# Signal Rectifier

1. Dec 13, 2016

### pierce15

I am studying The Art of Electronics and am confused about this particular part about a signal rectifier. In the first picture above, the first cap and resistor function differentiate the signal, and the signal passes the diode if the voltage is > .6V. The second circuit improves this by allowing any signal of above 0 V to pass.

I am confused about two things. First, why is R2 necessary? It seems like we could just measure the voltage coming out of the LED without the resistor there... Also, I do not understand how circuit 2 makes it so any signal > 0 V will pass. Can someone explain these two things?

2. Dec 13, 2016

### Staff: Mentor

In fig (1) the diode's anode sits at 0V DC so to forward bias it requires that the signal supply at least 0.5V.

In fig (2) the resistor network biases the diode at approx. 0.5V so then any rise/fall brought in by the coupling capacitor is superimposed on this DC bias.

3. Dec 13, 2016

### Baluncore

R2 is the output impedance of the circuit. Without R2 the output voltage could never fall.

4. Dec 13, 2016

### pierce15

OK i now see that the right side of the circuit needs to be completed but it cannot just be grounded because then the output voltage would always be 0V, hence the resistor to ground. I still don't understand how the part below the R1 in circuit 2 is biasing D1, however. I understand that it would be biased if there were a .6 V battery in front of D2 in circuit 1. However I don't understand how the new setup is achieving that.

5. Dec 13, 2016

### Baluncore

Current flows down through R3, then to ground through D1 which sets the anode voltage of D1 at about 0.6V.
That makes a voltage reference of one forward biassed diode voltage drop.
R1 then biasses the point in front of D2 from that 0.6V reference voltage.

6. Dec 15, 2016

### rbelli1

For accuracy keep the two diodes thermally coupled. You will get some self heating due to the bias current on D1. Higher frequencies or voltages may require buffering of the reference.

BoB

7. Dec 15, 2016

### pierce15

Alright I figured it out. The only way I am able to see how the bias effect works is by writing out the math. Let V be the voltage right after the cap and I the current through the cap. Then $I = -C \frac{d}{dt} (V - V_{in})$. Ignoring the load of D2 and R2 for now, the cap and R1 are in series, so the current through the resistor is the same as the current through the cap and it is $(V - .6)/R_1$. Hence $V = RC \frac{dV_{in}}{dt} + .6$, since $\frac{dV}{dt}$ is small by hypothesis. Adding the load, the signal $\frac{dV_{in}}{dt}$ will pass if it is $\ge 0$.