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I designed a front-end for the S5972 high-speed photodiode. See the following schematic.

Multisim schematic

In Multisim an AC sweep analysis gives the following Bode plot:

Multisim Bode plot

1. Is the presented front-end stable? If not, what part of the Bode plot tells me about the stability? What value do I need to adjust to stabilize the design?

From my understanding, we need constructive interference at the inverting input of the op-amp as well as a gain larger than unity. In the phase diagram of the Bode plot, the blue curve describes the difference between the voltage at the inverting input of the op-amp and the output of the op-amp. When the blue curve hits -180 degrees at about 230 MHz the constructive interference condition is satisfied, however, the magnitude of the frequency response is below unity, therefore, there is no positive feedback that makes the circuit ring.

2. How much of my understanding is correct?

What I believe I am missing is that usually the stability is defined in terms of noise gain. The reasoning behind using the noise instead of the signal gain (am I using the terms correctly in this context?) is that both gains can be different. For example, the standard inverting amplifier has a signal gain of -Rf/Ri while it has a noise gain of 1+Rf/Ri, thus, although if Rf/Ri falls we still have a unity amplification of the noise and the noise can be caught in a positive feedback loop.

3. Is this correct? How would I simulate the noise gain with Multisim?

Update 1:

As requested a transient analysis of a pulsed current source. The pulsed current source is configured to start at 0 A. After a delay of 100 ns the current source jumps in 1 ns rise time to a value of 440 uA. The pulse width is 50 ns. Fall time is 1 ns. The period between pulses is 300 ns.

transient analysis

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  • \$\begingroup\$ What is unity gain BW? What is Phase margin before 180 deg shift and positive feedback (unstable) at unity gain BW? Why is Ccm shorted out with -0 Ohm instead of matched R \$\endgroup\$ Jan 13, 2020 at 19:18
  • \$\begingroup\$ What is wire impedance? to PD? \$\endgroup\$ Jan 13, 2020 at 19:21
  • \$\begingroup\$ Punching your circuit and assuming ~3 pF parasitic input capacitance on the 847 (from past experience) into the TI calculators, it shouldn't oscillate. However, as a practical matter, getting 0.25 pF feedback capacitance is going to be very hard, so you are probably going to have to give up either some bandwidth or some gain. \$\endgroup\$ Jan 13, 2020 at 19:44
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    \$\begingroup\$ Do a transient analysis with a step input. \$\endgroup\$
    – Andy aka
    Jan 13, 2020 at 20:08
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    \$\begingroup\$ " When the blue curve hits 180 degrees at about 110 MHz "..." the magnitude of the frequency response is below unity" no, try again. f ~ 210 MHz and Av > 23dB \$\endgroup\$ Jan 13, 2020 at 23:12

2 Answers 2

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First of all, V_rev does nothing, it is connected in series with an ideal current source (which has infinite output impedance) and a capacitor (which does not couple DC), hence it might as well be 0V (or connected to gnd).

  1. Is the presented front-end stable? If not, what part of the Bode plot tells me about the stability? What value do I need to adjust to stabilize the design?

Yes it is stable, within the frequency range that you have sweept over.

For an amplifier with transfer function:

$$G(s)=\frac{V_o}{V_i}$$

If we put it in a feedback loop, where the feedback has a transfer function H(s), then the new transfer function with feedback becomes:

$$\frac{G(s)}{1+H(s)G(s)}$$

for an amplifier with feedback loop to be stable means the denumerator of this equation must not equal zero, this happens when:

$$H(s)G(s)=-1$$

This is the point that we are trying to avoid. If we hit this -1 point at any frequency then the whole system is unstable.

How stable your system is depends on how far it stays away from the -1 point, ie. what the minimal distance from the -1 the transfer function is at the frequency where it is closest.

  1. How much of my understanding is correct?

What you describe is reasonably well understood.

  1. Is this correct? How would I simulate the noise gain with Multisim?

No, to the best of my knowledge, this is not correct.

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  • \$\begingroup\$ What is Av-= Vout / Vin @ Vout, phase =180 deg? \$\endgroup\$ Jan 13, 2020 at 23:14
  • \$\begingroup\$ Why was this downvoted? I don't think the transfer function is correct in my case as we have a transimpedance (current-to-voltage) amplifier instead of an inverting amplifier configuration. \$\endgroup\$
    – bodokaiser
    Jan 14, 2020 at 8:24
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    \$\begingroup\$ I think it is very bad practice to downvote other peoples answers without giving an explanation why, I don't see anything at all wrong with my answer. @TonyStewartSunnyskyguyEE75 ?? What are you trying to say? \$\endgroup\$
    – user173292
    Jan 14, 2020 at 13:46
  • \$\begingroup\$ I asked what does your graph tell you about gain margin? \$\endgroup\$ Jan 15, 2020 at 18:25
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3: Noise gain

This is defined as the ratio between Vout, and the noise voltage internal to the amp: an apparent voltage in series with either/both inputs to an ideal differential amplifier.

Easiest way to simulate is add a voltage source in series with one input, set all others to zero amplitude, and sweep that.


The transient response looks fine, slightly peaky but still quickly settling. It looks about right, given the flat frequency response: note that a maximally-flat (Butterworth) frequency response still has some transient overshoot. There may be additional peaking going on, caused by excess phase shift (which ultimately limits how high you can tune the bandwidth, given some flatness/overshoot requirement), likely in the op-amp model. There may be additional phase shift in the real circuit, due to component impedances, trace lengths, stray inductance/capacitance, etc. Keep the layout compact.

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