Spike generator

[Andy Watson]

Won't work!
Consider the point labelled as "2" on the bridge rectifier, it can never get above one diode drop of "3", which is the other side of the capacitor that you're trying to charge.

Do you need full-wave rectification?

[Simon]

I need all the power I can get at 50V and less on the 100V. Should I just use a separate 2 diode voltage doubler or will it still not work ?

[Andy Watson]

Should I just use a separate 2 diode voltage doubler or will it still not work ?

Two separate diodes and an extra cap should work.

... and less on the 100V.

How much less? You could go to a whole bridge and two extra caps to get a bit more power, but even then, you'll be wanting reasonably large valued extra caps to realise any significant power output. You would effectively be A.C. coupling the output of the transformer through twice the impedance of the extra cap.

[Simon]

Well the 50V is dumping directly into the load. The 100V is dumping via a 0.5 ohm "source" resistance so less power will go through.

So you say make a completely separate voltage doubler, i still need to have a common negative though

[Simon]

my idea was that one of the diodes in the bridge would in effect be in parallel with one of the voltage doubler ones so why no commonize it and use the capacitors as 50V and as the lower half of the 100V

[Simon]

what is I just take the bridge out and use a voltage doubler instead, then I can take 50 or 100 volts at will, but I'd be inclined to have a larger capacitance on one side and what happens when take significant power from one side only, will the other be affected or will it affect the side I'm taking power from ?

[Andy Watson]

Do you have a figure for the amount of power? If all you want to do is charge a capacitor once in a blue moon, then a voltage doubler is the way to go. On the other hand, further up this thread, you seem to be suggesting that you want to drive 100V into 0.5R continuously! That would be entirely another level of engineering and not one that I would consider practical for a voltage doubler.

[Simon]

No i just need to dump the 100V for 50mS through the 0.5R resistor into the load under test. The dumps need to happen once every 1 second.

[Andy Watson]

No i just need to dump the 100V for 50mS through the 0.5R resistor into the load under test. The dumps need to happen once every 1 second.

An average of 10A, i.e. 1kW!  Do-able, with a voltage doubler, but if it were my problem I'd be looking at using another transformer - probably cheaper than the cost of the extra caps - and more reliable.

[Simon]

I don't think the idea is to supply constant 100V through 0.5R but just dump a load so it will peak at 100V and decay. details in the spec sheet are sketchy.

My concern now is switching that 100V, mosfet and BJT's are scare and expensive at those voltages and the BJT's have very poor gain, I'll end up needing more base current than the load is having dumped on it.

[Rufus]

I don't think the idea is to supply constant 100V through 0.5R but just dump a load so it will peak at 100V and decay. details in the spec sheet are sketchy.

It isn't sketchy. It is trying to simulate a load dump on a 28v dc system when the battery has also been disconnected and as I said on the first page of this thread

"The killer in that spec is the surge voltages unless your equipment can stand 100v for 10s of ms anyway."

The test set up in another similar standard is 100v of large lead-acid batteries and a contactor.

What your test set up needs to be depends on what you are testing - what current it is going to pull with 100v on the supply. If for example you have a 33v TVS diode on the input the theoretical supply current would be 134A or 13.4kW from the supply and 4.4kW in the TVS diode both for 50mS.

[Simon]

hm, I'll try not to point that out to the boss  He might not appreciate the requirement for something huge on a trolley. I think the main concern is the high voltage at 250V but as i already said to them it's more about the power.

The current power capabilites I have designed (680uF @ 100V) seems to work out at about 3.4J i have no idea what that is in the grand scheme of things.