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LM324 voltage follower with a single supply to rectify an AC signal?

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have a question that is mostly related to voltage input common mode ranges. The LM324 should be able to go pretty close to it's negative rails, as a voltage follower would the common mode voltage pretty much always be close to 0V?

My goal is to clip out the negative portions of a slow ac signal (10Hz, magnitude 2.5V centered at 0V) and then manipulate the now positive signal with a few stages of op-amp circuits. The following circuit is my first stage. The original circuit included a diode going from the ac signal to the non-inverting terminal of the op-amp which, accompanied by a large pull down resistor (1M+) functions, but attenuates the signal somewhat. If I remove the diode/resistor and plug the source directly into the op-amp, my output is exactly what I want: only the positive half of the signal comes through at the output (and the input actually reflects this as well).

All of the research I have done suggests not using inputs outside the rails of an op-amp but everything I've read says the problem is with unpredictable outputs, not necessarily the possibility of damaging the op-amp. My gut told me that the op-amp would simply clip the negative portion of the signal to the negative rail, and that is exactly what happened, but as reliability is very important for this circuit, I can't implement the following design without knowing exactly why it works and whether it will be reliable over time.


1. What is the common mode voltage of a voltage follower?

2. Can op-amp inputs safely exceed the rails if the common mode voltage is still within spec?

3. Is it safe to use a single-supply op-amp to half-wave rectify a slow ac signal?

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What are you designing this circuit for? What's the project?


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To directly answer your questions:

1. The common mode voltage of a buffer circuit is the DC offset you apply to the positive input. In your case, it's 0V.

2. You should look at all the specs in section 6.1 of the datasheet. It states the min/max input voltages that should be applied to the + and - inputs (-0.3V on the negative side).

3. IMO, it's not a safe practice to use an opamp in this way.

For further explanation:

In general it's a bad idea to exceed the limits of the OpAmp input specifications. Section 6.1, note (1)here states the maximum allowable negative input voltage is -0.3V. Stresses beyond this value may cause permanent damage to the device. If you're making a high-reliability product, you most definitely don't want to exceed these voltages.

Section 8.2 of that datasheet shows that you'll either reverse bias the input diodes or the NP junction of the input BJT. Doing this will clamp your signal like you want, but you're also pushing current through junctions that may not be able to handle the power you'll drop through them. Hence, voltages more negative than -0.3V may cause permanent damage to those inputs. In practice they can probably handle more than -0.3 for short periods, but it's not something I would risk.

You say that your original half-wave diode rectifier attenuates the signal... Are you talking purely about the voltage drop across the diode, or does it actually reduce the Vpeak voltage by more than the turn on voltage? I recommend finding a diode with an extremely small turn on voltage. A very quick search turns up something like this, with a turn on voltage of 0.16V. You'll lose any information from 0-0.16V, but if you just care about the peak of the signal, you can modify your voltage follower to a non-inverting amplifier and boost your signal back up.

You also may want to reduce the load resistor of your half-wave rectifier to something closer to the 10K-100K range as this will affect your output DC voltage as well as the forward voltage of the diode.

Hope this helps!

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I realized I mis-spoke slightly, the input diode will be forward biased, not reverse biased. It's most likely a protection diode with a 0.3V turn on voltage, which is where the absolute min -0.3V limit comes from.


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