# Increasing Line 120VAC to 340VDC

Dec 17, 2013
78
Xenon Flash Circuit (How to get 400 VDC)

Hello Everyone,

Project is a relatively high power xenon flash unit (200 joules/200wattsecond flash tube). Tube requires 300VDC minimum for flash tube, and 4kV+ for trigger. Ideally, I'd like to provide closer to 400VDC.

By my math, a 200ws/200 joule flashtube will need a bank of capacitors with a total capacitance of 2500uF at 400VDC.

Any recommendations for a power supply? How about recommendations for getting the 400VDC to charge that capacity bank?

I'd like to be able to be able to fire a third-power (1000uF) 400V flash once a second or so.

Any recommendations are greatly appreciated. Size and weight are somewhat important, as the finished unit will be on a tripod-stand.

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Dec 17, 2013
78
I'm considering a 19V 3.5A laptop power supply, because it's affordable, available, usable internationally, and it's common. Bonus points for size/weight as it can be used externally.

Thoughts on this power type of supply for this project?

#### duke37

Jan 9, 2011
5,364
200J each second = 200W

The power supply you suggest is only 66W

A boost converter to go from 19V to 400V may perhaps be purchased or made. Do not expect an efficiency of much over 60%.

Dec 17, 2013
78
Hi Duke,

You make a good point and it's one I've considered.

The flash tube is rated to 200 joules, but I'll almost never be using it at that capacity. It's far too bright. More often that not, I'll be using it at somewhere between 1/6th and 1/3 power.

To do this, I'll be dividing the capacitor bank into 6 identical caps in parallel and switched. I will engage/disengage each as I need them.

Does that make sense?

I have no experience with boost converters but at 60% efficiency, I think I'll need a bigger power supply. Do you have any recommendations? Ideally, the shorter recharge cycle the better, but as I said, only about 1/3rd power in most situations at 1s intervals.

#### OLIVE2222

Oct 2, 2011
690
You can have high current 340V DC quite easily with an 120/240V isolation transformer.When rectified and filtered you will have it.

#### duke37

Jan 9, 2011
5,364
Yes,

The mention of a laptop power supply threw my thinking off line.

If isolation is not required, then a voltage doubler from 120V mains would do without a heavy transformer. Excellent insulation would be needed.

The doubler would consist of a capacitor and two diodes only. It would have inherent current limiting which is necessary when the storage capacitor is discharged.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
You can have high current 340V DC quite easily with an 120/240V isolation transformer.When rectified and filtered you will have it.

I have done something very similar. I placed a light bulb in series with the input side of the transformer to limit the maximum current.

As a side effect, it tells you when the capacitors are charged

#### Fish4Fun

##### So long, and Thanks for all the Fish!
Aug 27, 2013
471
Hrrmmm, I understand 200J/S = 200W, and I understand that the OP stated he wanted to charge a bank of 2500uF capacitors to ~400V, AND I understand charging 2500uF to 400V = 1C; AND I get that it takes 400J to charge the Capacitor Bank, AND I get that 400J/Sec ==> 400W.....but there is an elephant in the room everyone seems to be missing:

The FLASH duration should be somewhere between the uS time domain and the mS time domain implying that the capacitor bank needs to be MUCH SMALLER.

The only real electrical limit to how short a pulse can be is the total-system inductance, including that of the capacitor, wires, and lamp itself. In contrast, changes in the input voltage or capacitance have no effect on discharge time. Short-pulse flashes require that all inductance be minimized. This is typically done using special capacitors, the shortest wires available, and electrical-leads with a lot of surface area but thin cross-sections. For extremely fast systems, low-inductance axial-leads, such as copper tubing, plastic-core wires, or even hollowed electrodes, may be used to decrease the total-system inductance. Dye lasers need very short pulses and sometimes use axial flashtubes, which have an annular shape with a large outer diameter, ring-shaped electrodes, and a hollow inner core, allowing both lower inductance and a dye cell to be placed like an axle through the center of the lamp.

The amount of power loading the glass can handle is the major mechanical limit. Even if the amount of energy (joules) that is used remains constant, electrical power (wattage) will increase in inverse proportion to a decrease in discharge time. Therefore, energy must be decreased along with the pulse duration, to keep the pulsed power levels from rising too high. Quartz glass (1 millimeter thick per 1 second discharge) can usually withstand a maximum of 160 watts per square centimeter of internal surface-area. Other glasses have a much lower threshold. Extremely fast systems, with inductance below critical damping (0.8 microhenries), usually require a shunt diode across the capacitor, to prevent current reversal (ringing) from destroying the lamp.

The limits to long pulse durations are the number of transferred electrons to the anode, sputter caused by ion bombardment at the cathode, and the temperature gradients of the glass. Pulses that are too long can vaporize large amounts of metal from the cathode, while overheating the glass will cause it to crack lengthwise. For continuous operation the cooling is the limit. Discharge durations for common flashtubes range from 1 microsecond to tens of milliseconds, and can have repetition rates of hundreds of hertz. Flash duration can be carefully controlled with the use of an inductor.[1][11]

The flash that emanates from a xenon flashtube may be so intense that it can ignite flammable materials within a short distance of the tube. Carbon nanotubes are particularly susceptible to this spontaneous ignition when exposed to the light from a flashtube.[20] Similar effects may be exploited for use in aesthetic or medical procedures known as intense pulsed light (IPL) treatments. IPL can be used for treatments such as hair removal and destroying lesions or moles.

Taken From: http://en.wikipedia.org/wiki/Flashtube Section Entitled: "Intensity and duration of flash"

The question that should be asked APPEARS to be, "how long a flash duration" is desired, and is the xenon lamp rated to handle this. How do you plan to TRIGGER the xenon lamp? etc. But @ the end of the day, I just can't see any Off-The-Shelf xenon tubes withstanding 200J/S to 400J/S for a FULL second.

That being said, if a single flash will suffice, and the duration chosen is 1mS with a re-charge time of < 999mS (to meet the OP's 1 second re-charge requirement), then: 400J/S * 0.001S = 0.4J/Second Average = 400mW, certainly within the capabilities of a 66W supply even @ < 1% efficiency.

Now, since Q = CV, and J = QV ==> For J = 0.4 and V = 400V --> Q = 0.001 so, 0.001 = 400 * C --> C = 0.001/400 = 25uF. A much more reasonable capacitance @ 400V. To charge a 25uF capacitor to to 400V with an input of 66W @ 1% efficiency (ie 6.6W @ 400V) implies an output current of 16.5mA (0.0165A). Since 1 Amp = 1 coulomb/second and we require 0.001 coulombs, we get --> 0.001q/0.0165q/s = 60.6mS to re-charge, WELL within the OP's specifications.

Now, the other elephants in the room seem to be the "trigger voltage" of 2kV-150kV, physical layout, stray inductance and cooling, not to mention the switching mechanism employed. While ALL of these can be solved with careful engineering, I question the viability of the project as a "DIY ** one-off **". A VERY nice flash can be purchased for < $500, and I can't see resolving any of the issues this project presents in less than 500 hours, suggesting that the OP's time has to be worth less than$1/hr before this project is even remotely fiscally viable, but then again, I frequently spend hundreds of hours on less-worthy projects, so who am I to throw stones? hehe.

Fish

#### duke37

Jan 9, 2011
5,364
I have only played with electric fencers which have pulses of around 1J, triggered at each second. That is plenty to deter me from touching the wire.
The capacitors are non-polarised pulse capacitors, electrolytic capacitors are not suitable.

I was under the impression that the energy in a capacitor is C*V*V/2.
If the capacitor is charged each second then the average input wattage equals the energy in joules. The instantaneous output wattage will be much higher since the time is much shorter.

Pulse transformers to trigger the tube are available, I have salvaged one out of a disposable camera.

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
The FLASH duration should be somewhere between the uS time domain and the mS time domain implying that the capacitor bank needs to be MUCH SMALLER.

It depends on what effect you want.

There is an interesting relationship between the pattern of light output and the voltage of the capacitor bank. See here. It is typical for standard photographic flash units to have an effective flash duration in the order of 1ms at some settings. For studio flash units, the duration can be significantly longer.

It is somewhat counter-intuitive that the time it takes for the light to fall to 10% of the peak is longer when the voltage is lower.

The choice of voltage and capacitance allows you to determine peak intensity, colour temperature, and also flash duration.

If this device is not being used for photography, the second may be of little importance, and the third (the duration) will effect the perception of the first (brightness).

If it is just being used for "effect" then a comparatively low voltage and a large capacitance will *possibly* seem brighter as our eyes tend to integrate the light over a period of time rather than respond to the peak intensity.

However, driving these things with an average power of 400W for any length of time is going to make things get *very* hot and the life of the tube (not to say that of the capacitors) is likely to be short.

I've heard the hiss of xenon escaping from an overheated flash tube and it is an expensive sound (or was for me).

#### Fish4Fun

##### So long, and Thanks for all the Fish!
Aug 27, 2013
471
@-->Steve,

Not arguing anything about flash duration, I simply picked 1mS as a calculation point. Your link suggests that 1mS is roughly 1/3 their "longest flash period" which (appears from the graphs) to be ~3mS (and in the literature 3.33mS). This does not fundamentally change my assertion that a 66W power supply SHOULD be more than adequate to "flash 1 time per second". I would think in a carefully designed circuit a 10W power supply would be more than adequate to exceed the OP's requirements.

hehe, I should also point out that the link's HIGHEST price flash is:

White Lightning™ X3200 Flash Unit

@ $549.95 Followed By: The Einstein™ E640 Flash Unit @$499.95, confirming my GUESS that a VERY NICE Flash wouldn't be much (if any) over $500. Again, not picking, just think this is a project that should likely end when the UPS truck arrives..... I was under the impression that the energy in a capacitor is C*V*V/2. YEP, and I missed a "0". Should have been 2.5uF, NOT 25uF. 0.4 = 0.0000025 * 400^2 Sorry. Fish #### adamq Dec 17, 2013 78 I have done something very similar. I placed a light bulb in series with the input side of the transformer to limit the maximum current. As a side effect, it tells you when the capacitors are charged This is very handy! there is an elephant in the room everyone seems to be missing: The FLASH duration should be somewhere between the uS time domain and the mS time domain implying that the capacitor bank needs to be MUCH SMALLER. The question that should be asked APPEARS to be, "how long a flash duration" is desired, and is the xenon lamp rated to handle this. The flash is salvaged from another photographic studio flash unit, and rated at 200ws. Specifications state full power flash is 1/300s, 1/64 power flash duration is1/9,500s. Am I right with these calculations: 200 watts per second = 200 j (1 second) If you cram 200J into the 1 second duration: 200j = 340VDC x 0.58A x 1 second Worst case scenario, if you cram the 200J into the max power flash duration of 1/300 second: 200j = 340VDC x 177A x 1/300 second 177A sounds like a lot of power, but consider that the duration is 33ms. Also consider that's the highest amperage for the longest duration this flash should ever see. More likely situations are in the area of 1/30th of that current: roughly 6amps? In either case, it's a surge of power through a circuit with little resistance from a bank of charged capacitors. Am I totally off here? How do you plan to TRIGGER the xenon lamp? Pulse transformers to trigger the tube are available, I have salvaged one out of a disposable camera. ^ This... But @ the end of the day, I just can't see any Off-The-Shelf xenon tubes withstanding 200J/S to 400J/S for a FULL second. This isn't an "off the shelf" from your local electronics store xenon strobe. This is a photographic strobe specifically designed to output high energy. Now, the other elephants in the room seem to be the "trigger voltage" of 2kV-150kV The trigger coil in a disposable camera easily generates >4kV from a 1.5v battery. not to mention the switching mechanism employed. I was thinking of using an octocoupler (MOC3020), HV diodes, a HV trigger coil cap, HV SCR, and a momentary type trigger switch. /While ALL of these can be solved with careful engineering, I question the viability of the project as a "DIY ** one-off **". A VERY nice flash can be purchased for <$500, and I can't see resolving any of the issues this project presents in less than 500 hours, suggesting that the OP's time has to be worth less than $1/hr before this project is even remotely fiscally viable, but then again, I frequently spend hundreds of hours on less-worthy projects, so who am I to throw stones? hehe. Fish I know the alienbees and Paul C Buff lights are affordable and good quality. I have used them extensively as a photog. This project is as much about proving I can do it at an even lower cost as it is about my need/want for a flash unit. Don't worry about the economics, that's my crux to bear It depends on what effect you want. There is an interesting relationship between the pattern of light output and the voltage of the capacitor bank. See here. It is typical for standard photographic flash units to have an effective flash duration in the order of 1ms at some settings. For studio flash units, the duration can be significantly longer. ... The choice of voltage and capacitance allows you to determine peak intensity, colour temperature, and also flash duration. If this device is not being used for photography, the second may be of little importance, and the third (the duration) will effect the perception of the first (brightness). It will be used as a photography flash. The voltage I will be using will be constant. The only control I have over the flash is the amount of power I choose to give it (capacitance). @$499.95, confirming my GUESS that a VERY NICE Flash wouldn't be much (if any) over \$500.

Again, not picking, just think this is a project that should likely end when the UPS truck arrives.....
Ah but that's no fun!!!

As an aside, I love all this back and forth. What a great forum! Already so much input and knowledge, problem solving and brain storming!

Dec 17, 2013
78
Here are the images reflecting flash output at varying voltages and power levels, and also shows how an IGBT can be used to shorten "full power" flash.

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Dec 17, 2013
78
A chart showing comparable flashtubes (200ws) requiring 2000uF capacitance for full discharge.

Dec 17, 2013
78
OK guys,

Based on all the feedback and suggestions here, and some research, this attached circuit is what I've come up with so far.

I've also created a project page for myself to track progress and share it with others.

Please give me your feedback if you have any. I'd really like to hear what you guys have to say.

Any design recommendations are greatly appreciated.

This is my first circuit so please keep that in mind

Please recommend diodes, fuses, resistors, etc wherever you think they should be

Some reasoning behind the circuit:

R1 is 75ohm, 50W. This is a power resister to limit recharge rate of capacitor bank.

IC1 is a MOC3020 octocoupler. Provides 7500V of isolation between trigger circuit and trigger switch.

SCR switches power from the 300VDC to the trigger coil via the closing of the trigger circuit.

T1 is a trigger coil transformer that I will either wind myself, or (more likely) I'll be salvaging from a disposable camera.

X1 is the 200wattsecond (joule) photo flashtube

The 240VAC is being supplied by an isolation step-up transformer (200W).

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#### Fish4Fun

##### So long, and Thanks for all the Fish!
Aug 27, 2013
471
Am I right with these calculations:
200 watts per second = 200 j (1 second)

If you cram 200J into the 1 second duration:
200j = 340VDC x 0.58A x 1 second

Worst case scenario, if you cram the 200J into the max power flash duration of 1/300 second:
200j = 340VDC x 177A x 1/300 second

177A sounds like a lot of power, but consider that the duration is 33ms. Also consider that's the highest amperage for the longest duration this flash should ever see.

More likely situations are in the area of 1/30th of that current: roughly 6amps?

In either case, it's a surge of power through a circuit with little resistance from a bank of charged capacitors.

Your MATH is correct, your equations are NOT representing what you want. First Watt-Second (ws) is a measure of power over time and is equivalent to Joules, similar to the kWh used on your power/hydro bill NOT Power per Unit of Time. @ Any Rate 1 Watt = 1J/1Sec. 1ws = 1j/s * 1s = 1J. So, 200ws = 200J.

As noted above Ej = C*V^2

Where:
Ej is Energy in Joules
V = Potential Difference in Volts

So IN THEORY:

C = 200 / 340^2 ==> C = 0.00173F or 1730uF

BUT this assumes the capacitor Starts @ 340V and Finishes @ 0V. IF the the tube stops conducting @ say 100V (I have NO IDEA if this occurs, it is just a supposition that there is some minimum Voltage required.) then we have to do some more figuring to get our total CONSUMED energy to 200J.

((100V)^2) * 2000uF = 20J

So we would actually need 220J @ the START of the discharge cycle.

220J / .002 = 110,000 = V^2 ==> V = 331V

Now, if the Flash Duration is 3.3mS (1/300) then we we have 200J/3.3mS (or more simply 200J * 300) --> 60,000W (60kW) with an average current during discharge of 60kW/((331V + 100V)/2) ==> 272A.

Now, ASSUMING you used the IGBT method of STOPPING the flash after a fixed interval of 1mS, but you still wanted FULL Power:

200J * 1000 = 200kW (BUT remember watts are a measure of Joules/Second, and here we only supply power for 1mS)

@ 1/9500s we are @ roughly 500uS (1/10000). ASS_U_MEing you still want the FULL POWER delivered (and I KNOW this is NOT the case) ==>

200J * 10,000 = 2Mw! (Again, the AVERAGE POWER is still only 200W, 2Mw is the PEAK POWER) And if the Voltage is the same as above, 2Mw/((331 + 100)/2) = 9070A!

As per charging, 200J refreshed once per second = 200W. NO getting around that. Using a line driven Step-Up Transformer would require a 200VA transformer @ the MINIMUM, and should likely be > 250VA. This is a fairly MASSIVE transformer @ mains frequencies of 50-60hz! As mentioned previously, using a voltage doubler to take 120Vac to 240Vac would give you peak voltages of 240Vac * 2^1/2 = 339V. BUT BE WARNED: Capacitive Coupling from MAINS to ANY CIRCUIT is DANGEROUS! MAKE SURE YOU ARE VERY CAREFUL! GALVANIC ISOLATION IS ALWAYS SAFER!

As a SIDE NOTE: Not all capacitors of the same voltage/capacitance value are EQUAL. You will need extremely LOW ESR with VERY HIGH current ratings. I certainly don't have any experience in the selection of capacitors like these; obviously there is a solution, I just don't have a clue about the "how".

As a quick example: the power dissipated in a purely resistive wire or PCB trace==>

Pd = I^2*R
Where:
I is the current in Amps
R = Resistance in Ohms

20ga copper wire has a resistance of ~10.15mOhms/ft. So a 3in 20ga wire leading from your capacitor bank to your Flash Tube would have a resistance of:

0.01015Ohms/ft * 3in * 1ft/12in = 2.5375mOhms.

And the Power Lost in the wire would be:

@ 272A ==> 272^2 * 0.0025375 = 187.7W
@ [email protected] ==> 9070^2 * 0.0025375 = 208.7kW!

Inductance in the same piece of STRAIGHT wire would be: 79nH and could certainly influence circuit behavior.

As I said @ the end of my last post, I cannot imagine this project being fiscally viable; HOWEVER, I Certainly understand wanting to play with it! JUST BE CAREFUL! You are playing with very dangerous voltages & currents. A small mistake could result in a life threatening outcome.

I Guess now I AM on the same page as everyone else. I had NO IDEA a flash consumed so much power. I am still having a hard time wrapping my head around it. Powering a flash from a battery pack seems highly improbable if this Flash is representative of MOST flashes. A 1AH battery pack @ 12V = 12Whr ==> 12Whr * 3600s/hr = 43,200J Assuming 20% efficiency in conversion, this would imply 8,640 delivered Joules, or roughly 43 "Flashes" per charge. "AA" Batteries have a 440mAH to 900mAH rating, so 4 * 1.5V = 6V, & 6V * 600mAH = 3.6Whr * 3600 = 12,960J. Assuming a "cheap" camera uses roughly 1/4 the power of the flash in question, and assuming the same 20% efficiency in power conversion: (12,960 * .2)/50 ~ 52 Flashes per battery set, which sounds optimistically high, but at least the right order of magnitude

Thanks! I have enjoyed playing with this

Fish

#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
Fish, the itty bitty flash on a disposable camera dumps about 10J of energy through the flash tube. My "good" on camera flash has about 60J at its disposal.

200J for a floor mounted studio flash is nothing, I've used several rated at 1000Ws, although you rarely used them at full power. These days everything is stuck into the head, and here is an example of a 1200Ws unit.

These things discharge the capacitors *FAST*. Even those old disposable cameras had current peaks in the order of 50A!

The boost converters are rather more than 20% efficient. Thy don't need to worry about regulation much, except as it allows them to drive a capacitive load.

Adamq, you're clearly looking at using an appropriate flash tube, and as long as you keep within the recommended voltages and do not exceed the recommended capacitance or dissipation limits, you will end up with a design which will both deliver you a correct colour temperature and not prematurely destroy the tube.

A further gotcha is that once the tube starts to conduct, it will do so until the voltage across it falls below some value that is somewhat variable and dependant on the temperature of the xenon etc. What can happen with a simple charging circuit is that the charger could be able to maintain this voltage and hence the tube will go into continuous conduction. You may not be able to see this and it can cause your tube to get quite hot. My rule of thumb was that my charger should not be able to charge the capacitors faster than about once per second or so. Incidentally, having a light bulb to limit the charging current is useful here as it will continue to glow in this case.

I recommend that your light bulb current limiter is placed before the bridge rectifier, not after it.

Dec 17, 2013
78
Adamq, you're clearly looking at using an appropriate flash tube, and as long as you keep within the recommended voltages and do not exceed the recommended capacitance or dissipation limits, you will end up with a design which will both deliver you a correct colour temperature and not prematurely destroy the tube.
Thanks for the kinds words! I'm excited.

I'm having difficulties triggering it right now, but it's pretty well assembled as it is. I'm picking up fresh parts tomorrow morning to try and finish the testing tomorrow.

A further gotcha is that once the tube starts to conduct, it will do so until the voltage across it falls below some value that is somewhat variable and dependant on the temperature of the xenon etc. What can happen with a simple charging circuit is that the charger could be able to maintain this voltage and hence the tube will go into continuous conduction. You may not be able to see this and it can cause your tube to get quite hot. My rule of thumb was that my charger should not be able to charge the capacitors faster than about once per second or so. Incidentally, having a light bulb to limit the charging current is useful here as it will continue to glow in this case.

I recommend that your light bulb current limiter is placed before the bridge rectifier, not after it.

That's a good and interesting point. Any specific recommendation for bulbs? I'm not sure what kind of bulb would be best.

Other than a visual indication (which is very valuable), does an indicator bulb provide any other benefit than the current power resistor I have in place (75R 50Watt)?

Could I put a small incandescent in series before the power resistor?

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#### (*steve*)

##### ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
Moderator
Jan 21, 2010
25,505
The advantage of a light bulb is that they are rated to place across the mains (you use a normal mains type light bulb) and that the resitance characteristic means that you can supply more current at lower differential voltages (it's not-linear n a protective way).

Also Lightbulbs don't fail because they get too hot.

Try a 75W light bulb.

Dec 17, 2013
78
The advantage of a light bulb is that they are rated to place across the mains (you use a normal mains type light bulb) and that the resitance characteristic means that you can supply more current at lower differential voltages (it's not-linear n a protective way).

Also Lightbulbs don't fail because they get too hot.

Try a 75W light bulb.

OK. I'll make use of standard 75W bulb for testing purposes until I'm sure the charging circuit isn't an issue. It'll be too big for the end product, but it'll be a great help for testing.

Thanks!

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