0
\$\begingroup\$

I have an op-amp based current-to-voltage converter, where the input current is summed from a number of LM13700 OTA's. The signal is audio, so the bandwidth of interest is up to \$20\mathrm{kHz}\$. The output gain I want is quite small, which naturally leads to a small feedback resistor, in this case \$1.5\mathrm k \Omega\$. Due to other constraints, I'd like to use the rather modest TL074, but the general rule of thumb seems to be that the load on the '07x should be at least \$3.3 \mathrm k \Omega \$ to avoid undue distortion.

My question is: is it really so that a low gain (i.e. small resistor) transimpedance amplifier distorts worse than a high gain one? Intuitively I'd expect that a small feedback resistor means less work for the op-amp, as the output voltage swing becomes smaller...

The circuit:

schematic

simulate this circuit – Schematic created using CircuitLab

Circuit usage details: the input current is very rarely the full \$5\mathrm{mA}\$, the typical case is more like \$0.5\mathrm{mA}\$. When the input is hitting full \$5\mathrm{mA}\$, the device is being intentionally driven to distortion (this is an effect-type device, not HiFi), so minor details matter less then anyway. However the normal case should not have obvious distortion.

\$\endgroup\$
1
  • \$\begingroup\$ A low feedback resistor means more work for the op amp, as it means higher output current. \$\endgroup\$
    – Hearth
    Commented Dec 8, 2022 at 16:00

2 Answers 2

1
\$\begingroup\$

is it really so that a low gain (i.e. small resistor) transimpedance amplifier distorts worse than a high gain one? Intuitively I'd expect that a small feedback resistor means less work for the op-amp, as the output voltage swing becomes smaller...

Due to negative feedback, the op-amp is trying to maintain its inverting input at a virtual ground potential (0 volts) so, in effect the feedback resistor is from output to 0 volts. If everything else was equal it means that the output amplitude is smaller for a lower value feedback resistor but, by no means is it less-work for the op-amp AND, when trying to drive bigger output signals, the output waveform inevitably clips at a lower level.

\$\endgroup\$
1
  • \$\begingroup\$ This is indeed how I understood it, it just seemed so silly that getting less gain is more difficult that I had to check. \$\endgroup\$
    – Timo
    Commented Dec 9, 2022 at 6:44
0
\$\begingroup\$

This distortion you are talking about might be from pure loading of your amplifier output.

You have to consider that your amplifier not only feeds current to your load, but also to your feedback network.

From the amplifier output point of view, your feedback resistor is effectively in parallel to your load (assuming your load is grounded). Therefore, your amplifier will distort if you make the parallel combination of your feedback and load resistor be so low that your output current sinking/sourcing capability cannot handle it.

The current through your feedback resistor is obviously dictated by your input signal. Therefore, that already demands maximum 5mA from the output of the amplifier. Looking into the TL074 datasheet, it says that if your load is between 2k-10k, then you can up to 10V, which leaves with a max. output current of 5mA (supply is +/-15V). If your load is higher than 10k then you can get up to 12V. This is probably due to output stage architecture.

In conclusion, consider also your load impedance and compute how much current you need for the voltage amplitude at the output you are looking for.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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