Explanations regarding the blocking of diodes in single phase rectifier

Question

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

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?

Thanks

Answer 1

This is not a correct schematic for ideal diode rectifier.

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

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.

Answer 2

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:

SPICE simulation of the circuit gives data for the following 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}
.ends

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.

Answer 3

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

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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 V_IN- = V_IN+. Since V_IN+ = 0 V both inputs can be considered to be at 0 V. (V_IN- is a "virtual ground".

Now consider the two cases for V_i:

1. When V_i > 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 V_F (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 V_O = - V_i.