In theory, for an op-amp in inverting design, the voltage gain is \$\dfrac{R_{f}}{R_{in}}\$.

For example: for a \$gain=10\$ we can use \$R_{f}=100\ \mathrm{k\Omega}\$ and \$R_{in} =10\ \mathrm{k\Omega}\$. The same gain can be obtained from \$R_{f}=1\ \mathrm{k\Omega}\$ and \$R_{in}=100\ \mathrm{\Omega}\$. What is the difference and which value is best suited?

A config like this:

enter image description here

| improve this question | | | | |
  • 1
    \$\begingroup\$ The AD811 and similar op-amps (Northon mode) operate internally very differently than the classical Voltage-input, Voltage-output op-amps that are taught in schools. While this is a valid question you really should constrain yourself to one type of OP-amp or the question won't be able to be answered in such a short format. referencing the AD811 here is unduly complicating. \$\endgroup\$ – placeholder Oct 9 '13 at 14:32
  • \$\begingroup\$ @rawbrawb Thanks! good point. The image edited. Just a question: Is Norton op-amp the same as Current Feedback OpAmp (CFO)? \$\endgroup\$ – Aug Oct 9 '13 at 17:02
  • 1
    \$\begingroup\$ Basically yes, although the AD811 also has one of the inputs having very low impedance. \$\endgroup\$ – placeholder Oct 9 '13 at 20:00
  • \$\begingroup\$ thanks for explaining. one question though - can the feedback resistor be 100 ohm ,because (as an example) in the datasheet for lt1028 feedback resistor of 100ohm can cause distortions is it valid for other opamps to ? whats the constant of it and how can we find minimum Rf ? \$\endgroup\$ – xslavic Dec 15 '13 at 11:40

If you want your op-amp to perform better at high frequencies you'll use lower value resistors to set the gain. Leakage capacitance around the feedback area might be in the order of 1pF due to circuit tracks and pads for components and this has an effect when resistor values are high.

If your feedback resistor were 100k ohm and your leakage capacitance were 1 pF, you'd find that at a frequency of: -

\$\dfrac{1}{2\pi RC} = \dfrac{1}{2\pi\times 100,000\times 1\times 10^{-12}} = 1.59MHz\$

The gain would be 3dB down on the dc gain i.e. if your dc gain is 10, at 1.59MHz the gain would be 7.07. If you need a flat response up to 32MHz then the biggest feedback resistor you can use is 5k ohm.

Op-amp data sheets are the best place to look to see what they recommend.

Taking the resistor values lower is fine but you will approach a point where the feedback resistor is starting to load the output circuits of your op-amp and you may get reduced amplitude swings and/or distortion.

But the bigger problem would be on the input resistor. To maintain a gain of ten with a feedback resistor of (say) 100 ohms means the input resistor is 10 ohms and this input resistor is the input impedance of your circuit. This would be seen by many circuits or signals as "too low" and can cause the inputted signals to distort or reduce in amplitude.

Typically you wouldn't go below 50 ohms for \$R_{IN}\$ and this means your feedback resistor is 500 ohms.

| improve this answer | | | | |
  • \$\begingroup\$ Thanks Andy! You answered some other questions of mine in other threads and I really like the way you explain details. Now I can understand why my op-amp datasheet (AD811) recommends RF=500 ohms and RG=50 ohms.Just another question: Can we say it also depends on the op-amp type (Current Feedback or Voltage Feedback Op-Amps?). As I get from your discussion, in Current Feedback Op-Amps that rely on input current instead of input voltage, I should concern about input impedance, and if so and I want a small RG, I may think about moving from AD811 which is a CFO to a VFO like AD826. \$\endgroup\$ – Aug Oct 9 '13 at 11:23
  • 1
    \$\begingroup\$ @Aug. Glad to help - please link to the question you refer to. Current feedback devices have less leakage capacitance at there input pins to ground and this means they perform better at high frequencies but I'd need a circuit really to see what you're proposing. \$\endgroup\$ – Andy aka Oct 9 '13 at 11:29
  • \$\begingroup\$ It is just a simple inverting op amp with input signal from a DDS (AD9850) with signal amplitude around 1 volt. RF connected between Out ant Vin- and RG between input signal and Vin- . Vin+ pin connected to the ground ( a very classic inverting config.) \$\endgroup\$ – Aug Oct 9 '13 at 11:34
  • \$\begingroup\$ A photo of the schematic added \$\endgroup\$ – Aug Oct 9 '13 at 11:37
  • 2
    \$\begingroup\$ The AD811 is fine but watch out at higher gains - also figure 19 tells you where to pitch the value of Rf to get best high frequency performance. Also fig 30 shows how the gain can droop at lower supply voltages. \$\endgroup\$ – Andy aka Oct 9 '13 at 11:42

This is a good general question and the general answer is: it depends.

Specifically, since you have two resistor values to find but only one specification, the gain, all you can constrain is the resistor value ratio; you need another constraint to fix individual values (or a bound on the them).

The most likely specification is the input resistance which, for this configuration, fixes the value of \$R_{in}\$ (or sets a lower bound) and then the value of \$R_F \$ follows.

| improve this answer | | | | |
  • \$\begingroup\$ And what specifies best input resistance? \$\endgroup\$ – Aug Oct 8 '13 at 22:52
  • 1
    \$\begingroup\$ The best input resistance is the one that fits the application. For example, for single-ended line-level audio signal applications, we typically want input resistance of around 10k or greater. Conversely, for a low-level, low-noise applications, the input resistance might need to be much lower. \$\endgroup\$ – Alfred Centauri Oct 8 '13 at 23:18
  • 1
    \$\begingroup\$ Each resistor is also a noise source. The bigger the resitance, the bigger the noise. In high gain systems, this may become important \$\endgroup\$ – Scott Seidman Oct 9 '13 at 1:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.