# OpAmp Precision Half Wave Rectifier

I am currently studying OpAmps and one of the topics that came up was the applications of OpAmp circuits. In particular, I was learning about precision rectifiers (using non-inverting version of it, thus the diode is pointing away from the "out" terminal of the OpAmp). The thing I cannot figure out is how does the OpAmp bypass the Vf limitations that regular full bridge rectifiers face. In other words, say for V_in = 0, the voltage at the inverting terminal is close to 0 as we assume an ideal device. Therefore, the voltage at the OpAmp output terminal is 0.7 volts higher than that of the inverting input. My assumption is that if V_in remains at 0, the voltage at the output terminal will stay at 0.7 volts.

To summarize the question, what properties of the OpAmp cause it to create and sustain that voltage difference that keeps the diode "on". Isn't it just easier for the OpAmp to establish 0 volts at the output and thus 0 volts at the inverting input node?

• The true output of the circuit is after the diode and not the opamp output. – Andy aka Apr 3 '15 at 8:48
• @Andy aka, I might have gotten some terminology wrong, can you suggest an edit to correct the question? – Rusag Apr 3 '15 at 15:08
• You said "the voltage at the output terminal is 0.7 volts higher than that of the inverting input" and in reality you meant "the voltage at the op-amp output is 0.7 volts higher than that of the inverting input" – Andy aka Apr 3 '15 at 15:43

Consider the simple precision half wave rectifier shown. Let the open loop gain of op-amp be $A$.

From the circuit, the voltage at cathode can be calculated as $$V_{anode} = A(V_{in} - V_{cathode})\tag1$$

Now the diode will conduct when $$V_{anode} > V_{cathode} + 0.7V\tag2$$

$$A(V_{in} - V_{cathode}) > V_{cathode} + 0.7V$$ $$V_{in} > \frac{A+1}{A}V_{cathode} + \frac{0.7}{A}$$

For A >> 1, we can write $$V_{in} > V_{cathode} + \frac{0.7}{A}\tag3$$

When the diode conducts, the circuit becomes a voltage follower and $V_{out}=V_{in}$. And when the diode does not conduct $V_{out}=0$. So the op-amp + diode can be replaced with a diode D1 with cut-in voltage 0.7/A. And as $A \rightarrow \infty$, this cut-in voltage $V_{D_1}\rightarrow 0$.

simulate this circuit – Schematic created using CircuitLab

Low amplitude voltage can make the precision diode forward biased because of the gain provided by the op-amp. So an ideal operational amplifier can make can make the non-ideal diode an ideal one.

EDIT: (Taken from OP's comment)

So ideally, for $V_{in} = 0$, $V_{out}$ and the voltage at the output node are both 0, but as soon as $V_{in}$ becomes just slightly greater than 0, assuming infinite gain, voltage at the output terminal will overcome $V_f$ needed to turn on the diode and the circuit will become a unity gain amplifier.

• Why does the diode has to conduct when V_in = 0? There is no current flowing since V_out is shorted to ground when it's equal to V_in = 0. Thus, wouldn't it be "easier" for the circuit to just set V_anode = V_cathode = 0? – Rusag Apr 3 '15 at 15:04
• @Rusag take an example. If A=100, the diode will conduct for Vin > 7mV. If A=1000, the diode will conduct for Vin > 0.7mV. Similarly if $A=\infty$, then diode will conduct for Vin > 0V. – nidhin Apr 3 '15 at 15:10
• So ideally, for V_in = 0, V_out and the voltage at the output node are both 0, but as soon as V_in becomes just slightly greater than 0, assuming infinite gain, voltage at the output terminal will overcome V_f needed to turn on the diode and the circuit will become a unity gain amplifier? – Rusag Apr 3 '15 at 15:23
• @Rusag exactly.. – nidhin Apr 3 '15 at 15:24
• I know that this is a little bit late, but can we using equation (3) find that the diode conducts if $V_{in}>0$ and vice versa for $V_{in}<0$ it seems to me that we need one more equation relating to $V_{cathode}$. – Essam Jan 20 at 10:28