# How exactly does a clamping zener diode pair in series work?

Discussion in 'Electronics Resources' started by CircuitZord, Oct 14, 2012.

1. ### CircuitZord Thread Starter Member

Oct 8, 2012
59
2

Why is the first case wrong but the second case correct? The website gives an explanation, but it's not clear to me exactly. I understand the reason for the resistor being there to prevent damage to the zener.

The way I see it, for the zeners in series. Say 12V is the signal coming out, and the zener voltage is 6.8V and the forward drop is 0.7V.

So, 12VAC comes along, the break down is -6.8V at the zener at the top, so if it's -12V it goes through the zener, so -12--6.8 = -5.2 + 0.7(other zener in forward drop) = -4.5V is lost through the zener. Hence -12--4.5 = -7.5V is the signal that goes through.

Is that the logic as to how it works?

2. ### crutschow Expert

Mar 14, 2008
13,496
3,373
The first case is wrong because the zener looks like a stardard silicon diode in the forward direction. Thus the voltage would be clipped at about ±0.7V instead of at the zener voltage as intended.

3. ### CircuitZord Thread Starter Member

Oct 8, 2012
59
2
Was my reasoning correct or wrong for the second case? I think I see now why the first case was wrong.

4. ### crutschow Expert

Mar 14, 2008
13,496
3,373
You reasoning is correct, but its typically stated as the "voltage drop across the two diodes is 7.5V" not "the signal that goes through", thus clipping the signal at ≈±7.5V