I am currently using a precision full wave rectifier as shown below. The issue I have is that the output look like a half wave signal. Images of the input and output signal of the circuit is shown below. The diode I am using is 1N4001G and the opamp I am using is LMH6611. I would like to know how to get a full wave signal on the output. Any help is appreciated.

enter image description here

Input Signal below Input Signal

Output Signal below Output Signal

Input vs Output Signal enter image description here

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    \$\begingroup\$ Your input signal swings positive and negative. Do your op amps have positive and negative voltage supply? \$\endgroup\$ Jul 30, 2021 at 20:23
  • \$\begingroup\$ Is Vcm near ground acceptable to IC spec \$\endgroup\$ Jul 30, 2021 at 20:29
  • \$\begingroup\$ Try adding bias to your signal or make sure you have a +ve and -ve power rails. \$\endgroup\$
    – NeuroEng
    Jul 30, 2021 at 20:29
  • \$\begingroup\$ Input bias is not enough, you have to bias the Vin+ references on both Op Amps too \$\endgroup\$ Jul 30, 2021 at 20:39
  • \$\begingroup\$ @JohnBirckhead The opamp is powered by battery; therefore, single supply currently. \$\endgroup\$
    – Sam
    Jul 30, 2021 at 20:41

3 Answers 3



simulate this circuit – Schematic created using CircuitLab

Figure 1. The first stage of the full-wave rectifier is a half-wave rectifier giving out the inverted positive half-cycles.

enter image description here

Figure 2. Input waveform in blue. Output in orange.

The problem is that you are using a single-ended supply and that means the op-amp can only output positive voltages.

enter image description here

Figure 1. Internals of the ancient 741 opamp. Source: Wikipedia.

From the internal schematic of the 741 op-amp it should be clear that the output can source current from the \$ V_{S+} \$ rail via Q14 or sink current to the \$ V_{S-} \$ rail via Q20. Nearly all (there are probably exceptions) op-amps will have a similar push-pull arrangement on the output.

It should be clear from the above that the lowest output voltage possible is when Q20 is turned fully on. The 741 is particularly bad and due to the proceeding stages the output can only get to within a few volts of VS-.

The LMH6611 you are using may be a lot better (I didn't check) but it can't possibly output a negative voltage if it is powered from a single 9 V cell.

The simplest solution is to add a second 9 V battery to provide the negative rail.

  • \$\begingroup\$ LM6611 is supplied from 2.7V to 11V max. \$\endgroup\$
    – Antonio51
    Jul 30, 2021 at 22:20

You are not biasing these bipolar Op Amps properly with a single supply.

You have two choices:

  1. Get a single supply OpAmp with PNP inputs that operate done to 0V on the input.
  2. Add a negative voltage to both Op Amps and keep input within the Vcommon mode range specs.


Since you are getting only the +ve out on one half cycle , most likely you are using a single supply not design for this operation. This requires a certain type with PNP or Pch FET input that work down to Vee.

To reduce power, raise all the resistors above 100k. then bias all the inputs above with low Rs >1k to Vbat/2.

  • \$\begingroup\$ I have added extra image for better clarification. I am only getting -ve out half cycle. \$\endgroup\$
    – Sam
    Jul 30, 2021 at 20:46
  • \$\begingroup\$ I already understood that. Did you understand my solution? \$\endgroup\$ Jul 30, 2021 at 20:57
  • \$\begingroup\$ I am sorry. I did not fully understand your solution; I did understand that i need to use higher resistor to reduce power. \$\endgroup\$
    – Sam
    Jul 30, 2021 at 21:01
  • 1
    \$\begingroup\$ @Sam Tony's first paragraph: "you are using a single supply". You should be using a dual supply, i.e. positive supply voltage connected to the positive op-amp power pin, negative supply voltage connected to the negative op-amp power pin. You can also mid-point bias (Tony's second paragraph) the circuit if you are bent on using a single supply, but this requires more design work and not suitable if you are just starting out playing with op-amps. Another note, use small signal diodes like 1N4148, not big slow rectifier diodes like the 1N4001. Don't worry about your resistor values right now. \$\endgroup\$
    – qrk
    Jul 30, 2021 at 21:32
  • \$\begingroup\$ If you don’t have the proper IC, the easiest solution is add a battery for Vee instead of 0V. Even if only -3V \$\endgroup\$ Jul 30, 2021 at 21:39

You are getting what you should. Try an experiment with just diodes and you will get the same waveform (voltages will be different). Look at the scope where the input vs the output. The output frequency is 2X the input and notice when the input swings negative the output goes positive as it should. It is working as it should.

  • \$\begingroup\$ I think you've missed something in your analysis. See my answer for a graph of the first stage output; it's the positive half-cycles inverted. This goes to an inverting summing amplifier with gains of -1 for the input signal and -2 for the output of the rectifying stage. The end result is a full-wave rectified signal. The problem is that the rectifying stage can't swing negative on a single-rail supply the use of which was admitted in the comments. \$\endgroup\$
    – Transistor
    Jul 30, 2021 at 22:18

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