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Hi,

I'm going to design an inverting voltage amplify circuit with a gain of -10. The simulation circuits are as below:

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1. Open-loop frequency response test, the result is consistent with the datasheet, with a 0dB cutoff frequency greater than 10MHz. And it's similar to a single-pole ideal opamp with -20dB/dec over a wide frequency range.  ---

2. A model I built for the feedback loop.

ADA4625-1's input capacitance is 20pF (11pF of CM, 9pF of DM), and I added a 10pF stray capacitance of PCB.

To compensate for the pole generated by Cin and Ri, I added a zero by feedback Capacitance, Cf = 0.1 * Cin. The AC result of Noise gain is plotted below marked as red lines.

And the frequency of the junction is about 1.25MHz, which I think is corresponded to the -3dB cut-off frequency of the closed-loop. ---

But the closed-loop simulation shows a much lower -3dB(20-3=17dB) cut-off frequency of 50kHz.  Which can't satisfy my need for closed-loop bandwidth.

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Please tell me what's the problem, is my simulation model not accurate, or is there some bug in ADA4625-1's SPICE model?

If my simulation circuit is wrong, please tell me how to fix it.

Thanks a lot!

Robert

correct a typo
[编辑人: RobertLiang 编辑时间: 19 Jan 2022 4:10 AM (GMT -5)]

## Top Replies

• Hi RobertLiang:  The 50kHz is set by Rf*Cf.  Reduce the value of Cf.

• Hi Glen,

You gave me a practical suggestion, and based on it, I thought a lot about its math principle.

I'd like to share the details of my thoughts here:

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In conclusion, for a standard inverting voltage amplifier circuit, Gclose=Avol/(1+Avol * 1/NoiseGain) is wrong! Because the input voltage from the voltage source is attenuated by Ri, Rf, Ci, and Cf at first, rather than minus feedback voltage directly.

Then the new block diagram specifically for my circuit is: ---

To verify that, I build a math model in Mathematica software, this is the result: And I calculated the 17dB(20-3dB) cut-off frequency is exactly equal to 53kHz, which is consistent with the LTSpice simulation result.

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Back to the closed-loop gain expression.

The new attenuation block(Labeled ''m") for inverting voltage amplify circuit contributes a new pole to the closed-loop gain's transfer function.

I calculated it in Mathematica: Its cut-off frequency is: Since Ri and Cf are 10 times smaller than Rf and Ci, this pole's frequency is approximatly .

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The new pole is dominated by Ri and Ci rather than Rf and Cf, so adjusting Cf is of limited help.

I'm trying to find some new methods (lowering Ri is not suitable for me).

Again, thanks for your professional help, Glen. You gave me a new direction.

Robert