12vdc to 470vdc converter for photoflashes

hello all, I am wanting to build a circuit to power my flashpoint 320m monolights from a 12v battery. these come factory ready for direct dc input. the battery pack transformer is 125$ from the factory. specs are as follows, each flash has a maximum output of 150ws, the factory version can run 2 flashes from 1 pack. The magic box outputs 470vdc at full charge. I am still learning but can build from a schematic even if I dont completely understand the magic inside. what is the best way to go about converting 12vdc up to 470vdc?

I was thinking a simple oscillator coupled to an appropriate transformer and a large cap to buffer the charging cycle of the internal caps with a hefty diode bridge to output the dc to the flash units.

Am I on the right track here?

You are on the right track. An circuit that uses the inductance of the transformer will be easier to make and will work better.

**broken link removed**

What recovery rate do you want?

Where did you come up with the 470 VDC number. I looked at the FP2PP unit they get about $124 for but I can't find what the thing actually outputs? I know it works around a NiMH battery but that was as good as it got.

They claim 400 flashes of 150 WS so with two heads (full power) that would be about 200 flashes and most reviews claim it won't meet the 400 flashes. So who knows on that note.

I did find a few references to using 12 volt to 120 VAC 60Hz inverters but this requires a pure sine inverter and not the typical modified square wave flavors. This 300 watt inverter seemed to be popular and showed up in this video. Then too, the inverter with battery would be large to lug around. I didn't price shop for the inverter. The Samlex 300 Watt Pure Sine Wave Inverter in the link was $159. More than the FP2PP battery pack sans battery and charger.

Then as Colin mentions:

What recovery rate do you want?

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If we assume the 470 volt range to be the correct DC input to the units, how much current for a good recycle time?

Ron

I am not too sure what the current is although I suppose I could figure it out as the flash will output 150 watt seconds of light with each flash at maximum output. I would like it to charge up in 2 seconds or less ideally when charging a single flash 5 seconds or so on two units. I got the 470 dc voltage output and pinout from a someone else's working unit.

I have a big inverter (1500 watt pure sine wave) I can use nearby the car, however it weighs in at about 100lbs. I would like the portable version to carry in to more remote locations using a smallish rechargeable battery.

Update>

Ok not sure if my math is correct but I come up with a hair over 1/3 amp at 470 v for a 150ws flash, so to be safe we could call it a full amp for two units.

Diver300 I went through that link you gave and it is very interesting, not that I was able to understand the whole thing. but using the collapsing field to generate the high voltage seems much more efficient, probably why when I was searching for a compatible transformer I couldnt find one with an appropriate coil count. I have tons of transformers in my junk box (ok its a small room). would a flyback from a crt work or will it generate too high of voltages?

I remembered this ic from jaimie's ring flash project LT3750 - Capacitor Charger Controller - Linear Technology
this looks like a workable solution for a simple charging circuit, my one question is can I replace the 1:10 transformer with a 1:15 and get the higher voltage, my math comes up with about 450 v.**broken link removed**

Hi. I'm not sure about your maths up there (1st post). To charge up a cap to 150 W.s in 2 seconds, you need 75W for those 2 seconds; if the conversion is 75% efficient, you would need 100W for those 2 seconds. For a 12V input this is 6.25 - 8.3A (based on 100% / 75% eff).

If 1:10 gives 300V out then 1:15 will give 450V out. There can be problems with the charge completion detection if depending on the construction of the transformer - it was thought (at somewhere I worked) that the leakage inductance was responsible for producing a spike on the primary causing premature charge termination.

You can try using a small ferrite core but just make sure that it doesn't saturate (at your maximum expected current) when you have the appropriate primary turn count wound. If it saturates, you can change the core, or maybe add an air gap and some extra turns (to make up for the decrease in permeability).

I did my calcs on the 470v side, I completely forgot the huge decrease in voltage would make a huge increase in amperage on the supply side. Ok so the ic probably wont work for my usage as it tops out at 6 amps. In searching for the size of the internal cap I just came across a link to the company that manufactures the unit and they are selling the battery pack/ inverter for 88$ Im not sure I can make one cheaper than that: ) any suggestions of a circuit schematic that will do what I need?

CalebG said:

k so the ic probably wont work for my usage as it tops out at 6 amps

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The IC doesn't top out; the limit current is specified by the 12 milliohm current sense resistor - just change that value to increase the value. Note that the average input current will be slightly less than half the peak current (due to the ramp-shaped current in the inductor). You will have to choose a suitable MOSFET for the current (> Ipk) and the voltage (> Vin+Vout/N).

Due to the high current and high switching speed, you may wish to use a multifilar (a number of individual insulated conductors - see 'bifilar' or litz wire as an example) primary winding.

CalebG said:

In searching for the size of the internal cap

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For 460V, 150W.s, the cap size would be ~1500uF.

As far as you making the whole thing for under $88, well that's up to you

470VDC@150WSec...

"I am still learning but can build from a schematic even if I dont completely understand the magic inside."...

Why do I get the premonition of an impending smell of burning protein...and a candidacy for a Darwin Award?

Ken

Ken , I work as a lineman so I dont put my fingers where they dont belong. No stranger to electricity, just dont completely understand all the ins and outs of how to design a circuit on my own and have it work properly. the same company that makes the chip above also makes this one
**broken link removed**

"The LT3751 is a high input voltage capable flyback controller designed to rapidly charge a large capacitor to a user-adjustable high target voltage set by the transformer turns ratio and three external resistors. Optionally, a feedback pin can be used to provide a low noise high voltage regulated output.

The LT3751 has an integrated rail-to-rail MOSFET gate driver that allows for efficient operation down to 4.75V. A low 106mV differential current sense threshold voltage accurately limits the peak switch current. Added protection is provided via user-selectable overvoltage and undervoltage lockouts for both VCC and VTRANS. A typical application can charge a 1000μF capacitor to 500V in less than one second.

The CHARGE pin is used to initiate a new charge cycle and provides ON/OFF control. The DONE pin indicates when the capacitor has reached its programmed value and the part has stopped charging. The FAULT pin indicates when the LT3751 has shut down due to either VCC or VTRANS voltage exceeding the user-programmed supply tolerances."

So if I have a 1500uf cap this should be able to charge it in 1.5 seconds according to the specs posted. Anyone think this would not work well?

CalebG said:

Anyone think this would not work well?

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Either IC should be capable of charging up the flash cap. The former is simpler to wire up but the output voltage is harder to control (you'll have to add or remove turns from the transformer to adjust). The latter requires you to work out a few more resistances, but the output voltage can be adjusted by changing a resistor value.