How to improve this simple circuit-- Darn Diodes!

Hi,
The following image that contains my circuit is simplified for this post. It works, but the losses are ridiculous due to the diodes! The
pulsed current through the inductor is around 50 amps, and lasts only for a few hundred microseconds at most. The inductance is very low. As you know, a MOSFET's Vsd (back voltage) is ridiculously low, and can't block much, hence the diode. Now I can place enough MOSFET's in- parallel to achieve low on-resistance, and the capacitance is low enough, but as far as placing diodes in-parallel it doesn't do much good beyond a few diodes.
The circuit:
http://apgpro.com/i/photo3.gif
Anyhow, is there a better circuit for doing this? Perhaps a way of replacing the diode with a MOSFET, but that doesn't provide a good ground path for the gate signal. I've tried IGBT instead of MOSFETs, but that didn't work well at all ... sorry, I forgot the reason. A mechanical relay would work if they didn't switch so slow.
Any ideas would be greatly appreciated!
Thanks! Paul
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What the hell is this? I make sense of nothing.
I see four N-channel JFETs, labelled IRF540 which is an enhancement mode MOSFET. I see ground, on the LEFT SIDE, not at the bottom, and current is flowing, up I guess? If I assume all transistors are common source to ground (I can't tell, because you show JFETs, which are symmetrical), then J1 will always be reverse-biased (i.e. body diode conducting) in series with V1, which I can only assume is supposed to be the power source and it's been drawn backwards?
Tim
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That's the MOSFET symbol used by LTspice. V1 & V2 are batteries. The ground symbol is for the signal ground.
There are three stages to the circuit -->
1. Pulse the inductor. 2. Short the inductor. 3. Unshort the inductor to allow the collapsing magnetic field charge V2.
Paul
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Paul wrote:

So as I see it, when OUT1 is turning on its MOSFET, V1 sends current through the inductor. Since there is no appreciable resistance, the only limit on current is the inductance of L1. So long as OUT1 is very brief (commensurate with the value of L1), that's seems okay.
But why bother with OUT2 and those three MOSFET?? The moment you turn off OUT1, the magnetic field of L1 will send current into V2 via D1 to charge V2.
By having the intermediate step of turning on the other three MOSFET via OUT2, you simply let the energy stored in the magnetic field of L1 dissipate in D2. The only energy you send into V2 is whatever is left over after turning off the OUT2 MOSFETs. So they don't seem to have any purpose except to waste energy in D2.
You could just eliminate the three MOSFETs controlled by OUT2 and the diode D2 completely. Shutting off OUT1 would 'pump' the current of L1 into V2. You just have to make sure that the MOSFET controlled by OUT1 is able to withstand the voltage across it from V1 + V2 (which it has to do anyway) plus the diode drop of D1 when forward conducting. The inductive 'kick' of L1 will be limited to V2 plus the diode drop of D1.
daestrom
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Paul wrote:

Why????
Makes no sense.
Ed

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1: how?

2: what does step 2 gain you?
you appear to have all the mosfets installed backwards, ( point 1) swap source and drain, replace D2 with a wire (or even leave it open circuit (point 2))
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wrote:

B+ | L | +---->|---+ | | D bat pulse---G | S gnd | gnd
I assume you're trying to punch a high current spike into a sulfated lead acid battery. The above circuit will do the same thing and is much simpler. You'll probably need to put a simple RC snubber from drain node to ground. You will need a decent electrolytic caps on B+ to handle the current pulse (United Chemicon KZE series or equiv). Put a resistor in series with the battery that supplies the B+ and you will have minimal stress on B+ battery.
--
Mark

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Tim Wiiliams this is not an Electronic Engineering Design Group as Troll Archimedes' putzit...Damn Flamers.
The Mighty WontVolt
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In article

You might as well write it up in Greek. As it stands, it is Greek to me.
Bill
--
Private Profit; Public Poop! Avoid collateral windfall!

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Paul wrote:

WHAT circuit? Get a wierd looking web page with a broken image placeholder.
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use a resistor?
what's it supposed to do? currently it looks like very elaborate heater.
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Hi Paul,
i think you made a mistake here, i understand what you are trying to do here, what you need to do is disconnect the right node of D2, and connect it to the right node of L1.
you are now draining the inductor current straight into ground, what you should do is create a path to source it from ground and put into V2.
you can search for topologies for buck or boost controller. use this as an example: http://www.analog.com/en/power-management/switching-controllers-external-switches/adp1821/products/product.html
if you can tell me what's your V1 and V2, and the current you need for charging the V2. i may come up with some solution.
yours Ren
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an follow up with my last reply. sorry, i didn't explain how you should correct it,
here's my drawing of what it should look like.
http://tomrei2000.googlepages.com/photo3.gif/photo3-full ;init:.gif
i didn't have the LTspice on this machine so i just did it with paint.
this is buck topology so V1 should always higher than V2.
when Out1 is on, Out2 is off, the V1 charges the L1, Cout supply the current to V2. when Out1 is off and Out2 is on, L1 supply the current to Cout and V2. (the current in inductor keeps flow in the same direction). be sure not to turn both FETs on in the same time. you have to synchronous the two FETs, so the controller will be called synchronous mode. the bottom FET can be replaced by Diode if the current isn't that big. (Paul, your 50A is because you discharged L1 to ground by turning Out2 on, that's why there's a big lost in your circuit) this mode you only need to control 1 FET, it's called asynchronous mode.
care must be taken to choose the capacitor and inductor values. you may also need to create circuits to compensate the zeros and poles, but that's a long story. you can find some more info on that Analog Devices' switching controller data sheet.
yours Ren
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wrote:

If you intend to apply a negative V1 pulse to L1, using the 'Out1' signal on it's respective IRF540n gate, then you're going to have to turn the fet around. The fet will then potentially see V1+V2 in the flyback period, when inductor energy can be delivered to V2.
1N5404 is not a schottky structure, so you may need to correct something in your documentation, if this is a simulation.
RL
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