How to improve this simple circuit-- Darn Diodes!

[Paul]

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

 

[Tim Williams]

The circuit:http://apgpro.com/i/photo3.gif

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

 

[Paul]

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

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

 

[daestrom]

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.

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

 

[Salmon Egg]

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

You might as well write it up in Greek. As it stands, it is Greek to me.

Bill

 

[ehsjr]

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.

Why????

Makes no sense.

Ed

 

[Jasen Betts]

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?

use a resistor?

what's it supposed to do? currently it looks like very elaborate heater.

 

[Jasen Betts]

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.

1: how?

2. Short the inductor.
3. Unshort the inductor to allow the collapsing magnetic field charge
V2.

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))

 

[tomrei]

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

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-mana...ternal-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

 

[qrk]

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

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.

 

[legg]

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

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

 

[tomrei]

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-exter...

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

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