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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?

enter image description here

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  • \$\begingroup\$ The true output of the circuit is after the diode and not the opamp output. \$\endgroup\$ – Andy aka Apr 3 '15 at 8:48
  • \$\begingroup\$ @Andy aka, I might have gotten some terminology wrong, can you suggest an edit to correct the question? \$\endgroup\$ – Rusag Apr 3 '15 at 15:08
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    \$\begingroup\$ 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" \$\endgroup\$ – Andy aka Apr 3 '15 at 15:43
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Consider the simple precision half wave rectifier shown. Let the open loop gain of op-amp be \$A\$.

enter image description here

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\$.

schematic

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.

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  • \$\begingroup\$ 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? \$\endgroup\$ – Rusag Apr 3 '15 at 15:04
  • \$\begingroup\$ @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. \$\endgroup\$ – nidhin Apr 3 '15 at 15:10
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    \$\begingroup\$ 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? \$\endgroup\$ – Rusag Apr 3 '15 at 15:23
  • \$\begingroup\$ @Rusag exactly.. \$\endgroup\$ – nidhin Apr 3 '15 at 15:24
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This is your precision rectifier.

schematic

simulate this circuit – Schematic created using CircuitLab

But before that, lets go back and review some opamp (ideal) basics.

  1. The opamp has no input current
  2. The output will do whatever it can do make sure the inverting and non inverting terminals are the same.

For your question, the first point does not matter. The second point is what you are interested in.

On the positive cycle, you get some positive voltage appearing at the V+ terminal. The opamp will do whatever it can to make sure that the voltage is between V+ and V- are zero. So if V- = V+ that means that the cathode of the diode is the same as V+. But in order for that to be true, them the opamp must output V+ + 0.7. This is how it overcomes the diode.

On the negative cycle, you have some negative voltage appearing at the V+ terminal. but because of the diode and its orientation, the output voltage from the opamp is going to go as far down as it possible can, because the diode has not turned on.

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