How make a dual +-12V supply from a 24V SMPS

up vote
12
down vote

favorite

I'm trying to power a home-made load cell transmitter using a 24V single SMPS. I need to make +12, 0 and -12 Volts that are capable of 50mA. I wish to power multiple channels of opamps and bridges.

I don't have much budget and availability of components in India.

I have an idea to use 1 LM7812 an 1 LM7912(negative) linear voltage regulators and a voltage divider setup to do this as per the circuit below.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Would this work? I've modified it from the suggestions and articles elsewhere.

Somebody suggested me one other circuit but I am concerned about the current capabilities of the opamp.

schematic

^{simulate this circuit}

Would this work? If yes, please suggest suitable op-amp.

Are there any other techniques that would do the job economically?

edited yesterday

JRE

20k43766

asked yesterday

Ohbhatt

708

Think about what will happen if you have mismatched loads between the rails.
– winny
yesterday

@winny That's what I am worried about.
– Ohbhatt
yesterday

How much current do you need? I have made such a contrapment for an audio circuit to prevent uneven clipping using an opamp and even divider like yours, but transistor+resisor buffered on the output. Wasted a lot of power and there are easier solutions. In your case, I would go for two switchmode converters or one isolated one.
– winny
yesterday

3

You mention you need to be able to supply 50mA of current. But I guess this is mainly through the +12 and -12 rails (e.g. powering dual-supply opamps). What exactly do you have to supply through the 0V rail? If the 0V rail simply serves as a reference and only goes to some opamp inputs or high-valued resistors, it means your current needs for the 0V rail are much much lower than 50mA, and the solution #2 is perfectly valid.
– dim
yesterday

Btw, both farnell.in and mouser.in ship in India. You would find pretty much any component on these(albeit a bit pricey ). Another option is ebay.com , but these ship from China and have quite long delivery times.
– Tejas Kale
7 hours ago

|
show 1 more comment

up vote
12
down vote

favorite

I'm trying to power a home-made load cell transmitter using a 24V single SMPS. I need to make +12, 0 and -12 Volts that are capable of 50mA. I wish to power multiple channels of opamps and bridges.

I don't have much budget and availability of components in India.

I have an idea to use 1 LM7812 an 1 LM7912(negative) linear voltage regulators and a voltage divider setup to do this as per the circuit below.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Would this work? I've modified it from the suggestions and articles elsewhere.

Somebody suggested me one other circuit but I am concerned about the current capabilities of the opamp.

schematic

^{simulate this circuit}

Would this work? If yes, please suggest suitable op-amp.

Are there any other techniques that would do the job economically?

edited yesterday

JRE

20k43766

asked yesterday

Ohbhatt

708

Think about what will happen if you have mismatched loads between the rails.
– winny
yesterday

@winny That's what I am worried about.
– Ohbhatt
yesterday

How much current do you need? I have made such a contrapment for an audio circuit to prevent uneven clipping using an opamp and even divider like yours, but transistor+resisor buffered on the output. Wasted a lot of power and there are easier solutions. In your case, I would go for two switchmode converters or one isolated one.
– winny
yesterday

3

You mention you need to be able to supply 50mA of current. But I guess this is mainly through the +12 and -12 rails (e.g. powering dual-supply opamps). What exactly do you have to supply through the 0V rail? If the 0V rail simply serves as a reference and only goes to some opamp inputs or high-valued resistors, it means your current needs for the 0V rail are much much lower than 50mA, and the solution #2 is perfectly valid.
– dim
yesterday

Btw, both farnell.in and mouser.in ship in India. You would find pretty much any component on these(albeit a bit pricey ). Another option is ebay.com , but these ship from China and have quite long delivery times.
– Tejas Kale
7 hours ago

|
show 1 more comment

up vote
12
down vote

favorite

I'm trying to power a home-made load cell transmitter using a 24V single SMPS. I need to make +12, 0 and -12 Volts that are capable of 50mA. I wish to power multiple channels of opamps and bridges.

I don't have much budget and availability of components in India.

I have an idea to use 1 LM7812 an 1 LM7912(negative) linear voltage regulators and a voltage divider setup to do this as per the circuit below.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Would this work? I've modified it from the suggestions and articles elsewhere.

Somebody suggested me one other circuit but I am concerned about the current capabilities of the opamp.

schematic

^{simulate this circuit}

Would this work? If yes, please suggest suitable op-amp.

Are there any other techniques that would do the job economically?

edited yesterday

JRE

20k43766

asked yesterday

Ohbhatt

708

I'm trying to power a home-made load cell transmitter using a 24V single SMPS. I need to make +12, 0 and -12 Volts that are capable of 50mA. I wish to power multiple channels of opamps and bridges.

I don't have much budget and availability of components in India.

I have an idea to use 1 LM7812 an 1 LM7912(negative) linear voltage regulators and a voltage divider setup to do this as per the circuit below.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Would this work? I've modified it from the suggestions and articles elsewhere.

Somebody suggested me one other circuit but I am concerned about the current capabilities of the opamp.

schematic

^{simulate this circuit}

Would this work? If yes, please suggest suitable op-amp.

Are there any other techniques that would do the job economically?

op-amp voltage-regulator voltage-divider 24v virtual-ground

edited yesterday

JRE

20k43766

asked yesterday

Ohbhatt

708

edited yesterday

JRE

20k43766

asked yesterday

Ohbhatt

708

edited yesterday

JRE

20k43766

edited yesterday

JRE

20k43766

edited yesterday

JRE

20k43766

asked yesterday

Ohbhatt

708

asked yesterday

Ohbhatt

708

asked yesterday

Ohbhatt

708

Think about what will happen if you have mismatched loads between the rails.
– winny
yesterday

@winny That's what I am worried about.
– Ohbhatt
yesterday

How much current do you need? I have made such a contrapment for an audio circuit to prevent uneven clipping using an opamp and even divider like yours, but transistor+resisor buffered on the output. Wasted a lot of power and there are easier solutions. In your case, I would go for two switchmode converters or one isolated one.
– winny
yesterday

3

You mention you need to be able to supply 50mA of current. But I guess this is mainly through the +12 and -12 rails (e.g. powering dual-supply opamps). What exactly do you have to supply through the 0V rail? If the 0V rail simply serves as a reference and only goes to some opamp inputs or high-valued resistors, it means your current needs for the 0V rail are much much lower than 50mA, and the solution #2 is perfectly valid.
– dim
yesterday

Btw, both farnell.in and mouser.in ship in India. You would find pretty much any component on these(albeit a bit pricey ). Another option is ebay.com , but these ship from China and have quite long delivery times.
– Tejas Kale
7 hours ago

|
show 1 more comment

Think about what will happen if you have mismatched loads between the rails.
– winny
yesterday

@winny That's what I am worried about.
– Ohbhatt
yesterday

How much current do you need? I have made such a contrapment for an audio circuit to prevent uneven clipping using an opamp and even divider like yours, but transistor+resisor buffered on the output. Wasted a lot of power and there are easier solutions. In your case, I would go for two switchmode converters or one isolated one.
– winny
yesterday

3

You mention you need to be able to supply 50mA of current. But I guess this is mainly through the +12 and -12 rails (e.g. powering dual-supply opamps). What exactly do you have to supply through the 0V rail? If the 0V rail simply serves as a reference and only goes to some opamp inputs or high-valued resistors, it means your current needs for the 0V rail are much much lower than 50mA, and the solution #2 is perfectly valid.
– dim
yesterday

Btw, both farnell.in and mouser.in ship in India. You would find pretty much any component on these(albeit a bit pricey ). Another option is ebay.com , but these ship from China and have quite long delivery times.
– Tejas Kale
7 hours ago

Think about what will happen if you have mismatched loads between the rails.
– winny
yesterday

@winny That's what I am worried about.
– Ohbhatt
yesterday

How much current do you need? I have made such a contrapment for an audio circuit to prevent uneven clipping using an opamp and even divider like yours, but transistor+resisor buffered on the output. Wasted a lot of power and there are easier solutions. In your case, I would go for two switchmode converters or one isolated one.
– winny
yesterday

You mention you need to be able to supply 50mA of current. But I guess this is mainly through the +12 and -12 rails (e.g. powering dual-supply opamps). What exactly do you have to supply through the 0V rail? If the 0V rail simply serves as a reference and only goes to some opamp inputs or high-valued resistors, it means your current needs for the 0V rail are much much lower than 50mA, and the solution #2 is perfectly valid.
– dim
yesterday

Btw, both farnell.in and mouser.in ship in India. You would find pretty much any component on these(albeit a bit pricey ). Another option is ebay.com , but these ship from China and have quite long delivery times.
– Tejas Kale
7 hours ago

|
show 1 more comment

7 Answers
7

active

oldest

votes

up vote
13
down vote

accepted

You first idea will not work at all.

Your second idea will work, but many OP-Amps aren't going to deliver more than a few mA on their output, which limits the current your circuit may draw from the virtual ground. There are Power-OP-Amps available which may deliver up to a few ampere, but if you cannot get your hands on one, you can use a PNP/NPN transistor pair to increase the output current:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though.

edited yesterday

answered yesterday

Janka

7,9991820

can you please suggest a Transistor suitable for 50 or 100mA.
– Ohbhatt
yesterday

1

You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application.
– Janka
yesterday

Okay. I will take a look at that. Thank you for the time.
– Ohbhatt
yesterday

Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails.
– Ohbhatt
17 hours ago

The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.)
– Janka
16 hours ago

|
show 1 more comment

up vote
7
down vote

You'd be better off using two 12V supplies, but if you insist...

#1 won't work.

#2 (given the very limited information you have supplied) might require the op-amp to dissipate as much as 600mW and stability would likely be an issue with capacitive loads. There are dedicated rail splitter chips which take stability seriously but they are not jellybean parts and, for example, the TLE2426 cannot handle the dissipation or current involved.

I suggest something more like this (assuming you have power to spare on your 12V supply:

enter image description here

This uses a ubiquitous TL431 shunt regulator and boosts it with a generic PNP power transistor.

The combination is like a precision power zener. Or just use a zener as below. Set Vo = 12V.

Then use this circuit:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

Note that if you excessively load the GND to -V the +V to GND voltage will increase to as much as 24V. Usually that's acceptable but take care about capacitor voltage rating and so on. You can add a higher voltage zener (say 14V) across R1 as a preventative measure.
R1 will dissipate less than 1W, under normal conditions, but the zener could dissipate as much as 1.3W if 50mA flows from +V to GND and there is no corresponding current from GND to -V.

You could use two 6.2V 1W zeners in series, for example. Keep the leads short, attach them to some PCB area and keep them apart so they run cooler.

edited yesterday

answered yesterday

Spehro Pefhany

200k4144397

I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though.
– Ohbhatt
yesterday

1

I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents.
– ThreePhaseEel
yesterday

add a comment |

up vote
5
down vote

Given your desire for as low power as possible, and my realization that this common problem is seldom approached this way. I came up with a self-oscillating switching solution just for the fun of it.

As with any switcher, single-tone emissions/ripple have to be considered (around 20kHz with these values). But if there is any significant ground current, I doubt you can be much more efficient (a more formal switcher with a separate oscillator can be made more efficient and could use a single inductor, but it would require more parts).

schematic

^{simulate this circuit – Schematic created using CircuitLab}

It is basically a relaxation oscillator that modulates the average current through L1 so that it oscillates around the required ground current. M1 and M2 are switched on and off relatively quickly (some acceleration capacitors would help with efficiency) and C12 provides positive feedback so that the opamp/comparator saturates on crossing the threshold (otherwise the load would damp the oscillator and it would become a linear regulator instead).

L3, C10, and C11 are there to filter the ripple and to isolate the oscillation from the load, so as to avoid dampening it too much. C10, and C11 also do double-duty as the regulator input capacitance. Excess energy in L1 and L2 would be returned to the required rail and stored in them. M1 and M2 source-drain diodes are conducting in this design.

R3,R4,R5, and R6 are chosen so as to keep M1 and M2 below threshold when there is no ground current. Unfortunately this also reduces the overall gain of the oscillator loop.

I have not done a very careful analysis of all of the implications of this design (particularly because of it being self-oscillating), so overall stability considerations on load changes might be an issue.

I don't think there are ICs for this type of configuration, which unnecessarily increases the part count and the design constraints. The only ones I know of are the DDR memory termination voltage regulators, but those are intended to work at very low voltages.

edited yesterday

answered yesterday

Edgar Brown

1,70116

+1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime.
– Janka
yesterday

1

Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components.
– Edgar Brown
yesterday

Yes, exactly the latter was my concern. What happens when the ground current changes direction.
– Janka
yesterday

1

@Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power.
– Edgar Brown
yesterday

1

@Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that..
– Edgar Brown
yesterday

|
show 4 more comments

up vote
4
down vote

The regulators won’t work. You have zero dropout allocated to them and your ground impedance is excessive.

The op amp is a better option, but it all depends on how much current you have going through ground. If the current is low enough you can just use a resistor divider with a couple of capacitors in parallel, if it is high you would need a hefty op amp.

You have a couple more options:

You could use two zeners with series resistors to reduce ground impedance

You could put together a class AB source follower with a few resistors and two transistors (basically what the op-amp is doing but with higher impedance)

If your ground current has a well-defined and consistent direction, you can use a positive or negative 12V regulator or even a transistor off one of the rails (making sure to put a bypass diode).

But regardless of what you do, any ground current will result in wasted power (unless you figure out how to design a ground-switcher regulator of course).

answered yesterday

Edgar Brown

1,70116

Thank you for the observation.
– Ohbhatt
yesterday

add a comment |

up vote
4
down vote

If your 24 V is well regulated you could just use a 7812 to create a mid point and call that your 0 volt rail.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This will only work if the 24 V is independent of whatever you're powering, and as per Edgar Brown's comment, positive linear regulators like the 7812 can't sink current.

edited yesterday

answered yesterday

Colin

2,4532920

That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify.
– Ohbhatt
yesterday

4

This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current.
– Edgar Brown
yesterday

That's a very valid point, thanks, @EdgarBrown, I've edited the answer.
– Colin
yesterday

3

@EdgarBrown Instead of a trusty 7812 you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive.
– crasic
yesterday

1

@Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design.
– immibis
yesterday

|
show 3 more comments

up vote
2
down vote

I think NJM4556A would work

you can draw current from negative and positive rails but there difference not to exceed the OP amp output current.

Note: I'm not experienced, i suggest you to read the following post

EEVBLOG - my-negative-voltage-rail-doesnt-work

answered yesterday

Hasan alattar

596

That is a good idea. I will took into that.
– Ohbhatt
yesterday

add a comment |

up vote
0
down vote

There are many low-cost methods. But switching method may help you with a minimum component that available everywhere.

you can use a flyback converter with a minimum circuit:

Edited: The main circuit:
mix circuit
Ref: a mix of two links (http://uzzors2k.4hv.org/index.phppage=flybacktransformerdrivers, https://wiki.analog.com/university/courses/electronics/text/chapter-6)

Component list:

Zener diode

555 IC

Mosfet

A toroid, the transformer can be made with wire and a toroid core

Diode in output

some capacitor

some resistor

some wire

Benefits:

you can generate any voltage in output even bigger than your first voltage

these components are available everywhere

you can generate any voltage even isolated voltage

you can increase your power by changing the Mosfet and selecting a bigger toroid.

The main references:

http://uzzors2k.4hv.org/index.php?page=flybacktransformerdrivers
enter image description here

Additionally, you need a Zener diode for 12-15volt and a 555 IC.( your coil feed with 24Volt but for 555, you should generate a 12-volt power with a Zener diode).

in the output, you need a diode with a capacitor.
link: https://wiki.analog.com/university/courses/electronics/text/chapter-6
enter image description here

It is dual polarity Full-wave rectifier using a center tapped transformer and 4 diodes

edited yesterday

answered yesterday

M KS

868

add a comment |

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7 Answers
7

active

oldest

votes

7 Answers
7

active

oldest

votes

up vote
13
down vote

accepted

You first idea will not work at all.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though.

edited yesterday

answered yesterday

Janka

7,9991820

can you please suggest a Transistor suitable for 50 or 100mA.
– Ohbhatt
yesterday

1

You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application.
– Janka
yesterday

Okay. I will take a look at that. Thank you for the time.
– Ohbhatt
yesterday

Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails.
– Ohbhatt
17 hours ago

The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.)
– Janka
16 hours ago

|
show 1 more comment

up vote
13
down vote

accepted

You first idea will not work at all.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though.

edited yesterday

answered yesterday

Janka

7,9991820

can you please suggest a Transistor suitable for 50 or 100mA.
– Ohbhatt
yesterday

1

You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application.
– Janka
yesterday

Okay. I will take a look at that. Thank you for the time.
– Ohbhatt
yesterday

Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails.
– Ohbhatt
17 hours ago

The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.)
– Janka
16 hours ago

|
show 1 more comment

up vote
13
down vote

accepted

You first idea will not work at all.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though.

edited yesterday

answered yesterday

Janka

7,9991820

You first idea will not work at all.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

The OP-Amp will take care of stabilizing the output so it matches the voltage set by the input voltage divider. Take care of capacitive loads, as Spehro noted in his answer, though.

edited yesterday

answered yesterday

Janka

7,9991820

edited yesterday

answered yesterday

Janka

7,9991820

answered yesterday

Janka

7,9991820

answered yesterday

Janka

7,9991820

can you please suggest a Transistor suitable for 50 or 100mA.
– Ohbhatt
yesterday

1

You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application.
– Janka
yesterday

Okay. I will take a look at that. Thank you for the time.
– Ohbhatt
yesterday

Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails.
– Ohbhatt
17 hours ago

The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.)
– Janka
16 hours ago

|
show 1 more comment

can you please suggest a Transistor suitable for 50 or 100mA.
– Ohbhatt
yesterday

1

You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application.
– Janka
yesterday

Okay. I will take a look at that. Thank you for the time.
– Ohbhatt
yesterday

Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails.
– Ohbhatt
17 hours ago

The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.)
– Janka
16 hours ago

can you please suggest a Transistor suitable for 50 or 100mA.
– Ohbhatt
yesterday

You have to look for the packages. For 100mA, each transistor had to dissipate 100mA*12V=1.2W, that's the limiting factor. Small signal transistors in TO-92 packages are typically limited to 500mW. There are exceptions as the SS8050/SS8550 pair from Fairchild which can dissipate 2W each. A far more conservative pair (also better available) would be BD233/BD234, BD235/BD236 or BD237/BD238. (Use transistors meant for audio applications, they are rated for linear region operation, as needed here.) Transistors in TO220 packages are overkill for your application.
– Janka
yesterday

Okay. I will take a look at that. Thank you for the time.
– Ohbhatt
yesterday

Would 2n2222(NPN) and 2N2907(PNP) Work for this? I am unable to avail any of the above pairs at acceptable cost or separately(I have to purchase in bulk otherwise). I crosschecked the power will not be above 500mW in any case between the reference and either + and - rails.
– Ohbhatt
17 hours ago

The 2N2222 can only dissipate 500mW, that's good for 41mA@12V. The 2N2907 can only dissipate 625mW, good for 52mA@12V. In general, transistors starting with BD are what you want (the 2N… prefix unfortunately gives no hint.)
– Janka
16 hours ago

|
show 1 more comment

up vote
7
down vote

You'd be better off using two 12V supplies, but if you insist...

#1 won't work.

I suggest something more like this (assuming you have power to spare on your 12V supply:

enter image description here

This uses a ubiquitous TL431 shunt regulator and boosts it with a generic PNP power transistor.

The combination is like a precision power zener. Or just use a zener as below. Set Vo = 12V.

Then use this circuit:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

You could use two 6.2V 1W zeners in series, for example. Keep the leads short, attach them to some PCB area and keep them apart so they run cooler.

edited yesterday

answered yesterday

Spehro Pefhany

200k4144397

I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though.
– Ohbhatt
yesterday

1

I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents.
– ThreePhaseEel
yesterday

add a comment |

up vote
7
down vote

You'd be better off using two 12V supplies, but if you insist...

#1 won't work.

I suggest something more like this (assuming you have power to spare on your 12V supply:

enter image description here

This uses a ubiquitous TL431 shunt regulator and boosts it with a generic PNP power transistor.

The combination is like a precision power zener. Or just use a zener as below. Set Vo = 12V.

Then use this circuit:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

You could use two 6.2V 1W zeners in series, for example. Keep the leads short, attach them to some PCB area and keep them apart so they run cooler.

edited yesterday

answered yesterday

Spehro Pefhany

200k4144397

I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though.
– Ohbhatt
yesterday

1

I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents.
– ThreePhaseEel
yesterday

add a comment |

up vote
7
down vote

You'd be better off using two 12V supplies, but if you insist...

#1 won't work.

I suggest something more like this (assuming you have power to spare on your 12V supply:

enter image description here

This uses a ubiquitous TL431 shunt regulator and boosts it with a generic PNP power transistor.

The combination is like a precision power zener. Or just use a zener as below. Set Vo = 12V.

Then use this circuit:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

You could use two 6.2V 1W zeners in series, for example. Keep the leads short, attach them to some PCB area and keep them apart so they run cooler.

edited yesterday

answered yesterday

Spehro Pefhany

200k4144397

You'd be better off using two 12V supplies, but if you insist...

#1 won't work.

I suggest something more like this (assuming you have power to spare on your 12V supply:

enter image description here

This uses a ubiquitous TL431 shunt regulator and boosts it with a generic PNP power transistor.

The combination is like a precision power zener. Or just use a zener as below. Set Vo = 12V.

Then use this circuit:

schematic

^{simulate this circuit – Schematic created using CircuitLab}

You could use two 6.2V 1W zeners in series, for example. Keep the leads short, attach them to some PCB area and keep them apart so they run cooler.

edited yesterday

answered yesterday

Spehro Pefhany

200k4144397

edited yesterday

answered yesterday

Spehro Pefhany

200k4144397

answered yesterday

Spehro Pefhany

200k4144397

answered yesterday

Spehro Pefhany

200k4144397

I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though.
– Ohbhatt
yesterday

1

I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents.
– ThreePhaseEel
yesterday

add a comment |

I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though.
– Ohbhatt
yesterday

1

I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents.
– ThreePhaseEel
yesterday

I have to keep minimal power consumption and I cant afford any change in the voltage. Thanks for the help though.
– Ohbhatt
yesterday

I wouldn't use a regulator approach to generating the virtual ground at all -- it's going to have problems either sourcing or sinking appropriate currents.
– ThreePhaseEel
yesterday

add a comment |

up vote
5
down vote

Given your desire for as low power as possible, and my realization that this common problem is seldom approached this way. I came up with a self-oscillating switching solution just for the fun of it.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

R3,R4,R5, and R6 are chosen so as to keep M1 and M2 below threshold when there is no ground current. Unfortunately this also reduces the overall gain of the oscillator loop.

edited yesterday

answered yesterday

Edgar Brown

1,70116

+1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime.
– Janka
yesterday

1

Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components.
– Edgar Brown
yesterday

Yes, exactly the latter was my concern. What happens when the ground current changes direction.
– Janka
yesterday

1

@Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power.
– Edgar Brown
yesterday

1

@Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that..
– Edgar Brown
yesterday

|
show 4 more comments

up vote
5
down vote

Given your desire for as low power as possible, and my realization that this common problem is seldom approached this way. I came up with a self-oscillating switching solution just for the fun of it.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

R3,R4,R5, and R6 are chosen so as to keep M1 and M2 below threshold when there is no ground current. Unfortunately this also reduces the overall gain of the oscillator loop.

edited yesterday

answered yesterday

Edgar Brown

1,70116

+1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime.
– Janka
yesterday

1

Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components.
– Edgar Brown
yesterday

Yes, exactly the latter was my concern. What happens when the ground current changes direction.
– Janka
yesterday

1

@Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power.
– Edgar Brown
yesterday

1

@Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that..
– Edgar Brown
yesterday

|
show 4 more comments

up vote
5
down vote

Given your desire for as low power as possible, and my realization that this common problem is seldom approached this way. I came up with a self-oscillating switching solution just for the fun of it.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

R3,R4,R5, and R6 are chosen so as to keep M1 and M2 below threshold when there is no ground current. Unfortunately this also reduces the overall gain of the oscillator loop.

edited yesterday

answered yesterday

Edgar Brown

1,70116

Given your desire for as low power as possible, and my realization that this common problem is seldom approached this way. I came up with a self-oscillating switching solution just for the fun of it.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

R3,R4,R5, and R6 are chosen so as to keep M1 and M2 below threshold when there is no ground current. Unfortunately this also reduces the overall gain of the oscillator loop.

edited yesterday

answered yesterday

Edgar Brown

1,70116

edited yesterday

answered yesterday

Edgar Brown

1,70116

answered yesterday

Edgar Brown

1,70116

answered yesterday

Edgar Brown

1,70116

+1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime.
– Janka
yesterday

1

Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components.
– Edgar Brown
yesterday

Yes, exactly the latter was my concern. What happens when the ground current changes direction.
– Janka
yesterday

1

@Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power.
– Edgar Brown
yesterday

1

@Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that..
– Edgar Brown
yesterday

|
show 4 more comments

+1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime.
– Janka
yesterday

1

Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components.
– Edgar Brown
yesterday

Yes, exactly the latter was my concern. What happens when the ground current changes direction.
– Janka
yesterday

1

@Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power.
– Edgar Brown
yesterday

1

@Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that..
– Edgar Brown
yesterday

+1, this is ingenious. But I think the reason it isn't too common is circuits which need split ground are most times for audio applications and we would certainly hear the chime.
– Janka
yesterday

Making a 400kHz-1MHz switcher is possible. You would not hear that at all!! And after all ground is the reference, it will be the rails that are moving... I am normally dealing with applications in which even 1µV of noise in high-impedance traces is an issue, we use switchers all over the place. Including driving variable analog power lines that run under those high-impedance traces. All it needs is good filtering, our only issues have arisen when the switcher control algorithms skip-beats and produce low-frequency components.
– Edgar Brown
yesterday

Yes, exactly the latter was my concern. What happens when the ground current changes direction.
– Janka
yesterday

@Janka In this architecture, assuming that it is fully stable, nothing of consequence will happen. The excess current will simply be steered via a MOSFET diode to the rail that is supplying it. Ideally zero waste power.
– Edgar Brown
yesterday

@Janka Oh, and regarding the tones, the issue only happened when during a design modification for an updated version of the product someone did not pay too much attention to a trace and made a long loop with it, which interacted with loops in the high-impedance traces. Even then the noise was barely above detection in the 2µV range. We fixed it by improving the control algorithm. In this architecture the switcher never has to go silent, small alternating pulses in both directions can ensure that. With this design that could be problematic to achieve, but a separate oscillator would fix that..
– Edgar Brown
yesterday

|
show 4 more comments

up vote
4
down vote

The regulators won’t work. You have zero dropout allocated to them and your ground impedance is excessive.

You have a couple more options:

You could use two zeners with series resistors to reduce ground impedance

You could put together a class AB source follower with a few resistors and two transistors (basically what the op-amp is doing but with higher impedance)

If your ground current has a well-defined and consistent direction, you can use a positive or negative 12V regulator or even a transistor off one of the rails (making sure to put a bypass diode).

But regardless of what you do, any ground current will result in wasted power (unless you figure out how to design a ground-switcher regulator of course).

answered yesterday

Edgar Brown

1,70116

Thank you for the observation.
– Ohbhatt
yesterday

add a comment |

up vote
4
down vote

The regulators won’t work. You have zero dropout allocated to them and your ground impedance is excessive.

You have a couple more options:

You could use two zeners with series resistors to reduce ground impedance

You could put together a class AB source follower with a few resistors and two transistors (basically what the op-amp is doing but with higher impedance)

If your ground current has a well-defined and consistent direction, you can use a positive or negative 12V regulator or even a transistor off one of the rails (making sure to put a bypass diode).

But regardless of what you do, any ground current will result in wasted power (unless you figure out how to design a ground-switcher regulator of course).

answered yesterday

Edgar Brown

1,70116

Thank you for the observation.
– Ohbhatt
yesterday

add a comment |

up vote
4
down vote

The regulators won’t work. You have zero dropout allocated to them and your ground impedance is excessive.

You have a couple more options:

You could use two zeners with series resistors to reduce ground impedance

You could put together a class AB source follower with a few resistors and two transistors (basically what the op-amp is doing but with higher impedance)

If your ground current has a well-defined and consistent direction, you can use a positive or negative 12V regulator or even a transistor off one of the rails (making sure to put a bypass diode).

But regardless of what you do, any ground current will result in wasted power (unless you figure out how to design a ground-switcher regulator of course).

answered yesterday

Edgar Brown

1,70116

The regulators won’t work. You have zero dropout allocated to them and your ground impedance is excessive.

You have a couple more options:

You could use two zeners with series resistors to reduce ground impedance

You could put together a class AB source follower with a few resistors and two transistors (basically what the op-amp is doing but with higher impedance)

If your ground current has a well-defined and consistent direction, you can use a positive or negative 12V regulator or even a transistor off one of the rails (making sure to put a bypass diode).

But regardless of what you do, any ground current will result in wasted power (unless you figure out how to design a ground-switcher regulator of course).

answered yesterday

Edgar Brown

1,70116

answered yesterday

Edgar Brown

1,70116

answered yesterday

Edgar Brown

1,70116

answered yesterday

Edgar Brown

1,70116

Thank you for the observation.
– Ohbhatt
yesterday

add a comment |

Thank you for the observation.
– Ohbhatt
yesterday

Thank you for the observation.
– Ohbhatt
yesterday

add a comment |

up vote
4
down vote

If your 24 V is well regulated you could just use a 7812 to create a mid point and call that your 0 volt rail.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This will only work if the 24 V is independent of whatever you're powering, and as per Edgar Brown's comment, positive linear regulators like the 7812 can't sink current.

edited yesterday

answered yesterday

Colin

2,4532920

That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify.
– Ohbhatt
yesterday

4

This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current.
– Edgar Brown
yesterday

That's a very valid point, thanks, @EdgarBrown, I've edited the answer.
– Colin
yesterday

3

@EdgarBrown Instead of a trusty 7812 you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive.
– crasic
yesterday

1

@Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design.
– immibis
yesterday

|
show 3 more comments

up vote
4
down vote

If your 24 V is well regulated you could just use a 7812 to create a mid point and call that your 0 volt rail.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This will only work if the 24 V is independent of whatever you're powering, and as per Edgar Brown's comment, positive linear regulators like the 7812 can't sink current.

edited yesterday

answered yesterday

Colin

2,4532920

That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify.
– Ohbhatt
yesterday

4

This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current.
– Edgar Brown
yesterday

That's a very valid point, thanks, @EdgarBrown, I've edited the answer.
– Colin
yesterday

3

@EdgarBrown Instead of a trusty 7812 you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive.
– crasic
yesterday

1

@Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design.
– immibis
yesterday

|
show 3 more comments

up vote
4
down vote

If your 24 V is well regulated you could just use a 7812 to create a mid point and call that your 0 volt rail.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This will only work if the 24 V is independent of whatever you're powering, and as per Edgar Brown's comment, positive linear regulators like the 7812 can't sink current.

edited yesterday

answered yesterday

Colin

2,4532920

If your 24 V is well regulated you could just use a 7812 to create a mid point and call that your 0 volt rail.

schematic

^{simulate this circuit – Schematic created using CircuitLab}

This will only work if the 24 V is independent of whatever you're powering, and as per Edgar Brown's comment, positive linear regulators like the 7812 can't sink current.

edited yesterday

answered yesterday

Colin

2,4532920

edited yesterday

answered yesterday

Colin

2,4532920

answered yesterday

Colin

2,4532920

answered yesterday

Colin

2,4532920

That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify.
– Ohbhatt
yesterday

4

This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current.
– Edgar Brown
yesterday

That's a very valid point, thanks, @EdgarBrown, I've edited the answer.
– Colin
yesterday

3

@EdgarBrown Instead of a trusty 7812 you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive.
– crasic
yesterday

1

@Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design.
– immibis
yesterday

|
show 3 more comments

That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify.
– Ohbhatt
yesterday

4

This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current.
– Edgar Brown
yesterday

That's a very valid point, thanks, @EdgarBrown, I've edited the answer.
– Colin
yesterday

3

@EdgarBrown Instead of a trusty 7812 you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive.
– crasic
yesterday

1

@Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design.
– immibis
yesterday

That is a fantastic solution. I don't have to invest in any expensive parts. But I still have to test this circuit to verify.
– Ohbhatt
yesterday

This will only work if your ground current is positive (it exits the regulator) normal regulators don’t sink current.
– Edgar Brown
yesterday

That's a very valid point, thanks, @EdgarBrown, I've edited the answer.
– Colin
yesterday

@EdgarBrown Instead of a trusty 7812 you can use an integrated switching regulator that will generally tolerate "reverse current", they are a little more money, but same simple implementation. I have used this in a design, granted in my case most of the system was running on the 24V rail, with only a small subset of components running off the virtual ground. In any case, this becomes a matter of component selection, and 1/2/3A switching regulators can be found with bidrectional current capability, the design is solid, but the BOM may be hard to find or expensive.
– crasic
yesterday

@Ohbhatt No, that does not eliminate any change. Imagine if you connected a resistor between +12V and 0V, the regulator would not handle that because the current would be trying to go into the regulator. But a resistor between 0V and -12V would be fine. That's why it depends on your circuit design.
– immibis
yesterday

|
show 3 more comments

up vote
2
down vote

I think NJM4556A would work

you can draw current from negative and positive rails but there difference not to exceed the OP amp output current.

Note: I'm not experienced, i suggest you to read the following post

EEVBLOG - my-negative-voltage-rail-doesnt-work

answered yesterday

Hasan alattar

596

That is a good idea. I will took into that.
– Ohbhatt
yesterday

add a comment |

up vote
2
down vote

I think NJM4556A would work

you can draw current from negative and positive rails but there difference not to exceed the OP amp output current.

Note: I'm not experienced, i suggest you to read the following post

EEVBLOG - my-negative-voltage-rail-doesnt-work

answered yesterday

Hasan alattar

596

That is a good idea. I will took into that.
– Ohbhatt
yesterday

add a comment |

up vote
2
down vote

I think NJM4556A would work

you can draw current from negative and positive rails but there difference not to exceed the OP amp output current.

Note: I'm not experienced, i suggest you to read the following post

EEVBLOG - my-negative-voltage-rail-doesnt-work

answered yesterday

Hasan alattar

596

I think NJM4556A would work

you can draw current from negative and positive rails but there difference not to exceed the OP amp output current.

Note: I'm not experienced, i suggest you to read the following post

EEVBLOG - my-negative-voltage-rail-doesnt-work

answered yesterday

Hasan alattar

596

answered yesterday

Hasan alattar

596

answered yesterday

Hasan alattar

596

answered yesterday

Hasan alattar

596

That is a good idea. I will took into that.
– Ohbhatt
yesterday

add a comment |

That is a good idea. I will took into that.
– Ohbhatt
yesterday

That is a good idea. I will took into that.
– Ohbhatt
yesterday

add a comment |

up vote
0
down vote

There are many low-cost methods. But switching method may help you with a minimum component that available everywhere.

you can use a flyback converter with a minimum circuit:

Component list:

Zener diode

555 IC

Mosfet

A toroid, the transformer can be made with wire and a toroid core

Diode in output

some capacitor

some resistor

some wire

Benefits:

you can generate any voltage in output even bigger than your first voltage

these components are available everywhere

you can generate any voltage even isolated voltage

you can increase your power by changing the Mosfet and selecting a bigger toroid.

The main references:

http://uzzors2k.4hv.org/index.php?page=flybacktransformerdrivers
enter image description here

Additionally, you need a Zener diode for 12-15volt and a 555 IC.( your coil feed with 24Volt but for 555, you should generate a 12-volt power with a Zener diode).

in the output, you need a diode with a capacitor.
link: https://wiki.analog.com/university/courses/electronics/text/chapter-6
enter image description here

It is dual polarity Full-wave rectifier using a center tapped transformer and 4 diodes

edited yesterday

answered yesterday

M KS

868

add a comment |

up vote
0
down vote

There are many low-cost methods. But switching method may help you with a minimum component that available everywhere.

you can use a flyback converter with a minimum circuit:

Component list:

Zener diode

555 IC

Mosfet

A toroid, the transformer can be made with wire and a toroid core

Diode in output

some capacitor

some resistor

some wire

Benefits:

you can generate any voltage in output even bigger than your first voltage

these components are available everywhere

you can generate any voltage even isolated voltage

you can increase your power by changing the Mosfet and selecting a bigger toroid.

The main references:

http://uzzors2k.4hv.org/index.php?page=flybacktransformerdrivers
enter image description here

Additionally, you need a Zener diode for 12-15volt and a 555 IC.( your coil feed with 24Volt but for 555, you should generate a 12-volt power with a Zener diode).

in the output, you need a diode with a capacitor.
link: https://wiki.analog.com/university/courses/electronics/text/chapter-6
enter image description here

It is dual polarity Full-wave rectifier using a center tapped transformer and 4 diodes

edited yesterday

answered yesterday

M KS

868

add a comment |

up vote
0
down vote

There are many low-cost methods. But switching method may help you with a minimum component that available everywhere.

you can use a flyback converter with a minimum circuit:

Component list:

Zener diode

555 IC

Mosfet

A toroid, the transformer can be made with wire and a toroid core

Diode in output

some capacitor

some resistor

some wire

Benefits:

you can generate any voltage in output even bigger than your first voltage

these components are available everywhere

you can generate any voltage even isolated voltage

you can increase your power by changing the Mosfet and selecting a bigger toroid.

The main references:

http://uzzors2k.4hv.org/index.php?page=flybacktransformerdrivers
enter image description here

Additionally, you need a Zener diode for 12-15volt and a 555 IC.( your coil feed with 24Volt but for 555, you should generate a 12-volt power with a Zener diode).

in the output, you need a diode with a capacitor.
link: https://wiki.analog.com/university/courses/electronics/text/chapter-6
enter image description here

It is dual polarity Full-wave rectifier using a center tapped transformer and 4 diodes

edited yesterday

answered yesterday

M KS

868

There are many low-cost methods. But switching method may help you with a minimum component that available everywhere.

you can use a flyback converter with a minimum circuit:

Component list:

Zener diode

555 IC

Mosfet

A toroid, the transformer can be made with wire and a toroid core

Diode in output

some capacitor

some resistor

some wire

Benefits:

you can generate any voltage in output even bigger than your first voltage

these components are available everywhere

you can generate any voltage even isolated voltage

you can increase your power by changing the Mosfet and selecting a bigger toroid.

The main references:

http://uzzors2k.4hv.org/index.php?page=flybacktransformerdrivers
enter image description here

Additionally, you need a Zener diode for 12-15volt and a 555 IC.( your coil feed with 24Volt but for 555, you should generate a 12-volt power with a Zener diode).

in the output, you need a diode with a capacitor.
link: https://wiki.analog.com/university/courses/electronics/text/chapter-6
enter image description here

It is dual polarity Full-wave rectifier using a center tapped transformer and 4 diodes

edited yesterday

answered yesterday

M KS

868

edited yesterday

answered yesterday

M KS

868

answered yesterday

M KS

868

answered yesterday

M KS

868

add a comment |

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