Voltage Double

[clel miller]

You guys are going to hate me.....
OK, I just got through reading a guys post (that had the same confusion as me...I think)...and I am still confused.

I can see how this schem would 2x the voltage...but not when there is a load connected.

Right now, I have two questions regards when a load is connected across the V Out:
1. As the AC flips polarity... does the non-charging cap discharge through the forward biased diode, and then through the load.?
2. If so, how much discharge is there, and what does that do to the 2x voltage.?

Thank You

 

[crutschow]

1. NO. When the circuit has operated for a few cycle, C1 is charged to the input positive peak voltage and C2 to the negative peak voltage. Any load will simply discharge the two capacitors, so the amount of voltage drop between the recharge from the AC input each cycle depends upon the load current and the value of the capacitors.

2. The ripple voltage drop between half cycles (for 60Hz power) is roughly Vdrop = (Iload * 8.3ms) / C (for C = C1 = C2).

Below is a simulation:

 

[clel miller]

I looked for some vids that suggested the path of travel I had in my head... and found this old Navy/Airforce video circa 1970.?
So is this guy wrong.?
At about 9:45 the pictorial shows just what I thought was taking place, and what made me wonder how the voltage doubles with a load present.
Thanks

 

[BobW]

These circuits have to be designed so that the diodes charge the capacitors much faster than the load can discharge them. The voltage is really only doubled when there is no load. As load is increased the voltage multiplication effect decreases. That's why these circuits are generally only used where the load current is very small.

 

[audioguru]

The voltage is almost doubled because the positive peak of the input charges C1 and the negative peak charges C2. The capacitors are in series so the output is almost double the peak voltage of the input sinewave. The load discharges the capacitors when they are not being charged so their values affect the average output voltage.

 

[clel miller]

These circuits have to be designed so that the diodes charge the capacitors much faster than the load can discharge them. The voltage is really only doubled when there is no load. As load is increased the voltage multiplication effect decreases. That's why these circuits are generally only used where the load current is very small.

Without knowing...I suppose this sums what I have been thinking.
Guess for most applications you would need to build some type of a regulated power supply.?
Thank You

 

[audioguru]

I have a cheap Chinese wall wart with an output of 9V at 200mA. Without a load its output is 17.5VDC and with a 200mA load the output is exactly 9.0VDC. If it is used with some circuits then a voltage regulator is needed.

I have a good quality wall wart with an output of 9VDC at 500mA. Without a load its output is 10.5VDC and with a 500mA load the output is exactly 9.0VDC. If it is used with ANY circuit then a voltage regulator IS NOT NEEDED.

 

[crutschow]

Without knowing...I suppose this sums what I have been thinking.
Guess for most applications you would need to build some type of a regulated power supply.?

The output has ripple similar to a normal rectifier/filter where the ripple increases with the load current. Normally you use a regulator after the filter to give a nice, steady DC voltage (with regulated output voltage below the minimum ripple voltage). You can do the same thing with a voltage doubler output.

 

[Dean Huster]

You aren't limited to doubters here. This is in a class called voltage multipliers. They're common in oscilloscopes, especially when encapsulated in the the form of the PDA multiplier. By adding more sections onto the basic doubled, yo u can have a voltage tripled, quadrupled, quadrupled, decoupled or more. I've made decoupled that would light a neon lamp when driven by the output of a 555 timer. As a family, they aren't expected to deliver a lot of current since they operate on the ability of the caps to hold a charge. The bigger the caps, the higher the current capability of the driving across supply, the heavier the load it will support.