3
\$\begingroup\$

I'm using a non inverting opamp circuit with a LM741 opamp to make an output signal have twice the amplitude as the input signal.

The problem I am having is the output is twice the amplitude upto around 7kHz and after this point it begins to decrease with increasing frequency, anything I can do to stop this happening or a counter measure to fix it?

\$\endgroup\$
  • \$\begingroup\$ Do you have a schematic? \$\endgroup\$ – Gustavo Litovsky Oct 18 '13 at 14:13
3
\$\begingroup\$

Even the 741 shouldn't be this bad but you could be running into slew rate limit problems. Firstly, for small signals, you should be able to get a gain of about 40dB. Here's the open loop gain response: -

enter image description here

Without any feedback at all, the 741 should still provide amplification of about 40dB at 7kHz. Shown on the graph above are two red lines that correspond to 7kHz and about 40dB. This is the small signal amplification that is possible with the 741. The graph is taken from the uA741 (because I couldn't locate an adequate picture of the LM741 so you need to do a bit of research to verify it is about the same. I believe it will be but please do check.

If you are trying to amplify with a gain of 2 (6dB) I'd expect a flat response up to about 300kHz (3dB point). You can also see this from the above graph by running your eye along from where 6dB is on the vertical scale and seeing where this crosses the thick black line and reading the frequency value on the X axis.

Slew rate limiting - this could be your problem - try reducing the signal amplitude to (say) 1Vp-p - does this improve things? Slew rate limits for the 741 are about 0.5V per microsecond and this means that if you are trying to produce an output that is 10Vp-p, the output rise time will be no faster than 20 microseconds. Correspondingly the fall time will also be 20 microseconds with the resultant approximate maximum frequency of \$\dfrac{1}{40\times 10^6}\$ = 25kHz. Again, this doesn't sound like the problem but please do check.

Finally, parasitic capacitances may be affecting your choice of feedback resistor. If Rf is in the order of 1M ohm and there is 10pF effectively across the 1M, the 3dB point of the op-amp circuit will only be around 15kHz.

In conclusion, I would start to suspect slew rate as your problem and maybe you are using resistors that are too high in value. Other than that you should get a reasonably flat frequency response to over 100kHz for a gain of 2.

\$\endgroup\$
  • \$\begingroup\$ Thanks for your reply. I should have noted that the gain is 28.29dB since the original signal is attenuated in another part of the project before it is fed to the amplifying circuit. From your comment I believe that 28.29dB still shouldn't cause problems. I tried reducing the resistors by an order of 1000 which made no difference. The input is also approximately 300mV and this amplifier is to increase it back to the original amplitude(or two times the amplitude if it doesn't cause issues which would be a gain of 35dB) \$\endgroup\$ – user27365 Oct 18 '13 at 22:59
  • \$\begingroup\$ there ya go... 35dbB gain and you are starting to attenuate at 7KHz a bit. This sounds like the problem. There are far better op amps available. \$\endgroup\$ – Andy aka Oct 18 '13 at 23:08
  • \$\begingroup\$ Unfortunately this is part of a university project and I can't use another opamp. Is my best option just to use a few non inverting amplifier circuits in series to get to the desired gain? \$\endgroup\$ – user27365 Oct 18 '13 at 23:16
  • \$\begingroup\$ Two op-amp gain stages in series would work;each with a smaller gain giving you the 35dB you need. \$\endgroup\$ – Andy aka Oct 18 '13 at 23:23
  • \$\begingroup\$ I'm currently using two non inverting circuits in series with the same amplification. When only the first circuit is connected the output is constant up to the value of 20kHz I need however when I connect the second the output of both combined is half the value it should be, any idea why? \$\endgroup\$ – user27365 Oct 21 '13 at 7:29

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.