# Ebook Volume3>Chapter3>Clamper circuits

#### BreachPattern

Joined Jul 18, 2010
2
Hi guys, I've been reading through these books and have been learning and understanding everything up to the clamper circuit section in the diodes chapter in the semiconductor book.

Some of the information doesn't add up in my mind, so I wanted to ask you generous guys and gals for some clarification.

There's this sentence:
"On the first positive half cycle, the diode conducts charging the capacator left end to +5 V."

I get that. The capacitor charges and has a ~5v drop across its two terminals. The left is +5 to ground and the right terminal is 0v to ground.

But then the next few sentences make my brain explode:
"This is -5v on the right end... The right end of the capacator is at -5v DC with respect to ground."

I cannot understand this. Isn't the right end of the cap at ground in the first half cycle? Isn't it more likely that the right end is -5v DC with respect to node 4 - not ground?

#### hgmjr

Joined Jan 28, 2005
9,027
One thing that you need to consider is that the voltage across the capacitor cannot change instantaneously. The result is that when the input AC voltage goes negative the capacitor acts briefly like a battery and thus the end of the diode attached to the diode ends up going negative. This is the concept associated with voltage doubling.

Don't forget to consider the forward voltage drop across the diode during the charge cycle.

hgmjr

#### BreachPattern

Joined Jul 18, 2010
2
I was thinking of the first half cycle as the first quarter cycle. I did this because reading the sentence ended with:

"on the first positive half cycle, the diode conducts charging the capacitor left end to +5v".

At the end of the first quarter cycle, the capacitor is charged, and the left end is +5v. At the end of the first half cycle, the left end is 0v and the right end is -5v.

I think it would make more sense to say:

"During first positive half cycle, the diode conducts, charging the capacitor to a voltage drop of +5V (4.3 V). At the end of the first positive half cycle, the right end of the capacitor is -5v (-4.3V) at v(1,4)".

This is a little more wordy, but is clearer. Saying that the capacator is charged by refering to one of the terminals makes sense, but saying the left end is +5 is not smart since that terminal isn't +5 at the frame of reference in which we are speaking.