Op-amp's input bias current
1
Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?
op-amp
edited 1 hour ago
Transistor
80.3k778173
asked 2 hours ago
Leo
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
1
Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?
op-amp
edited 1 hour ago
Transistor
80.3k778173
asked 2 hours ago
Leo
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
– Bimpelrekkie
2 hours ago
2
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
– Dave Tweed♦
2 hours ago
3
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
– Bimpelrekkie
2 hours ago
1
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
– WhatRoughBeast
1 hour ago
add a comment |
1
1
1
1
Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?
op-amp
edited 1 hour ago
Transistor
80.3k778173
asked 2 hours ago
Leo
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Could you please tell me how do you compute the impedance seen by the inverting input? The result is 100k//1M but according to me the voltage between the 100k and the 1M is not the same. So the resistors are not in parallel. This result is true in small signal analysis. But the input bias current is present at DC. Is there a way to resolve this problem with the thevenin's theorem ?
op-amp
op-amp
edited 1 hour ago
Transistor
80.3k778173
asked 2 hours ago
Leo
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Transistor
80.3k778173
asked 2 hours ago
Leo
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
Transistor
80.3k778173
edited 1 hour ago
Transistor
80.3k778173
edited 1 hour ago
Transistor
80.3k778173
80.3k778173
asked 2 hours ago
Leo
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Leo
61
asked 2 hours ago
Leo
61
61
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Leo is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
– Bimpelrekkie
2 hours ago
2
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
– Dave Tweed♦
2 hours ago
3
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
– Bimpelrekkie
2 hours ago
1
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
– WhatRoughBeast
1 hour ago
add a comment |
1
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
– Bimpelrekkie
2 hours ago
2
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
– Dave Tweed♦
2 hours ago
3
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
– Bimpelrekkie
2 hours ago
1
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
– WhatRoughBeast
1 hour ago
1
1
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
– Bimpelrekkie
2 hours ago
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
– Bimpelrekkie
2 hours ago
2
2
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
– Dave Tweed♦
2 hours ago
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
– Dave Tweed♦
2 hours ago
3
3
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
– Bimpelrekkie
2 hours ago
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
– Bimpelrekkie
2 hours ago
1
1
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
– WhatRoughBeast
1 hour ago
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
– WhatRoughBeast
1 hour ago
add a comment |
1 Answer
1
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oldest
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3
It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.
answered 2 hours ago
Dave Tweed♦
117k9145256
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
|
show 2 more comments
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It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.
answered 2 hours ago
Dave Tweed♦
117k9145256
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
|
show 2 more comments
3
It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.
answered 2 hours ago
Dave Tweed♦
117k9145256
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
|
show 2 more comments
3
3
3
It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.
answered 2 hours ago
Dave Tweed♦
117k9145256
It doesn't matter whether the two voltages are the same. Regardless of what they are, the two voltage sources and their resistors can be replaced with a single source and a single resistor (this is called the "Thévenin equivalent"). The Thévenin resistance is always simply the parallel combination of the original two resistors.
answered 2 hours ago
Dave Tweed♦
117k9145256
answered 2 hours ago
Dave Tweed♦
117k9145256
answered 2 hours ago
Dave Tweed♦
117k9145256
answered 2 hours ago
Dave Tweed♦
117k9145256
117k9145256
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
|
show 2 more comments
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
Although technically this does not answer the question, since bias currents are not necessarily perfectly proportional to input impedance. For instance, an op amp operating linearly with the inverting input within microvolts of ground can have considerable bias currents, at least for early models. The question is actually based on a mistaken assumption and is not answerable as stated.
– WhatRoughBeast
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: The question is, and I quote, "how do you compute the impedance seen by the inverting input?" and "Is there a way to resolve this problem with the thevenin's theorem?". This is precisely an answer to that. And the underlying problem is not about the magnitude of the bias current, but rather about how to mitigate its effect by balancing the source impedances. Again the assumptions are (yeah, I know) that the input structures are matched and that when the opamp is operating linearly, the input voltages are equal, and therefore, so are their bias currents.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
... The key concept here is that even if the bias current varies with voltage (or anything else), the matched source impedances make sure that this does not contribute to any offset at the output.
– Dave Tweed♦
1 hour ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
@WhatRoughBeast: Finally, there's nothing special about operating "within microvolts of ground", especially for "early models", which pretty much universally required bipolar power supplies.
– Dave Tweed♦
54 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
While I understand your point, the configuration implicitly assumes that bias current does not vary much in operation. It is intended, after all, to compensate for (essentially) DC offset voltages. While these vary (especially with temperature), any signal-dependent component is not referred to as offset voltage at all, but rather as a consequence of input resistance.
– WhatRoughBeast
51 mins ago
|
show 2 more comments
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1
The result is 100k//1M No it is not unless there is a voltage source to ground connected at the left of the 100k resistor. As drawn there is "nothing" so the 100 k resistor does nothing. How does an opamp's output behave? Does it have a low or high output impedance? You should first think about how this circuit normally operates, what will be connected to it, how is it used?
– Bimpelrekkie
2 hours ago
2
@Bimpelrekkie: In such circuits, it is generally understood that unknown inputs are voltage sources, unless otherwise specified.
– Dave Tweed♦
2 hours ago
3
@DaveTweed Sure, but I think that is a very bad habit! Nothing is specified at the input here. How much trouble is it to put a "Vin" there and then it would be clear. Very often I see assumptions being made and assumptions are the source of many errors and misunderstandings.
– Bimpelrekkie
2 hours ago
1
@Bimpelrekkie - I'm sure you are too polite to say it, but there is an old saying which goes, "When you 'assume', you make an 'ass' out of 'u' and 'me',"
– WhatRoughBeast
1 hour ago