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I have this circuit as a white noise source (with two outputs):-

schematic

So the 24V Zener diode makes noise due to a 60uA current. This is fed into two TL082 op amps. At the op amp input, the noise level is ~4Vp-p. However, at the op amps outputs, the noise level is ~11Vp-p. Both measured with an oscilloscope.

How can this be? Clearly something's wrong. The datasheet has clear examples of this chip being used in exactly this fashion as a simple voltage follower (Figs. 19, 26 & 27). This thing is, I don't recall seeing this weirdness when I bread boarded the circuit. It only seems to be happening on the PCB. When I probe the tip of the Zener, the noise seems to be at least in the order of 20MHz. I would expect the noise output to be lower than the input due to the op amp's slew rate not coping. What's going on?

The scope grab below shows the input noise signal:-

input signal

And this is the output. Avalanche noise should be a saw wave (as is the input). What are the highlighted peaks and where are they coming from? The PCB is a single sided one plus a ground plane.

output peaks

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    \$\begingroup\$ .Check your typing .Volts of noise seems very high . \$\endgroup\$
    – Autistic
    Commented Aug 10, 2017 at 1:35
  • \$\begingroup\$ @Autistic To the best of my ability it's correct. I've built these circuits several times. It's real avalanche noise with it's characteristic log normal distribution. I can provide a scope grab if necessary. If you plug it directly (buffered) into a small 8 ohm speaker you can hear it quite clearly. \$\endgroup\$
    – Paul Uszak
    Commented Aug 10, 2017 at 1:43
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    \$\begingroup\$ You have your probe set to 10x. Is that intentional? \$\endgroup\$
    – Blair Fonville
    Commented Aug 10, 2017 at 2:35
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    \$\begingroup\$ Paul you misunderstood my comment. You must use 10:1 with ground leads <=1cm for >+20MHz measurements otherwise probe coax resonance occurs from ground lead L and coax C. remove gnd wire and probe clip and use pins <1cm apart. Then capture both input and output at <0.5us /div and compare with 20MHz filter OFF. \$\endgroup\$
    – D.A.S.
    Commented Aug 10, 2017 at 13:14
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    \$\begingroup\$ @DiBosco en.m.wiktionary.org/wiki/metric I have heard it used quite a lot in the field of engineering when it comes to different ways to quantity something (in this case, amount of noise) \$\endgroup\$
    – Joren Vaes
    Commented Aug 10, 2017 at 15:36

1 Answer 1

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I would suggest that you place a load resistor, perhaps 2K along with a parallel capacitor of 100 pf on each of the outputs to ground. You are slewing the positive input faster than the output (and hence the negative input) can keep up with, which results in voltage difference on the op amp inputs. This type of op amp's push-pull output will behave better if you give it some load. Your breadboard probably had an (inadvertent) low-pass filter on the op amp positive input.

It's a little counter-intuitive, but you need high speed op-amps to provide adequate feedback to reduce higher-frequency open-loop amplification. Loading the output will help some, but I think you will still have a little overshoot on the output, although certainly not 100%.

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  • \$\begingroup\$ Excellent. If I wanted to read up on this, what would be a suitable search term please? \$\endgroup\$
    – Paul Uszak
    Commented Aug 10, 2017 at 20:58
  • \$\begingroup\$ I would try "op amp phase shift" or "op amp unity gain stability". You might also try "op amp output impedance" or "op amp output snubber" if you find that the trick I mentioned works. I tried and found Linear Technologies Application Note 148. Good luck and let us know what you find out! \$\endgroup\$
    – John Birckhead
    Commented Aug 10, 2017 at 21:21
  • \$\begingroup\$ Excellent2. All I have available tonight is a 1K resistor, 100nF capacitor and a hammer. It works though! I get about ~2Vp-p with a ~250kHz max. frequency with those components. No overshoot. The wave's turned into a sinusoidal type rather than the original saw tooth, but I can live with that. Thanks. \$\endgroup\$
    – Paul Uszak
    Commented Aug 10, 2017 at 22:41
  • \$\begingroup\$ R load invokes current slew rate limiting, hence large signal BW, on large signals when the real input is more like random shot impulses with C load. Examine all load reactance ( eg long cables) for regenerative feedback causing unity gain buffers to amplify with resonance , prone to oscillation with low Z out, and high R load with some LC reactance loading. So adding R reduces Q of output emitter follower spurious response. When since OA Zout rises with f due to reduction in gain, it tends towards an open loop emitter follower characteristic with ~ 300 Ohm , an oversimplified explanation \$\endgroup\$
    – D.A.S.
    Commented Aug 12, 2017 at 19:32

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