# Voltage Doubler problem

Discussion in 'General Electronics Discussion' started by kaijorg, Jun 15, 2016.

1. ### kaijorg

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Jun 15, 2016
Hi, I am building a xenon flash system for photography. The first part I need to build is a voltage doubler which will get it's power from standard 110v wall outlet. I have followed the schematic but upon testing it the 100ohm 1watt resistor burned up. I believe I have the diodes in correctly and the caps are correct I think (stripe on cap is neg?).There are three things I can think of but could use some help.
1-The schematic is not correct in specifying a 1watt 100ohm resistor
2-Diodes are in backwards
3-This circuit cannot be powered up without the rest of the circuit in place.
I wanted to build this portion and confirm I was getting 340volts dc before moving to the next parts which will include the charging circuit for the main capacitors and the trigger circuit. Attached is the schematic I am working from. Thank you.

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2. ### Harald KappModeratorModerator

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Nov 17, 2011
A 220µF capacitor at 60 Hz has an impedance of ~12Ω. In series with the 100Ω resistor the initial charge current will be at least 120V/112Ω~1A. 1A on 100Ω equals a peak power of P=I*R?100W. At 1W your resistor seems to be a bit underrated. Of course, the current will depend on the exact point on the input sine wave when you turn the circuit on and the current will also decrease as the capacitors get charged. By that time, however, the resistor will already have given his life away to the overloa(r)d

You will need at least 1kΩ or more at the cost of a longer timespan until the capacitors are fully charged. Or use a much more powerful resistor.

3. ### hevans1944Hop - AC8NS

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Jun 21, 2012
This is a standard half-wave voltage doubler circuit. The diodes and capacitors are drawn correctly. The purpose of the resistor is to limit the charging current of the two capacitors. Each diode conducts alternately on the positive and then the negative half-cycles of the power line, charging the capacitors to the peak of the line voltage or approximately 170 V DC across each capacitor. The capacitors are connected in series, so the total voltage is 170 + 170 = 340 V DC. Since you plan to use this with a xenon flash system for still (single frame) photography, the time to charge the capacitors is not very important. You could increase the resistor value to decrease the initial charging current and the power dissipated in the resistor.

If this results in too long a time to re-charge the capacitors between flashes, decrease the value and perhaps increase the wattage rating of the resistor. Most commonly, a resistor is not needed at all, but that means a large "surge" current will occur through the 1N4007 diodes which are rated for 30 A for a one-cycle surge, so assuming the capacitors will fully charge in one half-cycle of the AC line voltage, the current-limiting resistor could be as small as 170 / 30 = 6 ohms, dissipating an instantaneous power of about 5400 watts for a duration of less than 8 ms at 60 Hz line frequency... about 43 J of energy. Note this is worst case. An uncharged capacitor has zero apparent resistance, so the instantaneous current can be as much as the peak AC line voltage (170 V) divided by the value of the current-limiting resistor if power is applied at the peak of the AC line cycle.

Unfortunately, connecting this circuit directly to the mains power means there is virtually no way to limit the surge current without using a resistor. It's possible that the resistor should have been specified at 1000 ohms at 1 watt. This would limit the current to 170 /1000 = 0.17 A and an instantaneous power dissipation of about 29 watts during the half-cycle or so required to charge the capacitor... less than a quarter joule of energy to dissipate.

Once the capacitors are charged, no further current should flow through the current-limiting resistor, so power dissipation will be nil until the flash-tube discharges the capacitors and the power line voltage begins to re-charge them again. If you make the resistor too large, it will require more than one half-cycle to charge the capacitor, but the peak current always occurs first and decreases on subsequent half-cycles. Try using a 1 kΩ, 1 W resistor.

This circuit is dangerous if operated directly from mains power. You should use a 1:1 isolation transformer to provide line voltage to the diodes. If you do that you won't need the current-limiting resistor.

Last edited by a moderator: Jun 15, 2016
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4. ### davennModerator

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Sep 5, 2009
@kaijorg

please note very carefully the last statement hevans1944's post
I have hilited and bolded all of it

( Hop ... hope you don't mind )

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5. ### BobK

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Jan 5, 2010
The watt rating of a resistor is for continuous operation. They can survive much higher power for short periods. I would not assume that the resistor is really burning in this circuit if it is built correctly, I would look for a short.

Bob

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6. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Yeah, most common is installing the electrolytic capacitors with their polarity reversed. That will do it every time and may cause the capacitors to explode if current-limiting resistor doesn't burn out first, acting somewhat as a fuse. Please don't power this circuit directly from the AC mains.

7. ### kaijorg

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Jun 15, 2016
Thank you for the input. After going over my build I discovered that I had installed the diodes backwards. The capacitors were installed correctly and no short in the wiring. After reversing the diodes and installing a new resistor (same rating as before) no problem with my test of about 1 minute.

I would like to add the 1:1 isolation transformer but could use some guidance on getting the correct one. Any help with where and what specs will be of great help. The transformer needs to be as small and light as possible and meet the requirements.

Thank you again, this has been a big help.

8. ### (*steve*)¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥdModerator

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Jan 21, 2010
instead of a resistor, use a light bulb. When the light goes off the capacitors are charged.

I would recommend you use an isolating transformer so your circuit changes from scarily lethal to plain old lethal.

scary lethal means you can be killed by coming into contact with any part of the circuit.

plain old lethal means it won't kill you unless you come into contact with at least two parts of the circuit.

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9. ### CDRIVEHauling 10' pipe on a Trek Shift3

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May 8, 2012
I would think there are safer low voltage (source) methods to achieve the high DC voltage for xenon flash systems. Cheaper than finding a suitably small 1:1 120V transformer too.

Chris

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10. ### hevans1944Hop - AC8NS

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Jun 21, 2012
Correcto! However, a mains-operated voltage doubler will allow incredibly fast xenon flashing rates, especially if the follow-on circuit has inductor/SCR energy management of the xenon tube discharge. It depends on the minimum cycle time requirement, which was not specified.

I once built a high-current DC power supply for a short-arc, high-pressure, xenon lamp that provided about 300 watts continuous power to the forced-air cooled lamp. Fun project. Hardest part was initially triggering the lamp arc by using a series-connected pulse transformer. Most short-arc high-pressure lamps don't have a trigger electrode. So we found a pulse transformer with a low-impedance secondary and made a relaxation oscillator to pulse current through the primary until the arc formed. Takes a few seconds to do that, and the arc voltage changes as the lamp heats up and the pressure inside builds up, so some fancy feedback required to control the lamp current. After building it, I placed the lamp and its collimating optics in a fourth-floor window of our electronics lab and pointed it toward Stewart Street, which I drove down every day to work. Let it run continuously for about a week to see if anyone would notice and comment on it, but AFAIK no one did.