Mixing
Help With Resistor Type? [closed]
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
asked Nov 20 '18 at 20:58
G.Medina
574
closed as off-topic by Dave Tweed♦ Nov 21 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
asked Nov 20 '18 at 20:58
G.Medina
574
closed as off-topic by Dave Tweed♦ Nov 21 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
Nov 20 '18 at 21:01
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
Nov 20 '18 at 22:07
add a comment |
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
asked Nov 20 '18 at 20:58
G.Medina
574
Im having a hard time finding the spects for this piece
I know is a metal film resistor, according to the colors its a 470? is this correct, does anybody know where can I purchase a ton of this?
resistors
resistors
asked Nov 20 '18 at 20:58
G.Medina
574
asked Nov 20 '18 at 20:58
G.Medina
574
asked Nov 20 '18 at 20:58
G.Medina
574
asked Nov 20 '18 at 20:58
G.Medina
574
asked Nov 20 '18 at 20:58
G.Medina
574
574
closed as off-topic by Dave Tweed♦ Nov 21 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Dave Tweed♦ Nov 21 '18 at 15:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "Questions seeking recommendations for specific products or places to purchase them are off-topic as they are rarely useful to others and quickly obsolete. Instead, describe your situation and the specific problem you're trying to solve." – Dave Tweed
If this question can be reworded to fit the rules in the help center, please edit the question.
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
Nov 20 '18 at 21:01
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
Nov 20 '18 at 22:07
add a comment |
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
Nov 20 '18 at 21:01
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
Nov 20 '18 at 22:07
8
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
Nov 20 '18 at 21:01
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
Nov 20 '18 at 21:01
7
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
Nov 20 '18 at 22:07
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
Nov 20 '18 at 22:07
add a comment |
2 Answers
2
active
oldest
votes
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
16
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
3
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
|
show 1 more comment
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered Nov 20 '18 at 21:04
Electron
979213
2
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
2
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
1
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
1
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
|
show 2 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
16
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
3
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
|
show 1 more comment
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
16
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
3
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
|
show 1 more comment
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
So, that's clearly an inductor (see the labeling "L1"). Probably a Bourns 78F470J.
The hard part really is
where can I purchase a ton of this
because, well, I don't know why you need a certain mass of this, but considering the weight according to mouser is 185 mg, and 1 ton = 1 Mg, you'll need 5,405,406 of these.
That's a real problem right there. Even the largest known inventory (according to octopart.com) doesn't come close to that number (more like 70,000-ish); combining all sources together, you might not even hit one tenth of that. Also, your purchase would probably instantly increase the price for wirewound axial inductors.
So, you probably have to live with the 6 weeks lead time of Bourns, at the very least. Good news is that you can probably just call them right now and place the order. At this quantity for this class of device, they will certainly have an open ear for your customer needs.
I really don't know the distributor markup on these, but it's realistic they'll demand no less than $0.03 apiece, so that your order would have a volume of around 162 k$.
A bit of interesting knowledge: The maximum specified DC current through this inductor is 205 mA. The energy stored in a magnetic field is
$$E=frac12 LI^2text,$$
leading to a total maximum energy stored in your ton of inductors of
begin{align}
E_{tot}&=Ncdotfrac12 LI^2\
&=leftlceilfrac{1,text{Mg}}{185,text{mg}} rightrceilcdot frac12 47,mathrmmutext{H},left(205,text{mA}right)^2\
&approx 5.4cdot 10^6,cdot, 2.4cdot10^{-6} ,text{kg},text{m}^2,text{s}^{−2},text{A}^{−2},cdot ,4.2cdot10^{-3},text{A}^2\
&=5.4cdot2.4cdot4.2cdot10^{-3},text{N·m}\
&approx 54 ,text{J.}
end{align}
That is, disappointingly, the energy that it takes to trigger your xenon camera flash (not your smartphone's "flash" LED, a proper flash) five times.
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
edited Nov 20 '18 at 22:46
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
answered Nov 20 '18 at 21:39
Marcus Müller
31.7k35794
31.7k35794
16
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
3
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
|
show 1 more comment
16
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
3
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
16
16
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
Again with the dry comedy!
– winny
Nov 20 '18 at 22:09
2
2
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
Thanks, it's a much more realistic answer now! Nice touch with the stored energy calculation, that's disappointingly small. Though I'd like to see the circuit board where all 5M of these are wired in parallel.
– Johnny
Nov 20 '18 at 22:55
3
3
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
I can't tell you what that'll look like, but I can give you a lower boundary for its weight ;)
– Marcus Müller
Nov 20 '18 at 22:59
2
2
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
@Johnny - Even better, the one where they're all wired in series.
– WhatRoughBeast
Nov 21 '18 at 1:12
1
1
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
+1 Nice. Probably can get a metric ton of these in a few weeks in China, for between 50 and 100K.
– Spehro Pefhany
Nov 21 '18 at 2:00
|
show 1 more comment
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered Nov 20 '18 at 21:04
Electron
979213
2
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
2
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
1
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
1
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
|
show 2 more comments
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered Nov 20 '18 at 21:04
Electron
979213
2
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
2
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
1
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
1
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
|
show 2 more comments
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered Nov 20 '18 at 21:04
Electron
979213
NO, it is an inductor -- you can see on PCB they define it as L1
the colour code is YELLOW VIOLET BROWN SILVER = 470uH +/- 10%
answered Nov 20 '18 at 21:04
Electron
979213
answered Nov 20 '18 at 21:04
Electron
979213
answered Nov 20 '18 at 21:04
Electron
979213
answered Nov 20 '18 at 21:04
Electron
979213
979213
2
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
2
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
1
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
1
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
|
show 2 more comments
2
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
2
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
1
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
1
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
2
2
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
I think you mean 47uH, right?
– Spehro Pefhany
Nov 20 '18 at 21:08
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
@SpehroPefhany: Hmm, do you think it black not brown?
– Rev1.0
Nov 20 '18 at 21:09
2
2
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
It looks black on my monitor but if it's brown then 470uH.
– Spehro Pefhany
Nov 20 '18 at 21:10
1
1
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
@SpehroPefhany, its brown actually
– Electron
Nov 20 '18 at 21:11
1
1
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
Definitely black. There are no brown components what so ever in that band. Open it up in an image analysis software and check.
– pipe
Nov 21 '18 at 10:56
|
show 2 more comments
8
Since the reference designator is L1, I suspect that it is an inductor, not a resistor.
– Peter Bennett
Nov 20 '18 at 21:01
7
+1 for a good picture, cropped, with a clear indication of the part in question. And (incidentally?) the L1 designator is visible too ;)
– Wouter van Ooijen
Nov 20 '18 at 22:07