Mixing
Understanding class A amplifier waveforms
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I am working on an audio amplifier but got stuck on to a basic of npn transistor circuit.
Attaching a class-A amplifier circuit. Take a look here or here. I tried to simulate the circuit but as everywhere on net it is said that this can amplify whole sine wave , in my P Spice simulation it is not. Only one half of input signal is getting amplified.
My question is how I can get full sine wave (as all the schematic of class a amplifier over the internet claims)? Also please explain why exactly I am getting attached waveform in my circuit.
transistors amplifier bjt
edited Jan 25 at 18:10
JRE
22.3k53773
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
$endgroup$
add a comment |
$begingroup$
I am working on an audio amplifier but got stuck on to a basic of npn transistor circuit.
Attaching a class-A amplifier circuit. Take a look here or here. I tried to simulate the circuit but as everywhere on net it is said that this can amplify whole sine wave , in my P Spice simulation it is not. Only one half of input signal is getting amplified.
My question is how I can get full sine wave (as all the schematic of class a amplifier over the internet claims)? Also please explain why exactly I am getting attached waveform in my circuit.
transistors amplifier bjt
edited Jan 25 at 18:10
JRE
22.3k53773
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
$endgroup$
2
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. Another problem.
$endgroup$
– jonk
Jan 25 at 13:16
add a comment |
$begingroup$
I am working on an audio amplifier but got stuck on to a basic of npn transistor circuit.
Attaching a class-A amplifier circuit. Take a look here or here. I tried to simulate the circuit but as everywhere on net it is said that this can amplify whole sine wave , in my P Spice simulation it is not. Only one half of input signal is getting amplified.
My question is how I can get full sine wave (as all the schematic of class a amplifier over the internet claims)? Also please explain why exactly I am getting attached waveform in my circuit.
transistors amplifier bjt
edited Jan 25 at 18:10
JRE
22.3k53773
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
$endgroup$
I am working on an audio amplifier but got stuck on to a basic of npn transistor circuit.
Attaching a class-A amplifier circuit. Take a look here or here. I tried to simulate the circuit but as everywhere on net it is said that this can amplify whole sine wave , in my P Spice simulation it is not. Only one half of input signal is getting amplified.
My question is how I can get full sine wave (as all the schematic of class a amplifier over the internet claims)? Also please explain why exactly I am getting attached waveform in my circuit.
transistors amplifier bjt
transistors amplifier bjt
edited Jan 25 at 18:10
JRE
22.3k53773
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
edited Jan 25 at 18:10
JRE
22.3k53773
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
edited Jan 25 at 18:10
JRE
22.3k53773
edited Jan 25 at 18:10
JRE
22.3k53773
edited Jan 25 at 18:10
JRE
22.3k53773
22.3k53773
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
asked Jan 25 at 11:48
Ashish JhaAshish Jha
285
285
2
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. Another problem.
$endgroup$
– jonk
Jan 25 at 13:16
add a comment |
2
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. Another problem.
$endgroup$
– jonk
Jan 25 at 13:16
2
2
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. Another problem.
$endgroup$
– jonk
Jan 25 at 13:16
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. Another problem.
$endgroup$
– jonk
Jan 25 at 13:16
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The 10 K and 1 K resistors you have connected to the base of the transistor are for biasing, however, the voltage source you have for the oscillator is preventing that from happening. If you AC couple the oscillator instead (add a capacitor between the signal generator and transistor) you should see correct operation. With the voltages you have you will see distortion (clipping), try reducing the oscillator to 1 V.
You can experiment with the ratio of the biasing resistors to how that influences the output waveform.
answered Jan 25 at 12:09
ColinColin
2,98121123
$endgroup$
3
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
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Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
add a comment |
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. That's another serious problem.
I'll discuss an approach and a schematic (below) that will show you how to approach a more proper design for a BJT stage like this.
Looks like you want a gain of 13, just quickly glancing. Obviously, you can tolerate an output impedance of $13:text{k}Omega$, too. I'll choose a different gain and a different output impedance, but not too far away.
Let's say the voltage gain is to be $A_V=10$ and I'll keep your existing power supply rail of $V_text{CC}=10:text{V}$. Here's an approach. (There are many such, not just one. But I'm not going to go through more than one for you. You can pick up others when other folks tell you about them.)
- The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) With $A_V=10$, this means $I_{text{C}_text{Q}}gt 250:mutext{A}$. I'd like twice that much, if possible. So let's set $I_{text{C}_text{Q}}=500:mutext{A}$.
- Given $I_{text{C}_text{Q}}=500:mutext{A}$ and typical small-signal BJTs, it is reasonable to conclude that the quiescent base-emitter voltage is about $V_{text{BE}_text{Q}}approx 660:text{mV}$.
- I like to reserve about $2:text{V}$ for the minimum $V_text{CE}$ of the BJT, in order to keep it well away from saturation, to help deal with BJT variations, and to slightly reduce the impact of the Early Effect.
- I like to reserve at least $1:text{V}$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
- With $V_text{CC}=10:text{V}$ and subtracting the above two margins I just reserved, this means there is about $7:text{V}$ left over for the collector swing. But I also want to leave about $2:text{V}$ margin at the top end of the collector swing (limiting distortion due to gain variation and mitigating Early Effect.) So I don't want the collector to move any higher than $8:text{V}$. So this leaves only $5:text{V}$ for the collector swing (max.)
- Therefore, the quiescent collector voltage will be $V_{text{C}_text{Q}}=10:text{V}-2:text{V}-frac{5:text{V}}{2}=5.5:text{V}=1:text{V}+2:text{V}+frac{5:text{V}}{2}$. In short, $V_{text{C}_text{Q}}=5.5:text{V}$.
- From (1) and (6), I can compute a collector resistor of $R_{text{C}}=frac{10:text{V}-5.5:text{V}}{500:mutext{A}}=9:text{k}Omega$. Set this to the nearby 5% precision value of $R_{text{C}}=9.1:text{k}Omega$.
- From (1) and (4), I can compute a DC emitter resistor of $R_{text{E}_1}=frac{1:text{V}}{500:mutext{A}}=2:text{k}Omega$. That's a standard 5% value, so keep it.
- From (2) and (4), I know that the quiescent DC base voltage should be $V_{text{B}_text{Q}}=1:text{V}+660:text{mV}=1.66:text{V}$.
- To be conservative, I'll assume that the base current of the BJT will be no more than about $I_{text{B}_text{Q}}=frac{I_{text{C}_text{Q}}=500:mutext{A}}{beta=100}=5:mutext{A}$.
- To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current due to signal variations), I know that the current through the two base divider resistors should be about $frac1{10}$th the quiescent collector current [or 10 times the current calculated in (10) above.] So this means about $50:mutext{A}$ for the base's resistor divider pair; used in steps (12) and (13) next.
- The divider resistor, from base to ground, is then $R_2=frac{1.66:text{V}}{50:mutext{A}}=33.2:text{k}Omega$. Use the nearby 5% value of $R_2=33:text{k}Omega$.
- The divider resistor, from base to the supply rail, is then $R_1=frac{10:text{V}-1.66:text{V}}{50:mutext{A}+5:mutext{A}}=151.6:text{k}Omega$. Use the nearby 5% value of $R_1=150:text{k}Omega$.
- To get the gain, I need the total AC emitter resistance to be $frac{R_text{C}=9.1:text{k}Omega}{A_V=10}-frac{V_T=26:text{mV}}{I_{text{C}_text{Q}}=500:mutext{A}}approx 858:Omega$. However, as you will soon see below, there is already a $2:text{k}Omega$ emitter resistor for the DC operating point computed in (8) above. So I need a new AC resistor value of $R_{text{E}_2}=frac{2:text{k}Omega,cdot, 858:Omega}{2:text{k}Omega-858:Omega}approx 1503:Omega$. I'll use the nearby 5% value of $R_{text{E}_2}=1.5:text{k}Omega$.
So here is the resulting design using standard resistor values:
simulate this circuit – Schematic created using CircuitLab
The above should take a maximum of a $500:text{mV}_text{PP}$ input signal and generate a maximum $5:text{V}_text{PP}$ output signal.
Feel free to ask questions, now. But hopefully that provides an approach to similar design questions.
To make the design still more bullet-proof to BJT variations, reduce the allowable maximum collector swing to $4:text{V}$ (or even less) and follow step (6) to set $V_{text{C}_text{Q}}=6:text{V}$ (or even a little bit higher, perhaps to $V_{text{C}_text{Q}}=6.3:text{V}$) and then recalculate the rest from there.
There's an issue with the design. It probably needs something to reduce its gain at higher frequencies. (Given the above-designed collector resistor, a $470:text{pF}$ capacitor across $R_c$ might be added, for example.) But I'm not going to address that issue any further, here.
answered Jan 25 at 14:04
jonkjonk
33.9k12874
$endgroup$
3
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As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
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@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
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Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
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– Ashish Jha
Jan 27 at 7:13
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@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
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– jonk
Jan 27 at 8:26
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@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
add a comment |
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2 Answers
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votes
2 Answers
2
active
oldest
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active
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$begingroup$
The 10 K and 1 K resistors you have connected to the base of the transistor are for biasing, however, the voltage source you have for the oscillator is preventing that from happening. If you AC couple the oscillator instead (add a capacitor between the signal generator and transistor) you should see correct operation. With the voltages you have you will see distortion (clipping), try reducing the oscillator to 1 V.
You can experiment with the ratio of the biasing resistors to how that influences the output waveform.
answered Jan 25 at 12:09
ColinColin
2,98121123
$endgroup$
3
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
$begingroup$
Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
add a comment |
$begingroup$
The 10 K and 1 K resistors you have connected to the base of the transistor are for biasing, however, the voltage source you have for the oscillator is preventing that from happening. If you AC couple the oscillator instead (add a capacitor between the signal generator and transistor) you should see correct operation. With the voltages you have you will see distortion (clipping), try reducing the oscillator to 1 V.
You can experiment with the ratio of the biasing resistors to how that influences the output waveform.
answered Jan 25 at 12:09
ColinColin
2,98121123
$endgroup$
3
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
$begingroup$
Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
add a comment |
$begingroup$
The 10 K and 1 K resistors you have connected to the base of the transistor are for biasing, however, the voltage source you have for the oscillator is preventing that from happening. If you AC couple the oscillator instead (add a capacitor between the signal generator and transistor) you should see correct operation. With the voltages you have you will see distortion (clipping), try reducing the oscillator to 1 V.
You can experiment with the ratio of the biasing resistors to how that influences the output waveform.
answered Jan 25 at 12:09
ColinColin
2,98121123
$endgroup$
The 10 K and 1 K resistors you have connected to the base of the transistor are for biasing, however, the voltage source you have for the oscillator is preventing that from happening. If you AC couple the oscillator instead (add a capacitor between the signal generator and transistor) you should see correct operation. With the voltages you have you will see distortion (clipping), try reducing the oscillator to 1 V.
You can experiment with the ratio of the biasing resistors to how that influences the output waveform.
answered Jan 25 at 12:09
ColinColin
2,98121123
edited Jan 25 at 12:22
answered Jan 25 at 12:09
ColinColin
2,98121123
answered Jan 25 at 12:09
ColinColin
2,98121123
answered Jan 25 at 12:09
ColinColin
2,98121123
2,98121123
3
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
$begingroup$
Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
add a comment |
3
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
$begingroup$
Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
3
3
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
$begingroup$
And likewise, a capacitor on the output, to remove the DC offset of the output.
$endgroup$
– rdtsc
Jan 25 at 12:18
$begingroup$
Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
$begingroup$
Hi Colin, yes it worked and i now understand why, thanks. Can you explain math of selecting the value of that coupling capacitor.
$endgroup$
– Ashish Jha
Jan 27 at 3:53
add a comment |
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. That's another serious problem.
I'll discuss an approach and a schematic (below) that will show you how to approach a more proper design for a BJT stage like this.
Looks like you want a gain of 13, just quickly glancing. Obviously, you can tolerate an output impedance of $13:text{k}Omega$, too. I'll choose a different gain and a different output impedance, but not too far away.
Let's say the voltage gain is to be $A_V=10$ and I'll keep your existing power supply rail of $V_text{CC}=10:text{V}$. Here's an approach. (There are many such, not just one. But I'm not going to go through more than one for you. You can pick up others when other folks tell you about them.)
- The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) With $A_V=10$, this means $I_{text{C}_text{Q}}gt 250:mutext{A}$. I'd like twice that much, if possible. So let's set $I_{text{C}_text{Q}}=500:mutext{A}$.
- Given $I_{text{C}_text{Q}}=500:mutext{A}$ and typical small-signal BJTs, it is reasonable to conclude that the quiescent base-emitter voltage is about $V_{text{BE}_text{Q}}approx 660:text{mV}$.
- I like to reserve about $2:text{V}$ for the minimum $V_text{CE}$ of the BJT, in order to keep it well away from saturation, to help deal with BJT variations, and to slightly reduce the impact of the Early Effect.
- I like to reserve at least $1:text{V}$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
- With $V_text{CC}=10:text{V}$ and subtracting the above two margins I just reserved, this means there is about $7:text{V}$ left over for the collector swing. But I also want to leave about $2:text{V}$ margin at the top end of the collector swing (limiting distortion due to gain variation and mitigating Early Effect.) So I don't want the collector to move any higher than $8:text{V}$. So this leaves only $5:text{V}$ for the collector swing (max.)
- Therefore, the quiescent collector voltage will be $V_{text{C}_text{Q}}=10:text{V}-2:text{V}-frac{5:text{V}}{2}=5.5:text{V}=1:text{V}+2:text{V}+frac{5:text{V}}{2}$. In short, $V_{text{C}_text{Q}}=5.5:text{V}$.
- From (1) and (6), I can compute a collector resistor of $R_{text{C}}=frac{10:text{V}-5.5:text{V}}{500:mutext{A}}=9:text{k}Omega$. Set this to the nearby 5% precision value of $R_{text{C}}=9.1:text{k}Omega$.
- From (1) and (4), I can compute a DC emitter resistor of $R_{text{E}_1}=frac{1:text{V}}{500:mutext{A}}=2:text{k}Omega$. That's a standard 5% value, so keep it.
- From (2) and (4), I know that the quiescent DC base voltage should be $V_{text{B}_text{Q}}=1:text{V}+660:text{mV}=1.66:text{V}$.
- To be conservative, I'll assume that the base current of the BJT will be no more than about $I_{text{B}_text{Q}}=frac{I_{text{C}_text{Q}}=500:mutext{A}}{beta=100}=5:mutext{A}$.
- To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current due to signal variations), I know that the current through the two base divider resistors should be about $frac1{10}$th the quiescent collector current [or 10 times the current calculated in (10) above.] So this means about $50:mutext{A}$ for the base's resistor divider pair; used in steps (12) and (13) next.
- The divider resistor, from base to ground, is then $R_2=frac{1.66:text{V}}{50:mutext{A}}=33.2:text{k}Omega$. Use the nearby 5% value of $R_2=33:text{k}Omega$.
- The divider resistor, from base to the supply rail, is then $R_1=frac{10:text{V}-1.66:text{V}}{50:mutext{A}+5:mutext{A}}=151.6:text{k}Omega$. Use the nearby 5% value of $R_1=150:text{k}Omega$.
- To get the gain, I need the total AC emitter resistance to be $frac{R_text{C}=9.1:text{k}Omega}{A_V=10}-frac{V_T=26:text{mV}}{I_{text{C}_text{Q}}=500:mutext{A}}approx 858:Omega$. However, as you will soon see below, there is already a $2:text{k}Omega$ emitter resistor for the DC operating point computed in (8) above. So I need a new AC resistor value of $R_{text{E}_2}=frac{2:text{k}Omega,cdot, 858:Omega}{2:text{k}Omega-858:Omega}approx 1503:Omega$. I'll use the nearby 5% value of $R_{text{E}_2}=1.5:text{k}Omega$.
So here is the resulting design using standard resistor values:
simulate this circuit – Schematic created using CircuitLab
The above should take a maximum of a $500:text{mV}_text{PP}$ input signal and generate a maximum $5:text{V}_text{PP}$ output signal.
Feel free to ask questions, now. But hopefully that provides an approach to similar design questions.
To make the design still more bullet-proof to BJT variations, reduce the allowable maximum collector swing to $4:text{V}$ (or even less) and follow step (6) to set $V_{text{C}_text{Q}}=6:text{V}$ (or even a little bit higher, perhaps to $V_{text{C}_text{Q}}=6.3:text{V}$) and then recalculate the rest from there.
There's an issue with the design. It probably needs something to reduce its gain at higher frequencies. (Given the above-designed collector resistor, a $470:text{pF}$ capacitor across $R_c$ might be added, for example.) But I'm not going to address that issue any further, here.
answered Jan 25 at 14:04
jonkjonk
33.9k12874
$endgroup$
3
$begingroup$
As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
$begingroup$
@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
$begingroup$
Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
$endgroup$
– Ashish Jha
Jan 27 at 7:13
$begingroup$
@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
$endgroup$
– jonk
Jan 27 at 8:26
$begingroup$
@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
add a comment |
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. That's another serious problem.
I'll discuss an approach and a schematic (below) that will show you how to approach a more proper design for a BJT stage like this.
Looks like you want a gain of 13, just quickly glancing. Obviously, you can tolerate an output impedance of $13:text{k}Omega$, too. I'll choose a different gain and a different output impedance, but not too far away.
Let's say the voltage gain is to be $A_V=10$ and I'll keep your existing power supply rail of $V_text{CC}=10:text{V}$. Here's an approach. (There are many such, not just one. But I'm not going to go through more than one for you. You can pick up others when other folks tell you about them.)
- The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) With $A_V=10$, this means $I_{text{C}_text{Q}}gt 250:mutext{A}$. I'd like twice that much, if possible. So let's set $I_{text{C}_text{Q}}=500:mutext{A}$.
- Given $I_{text{C}_text{Q}}=500:mutext{A}$ and typical small-signal BJTs, it is reasonable to conclude that the quiescent base-emitter voltage is about $V_{text{BE}_text{Q}}approx 660:text{mV}$.
- I like to reserve about $2:text{V}$ for the minimum $V_text{CE}$ of the BJT, in order to keep it well away from saturation, to help deal with BJT variations, and to slightly reduce the impact of the Early Effect.
- I like to reserve at least $1:text{V}$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
- With $V_text{CC}=10:text{V}$ and subtracting the above two margins I just reserved, this means there is about $7:text{V}$ left over for the collector swing. But I also want to leave about $2:text{V}$ margin at the top end of the collector swing (limiting distortion due to gain variation and mitigating Early Effect.) So I don't want the collector to move any higher than $8:text{V}$. So this leaves only $5:text{V}$ for the collector swing (max.)
- Therefore, the quiescent collector voltage will be $V_{text{C}_text{Q}}=10:text{V}-2:text{V}-frac{5:text{V}}{2}=5.5:text{V}=1:text{V}+2:text{V}+frac{5:text{V}}{2}$. In short, $V_{text{C}_text{Q}}=5.5:text{V}$.
- From (1) and (6), I can compute a collector resistor of $R_{text{C}}=frac{10:text{V}-5.5:text{V}}{500:mutext{A}}=9:text{k}Omega$. Set this to the nearby 5% precision value of $R_{text{C}}=9.1:text{k}Omega$.
- From (1) and (4), I can compute a DC emitter resistor of $R_{text{E}_1}=frac{1:text{V}}{500:mutext{A}}=2:text{k}Omega$. That's a standard 5% value, so keep it.
- From (2) and (4), I know that the quiescent DC base voltage should be $V_{text{B}_text{Q}}=1:text{V}+660:text{mV}=1.66:text{V}$.
- To be conservative, I'll assume that the base current of the BJT will be no more than about $I_{text{B}_text{Q}}=frac{I_{text{C}_text{Q}}=500:mutext{A}}{beta=100}=5:mutext{A}$.
- To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current due to signal variations), I know that the current through the two base divider resistors should be about $frac1{10}$th the quiescent collector current [or 10 times the current calculated in (10) above.] So this means about $50:mutext{A}$ for the base's resistor divider pair; used in steps (12) and (13) next.
- The divider resistor, from base to ground, is then $R_2=frac{1.66:text{V}}{50:mutext{A}}=33.2:text{k}Omega$. Use the nearby 5% value of $R_2=33:text{k}Omega$.
- The divider resistor, from base to the supply rail, is then $R_1=frac{10:text{V}-1.66:text{V}}{50:mutext{A}+5:mutext{A}}=151.6:text{k}Omega$. Use the nearby 5% value of $R_1=150:text{k}Omega$.
- To get the gain, I need the total AC emitter resistance to be $frac{R_text{C}=9.1:text{k}Omega}{A_V=10}-frac{V_T=26:text{mV}}{I_{text{C}_text{Q}}=500:mutext{A}}approx 858:Omega$. However, as you will soon see below, there is already a $2:text{k}Omega$ emitter resistor for the DC operating point computed in (8) above. So I need a new AC resistor value of $R_{text{E}_2}=frac{2:text{k}Omega,cdot, 858:Omega}{2:text{k}Omega-858:Omega}approx 1503:Omega$. I'll use the nearby 5% value of $R_{text{E}_2}=1.5:text{k}Omega$.
So here is the resulting design using standard resistor values:
simulate this circuit – Schematic created using CircuitLab
The above should take a maximum of a $500:text{mV}_text{PP}$ input signal and generate a maximum $5:text{V}_text{PP}$ output signal.
Feel free to ask questions, now. But hopefully that provides an approach to similar design questions.
To make the design still more bullet-proof to BJT variations, reduce the allowable maximum collector swing to $4:text{V}$ (or even less) and follow step (6) to set $V_{text{C}_text{Q}}=6:text{V}$ (or even a little bit higher, perhaps to $V_{text{C}_text{Q}}=6.3:text{V}$) and then recalculate the rest from there.
There's an issue with the design. It probably needs something to reduce its gain at higher frequencies. (Given the above-designed collector resistor, a $470:text{pF}$ capacitor across $R_c$ might be added, for example.) But I'm not going to address that issue any further, here.
answered Jan 25 at 14:04
jonkjonk
33.9k12874
$endgroup$
3
$begingroup$
As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
$begingroup$
@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
$begingroup$
Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
$endgroup$
– Ashish Jha
Jan 27 at 7:13
$begingroup$
@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
$endgroup$
– jonk
Jan 27 at 8:26
$begingroup$
@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
add a comment |
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. That's another serious problem.
I'll discuss an approach and a schematic (below) that will show you how to approach a more proper design for a BJT stage like this.
Looks like you want a gain of 13, just quickly glancing. Obviously, you can tolerate an output impedance of $13:text{k}Omega$, too. I'll choose a different gain and a different output impedance, but not too far away.
Let's say the voltage gain is to be $A_V=10$ and I'll keep your existing power supply rail of $V_text{CC}=10:text{V}$. Here's an approach. (There are many such, not just one. But I'm not going to go through more than one for you. You can pick up others when other folks tell you about them.)
- The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) With $A_V=10$, this means $I_{text{C}_text{Q}}gt 250:mutext{A}$. I'd like twice that much, if possible. So let's set $I_{text{C}_text{Q}}=500:mutext{A}$.
- Given $I_{text{C}_text{Q}}=500:mutext{A}$ and typical small-signal BJTs, it is reasonable to conclude that the quiescent base-emitter voltage is about $V_{text{BE}_text{Q}}approx 660:text{mV}$.
- I like to reserve about $2:text{V}$ for the minimum $V_text{CE}$ of the BJT, in order to keep it well away from saturation, to help deal with BJT variations, and to slightly reduce the impact of the Early Effect.
- I like to reserve at least $1:text{V}$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
- With $V_text{CC}=10:text{V}$ and subtracting the above two margins I just reserved, this means there is about $7:text{V}$ left over for the collector swing. But I also want to leave about $2:text{V}$ margin at the top end of the collector swing (limiting distortion due to gain variation and mitigating Early Effect.) So I don't want the collector to move any higher than $8:text{V}$. So this leaves only $5:text{V}$ for the collector swing (max.)
- Therefore, the quiescent collector voltage will be $V_{text{C}_text{Q}}=10:text{V}-2:text{V}-frac{5:text{V}}{2}=5.5:text{V}=1:text{V}+2:text{V}+frac{5:text{V}}{2}$. In short, $V_{text{C}_text{Q}}=5.5:text{V}$.
- From (1) and (6), I can compute a collector resistor of $R_{text{C}}=frac{10:text{V}-5.5:text{V}}{500:mutext{A}}=9:text{k}Omega$. Set this to the nearby 5% precision value of $R_{text{C}}=9.1:text{k}Omega$.
- From (1) and (4), I can compute a DC emitter resistor of $R_{text{E}_1}=frac{1:text{V}}{500:mutext{A}}=2:text{k}Omega$. That's a standard 5% value, so keep it.
- From (2) and (4), I know that the quiescent DC base voltage should be $V_{text{B}_text{Q}}=1:text{V}+660:text{mV}=1.66:text{V}$.
- To be conservative, I'll assume that the base current of the BJT will be no more than about $I_{text{B}_text{Q}}=frac{I_{text{C}_text{Q}}=500:mutext{A}}{beta=100}=5:mutext{A}$.
- To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current due to signal variations), I know that the current through the two base divider resistors should be about $frac1{10}$th the quiescent collector current [or 10 times the current calculated in (10) above.] So this means about $50:mutext{A}$ for the base's resistor divider pair; used in steps (12) and (13) next.
- The divider resistor, from base to ground, is then $R_2=frac{1.66:text{V}}{50:mutext{A}}=33.2:text{k}Omega$. Use the nearby 5% value of $R_2=33:text{k}Omega$.
- The divider resistor, from base to the supply rail, is then $R_1=frac{10:text{V}-1.66:text{V}}{50:mutext{A}+5:mutext{A}}=151.6:text{k}Omega$. Use the nearby 5% value of $R_1=150:text{k}Omega$.
- To get the gain, I need the total AC emitter resistance to be $frac{R_text{C}=9.1:text{k}Omega}{A_V=10}-frac{V_T=26:text{mV}}{I_{text{C}_text{Q}}=500:mutext{A}}approx 858:Omega$. However, as you will soon see below, there is already a $2:text{k}Omega$ emitter resistor for the DC operating point computed in (8) above. So I need a new AC resistor value of $R_{text{E}_2}=frac{2:text{k}Omega,cdot, 858:Omega}{2:text{k}Omega-858:Omega}approx 1503:Omega$. I'll use the nearby 5% value of $R_{text{E}_2}=1.5:text{k}Omega$.
So here is the resulting design using standard resistor values:
simulate this circuit – Schematic created using CircuitLab
The above should take a maximum of a $500:text{mV}_text{PP}$ input signal and generate a maximum $5:text{V}_text{PP}$ output signal.
Feel free to ask questions, now. But hopefully that provides an approach to similar design questions.
To make the design still more bullet-proof to BJT variations, reduce the allowable maximum collector swing to $4:text{V}$ (or even less) and follow step (6) to set $V_{text{C}_text{Q}}=6:text{V}$ (or even a little bit higher, perhaps to $V_{text{C}_text{Q}}=6.3:text{V}$) and then recalculate the rest from there.
There's an issue with the design. It probably needs something to reduce its gain at higher frequencies. (Given the above-designed collector resistor, a $470:text{pF}$ capacitor across $R_c$ might be added, for example.) But I'm not going to address that issue any further, here.
answered Jan 25 at 14:04
jonkjonk
33.9k12874
$endgroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. That's another serious problem.
I'll discuss an approach and a schematic (below) that will show you how to approach a more proper design for a BJT stage like this.
Looks like you want a gain of 13, just quickly glancing. Obviously, you can tolerate an output impedance of $13:text{k}Omega$, too. I'll choose a different gain and a different output impedance, but not too far away.
Let's say the voltage gain is to be $A_V=10$ and I'll keep your existing power supply rail of $V_text{CC}=10:text{V}$. Here's an approach. (There are many such, not just one. But I'm not going to go through more than one for you. You can pick up others when other folks tell you about them.)
- The maximum voltage gain is about 40 times the quiescent collector current (in millamps.) With $A_V=10$, this means $I_{text{C}_text{Q}}gt 250:mutext{A}$. I'd like twice that much, if possible. So let's set $I_{text{C}_text{Q}}=500:mutext{A}$.
- Given $I_{text{C}_text{Q}}=500:mutext{A}$ and typical small-signal BJTs, it is reasonable to conclude that the quiescent base-emitter voltage is about $V_{text{BE}_text{Q}}approx 660:text{mV}$.
- I like to reserve about $2:text{V}$ for the minimum $V_text{CE}$ of the BJT, in order to keep it well away from saturation, to help deal with BJT variations, and to slightly reduce the impact of the Early Effect.
- I like to reserve at least $1:text{V}$ for the quiescent emitter voltage for a variety of reasons, but importantly because I would like to place temperature and part variation issues under management.
- With $V_text{CC}=10:text{V}$ and subtracting the above two margins I just reserved, this means there is about $7:text{V}$ left over for the collector swing. But I also want to leave about $2:text{V}$ margin at the top end of the collector swing (limiting distortion due to gain variation and mitigating Early Effect.) So I don't want the collector to move any higher than $8:text{V}$. So this leaves only $5:text{V}$ for the collector swing (max.)
- Therefore, the quiescent collector voltage will be $V_{text{C}_text{Q}}=10:text{V}-2:text{V}-frac{5:text{V}}{2}=5.5:text{V}=1:text{V}+2:text{V}+frac{5:text{V}}{2}$. In short, $V_{text{C}_text{Q}}=5.5:text{V}$.
- From (1) and (6), I can compute a collector resistor of $R_{text{C}}=frac{10:text{V}-5.5:text{V}}{500:mutext{A}}=9:text{k}Omega$. Set this to the nearby 5% precision value of $R_{text{C}}=9.1:text{k}Omega$.
- From (1) and (4), I can compute a DC emitter resistor of $R_{text{E}_1}=frac{1:text{V}}{500:mutext{A}}=2:text{k}Omega$. That's a standard 5% value, so keep it.
- From (2) and (4), I know that the quiescent DC base voltage should be $V_{text{B}_text{Q}}=1:text{V}+660:text{mV}=1.66:text{V}$.
- To be conservative, I'll assume that the base current of the BJT will be no more than about $I_{text{B}_text{Q}}=frac{I_{text{C}_text{Q}}=500:mutext{A}}{beta=100}=5:mutext{A}$.
- To make a "stiff" resistor divider (in the sense that it is relatively unaffected by variations in the required base current due to signal variations), I know that the current through the two base divider resistors should be about $frac1{10}$th the quiescent collector current [or 10 times the current calculated in (10) above.] So this means about $50:mutext{A}$ for the base's resistor divider pair; used in steps (12) and (13) next.
- The divider resistor, from base to ground, is then $R_2=frac{1.66:text{V}}{50:mutext{A}}=33.2:text{k}Omega$. Use the nearby 5% value of $R_2=33:text{k}Omega$.
- The divider resistor, from base to the supply rail, is then $R_1=frac{10:text{V}-1.66:text{V}}{50:mutext{A}+5:mutext{A}}=151.6:text{k}Omega$. Use the nearby 5% value of $R_1=150:text{k}Omega$.
- To get the gain, I need the total AC emitter resistance to be $frac{R_text{C}=9.1:text{k}Omega}{A_V=10}-frac{V_T=26:text{mV}}{I_{text{C}_text{Q}}=500:mutext{A}}approx 858:Omega$. However, as you will soon see below, there is already a $2:text{k}Omega$ emitter resistor for the DC operating point computed in (8) above. So I need a new AC resistor value of $R_{text{E}_2}=frac{2:text{k}Omega,cdot, 858:Omega}{2:text{k}Omega-858:Omega}approx 1503:Omega$. I'll use the nearby 5% value of $R_{text{E}_2}=1.5:text{k}Omega$.
So here is the resulting design using standard resistor values:
simulate this circuit – Schematic created using CircuitLab
The above should take a maximum of a $500:text{mV}_text{PP}$ input signal and generate a maximum $5:text{V}_text{PP}$ output signal.
Feel free to ask questions, now. But hopefully that provides an approach to similar design questions.
To make the design still more bullet-proof to BJT variations, reduce the allowable maximum collector swing to $4:text{V}$ (or even less) and follow step (6) to set $V_{text{C}_text{Q}}=6:text{V}$ (or even a little bit higher, perhaps to $V_{text{C}_text{Q}}=6.3:text{V}$) and then recalculate the rest from there.
There's an issue with the design. It probably needs something to reduce its gain at higher frequencies. (Given the above-designed collector resistor, a $470:text{pF}$ capacitor across $R_c$ might be added, for example.) But I'm not going to address that issue any further, here.
answered Jan 25 at 14:04
jonkjonk
33.9k12874
edited Jan 26 at 6:02
answered Jan 25 at 14:04
jonkjonk
33.9k12874
answered Jan 25 at 14:04
jonkjonk
33.9k12874
answered Jan 25 at 14:04
jonkjonk
33.9k12874
33.9k12874
3
$begingroup$
As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
$begingroup$
@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
$begingroup$
Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
$endgroup$
– Ashish Jha
Jan 27 at 7:13
$begingroup$
@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
$endgroup$
– jonk
Jan 27 at 8:26
$begingroup$
@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
add a comment |
3
$begingroup$
As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
$begingroup$
@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
$begingroup$
Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
$endgroup$
– Ashish Jha
Jan 27 at 7:13
$begingroup$
@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
$endgroup$
– jonk
Jan 27 at 8:26
$begingroup$
@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
3
3
$begingroup$
As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
$begingroup$
As always, @jonk, your answer goes above and beyond. Always a pleasure to read, thanks!
$endgroup$
– Colin
Jan 25 at 14:26
$begingroup$
@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
$begingroup$
@Colin Thanks! I appreciate the kind words, very much. :)
$endgroup$
– jonk
Jan 25 at 14:28
$begingroup$
Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
$endgroup$
– Ashish Jha
Jan 27 at 7:13
$begingroup$
Thank you sir for detailed reply. I understood your calculations and approach.Sir, Step 1,11 and 14 is little unclear to me. In step 1 how you decided that Vgain should be at least 40 times higher than Ic. Is this a rule of thumb?In step 11 ,Why current through voltage divider should be 10 times more? In step 14 i didn't get AC emmiter resistance calculation. And, sir with this circuit when i calculated beta(Ic/Ib) is coming 86 instead of 100 in PSpice(in Circuit lab its coming 100). If bjt is in active region beta should not vary, right? or there is some another reason?
$endgroup$
– Ashish Jha
Jan 27 at 7:13
$begingroup$
@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
$endgroup$
– jonk
Jan 27 at 8:26
$begingroup$
@AshishJha It's easy to derive (1). Look for the terms $r_e$ and $r_pi$, for example. But you've selected an answer and I've written an overly complete answer already.
$endgroup$
– jonk
Jan 27 at 8:26
$begingroup$
@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
$begingroup$
@jonk , sir I am searching over the net. It would be helpful if you can send some relative documents or link. my mail id- ashish.wityliti@gmail.com
$endgroup$
– Ashish Jha
Jan 29 at 5:15
add a comment |
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2
$begingroup$
In active mode operation, the BJT collector will be 180 degrees out of phase with the base. But when the BJT is saturated, its collector will tend to follow the base when the emitter resistor is present. So the interpretation here is that for part of the cycle your BJT is saturated and for the rest of it you have way too much gain for the signal, winding up with an almost-square wave as the output result. Of course, as already pointed out your base bias resistors are being over-ridden by an ideal voltage source at the base. Another problem.
$endgroup$
– jonk
Jan 25 at 13:16