Need Clever MOSFET Drive

[David Russell]

All:
I'm cobbling together a capacitive discharge circuit, starts at 34V and
discharges to about 20V, using a P-chan mosfet hooked to the positive rail,
discharging through a grounded load. The problem is I want a current limit
of about 5A. It can be less than 5A, but not more, so it's not a constant
current design. I noticed from the specs of the fet I'm using, that if I
can keep the source-gate voltage differential to 5V, it should self-limit,
but the trick is to keep the gate 5V below the source as the source voltage
changes. Anyone got a clever idea? I'm stumped.

Thanks

 

[Robert Monsen]

All:
I'm cobbling together a capacitive discharge circuit, starts at 34V and
discharges to about 20V, using a P-chan mosfet hooked to the positive rail,
discharging through a grounded load. The problem is I want a current limit
of about 5A. It can be less than 5A, but not more, so it's not a constant
current design. I noticed from the specs of the fet I'm using, that if I
can keep the source-gate voltage differential to 5V, it should self-limit,
but the trick is to keep the gate 5V below the source as the source voltage
changes. Anyone got a clever idea? I'm stumped.

Thanks

You could also just use a voltage divider from the 34V rail to ground to set
the gate voltage. As the thing drains, the gate voltage will rise relative
to the source, which will further limit the current.

Regards,
Bob Monsen

 

[Kevin McMurtrie]

All:
I'm cobbling together a capacitive discharge circuit, starts at 34V and
discharges to about 20V, using a P-chan mosfet hooked to the positive rail,
discharging through a grounded load. The problem is I want a current limit
of about 5A. It can be less than 5A, but not more, so it's not a constant
current design. I noticed from the specs of the fet I'm using, that if I
can keep the source-gate voltage differential to 5V, it should self-limit,
but the trick is to keep the gate 5V below the source as the source voltage
changes. Anyone got a clever idea? I'm stumped.

Thanks

Pick up an electronics book and look up constant current sources. I
don't know what your current setup is, but probably need nothing more
than a resistor.

 

[Tim Jackson]

All:
I'm cobbling together a capacitive discharge circuit, starts at 34V and
discharges to about 20V, using a P-chan mosfet hooked to the positive rail,
discharging through a grounded load. The problem is I want a current limit
of about 5A. It can be less than 5A, but not more, so it's not a constant
current design. I noticed from the specs of the fet I'm using, that if I
can keep the source-gate voltage differential to 5V, it should self-limit,
but the trick is to keep the gate 5V below the source as the source voltage
changes. Anyone got a clever idea? I'm stumped.

Thanks

The usual trick is to put in a series resistor, in this case about 0.14 ohms
with the emitter-base junction of a transistor across it and arrange for
increasing collector current to depress the gate-source voltage of the main
device. This will rapidly turn off the device as the current exceeds 5A.
Of course the FET must have suitable cooling to handle the power it will
dissipate running with a significant voltage drop.

Relying on a fixed gate voltage is tricky as the voltage tends to vary from
device to device and with temperature.

Tim Jackson

 

[CBarn24050]

Why use a mosfet? you could use a pnp transistor with current limiting. Look up
linear power supplies to get a circuit.

 

[David Russell]

Yeah, I thought of that. The problem is on the voltage divider the Vo to
the gate is a constant percentage of the Vi, and that will vary considerably
from 34 to 20V. In fact, it's enough that if you set the R values to get a
5A current limit at 20V, by the time you're back up to 34 it's up to 18A.

Thanks, though.
daver

 

[David Russell]

But it's not a constant current, it's a current limit. It can and will be
less than 5A depending on the load, but can't be more than 5A.

Thanks for the reply, I appreciate it.
daver

 

[David Russell]

This is close... The problem is I don't want to turn the device completely
off, nor additional resistance in the circuit. The mosfet's ON resistance is
..3 ohms, resulting in a drop of 1.5V at 5A. Could I use that as the current
sensor? I'd need a transistor in there somewhere to regulate the gate
voltage? You seem to have this one, could you give me a few more specifics?

Thanks, I really appreciate the help.
daver

 

[David Russell]

Thanks, I'll double check that.
daver

 

[Jeff]

Slap a 5.1V ziner across the divider?

 

[Tim Jackson]

This is close... The problem is I don't want to turn the device completely
off, nor additional resistance in the circuit. The mosfet's ON resistance is
.3 ohms, resulting in a drop of 1.5V at 5A. Could I use that as the current
sensor? I'd need a transistor in there somewhere to regulate the gate
voltage? You seem to have this one, could you give me a few more specifics?

Thanks, I really appreciate the help.
daver

It doesn't turn off completely, it raises the FET resistance as high as is
needed to hold the current down to 5A. That was what you wanted wasn't it?
How else did you aim to hold off the current? Anything cold is a lot more
complicated - you would be building a switch-mode voltage-bucking regulator.
I'm not about to describe that circuit without diagrams.

When active this is a linear negative feedback loop, the FET is in common
drain mode, the transistor is in common emitter. One inversion, and Vbe is
the set point for the control loop. As only one transistor is providing
voltage gain it is (almost) guaranteed to be inherently stable whatever
parts you chose.

You can't use the FET's resistance to regulate the current because that will
shoot up as soon as it starts to throttle, and you'd get a current trip
rather than a current limiter. That works like a thyristor. You do need a
low value real resistor for the current shunt. If 0.7V at full current is
too much drop then you have to start getting fancy. You could use an even
lower resistor and an op-amp to detect the voltage drop, but then your
inherent stability goes out of the window and you'd need some gain control
components. You could use a germanium transistor if you can find one!

Another trick to avoid voltage drop would be to use a hall effect magnetic
field sensor driven by a few turns of wire, in place of the transistor. A
dust-iron ring core is easy to slit and insert the sensor, but retains a
little (probably irrelevant) residual field at zero current. Ferrites don't
but are hard to cut - if you don't have a diamond slitting saw the simplest
way is to get a pair of C cores and grind one leg down..

What's the big deal about voltage drop anyway. One minute you're
complaining about too much current, the next it's too much voltage drop.
You can't have both surely.

Tim

 

[Robert Monsen]

Yeah, I thought of that. The problem is on the voltage divider the Vo to
the gate is a constant percentage of the Vi, and that will vary considerably
from 34 to 20V. In fact, it's enough that if you set the R values to get a
5A current limit at 20V, by the time you're back up to 34 it's up to 18A.

Thanks, though.
daver

No, set the divider so it sets the gate to 29V when the cap has 34V across
it. Then, when it drops to 20V, you have a 3V gate drop, which simply limits
the current a bit more. You stated initially that you didn't care if the
current was less than 5A.

Regards,
Bob Monsen

 

[Robert Monsen]

Slap a 5.1V ziner across the divider?

Yes, that works well. Then, you have 5.1V across your mosfet no matter what
the source voltage is.

Regards
Bob Monsen

 

[Tony Williams]

I'm cobbling together a capacitive discharge circuit, starts at
34V and discharges to about 20V, using a P-chan mosfet hooked to
the positive rail, discharging through a grounded load. The
problem is I want a current limit of about 5A. It can be less
than 5A, but not more, so it's not a constant current design.

Perhaps something along the lines of the sketch below.


V-capacitor, 34V to 20V --> +-----+--------+-------+
| | | |
+Vs -------------------+ | | |/ |
| / | +--| |
Control>--/\/\-+ | \6k | | |\ |s
| | / | | | |--+
e\| | +---------+/\/\+---||p-type
pnp|-+-0v | | | | | |--+
/| | | | | | |/ |
| \ | | | +--|pnp |
| / | | ===C |\ |
| \ | | | | +
FET Off= +0.7V | | |/ \| | | |
On= -0.25V +---+----| npn |---+ | LOAD
| |\e e/| | | |
| | | \ | +
| +-+-+ /0.05ohm | |
| | \ | |
\ \ | | |
/ / 0V+--------+-------+
\ \
| \|/ 2mA?
-Vs -----------+--------+

The basic idea is to sense the output current in the 0V leg
of the C, with a 0.05 ohm current shunt. During discharge
this voltage will reach -0.25V at 5A.

There is a long-tail pair, with about 2mA of tail current.
The r.hand Tr of the pair drives up through a 6k resistor
hanging off the cap voltage, into a pnp/npn emitter
follower, which drives the gate of the P-type MOSFET.

If the input Control voltage is high, then in input pnp
is On, and the base of the l.hand npn is at +0.7V. The
l.hand npn takes all the tail current, so the MOSFET is Off.

If the input Control voltage is 0v, then the input pnp
is Off, and the base of the l.hand npn is at -0.25V. The
r.hand npn takes all the 2mA tail current, developing 12V
across the 6k. The MOSFET is fully ON.

However, if the voltage across the 0.05 ohm shunt approaches
-0.25V then the lt-pair starts to reduce the MOSFET gate
voltage.

I sketched discrete transistors for clarity, but the idea
could be realised in other ways, using opamps, perhaps
using the latch+comparator facilities of a 555 timer as the
main discharge controller.

 

[Ian Buckner]

But it's not a constant current, it's a current limit. It can and will be
less than 5A depending on the load, but can't be more than 5A.

Thanks for the reply, I appreciate it.
daver

So, why can't it be a constant current?

I'd be inclined to look at a voltage regulator with a shut down
feature.
Hook the input to your capacitor, output to your load dump R (or
just ground). Turn it on when needed, off when the C voltage gets
where you need it using the shutdown. Although it might seem
complicated, it will actually be simpler than the alternatives once
you add in current and thermal limiting.

Regards
Ian