# Half-Wave Voltage Doubler

Discussion in 'General Electronics Chat' started by epsilonjon, Aug 19, 2011.

1. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Hi.

I'm having a bit of trouble understanding the haf-wave voltage doubler. I've copied and pasted the section from the book i'm reading to save typing it out:

I understand the first part where C1 is charging up. What I don't understand is how C2 can charge without C1 discharging. The book says that C1 cannot discharge (I guess because there is no resistor for it to discharge through), but any current which goes through C2 has to be going through C1 too. So is C1 not discharging at the same rate that C2 is charging?

If someone can explain it I'd be most grateful!

Cheers,
Jon.

2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,993
1,116
Yes you are right C1 is discharging at the same rate that C2 is charging but you book assume steady state situation.

3. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
C1 discharges so that C2 can get charged.The peak at which the C2 is charged is
VC2 = Vp + VC1 and VC1 is Vp - forward voltage drop of diode i.e.. approx Vp
So VC2 approx Vp + Vp.

The e-book by Tony R. Kuphaldt provided in this site could help you a bit more...

Good Luck

4. ### epsilonjon Thread Starter Member

Feb 15, 2011
65
1
Thanks for the replies guys. So tell me if i'm understanding correctly:

1. During the first positive quarter cycle D1 is forward biased and D2 is reverse biased. C1 charges up to the maximum input voltage minus the diode drop.

2. As the input voltage starts to decrease, C1 maintains its charge which means that D1 becomes reverse biased and D2 becomes forward biased. C1 discharges and C2 charges a bit, but not yet to the maximum.

3. The input voltage eventually returns back to the positive part of its cycle and this, combined with the voltages now across C1 and C2, causes D2 to become reverse biased and D1 to become forward biased.

4. C2 maintains its charge and C1 again charges back up to the max input voltage minus the diode drop.

5. The above cycle repeats, each time with C2 getting more and more charged. Eventually D2 only becomes forward biased for a very short period of the cycle (when the input voltage is at its negative peak). Likewise, D1 only becomes forward biased for the short period when the input voltage is at its positive peak. Consequently, the capacitors are no longer going through this charging/discharging cycle, and maintain steady voltages.

6. When this steady-state has occured we see that for the short time when D2 is forward biased we must have the voltage across C2 being approximately 2 times the maximum input voltage. And since it's in steady-state, C2 stays at this voltage forever after.

Is that roughly how it goes?

Jon.

5. ### Hi-Z Member

Jul 31, 2011
157
17
Yes, that sounds about right. In the absence of any load current, C2 will eventually (actually rather quickly) be pumped up to about 2Vp. You could make C1 larger than C2 to speed up the process.

Feb 15, 2011
65
1
Thanks