(solved) newbie question - voltage tripler

Can anyone tell me how the principal of a voltage tripler works - step by step, what is really happening? I don't really get it and cannot find something useful in the internet about that.

Thank you so much!
Christian

fatheaddrummer said:

Can anyone tell me how the principal of a voltage tripler works - step by step, what is really happening? I don't really get it and cannot find something useful in the internet about that.

Click to expand...

That is quite surprising; I have googled "voltage tripler" and on the first page I get at least two pages that give a quite elaborate answer.

Good training for other newbies to answer....

JR

Hi,

as a very newbie, i'll try to explain. Not sure i got it right.

In The first Half-Cycle, Vin is positive, so C1 Charges through D2 to V(C1) = Vin(peak)

In second Half-Cycle, Vin is negative so C2 charge through D1 and to V(C2) =  Vin(peak) + V(C1) = 2Vin(peak)

In third half-cycle, Vin is positive , so, C1 charges back to Vin plus, it allows C3 to charge through D3 to V(C3) = Vin(peak) + V(C2) - V(C1) = 2Vin(peak)

after 3 half-cycles, Vout = V(C1)+ V(C3) = 2 3Vin(peak).

This is all about Kirchhoff's voltage law.

Am i correct?

Thomas

its a tripler so answer should be 3x

JR

Of course, my math gone wrong,

Vout = V(C1) + V(C3)

since V(C1)= Vin(peak) and V(C3) = 2Vin(peak) it's obviously Vout = 3Vin (peak)

Sorry, it's pretty late here 

Thomas

fatheaddrummer said:

Can anyone tell me how the principal of a voltage tripler works - step by step, what is really happening? I don't really get it and cannot find something useful in the internet about that.

Thank you so much!
Christian

Click to expand...

OK step by step...

#1 understand the basic components.

A diode only passes current in one direction. A (relatively large) capacitor passes AC current in both directions.

#2 divide and conquer.

First stage C1 and D2.... The input labelled + is really an AC signal that swings + and -. The unlabeled input lead connected to the cathode of D2 is ground or 0V.  As the AC signal at the input swings + and - the cap output tries to follow, but when the right or negative end of C1 tries to swing positive, D2 conducts and clamps that terminal to approximately 0.5V positive.  However in the negative direction the diode does not conduct, so the cap swings freely in the negative direction. After a few cycles the output side of C1 is swinging the same AC voltage just shifted negative by a DC voltage equal to the peak input . 

The next stage is C2 and  D1. The left positive end of C2 is connected to 0V so does not change. The negative side of C2 is connected through diode D1 to the output of the first stage. Since diode D1 conducts when the output of the first stage swings negative, capacitor C2 gets discharged down to -2x the input peak voltage.

The final stage C3 and D3 has C3 connected to the output of the first stage that swings the full AC voltage just shift down by the rectified DC voltage. Diode D3 biases the output side of the cap C3 down 2x the peak DC voltage so 2x DC bias + 1x from the AC swing generates the 3x voltage tripler.

The output of this tripler is still an AC voltage so you need another cap and diode (like the second stage) to extract a DC voltage from this.  In fact you can stack any number of stages to generate Nx multipliers.

There are a number of other practical considerations when trying to pull significant output current since the simple tripler will pull 3x the output current in the input stage. 

These circuits are popular for making low current higher voltage supplies like 48V phantom  that only draw a few mA.

JR 

> after 3 half-cycles, Vout = V(C1)+ V(C3) = . . . . . . .

Without actually looking at the reference (late here also):

Usually the D-C string stands-on one side of the input, so at the third step:

Vout = V(in) + V(C1) + V(C3) =

Which is roughly "3" in most lands.

Note also that this is 3 times the Peak input voltage. A "10V" winding is usually Sine RMS (because that is what the electric company bills us for lighting and heating). The diode/capacitor networks grab the Peak, 1.414 times higher than the RMS. So far, a "10V (rms)" winding gives 42.4V out.

There's three diode drops, passing large surges, so expect 1V drop each. A bit under 40V.

These large surges also cause large drops in the transformer resistance. For small iron this can be a LOT.

The caps sag between surges, the lowest point of the sag is critical to a regulator, and there's no good simple way (AFAIK) to pick these sizes.

If your "tripler" actually gives 3X the *RMS* under load, you are doing good. If it does not give 3X RMS, it has real problems which could include overheating or surge-damage.

Thank you all very very much - got it now!!

Best,
Christian