Reason for small-valued feedback resistors in low noise Op Amp
$begingroup$
I am wondering why the feedback resistor values specified in the datasheet of the AD797 are so small. My understanding is that the low feedback resistances keep the noise small, but isn't it also not ideal to have large currents flowing through the feedback network? My understanding was that feedback resistors should be >1k.
Here is a link to the datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/ad797.pdf
And a picture of an example application:
26.1 ohms seems like far too small a value for feedback resistors.
op-amp feedback
$endgroup$
add a comment |
$begingroup$
I am wondering why the feedback resistor values specified in the datasheet of the AD797 are so small. My understanding is that the low feedback resistances keep the noise small, but isn't it also not ideal to have large currents flowing through the feedback network? My understanding was that feedback resistors should be >1k.
Here is a link to the datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/ad797.pdf
And a picture of an example application:
26.1 ohms seems like far too small a value for feedback resistors.
op-amp feedback
$endgroup$
add a comment |
$begingroup$
I am wondering why the feedback resistor values specified in the datasheet of the AD797 are so small. My understanding is that the low feedback resistances keep the noise small, but isn't it also not ideal to have large currents flowing through the feedback network? My understanding was that feedback resistors should be >1k.
Here is a link to the datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/ad797.pdf
And a picture of an example application:
26.1 ohms seems like far too small a value for feedback resistors.
op-amp feedback
$endgroup$
I am wondering why the feedback resistor values specified in the datasheet of the AD797 are so small. My understanding is that the low feedback resistances keep the noise small, but isn't it also not ideal to have large currents flowing through the feedback network? My understanding was that feedback resistors should be >1k.
Here is a link to the datasheet: https://www.analog.com/media/en/technical-documentation/data-sheets/ad797.pdf
And a picture of an example application:
26.1 ohms seems like far too small a value for feedback resistors.
op-amp feedback
op-amp feedback
asked 4 hours ago
SaundersSaunders
565
565
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4 Answers
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votes
$begingroup$
This op-amp boasts input noise of 0.9nV/rt.Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor over the human audio bandwidth. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper.
$endgroup$
add a comment |
$begingroup$
If you refer to figure 34, the total noise (assumed to be the white noise region for both en and in) is approximately 8x better at 10 ohms source resistance compared to 1K.
Remember that it’s not only the noise voltage and the Johnson-Nyquist noise from the feedback resistors but the input noise current multiplied by the resistance seen looking out from the inverting input.
The transistors at the input are run at very high currents so the voltage noise is quite low, but there is a cost in current noise.
They all add in quadrature, of course, since they are typically uncorrelated.
The resistors as shown are tolerable for smallish output swings and if you don’t care too much about accuracy of gain.
$endgroup$
add a comment |
$begingroup$
With honor to Walt Jung (of ADI, et al), low values of resistors can cause detectable THERMAL distortion. Drive the opamp with 20Hz and 2,000Hz; use a spectrum analyzer, and you'll see the 2,000Hz output with some 20Hz sidebands.
Which means what? Use physically larger resistors. Or experiment with resistors having different resistive element structures? The very thin metal-file-spiral-trimmed has a very fast timeconstant for heating/cooling, as heat is dumped into the ceramic/clay core.
And the opamp may need a BUFFER, to avoid generating thermal distortion as the silicon undergoes transient heating, as the UP transistors turn off and the DOWN transistors turn off.
==============================================
Separate topic: the opamp has only 70dB PSRR at 10KHz. 70dB is 3,000:1.
So what? Will a thermally-noise VDD-regulator be a problem? Some LDOs have internal equivalent Rnoise of 10,000,000 ohms (Often in poly-silicon servo-feedback resistors, and in diffpairs operating in subthreshold at 100nanoAmp
currents). This produces 1 microvolt per rootHertz random thermal noise on the "clean" VDD rail. Is this a risk?
If you have 60dB PSRR at 10KHz, that 1microVolts becomes 1 nanoVolt Referred to Input, which is a 3dB increase of the opamp's noise floor. And at 100KHz, the opamp has only 50dB PSRR (from a datasheet plot).
Summary: pay attention to the VDD rail random noise. And don't think about using switchRegs in these systems, unless you
---- use magnetic shielding
---- use electric-field shielding
---- pay attention to building "local batteries" for the opamp's two rails
---- design the Ground, with slits, etc to keep trash away from the opamp
$endgroup$
add a comment |
$begingroup$
A graph in the datasheet of the opamp shows it clipping with an output voltage swing of only plus and minus 2V into the 26.1 ohm resistors which might not be enough.
$endgroup$
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
1
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
add a comment |
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4 Answers
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active
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4 Answers
4
active
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$begingroup$
This op-amp boasts input noise of 0.9nV/rt.Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor over the human audio bandwidth. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper.
$endgroup$
add a comment |
$begingroup$
This op-amp boasts input noise of 0.9nV/rt.Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor over the human audio bandwidth. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper.
$endgroup$
add a comment |
$begingroup$
This op-amp boasts input noise of 0.9nV/rt.Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor over the human audio bandwidth. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper.
$endgroup$
This op-amp boasts input noise of 0.9nV/rt.Hz, which is roughly equal to the Johnson noise of a 50 Ω resistor over the human audio bandwidth. If you aren't putting resistors smaller than that around it, you're wasting some of this op-amp's performance, and probably should be buying something cheaper.
answered 1 hour ago
Warren YoungWarren Young
3,3771628
3,3771628
add a comment |
add a comment |
$begingroup$
If you refer to figure 34, the total noise (assumed to be the white noise region for both en and in) is approximately 8x better at 10 ohms source resistance compared to 1K.
Remember that it’s not only the noise voltage and the Johnson-Nyquist noise from the feedback resistors but the input noise current multiplied by the resistance seen looking out from the inverting input.
The transistors at the input are run at very high currents so the voltage noise is quite low, but there is a cost in current noise.
They all add in quadrature, of course, since they are typically uncorrelated.
The resistors as shown are tolerable for smallish output swings and if you don’t care too much about accuracy of gain.
$endgroup$
add a comment |
$begingroup$
If you refer to figure 34, the total noise (assumed to be the white noise region for both en and in) is approximately 8x better at 10 ohms source resistance compared to 1K.
Remember that it’s not only the noise voltage and the Johnson-Nyquist noise from the feedback resistors but the input noise current multiplied by the resistance seen looking out from the inverting input.
The transistors at the input are run at very high currents so the voltage noise is quite low, but there is a cost in current noise.
They all add in quadrature, of course, since they are typically uncorrelated.
The resistors as shown are tolerable for smallish output swings and if you don’t care too much about accuracy of gain.
$endgroup$
add a comment |
$begingroup$
If you refer to figure 34, the total noise (assumed to be the white noise region for both en and in) is approximately 8x better at 10 ohms source resistance compared to 1K.
Remember that it’s not only the noise voltage and the Johnson-Nyquist noise from the feedback resistors but the input noise current multiplied by the resistance seen looking out from the inverting input.
The transistors at the input are run at very high currents so the voltage noise is quite low, but there is a cost in current noise.
They all add in quadrature, of course, since they are typically uncorrelated.
The resistors as shown are tolerable for smallish output swings and if you don’t care too much about accuracy of gain.
$endgroup$
If you refer to figure 34, the total noise (assumed to be the white noise region for both en and in) is approximately 8x better at 10 ohms source resistance compared to 1K.
Remember that it’s not only the noise voltage and the Johnson-Nyquist noise from the feedback resistors but the input noise current multiplied by the resistance seen looking out from the inverting input.
The transistors at the input are run at very high currents so the voltage noise is quite low, but there is a cost in current noise.
They all add in quadrature, of course, since they are typically uncorrelated.
The resistors as shown are tolerable for smallish output swings and if you don’t care too much about accuracy of gain.
answered 2 hours ago
Spehro PefhanySpehro Pefhany
208k5159421
208k5159421
add a comment |
add a comment |
$begingroup$
With honor to Walt Jung (of ADI, et al), low values of resistors can cause detectable THERMAL distortion. Drive the opamp with 20Hz and 2,000Hz; use a spectrum analyzer, and you'll see the 2,000Hz output with some 20Hz sidebands.
Which means what? Use physically larger resistors. Or experiment with resistors having different resistive element structures? The very thin metal-file-spiral-trimmed has a very fast timeconstant for heating/cooling, as heat is dumped into the ceramic/clay core.
And the opamp may need a BUFFER, to avoid generating thermal distortion as the silicon undergoes transient heating, as the UP transistors turn off and the DOWN transistors turn off.
==============================================
Separate topic: the opamp has only 70dB PSRR at 10KHz. 70dB is 3,000:1.
So what? Will a thermally-noise VDD-regulator be a problem? Some LDOs have internal equivalent Rnoise of 10,000,000 ohms (Often in poly-silicon servo-feedback resistors, and in diffpairs operating in subthreshold at 100nanoAmp
currents). This produces 1 microvolt per rootHertz random thermal noise on the "clean" VDD rail. Is this a risk?
If you have 60dB PSRR at 10KHz, that 1microVolts becomes 1 nanoVolt Referred to Input, which is a 3dB increase of the opamp's noise floor. And at 100KHz, the opamp has only 50dB PSRR (from a datasheet plot).
Summary: pay attention to the VDD rail random noise. And don't think about using switchRegs in these systems, unless you
---- use magnetic shielding
---- use electric-field shielding
---- pay attention to building "local batteries" for the opamp's two rails
---- design the Ground, with slits, etc to keep trash away from the opamp
$endgroup$
add a comment |
$begingroup$
With honor to Walt Jung (of ADI, et al), low values of resistors can cause detectable THERMAL distortion. Drive the opamp with 20Hz and 2,000Hz; use a spectrum analyzer, and you'll see the 2,000Hz output with some 20Hz sidebands.
Which means what? Use physically larger resistors. Or experiment with resistors having different resistive element structures? The very thin metal-file-spiral-trimmed has a very fast timeconstant for heating/cooling, as heat is dumped into the ceramic/clay core.
And the opamp may need a BUFFER, to avoid generating thermal distortion as the silicon undergoes transient heating, as the UP transistors turn off and the DOWN transistors turn off.
==============================================
Separate topic: the opamp has only 70dB PSRR at 10KHz. 70dB is 3,000:1.
So what? Will a thermally-noise VDD-regulator be a problem? Some LDOs have internal equivalent Rnoise of 10,000,000 ohms (Often in poly-silicon servo-feedback resistors, and in diffpairs operating in subthreshold at 100nanoAmp
currents). This produces 1 microvolt per rootHertz random thermal noise on the "clean" VDD rail. Is this a risk?
If you have 60dB PSRR at 10KHz, that 1microVolts becomes 1 nanoVolt Referred to Input, which is a 3dB increase of the opamp's noise floor. And at 100KHz, the opamp has only 50dB PSRR (from a datasheet plot).
Summary: pay attention to the VDD rail random noise. And don't think about using switchRegs in these systems, unless you
---- use magnetic shielding
---- use electric-field shielding
---- pay attention to building "local batteries" for the opamp's two rails
---- design the Ground, with slits, etc to keep trash away from the opamp
$endgroup$
add a comment |
$begingroup$
With honor to Walt Jung (of ADI, et al), low values of resistors can cause detectable THERMAL distortion. Drive the opamp with 20Hz and 2,000Hz; use a spectrum analyzer, and you'll see the 2,000Hz output with some 20Hz sidebands.
Which means what? Use physically larger resistors. Or experiment with resistors having different resistive element structures? The very thin metal-file-spiral-trimmed has a very fast timeconstant for heating/cooling, as heat is dumped into the ceramic/clay core.
And the opamp may need a BUFFER, to avoid generating thermal distortion as the silicon undergoes transient heating, as the UP transistors turn off and the DOWN transistors turn off.
==============================================
Separate topic: the opamp has only 70dB PSRR at 10KHz. 70dB is 3,000:1.
So what? Will a thermally-noise VDD-regulator be a problem? Some LDOs have internal equivalent Rnoise of 10,000,000 ohms (Often in poly-silicon servo-feedback resistors, and in diffpairs operating in subthreshold at 100nanoAmp
currents). This produces 1 microvolt per rootHertz random thermal noise on the "clean" VDD rail. Is this a risk?
If you have 60dB PSRR at 10KHz, that 1microVolts becomes 1 nanoVolt Referred to Input, which is a 3dB increase of the opamp's noise floor. And at 100KHz, the opamp has only 50dB PSRR (from a datasheet plot).
Summary: pay attention to the VDD rail random noise. And don't think about using switchRegs in these systems, unless you
---- use magnetic shielding
---- use electric-field shielding
---- pay attention to building "local batteries" for the opamp's two rails
---- design the Ground, with slits, etc to keep trash away from the opamp
$endgroup$
With honor to Walt Jung (of ADI, et al), low values of resistors can cause detectable THERMAL distortion. Drive the opamp with 20Hz and 2,000Hz; use a spectrum analyzer, and you'll see the 2,000Hz output with some 20Hz sidebands.
Which means what? Use physically larger resistors. Or experiment with resistors having different resistive element structures? The very thin metal-file-spiral-trimmed has a very fast timeconstant for heating/cooling, as heat is dumped into the ceramic/clay core.
And the opamp may need a BUFFER, to avoid generating thermal distortion as the silicon undergoes transient heating, as the UP transistors turn off and the DOWN transistors turn off.
==============================================
Separate topic: the opamp has only 70dB PSRR at 10KHz. 70dB is 3,000:1.
So what? Will a thermally-noise VDD-regulator be a problem? Some LDOs have internal equivalent Rnoise of 10,000,000 ohms (Often in poly-silicon servo-feedback resistors, and in diffpairs operating in subthreshold at 100nanoAmp
currents). This produces 1 microvolt per rootHertz random thermal noise on the "clean" VDD rail. Is this a risk?
If you have 60dB PSRR at 10KHz, that 1microVolts becomes 1 nanoVolt Referred to Input, which is a 3dB increase of the opamp's noise floor. And at 100KHz, the opamp has only 50dB PSRR (from a datasheet plot).
Summary: pay attention to the VDD rail random noise. And don't think about using switchRegs in these systems, unless you
---- use magnetic shielding
---- use electric-field shielding
---- pay attention to building "local batteries" for the opamp's two rails
---- design the Ground, with slits, etc to keep trash away from the opamp
edited 21 mins ago
answered 41 mins ago
analogsystemsrfanalogsystemsrf
14.7k2718
14.7k2718
add a comment |
add a comment |
$begingroup$
A graph in the datasheet of the opamp shows it clipping with an output voltage swing of only plus and minus 2V into the 26.1 ohm resistors which might not be enough.
$endgroup$
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
1
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
add a comment |
$begingroup$
A graph in the datasheet of the opamp shows it clipping with an output voltage swing of only plus and minus 2V into the 26.1 ohm resistors which might not be enough.
$endgroup$
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
1
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
add a comment |
$begingroup$
A graph in the datasheet of the opamp shows it clipping with an output voltage swing of only plus and minus 2V into the 26.1 ohm resistors which might not be enough.
$endgroup$
A graph in the datasheet of the opamp shows it clipping with an output voltage swing of only plus and minus 2V into the 26.1 ohm resistors which might not be enough.
answered 3 hours ago
AudioguruAudioguru
42413
42413
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
1
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
add a comment |
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
1
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
$begingroup$
But the graph has it rated down to about 20 ohms, with a weak output as you indicated. Up to the designer to keep the load over 50 ohms if possible. At 200 ohm loads you get full output swing.
$endgroup$
– Sparky256
3 hours ago
1
1
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
$begingroup$
Thank you for making an intelligent comment. In the future, please use the comment section when you want to make an intelligent comment. I believe you have enough reputation to do so. We try to reserve the Answer section for answers to the question.
$endgroup$
– mkeith
2 hours ago
add a comment |
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