How does this rectifier work, since the op-amp input is 0 V?

Question

Consider the rectifier below. Both inputs for the op-amp are 0 V. There are two diodes, that only allow current if the potential difference over them is above 0.7 V. Also, V1 is a sinusoid.

I don't quite understand how V3 and V2 vary for v1 > 0 and v1 < 0 since the op-amp input is always 0 V and therefore no current should pass through the diodes. V2 is not an input source, so it should also be 0 V because no voltage is getting there (it is being cut by the virtual ground from the op-amp).

What am I understanding wrong here?

Answer 1

What am I understanding wrong here?

An op-amp circuit that employs a virtual ground maintains the inverting input at virtually the same voltage as the non-inverting input. It achieves this by adjusting the op-amp output (\$V_3\$). Then, for convenience, we can approximate the term virtual to exactly. This makes the calculations simpler and, at DC and low frequencies, that approximation has almost negligible effect on accuracy.

I don't quite understand how V3 and V2 vary for v1 > 0 and v1 < 0

So, in your particular situation, when the input voltage (\$V_1\$) is positive, the virtual ground is maintained by a negative voltage on the op-amp output (\$V_3\$) and, current flows through \$R_3\$ from left to right.

When the input voltage (\$V_1\$) is negative, the virtual ground is maintained by a positive voltage on the op-amp output (\$V_3\$) and, current flows through \$R_2\$ from right to left.

When the op-amp output is negative, diode \$D_1\$ is reversed biased thus preventing that negative voltage at \$V_3\$ influencing the true circuit output (\$V_2\$). When the op-amp output is positive, \$V_2\$ becomes that voltage minus one diode drop.

Hence, the circuit is an inverting, precision half-wave rectifier.

V2 is not an input source, so it should also be 0 V because no voltage is getting there (it is being cut by the virtual ground from the op-amp).

No, the inverting input is virtually 0 volts and, that might mean it's a few microvolts higher or lower than the non-inverting terminal and, this is enough to drive the output (\$V_2\$ or \$V_3\$) towards the power rails because, a typical op-amp open loop gain might be around 1,000,000.

Answer 2

How do we invent circuits?

In addition to revealing the idea behind this clever circuit, answering the question gave me the unique opportunity to show a somewhat little paradoxical way to invent electronic circuits by finding a useful application of a harmful phenomenon. I do this with the belief that no less important than the specific circuit explanations is to develop skills in how to understand, explain and invent circuits.

Basic idea

Inverting amplifier viewpoint

The circuit can be considered as an inverting amplifier in which the negative feedback network is split into two parts - one (D1-R2) operating during the negative input half-waves and the other (D2-R3) during the positive half-waves. The switching of the two parts is done with the help of the diodes D1 and D2. The two output signals are taken after the diodes; therefore the unwanted forward voltage drops across them are compensated by the op-amp.

The role of virtual ground

Thus, the circuit has two outputs - one for the positive output half-waves and the other for the negative half-waves... and we use one of them, in this case the positive one. Quite rightly, the question arises, "And what the hell is the second output after we do not use it?"

The role of this part of the negative feedback network is to maintain the virtual ground at the inverting input during the negative input half waves. If it were not there would be no virtual ground and the input signal would be transferred through resistors R1 and R2 to the circuit output (V2)... the feedback would become "feedforward":-)

The op-amp function

It is the same as our role in life when we realize our goals.

Op-amp follower. So, although this is a complex negative feedback circuit, the op-amp work is very simple - by varying its output voltage, it keeps the inverting input voltage equal to the non-inverting input voltage (zero); this is known as the "golden rule" of H&H. In this sense, any op-amp circuit with negative feedback is a follower where the voltage of the non-inverting input is the input and the voltage of the inverting input is the output. The simplest way to make such a follower is to connect the op-amp output to its inverting input. Then the op-amp will do its job without any effort.

Disturbed follower. But when the op-amp encounters obstacles (voltage drops across diodes and resistors), it starts increasing its output voltage more than usual in order to compensate for these disturbances. And as strange as it sounds, the primary disturbance is the external input voltage Vin... but it is a useful "disturbance".

D1 forward voltage is an undesired disturbance here; that is why, the output is taken after it and the disturbance is eliminated. The voltage divider R1-R2 is a useful "disturbance"; so the output is taken before it.

Implementation

STEP 1: Perfect half wave rectifier

^{simulate this circuit – Schematic created using CircuitLab}

Zero input voltage. The op-amp effortlessly maintains zero voltage at its inverting input since no currents flow, no disturbing voltages; therefore its output voltage is zero... it just idles.

Negative input voltage. In keeping with our negative feedback philosophy above, this voltage change is a "disturbance" for the op-amp which it must eliminate. In the first instant, the op-amp fails to respond (as we do in life) and its output voltage is zero. The negative input voltage is transferred through the resistor R1 to the inverting input. The op-amp senses this change and begins to compensate by increasing its output voltage in a positive direction. With an excess of 0.7 V, it compensates the D1 forward voltage drop and starts passing current through the circuit R2 -> R1 -> Vin with the idea of ​​restoring zero voltage to the virtual ground. Figuratively speaking, R2 acts as a pull-up resistor ("pulled up" by the op-amp) and R1 as a pull-down resistor ("pulled down" by Vin).

Positive input voltage. Now, the op-amp senses a positive change at the inverting input ("disturbance") and begins to compensate by increasing its output voltage in a negative direction. With an excess of 0.7 V, it compensates the D2 forward voltage drop and starts sinking current through the circuit Vin -> R1 -> R3 to ​​restore the zero voltage of the virtual ground. Now R3 acts as a pull-down resistor and R1 as a pull-up resistor.

And here is the role of the negative feedback in this case - the virtual ground does not allow the transfer of the input voltage through the circuit R1 -> R2 to the load (more precisely, the op amp prevents this by drawing current from that point).

STEP 2: Imperfect half wave rectifier

The best way to convince ourselves of the need for a second negative feedback network is to see what happens if we remove it. Let's do it.

^{simulate this circuit}

Only the case with a positive input voltage is of interest. Now the diode D1 is off and the op-amp can do nothing - it reaches the negative supply rail voltage and "stops". The input voltage is freely transmitted through the R1 -> R2 circuit to the load. A resistive voltage divider is formed and 1/3 of the input voltage is applied to the load.

STEP 3: Unloaded half wave rectifier

Apparently the load is the reason for this, we think; then let's remove it! But it gets even worse - now all the input voltage is passed to the circuit output!

^{simulate this circuit}

STEP 4: Perfect full wave rectifier

And here we realize that we most often need a full-wave rectifier... and somehow, it worked itself out (I myself realized this idea a month ago when I answered a similar question).

It remains only to include a voltage follower (OA2) between the circuit and the load.

^{simulate this circuit}

STEP 5: Deriving another principle

What have we done here? We have found a useful application of a harmful circuit phenomenon. This is a very powerful principle for solving inventive problems that can help us in other cases as well.

More examples of this principle are the use of the "harmful" voltage drop across a forward-biased diode as a useful voltage (e.g., in bias circuits)... or the "harmful" voltage drop across a resistor (current-to-voltage converter), capacitor (integrator), etc.

Answer 3

Ohm’s Law indicates that the current through \$R_1\$ is:

$$i_{R_{1}}=\frac{v_1}{R_1}$$ since \$v_-=0\$.

The op-amp defining relations state that the current into the op-amp inputs is zero. So where does the current go?

It must go through the feedback network via \$R_2\$ or \$R_3\$.

The misunderstanding is that zero volts at a node implies zero current. Voltage across a device indicates current.