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How to calculate the min. feedback resistor value in op-amps.
How to calculate the min. feedback resistor value in op-amps.
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Old 1st December 2019, 12:12 PM   #1
Piotr is offline Piotr  Poland
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Default How to calculate the min. feedback resistor value in op-amps.

Hi!
This topic bothers me lately. I would like to use as small resistors as possible in the feedback, but I do not want to overdo it. As a presumption, do not exceed a certain power or current per resistor and it results from the maximum op-amp output voltage that I have?? How to calculate it? I mean this current/power per resistor for a specific op-amp.
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Old 1st December 2019, 01:09 PM   #2
Mark Tillotson is offline Mark Tillotson
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Each opamp has a minimum load impedance it can comfortably drive without increase distortion - usually this is clear from the datasheet graphs where they often show curves for different loadings, the lowest load resistance shown is likely to be the limit for best performance (manufacturers like to show off a chip at its best).


Any random opamp is likely to be able to drive 2k without issue. Some are happy down to 1k or 600 ohms, but not all.


The output swing will be diminished with low impedance loads (even if the distortion doesn't increase). What is touted as a rail-to-rail opamp will not be when loaded to 600 ohms - in fact often to get reasonably-close-to-the-rails output swing you'd limit the output load to 10k minimum impedance.
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Old 1st December 2019, 01:25 PM   #3
FauxFrench is online now FauxFrench  France
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I can read the question in more ways. Mark did one good interpretation.
One way to read the question is: what is the maximum power loss in the feed-back resistor of an OP-AMP, going from the output of the OP-AMP to the inverting input?

We call the resistor going to ground (non-inverting OP-AMP coupling) or to the input (inverting OP-AMP coupling) R1. We call the feed-back resistor R2. The signal across the feedback resistor is then the amplitude of the input signal multiplied with R2/R1. The power loss in the feedback resistor is then found by the RMS value at the input multiplied with R2/R1, then squaring the result and finally dividing the squared result with the value of the feedback resistor. For the max. power loss in the feedback resistor, use the (RMS value) of the max. input amplitude.

Last edited by FauxFrench; 1st December 2019 at 01:28 PM.
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Old 1st December 2019, 01:48 PM   #4
jackinnj is offline jackinnj  United States
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How to calculate the min. feedback resistor value in op-amps.
Bob Pease wrote an application note on the topic: http://www.ti.com/lit/an/snoa471b/snoa471b.pdf

Application note 1485
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Old 1st December 2019, 04:04 PM   #5
Piotr is offline Piotr  Poland
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Thanks! I did not read the TI application yet. It looks like a good piece of thing.


@Mark Tillotson


What about power amplifier chips like LM3886. It can be loaded with 4ohms, or lets say 6 amps, but datasheet still recommending R2=20k and R1=1k. Why resistors that big? You can set 23V RMS at the output for the 8ohm load, this is about 25miliwatts at 20k resistor, only 25miliwatts. And 1,6mA at the peak of the current swing.

Last edited by Piotr; 1st December 2019 at 04:06 PM.
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Old 1st December 2019, 04:51 PM   #6
FauxFrench is online now FauxFrench  France
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Quote:
Originally Posted by Piotr View Post
Thanks! I did not read the TI application yet. It looks like a good piece of thing.


@Mark Tillotson


What about power amplifier chips like LM3886. It can be loaded with 4ohms, or lets say 6 amps, but datasheet still recommending R2=20k and R1=1k. Why resistors that big? You can set 23V RMS at the output for the 8ohm load, this is about 25miliwatts at 20k resistor, only 25miliwatts. And 1,6mA at the peak of the current swing.

OK, you meant resistance value.
Why would it be an advantage to reduce the feedback voltage resistor divider below what would make a difference for the chip? You would only require larger power rating of the resistors. If you lower the resistor values and use a capacitor in the line to ground (non-inverting OP-AMP coupling) to reduce gain at very low frequencies and DC, the capacitor value will need to be increased for the same cut-off frequency.

Last edited by FauxFrench; 1st December 2019 at 04:53 PM.
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Old 1st December 2019, 05:18 PM   #7
Piotr is offline Piotr  Poland
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Quote:
Originally Posted by FauxFrench View Post
OK, you meant resistance value.
Why would it be an advantage to reduce the feedback voltage resistor divider below what would make a difference for the chip? You would only require larger power rating of the resistors. If you lower the resistor values and use a capacitor in the line to ground (non-inverting OP-AMP coupling) to reduce gain at very low frequencies and DC, the capacitor value will need to be increased for the same cut-off frequency.

I never use this HPF capacitor under R1. I only use the coupling serial capacitor on the input. Lower resistor values mean less noise. The AD797 datasheet boasts low noise thanks to the use of small resistors. They also boast that AD797 works well with small resistors (it has 50mA output current capacity)



https://www.analog.com/media/en/tech...eets/AD797.pdf
Page 13, Table 6
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Old 1st December 2019, 05:26 PM   #8
dmills is offline dmills  United Kingdom
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In power amps, lower feedback resistors help to lower the noise figure, as well as reducing the effect of non linearity in the input impedance of the LTP.

You can take this way too far of course, and in a poorly thought out power stage it is easy to go past the point at which the resistors are no longer a significant contributor to the system noise, but it is surprising how often the feedback network dominates the noise performance of a power stage. Much more of an issue when you have a high power amp that spends 99% of its time 30 to 60dB below peak output of course.

For most audio opamps (excluding the low current stuff like {06/07/08}x and the cmos low supply rail stuff, 1k or so is about as low as I like to go unless I am doing really noise critical things in which case sums are indicated (Remember, Voltage noise is NOT the only noise). In a big power stage going that low can cause an annoying amount of dissipation, so you sometimes need to go higher to avoid needing a 10W resistor in the feedback path (Which is just silly).

Last edited by dmills; 1st December 2019 at 05:31 PM.
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Old 1st December 2019, 05:28 PM   #9
FauxFrench is online now FauxFrench  France
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You mention the LM3886 yourself. I am convinced that you can replace the 20K with 2K and the 1K with 100 Ohm. The capacitor is no problem, then. I doubt you can really hear a difference in noise.
Your output offset may increase as the DC impedance level difference seen between the two inputs increases.
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Old 1st December 2019, 05:43 PM   #10
Piotr is offline Piotr  Poland
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Quote:
Originally Posted by dmills View Post
In power amps, lower feedback resistors help to lower the noise figure, as well as reducing the effect of non linearity in the input impedance of the LTP.

You can take this way too far of course, and in a poorly thought out power stage it is easy to go past the point at which the resistors are no longer a significant contributor to the system noise, but it is surprising how often the feedback network dominates the noise performance of a power stage. Much more of an issue when you have a high power amp that spends 99% of its time 30 to 60dB below peak output of course.

For most audio opamps (excluding the low current stuff like {06/07/08}x and the cmos low supply rail stuff, 1k or so is about as low as I like to go unless I am doing really noise critical things in which case sums are indicated (Remember, Voltage noise is NOT the only noise). In a big power stage going that low can cause an annoying amount of dissipation, so you sometimes need to go higher to avoid needing a 10W resistor in the feedback path (Which is just silly).

In the signal, if I have to use a 500R resistor, I do not care, but above 1000R I am already starting to think if it can't be made smaller. As you say, we often listen to integrated or headphone amplifiers below 30 or 40dB from full scale. Then the noise gives its character to the sound, in a word it belongs to audible distortions.
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