0
\$\begingroup\$

So I had to construct this (practical differentiation) and deduce the bandwidth for which the differentiation occurs. We also had to deduce for what higher frequencies it turns into an inverting amplifier. I am very confused by what I am looking at because it doesnt look correct. The top is not anywhere near flat enough. However, I was playing around with it, and changing the capacitor to 100uF corrected this. The OP amp in question is LM324, with a unity gain of 1Mhz.

enter image description here

This is the frequency response of the circuit. The amplification is in DB. Am I correct in thinking that the differentiation occurs from 112.179 hz to 27.295k Hz? And that the max amplification is at 1.7507k Hz. Lastly, that the bandwidth is 27.295k - 112.179 = 27182.821 hz? It turns into an inverting amplifier at frequencies greater than 27.295k.

enter image description here

and without R3 ...

enter image description here

|improve this question
\$\endgroup\$
  • \$\begingroup\$ I dont understand what you are asking. \$\endgroup\$ – MathWannaBe456 Dec 9 '16 at 22:04
  • 1
    \$\begingroup\$ R3 is 1 ohm, why such a strange value? \$\endgroup\$ – Neil_UK Dec 9 '16 at 22:08
  • \$\begingroup\$ That was left over when I was toying around trying to figure out how to deduce the bandwidths. When I was setting up the simulation profile wrong. So, removal of that caused the weirdness at the top to go away. Is my thought process for the bandwidth correct? I tried to position the trace as close to -3dB as possible. \$\endgroup\$ – MathWannaBe456 Dec 9 '16 at 22:10
  • 1
    \$\begingroup\$ remove it, then simulate again. It's like wondering why your car tyres are smoking, when you're spinning the wheels tethered to a concrete block. Let the amplifier behave like an amplifier, then you'll be able to see what it's doing. \$\endgroup\$ – Neil_UK Dec 9 '16 at 22:18
  • 1
    \$\begingroup\$ I'm glad you can see it. I can't. try editing your question to include it. \$\endgroup\$ – Neil_UK Dec 9 '16 at 22:22
2
\$\begingroup\$

Am I correct in thinking that the differentiation occurs from 112.179 hz to 27.295k Hz?

  • No

    • by differentiation, we generally mean the sinusoidal phase current lags voltage by 90 deg. In this circuit, everything at least an octave below the input breakpoint in the response curve meets criteria, even though you have not plotted the phase response.
    • Since the input RC breakpoint T=0.01uF*1k6 = 1.6e-5 or f=10kHz (approx) we can say everything below 10kHZ is a derivative although it needs at least an octave to get close to -90 deg

  • This starts at too high a frequency for the plateau response to work

    • open loop gain @10kHz.= 100 and desired plateau gain =60
    • with |Zc|=R1=1k6 the impedance feedback ratio,@10kHz = 100k/3k2=31

And that the max amplification is at 1.7507k Hz.

  • No .. maybe you meant 10.75 kHz
    • the Op Amp is an integrator with -1 slope and the peak gain occurs at the intersection of the +1 differentiator slope just above 10kHz where the impedance ratio increases from 31 towards 60 while open loop gain of 100 drops 50% per octave above 10kHz, so it will be around 11kHz at peak gain

bandwidth is 27.295k - 112.179 = 27182.821 hz?

  • We could call this a Bandpass filter with order +1 diff on low side and -1 on high side, and use the definition of -3dB spread to define BW which looks like +/-2kHz or 4kHz BW on your graph at peak minus -3dB on either side.

It turns into an inverting amplifier at frequencies greater than 27.295

Since it uses the inverting feedback, the output is always inverting except it has voltage 90 deg phase lead up to center then phase lag by 90 deg. so the net phase is -90 then rapid phase shift from 8 to 12kHz starting at -90deg to -180 (@11kHZ) to -270 deg for increasing frequency.

p.s. try to follow conventions for orientation on power supplies

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ Thank, and I will make it a habit to follow common conventions. \$\endgroup\$ – MathWannaBe456 Dec 10 '16 at 20:03
2
\$\begingroup\$

Your second picture shows a nice straight line from low frequency to approaching 10kHz, which is what one would expect from the circuit.

A differentiator responds to the rate of change of a signal.

A convenient way to make a differentiator is to use a capacitor input to a virtual ground amplifier. That way, the output responds to the current through the capacitor.

As a differentiator has a rising response, which can amplify high frequency noise excessively, and demand a very high gain amplifier, and present a very low impedance to the circuit driving it, a convenient way to make a practical differentiator is to put a small R in series with the C, to limit the differentiation bandwidth.

In your schematic, you have a virtual ground amplifier, made of R1 and UA1. Your input is a series RC, with a 10kHz (16uS) time constant. With an ideal amplifier, you would expect differentiation behaviour at low frequencies where C1 dominates, and it to be 'running out of steam' as it approaches 10kHz where the R and C impedances become equal in magnitude. You would expect flat behaviour above that, where R2 dominates. The gain at 10kHz would be 3dB below the intersection of the linear and the differentiation asymptotes.

However, you are asking for a flat gain of 60 from your opamp (R1/R2), and it has only 1MHz gain bandwidth. Therefore your flat gain will be 3dB into rolloff by 16kHz. The picture you linked to, and I edited in, shows the gain peaking around 10kHz. This is consistent with the amplifier running out of gain.

When you load the output with 1 ohm as in your first picture, you reduce the amplifier gain even further, so it rolls off at a lower frequency.

|improve this answer
\$\endgroup\$
  • \$\begingroup\$ what do you mean by "3dB into rolloff by 16kHz"? You mean the frequencys for which the rate of change becomes negative? And we're losing amplification? \$\endgroup\$ – MathWannaBe456 Dec 11 '16 at 0:30
  • \$\begingroup\$ Also, the absolute gain is supposed to be R1/R2, which is 62.5. Why does my amplifier have a max gain of only 35.838. There is much better "flat lining" for higher values of C. I ran a parametric sweep, using .01uF as initial value, until 100uF, with 25uF jumps. You can see the increasing values of the bandwidth. Or by absolute gain, is it meant the output voltage? As if I plot the voltage (out) by frequency, it does hit the max voltage. The point of this max voltage, this corresponds to when the roll off starts occurring. \$\endgroup\$ – MathWannaBe456 Dec 11 '16 at 0:47
  • \$\begingroup\$ Here is the graph, the colors are read from the bottom, in increasing order. So, orange corresponds to voltage with .01 uF capacitor and so on.imgur.com/Bb3frlZ \$\endgroup\$ – MathWannaBe456 Dec 11 '16 at 0:49
  • 1
    \$\begingroup\$ Are you trying to make a differentiaitor, or an inverting amplifier? If a the first, you have the right value C1 and R1, to get the widest bandwidth out of the amp you've chosen, given its lousy GBW and the high gain of R1/R2. If you want to make an inverter, then C1 should be 'large', large enough to push your flat frequency response down towards DC. If you want a wider bandwidth differentiator, then you want a smaller R1, to extend your frequency upwards, or a faster amplifier and a smaller R2. What do you want? We don't use amplifiers >GBW, it's unrepeatable, poorly controlled. \$\endgroup\$ – Neil_UK Dec 11 '16 at 8:31
  • \$\begingroup\$ Im just trying to fully understand the topic for my lab. We've barely gotten into the phasor domain and we're told to test a bunch of different op amp configurations, one being the practical differentiator. \$\endgroup\$ – MathWannaBe456 Dec 11 '16 at 15:04

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.