2
\$\begingroup\$

I'm trying to amplify a low signal with square pulses at 20 mV using op amps. According to simulation, rise and fall time of pulses increase too much after amplification, approximately 60 us. The component which will receive the signal accepts maximum 300 ns, the datasheet says so.

What kind of things can be done to reduce rise/fall time after amplification or to solve this problem? If I should choose another op amp, which is the best for this purpose?

I have another solution in my mind but I'm not sure. If I shift reference clock a little bit so that rising edges of clock don't meet the rising edges of signal(which are too long), will it work properly?

To see the details of application I'm working on it, visit the following question: How to amplify a signal containing square pulses at milivolts to a couple volts

\$\endgroup\$
5
  • \$\begingroup\$ an opamp ideally does not change the rise/fall times, so it is the wrong tool for changing them. \$\endgroup\$
    – PlasmaHH
    Commented May 21, 2015 at 8:28
  • \$\begingroup\$ @PlasmaHH He's trying to avoid changing them, but the slew rate of the opamp is causing them to slow. \$\endgroup\$
    – Nick Johnson
    Commented May 21, 2015 at 8:47
  • \$\begingroup\$ Safa, are you sure an opamp is the right tool for the job? Perhaps a comparator is a better match here? \$\endgroup\$
    – Nick Johnson
    Commented May 21, 2015 at 8:47
  • \$\begingroup\$ @NickJohnson: The question for me reads that he wants to aplify and reduce rise/fall time at the same step. "reduce [...] while amplifying" \$\endgroup\$
    – PlasmaHH
    Commented May 21, 2015 at 8:49
  • 1
    \$\begingroup\$ @PlasmaHH "According to simulation, rise and fall time of pulses increase too much after amplification" - he's clearly suffering from slew-rate issues, not trying to speed up the original pulse. \$\endgroup\$
    – Nick Johnson
    Commented May 21, 2015 at 8:54

3 Answers 3

Reset to default
1
\$\begingroup\$

An ideal op-amp will amplify and not distort. Unfortunately op-amps are not perfect and to amplify edges you need quite a high gain-bandwidth-product (GBP). GBP tells you about the speed of the op-amp and can be found in the data sheet. Also, an op-amp has another parameter called "slew rate". This tells you how fast the op-amp output can move in volts per microsecond.

An op-amp with a slew rate of 1V/us will easily cope with a small signal but at higher levels of amplification the edges will begin to have finite ramp speeds and eventually a square wave will look more like a triangle wave.

Choose your op-amp carefully.

\$\endgroup\$
1
\$\begingroup\$

Posted as a comment in the question you refer to, the suggestion was made to use a voltage comparator.

The suggestion still holds, and by picking the proper comparator you can get blindingly fast edges.

For example:

enter image description here

and here's the LTspice circuit list so you can simulate it if you want to.

\$\endgroup\$
1
  • \$\begingroup\$ You might also consider hooking the negative input of the comparator up to an RC filtered version of the input; this will allow your output to adapt to the DC level of the input signal. \$\endgroup\$
    – Nick Johnson
    Commented May 21, 2015 at 9:53
0
\$\begingroup\$

What is the gain value and what is the output level?

  • If the rise time improves (smaller) while drastically reducing the input (and output) level the cause is the limited slew rate (large signal effect).
  • Otherwise, it is the limited small-signal bandwidth. In this case, the rise time should improve while reducing the gain value (more feedback, wider bandwidth).
\$\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.