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enter image description here

I am simulating this circuit for personal development. The gain of the integrator can be calculated using the following formula:

Gain = -Rf/R1 (1/(1+(2 x pi x f x C1 x Rf)))

How can I calculate the gain of this circuit when diodes are introduced for rectification?

I tried for some time, but couldn't get the result I see on simulation. How can I calculate the gain of this circuit and the output voltage (both Vrms and Vdc) with that gain?

It would also help if someone could explain how this circuit works.

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    \$\begingroup\$ Ignore the capacitors. Analyze the circuit assuming the op amp output is positive. Analyze the circuit assuming the op amp output is negative. The op amp negative terminal is always at ground potential. Assume the diodes are shorts when forward biased and opens when reversed biased. \$\endgroup\$
    – user69795
    Dec 10, 2020 at 5:07
  • \$\begingroup\$ @user69795, can you be more specific? \$\endgroup\$
    – CNA
    Dec 10, 2020 at 12:59
  • \$\begingroup\$ the circuit as far as R5 is called a precision half wave rectifier. You can search on that and find explanations,it is a well known circuit. Then you have a simple CR circuit on its output. \$\endgroup\$
    – danmcb
    Feb 6 at 15:06

2 Answers 2

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When the input is negative, the op amp output is positive. D3 is ON. The circuit is an inverting amp with a gain of 1.4. When the input is positive, the output is negative. D2 is ON. The circuit has no gain. The op amp output is then a half wave rectified signal. R5 and C3 are a low pass filter. C2 also acts as a low pass, when the op amp has positive outputs.

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  • \$\begingroup\$ If so, how should i calculate the output voltage? \$\endgroup\$
    – CNA
    Dec 13, 2020 at 4:03
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When we speak about the "gain" of an active circuit, there must be one fixed DC operating point around which a small signal can swing. Because the input signal has no DC component, the only possible operating point is a DC voltage at the opamps output node of Vout=0 (we must asssume an ideal amplifier because there is no other information).

Because of the small-signal requirement ("gain" is defined for linearized transfer functions only) both diodes are not "open" and there will be no feedback.

Therefore, the circuit operates with the opamps open-loop gain - referred to the opamp output node.

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