I have a simple full-wave precision rectifier made of two op-amps integrated into the OPA2810. The gain of the full-wave precision rectifier is 1. The circuit is the following:

enter image description here

The positive supply is 5 V and the negative supply is 5 V. The decoupling capacitors are each equal to 10 nF. All the resistors are 10 kΩ.

I was measuring the voltage with a standard probe referenced to ground.

The output of the circuit is then followed by a differential op-amp.

Here is the picture of the output. The green line is the input multiplied by a gain of 20, but the gain is not linked to the full-wave rectifier; it is linked to a differential op-amp, but the input of the full-wave rectifier is clean.

enter image description here

Do you know where the noise is coming from?

-------------E D I T -------------------------------------------------- Here are some precisions.

I isolated the circuit form the rest of the circuit. The supplies are provided by a flyback which works around 350 kHz. I used a probe with a very short ground for the measurement but it still remains some noise coming from the flyback. In any case there is an other source of noise which I think could come from the diodes. I will take a look on it. The diodes are BAS16LD which I think are not very fast.

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  • \$\begingroup\$ The output of the circuit is then followed by a differential op-amp <-- please post the full exact circuit (with resistor values) and indicate what supply de-couplers you used. Maybe a PCB layout image might be required. Don't try this circuit on breadboard and make sure you have met the PCB design layout standards in the data sheet (section 11). It's a very fast op-amp and incorrect layout could easily cause problems. \$\endgroup\$
    – Andy aka
    Jan 9, 2023 at 19:08
  • \$\begingroup\$ That'a a pretty high GBW op-amp. Maybe try reducing the resistor value by an order of magnitude. \$\endgroup\$ Jan 11, 2023 at 19:52
  • \$\begingroup\$ I'm trying to figure out how the circuit's input gains are equalized for the two half-waves. I think they are 1 and -3/2. \$\endgroup\$ Jan 11, 2023 at 20:33
  • \$\begingroup\$ What diodes are you using? \$\endgroup\$ Jan 12, 2023 at 8:41
  • \$\begingroup\$ Mind boggling. If just wanting a precision rectifier, use a standard topology. \$\endgroup\$
    – greybeard
    Jan 12, 2023 at 9:20

2 Answers 2


Don't know if this can help for reducing noise.

I would add a resistor (R6) which seems not very "useful".
It should be checked at the lab. But it really should help when D5 is OFF.

NB1: X1 is always a "closed loop" opamp, except when both diodes are OFF (a very short time). NB2: from the point of view of Vd1 and Vd2 ...
Seems that:
Vo1=Vd2-Vd1 (first phase with Vd2=0) and Vo2= Vd2*(1+1/2) for the second phase.

enter image description here

UPDATE: note that with some opamps, oscillations may occur.
Tested at 300 kHz with OP37C, parasitic capacitors at inputs as added.

enter image description here

When oscillation occur only at a one-half sinusoid, just lower resistors (use rr => 1 kOhm). Oscillation disappear.

enter image description here


The circuit idea

The circuit's gains for the two half-waves are not equalized. Let's see why.

The first stage of the circuit is a precision half-wave rectifier with two outputs - one for the positive input half-wave and the other for the negative half-wave. Actually, it is an inverting amplifier implemented through an op-amp (A1) with two separate negative feedback networks - one (RD1) for the positive input half-wave and the other (RD2) for the negative half-wave. The diodes are put in the negative feedbacks so their VF are eliminated.

The rectifier's outputs are connected to the second stage of the circuit (A2) that is a "bad differential amplifier" with unbalanced gains. It can be considered as a combination of an inverting amplifier with gain of -R/R = -1 and a non-inverting amplifier with gain of 3R/2R = 3/2.

So the positive input half-waves are amplified with gain of 1 and the negative input half-waves - with gain of -3/2.

EDIT 1: An answer to @Spehro Pefhany's comments

Actually it does work, if you calculate the output voltage correctly. Gain is +1 for positive inputs and -1 for negative inputs... Part of the feedback goes through both both op-amps so one might be suspicious of stability unless the op-amps have a lot of phase margin to start with.

Apparently op-amp circuits hide more secrets from me... Now I will try to figure this one out...

Let's focus on the input negative half-wave to see why then the gain is not -3/2 but exactly -1. We are starting to think...

A2 must be plugged into the global feedback... so it must be facing its output back to A1's input... but it is not; instead, it is oriented with its inverting input towards the A1's inverting input. What the hell is that?

Voltage follower. Here I am reminded of an observation of mine about the inverting amplifier: It has a connection (via the resistor R2) between its output and inverting input... and through this circuit it controlls the voltage of the inverting input so that it is always equal to the voltage of the non-inverting input (H&H's Golden Rule). So, we can make the paradoxical conclusion that A2 and its R2 act as a voltage follower (more precusely, a "disturbed follower")... that is connected in A1's negative feedback!

It turns out that the circuit of the two top upper resistors in series is connected to a point with the same potential as the D2's cathode... as though it appears in parallel with the bottom resistor connected to D2! Figuratively speaking, we can replace the voltage follower by a piece of wire. Let's start calculating then... although it is obvious:

First, there is an inverting amplifier made by A1; its R1 = R and R2 = 2R||R so its gain is -R2/R1 = -2/3. Then, there is a non-inverting amplifier (made by A2); its R1 = 2R and R2 = R so its gain is 1 + R2/R1 = 3/2. The overall gain is -2R.R/(2R + R)/R.(2R + R)/2R = -2/3.3/2 = -1.


The two input gains of the "imperfect differential amplifier" (A2) are equalized by connecting an "inverting follower" before the inverting input and "inverting attenuator" before the non-inverting input. See more about this technique in another answer of mine.

EDIT 2: The source of noise

A possible source of noise can be A1's output. When its output voltage crosses the zero voltage region (-0.7 V ÷ 0.7 V), both diodes are "off". So there is no feedback and oscillations with maximum magnitude ("noise") appear at the output.

  • \$\begingroup\$ Actually it does work, if you calculate the output voltage correctly. Gain is +1 for positive inputs and -1 for negative inputs. \$\endgroup\$ Jan 11, 2023 at 23:13
  • \$\begingroup\$ @Spehro Pefhany, It's already 1 past midnight here and I'm writing in the dark falling asleep in bed... Maybe that's why I don't see the obvious solution, just a non-inverting amplifier with 2k and 1k resistors :-) \$\endgroup\$ Jan 11, 2023 at 23:23
  • 1
    \$\begingroup\$ @Spehro Pefhany, Very intriguing... Tomorrow I will try to figure out it... \$\endgroup\$ Jan 11, 2023 at 23:35
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    \$\begingroup\$ @Spehro Pefhany, I was excited anticipating the clever trick and couldn't sleep...but I finally saw it. Now I will describe it in a hurry at the bottom of my answer… even though it is already 2 past midnight... \$\endgroup\$ Jan 12, 2023 at 0:13
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    \$\begingroup\$ Re. gains, stare hard at the oscillograph in Antonio51's answer, esp. \$V_{d1}\$ & \$V_{d2}\$. \$\endgroup\$
    – greybeard
    Jan 12, 2023 at 9:28

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