I have started an analog circuits course and have trouble understanding the behavior of such opAmps and rectifiers. enter image description here

This is the circuit given as example, however there's no extra explanations provided regarding the functioning of diodes and how the amp distributes the current.

LE: Does the opening/closing of diodes depends on the direction of current? Am i misunderstanding that the direction of current changes when Vi changes sign?


  • \$\begingroup\$ There is a much simpler way to build a half wave voltage rectifier. \$\endgroup\$ Aug 4, 2020 at 19:41
  • \$\begingroup\$ @Helena, this isn't just a half-wave rectifier. It's an inverting precision half-wave rectifier (with the inputs accidentally swapped). OP isn't looking for a simpler half-wave rectifier anyway so it's not clear why you commented. \$\endgroup\$
    – Transistor
    Aug 4, 2020 at 19:45
  • \$\begingroup\$ Hmm why use op amps when you can use something less hard and fun?And OK why isn't he/she using an inverter and the voltage of the base is controlled by an AC source? See, easy like cake. \$\endgroup\$ Aug 4, 2020 at 19:47
  • \$\begingroup\$ @HelenaWells It's for when you really need to not have the forward voltage drop of the diode be zero. More for signal applications than power applications. \$\endgroup\$
    – DKNguyen
    Aug 4, 2020 at 21:22
  • \$\begingroup\$ @Helena, can you show an example of the circuit you are thinking of. I suspect that you are missing some details. \$\endgroup\$
    – Transistor
    Aug 5, 2020 at 18:46

3 Answers 3


This is not a correct schematic for ideal diode rectifier. rect-error

The feedback must go to the negative input of opamp. The negative feedback will eliminate the voltage drop of the diode D2. ideal diode

This is an "ideal diode" single phase rectifier.

I've added a test point at the opamp out and smaller input voltage for better understanding. rect3

  • \$\begingroup\$ the behavior is the same regardless of sign of VG1? \$\endgroup\$
    – xampierre
    Aug 4, 2020 at 8:26
  • \$\begingroup\$ At positive inputs the output of opamp will at voltage approx. -0.5V, the forward voltage of the diode D1. At negative inputs the amplifikation is -1 (R1=R2) regardless of diodes. The amplifier output voltage will be -Uin+voltage drop of the D2. \$\endgroup\$
    – csabahu
    Aug 4, 2020 at 8:51
  • \$\begingroup\$ I understand now, thanks a lot! \$\endgroup\$
    – xampierre
    Aug 5, 2020 at 9:31

As has been mentioned, the schematic should be changed so that the feedback is to the negative terminal of the op amp rather than the positive terminal. Here is my updated circuit:

Ideal diode rectifier

SPICE simulation of the circuit gives data for the following plot:

Simulation plot

I used the following model for a mostly ideal op amp:

* Mostly ideal op amp
* in+ in- vcc vdd out
.subckt opamp 1 2 3 4 5
Rin 1 2 10G
Cin 1 2 100pF
Rout 6 0 0.1
* Ignores power rails (more ideal)
* Bout 3 6 v={100k*v(1,2)}
* Has less amplification for low voltages and approaches supply voltages
Bout 3 6 v={(v(4)+v(5))/2+v(4,5)*atan(100k*v(1,2))/pi}

As to a simple, intuitive explanation as to why this circuit works, remember that:

  1. A diode may be loosely approximated by a small voltage source (usually about 0.7V) and a small resistor which allows current flow in only one direction.
  2. An ideal op amp with its output connected back to its negative input will try to output whatever voltage causes its inputs to be the same (i.e. it will make the negative input voltage match the positive input voltage).

When Vin is positive, current will want to flow to the right through R1. Nearly all of this current will flow through D1 (low resistance), causing Vx to be about -0.7V (since the input is brought to ground and the diode drops about 0.7V). Since nearly all of the current will flow through D1, no current flows through R2 and the output voltage is nearly the same as the input voltage (0V).

When Vin is negative, current will want to flow to the left through R1. Since almost no current will flow backwards through D1, this current has to travel through R2. By Ohm's law, the current will be Vin/R1 giving an output voltage of -(Vin/R1)*R2=-(R2/R1)*Vin. This voltage will be positive since Vin is negative.

  • \$\begingroup\$ Thanks a lot for the explanations! \$\endgroup\$
    – xampierre
    Aug 5, 2020 at 9:31

The schematic is incorrect. The inputs have been swapped. See imgur.com/a/RAKtLwR for further slides from the tutorial for example.

It pretty straight-forward. Remember that:

  • The op-amp has very high input impedance so usually we can ignore any current in or out of that pin.
  • With negative feedback the op-amp will settle down with VIN- = VIN+. Since VIN+ = 0 V both inputs can be considered to be at 0 V. (VIN- is a "virtual ground".

Now consider the two cases for Vi:

  1. When Vi > 0 current through R2 can only go through D2 as D1 is reverse biased. The op-amp output will settle at about -0.7 V or so (depending on VF (the forward voltage) of D2.
  2. When i < 0 current can only come back through R2 via D1 as D2 is reverse biased. Since R1 = R2 = R the gain of the circuit is -1 so VO = - Vi.
  • 2
    \$\begingroup\$ Just a heads up. The inverting input is not being used as it should. Not a correct schematic where - input is a virtual ground. \$\endgroup\$
    – user105652
    Aug 4, 2020 at 6:39
  • \$\begingroup\$ Can it be a negative feedback if the output is fed back to V+? \$\endgroup\$
    – xampierre
    Aug 4, 2020 at 6:45
  • \$\begingroup\$ Oops! I didn't spot that it's positive feedback. Is the schematic correct? \$\endgroup\$
    – Transistor
    Aug 4, 2020 at 8:14
  • \$\begingroup\$ the image is an excerpt from the course and maybe the opamp was placed by mistake. few slides further this is presented imgur.com/a/RAKtLwR \$\endgroup\$
    – xampierre
    Aug 4, 2020 at 8:22
  • \$\begingroup\$ OK. My answer stands then. \$\endgroup\$
    – Transistor
    Aug 4, 2020 at 8:59

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