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Precision active half-wave rectifier

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In Figure 1 of the above link (Basic Precision Halfwave Rectifier), how would I analyse this circuit? At first, I thought R1 and R2 didn't really do anything because it seemed like they weren't part of any feedback networks, but when I ran a simulation without them, the circuit did not rectify.

Any tips?

I initially analyzed the circuit by replacing the diode with a voltage drop. I assumed 0.7V, so I wrote my output as Vout = Immediate voltage output of opamp - 0.7 V. Then, set Vout equal to my Vin. That's why I had the impression that R1 and R2 did not do anything.

Thanks!

Reply to
MRW
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R1 shouldn't really be needed. It's just defining the input voltage as being wrt 0V. R2 in reality wouldn't normally be explicity shown as part of the circuit. Some kind of lowish value is needed though in order to hold the output voltage near 0V when the opamp and it's reverse biased diode has flopped down to near the negative rail (any negative input voltage).

Without the resistor your sim will be measuring the output voltage with some kind of infinite resistance voltmeter and the multi megohms due to a reversed biased diode, would look essentially like a short circuit. john

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Reply to
john jardine

Download LTSpice from linear.com. Then you can see the waveforms everywhere in the circuit.

- Tom Gootee

Reply to
tomg

Do you actually want the half wave rectified sine wave, or a DC voltage equal to the peak?

Tam

Reply to
Tam/WB2TT

R1 is there for capacitive/inductive coupling., the input of the OP-amp could be bipolar, thus you need to rid that diode effect and give some in range impedance for the AC source. the R2 is needed to create load on the OP-amp since, at the zero state, you will not have any load coupling for the next stage. A Oscope with AC input set for example, would not work very well because of the diode in line. there are other more in death reasons for the state of the op-amp, but i won't get into that.

That is about as simple as i can explain it.

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Reply to
Jamie

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