Current limiting with a mosfet?...

[Chris Jones]

On 15/06/2023 4:37 am, Commander Kinsey wrote:

On Wed, 14 Jun 2023 18:31:09 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 18:07:05 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 17:59:32 +0100, Commander Kinsey <CK1@nospam.com
wrote:

Is it easy to make a simple circuit to limit a 12V DC current to 50A
using a mosfet?

Can I just give it a variable voltage to turn it on a certain
amount?  Or is it not that easy?

If it\'s simpler, a potentiometer to vary the voltage drop across the
mosfet would do.

I\'m trying to balance a few power supplies all running in parallel,
so they do the same work each.  I don\'t mind tweaking a potentiometer
on each one to make the current about equal on each.


You could make a circuit that shuts the FET   OFF   when you hit 51
amps.    If you want to limit current by lowering the output voltage
using that FET, then you are asking for  blown up FET I think.

You\'d need a huge amount of dissipation to make that work and maybe
you don\'t want to reduce the output voltage.

But just shutting off could work fairly easily.

Shutting it off would worsen the problem - the other supplies would then
be likely to hit their limits.

Surely the TO-247 MOSFETs can take a high current?  Or is that only when
turned fully on?  I guess it would be complicated to make them pulse,
and probably upset the SMPS it\'s adjusting.

At the moment I\'m just going to sort out the biggest imbalance - two of
the supplies being 12.85V, two being 12.35V, and one adjustable.  The
two high voltage ones I can put big TO-220 schottkys on to drop half a
volt.

Simplest method:
Open up the power supplies and find the resistive divider in the voltage
sensing circuit that sets the output voltage. Put a trimpot with series
resistor across one of the resistors, to make it slightly adjustable,
but with hardly any more range than needed to reach the desired voltage.
Then adjust each of the SMPS to the same output voltage. Then, rather
than paralleling all of the SMPS with fat wire and running longer wire
from there to your load, instead run separate, fairly long (maybe longer
than physically necessary) (and equally long for each PSU) wires, no
thicker than necessary, from each SMPS separately to the load. This will
provide a small ballast resistor between each SMPS and the load. You can
still twist the positive and negative output wires of each SMPS, to
minimise the loop area between them and thereby minimise inductance. If
the inductance of the long wires is still a problem, you could add lots
more electrolytic capacitors at the load end.

Proper method:
Buy SMPS designed for parallel operation.

 

[Commander Kinsey]

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

> Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the source-drain path.

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

it will not get very hot in such a case at 50 A (but still calculate it
and use a heatsink, for example 50 mOhm Rds_on at 50 A, gives P = i^2 x R so 50 * 50 * .05 still makes 125 Watt!

The 1 milliohm one at 50A will produce only 2.5W when fully on, and 25W when dropping 0.5V.

If you just want to limit to 50 A as in blowing a fuse
then you need a sensing shunt and a trigger circuit to quickly switch the MOSFET off.

As a switch you could use the MOSFET upside down like I do here for example to get enough gate drive:
https://panteltje.nl/panteltje/cb/tx_power_switch2.jpg
but even that is 30 A peak, not continous...

So maybe simpler to just use a fuse?

No, I definitely don\'t want it cutting out. Then the other supplies would be even more overloaded, they would cut out, and I\'d have no power.

There is more to it.
Else just get a few power MOSFETs and play with those in some test circuit to get the hang of it.

I have ordered some and will play around with them and some lightbulbs as a load and see what voltages I can get.

 

[Commander Kinsey]

On Thu, 15 Jun 2023 13:18:28 +0100, Chris Jones <lugnut808@spam.yahoo.com> wrote:

On 15/06/2023 4:37 am, Commander Kinsey wrote:
On Wed, 14 Jun 2023 18:31:09 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 18:07:05 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 17:59:32 +0100, Commander Kinsey <CK1@nospam.com
wrote:

Is it easy to make a simple circuit to limit a 12V DC current to 50A
using a mosfet?

Can I just give it a variable voltage to turn it on a certain
amount? Or is it not that easy?

If it\'s simpler, a potentiometer to vary the voltage drop across the
mosfet would do.

I\'m trying to balance a few power supplies all running in parallel,
so they do the same work each. I don\'t mind tweaking a potentiometer
on each one to make the current about equal on each.


You could make a circuit that shuts the FET OFF when you hit 51
amps. If you want to limit current by lowering the output voltage
using that FET, then you are asking for blown up FET I think.

You\'d need a huge amount of dissipation to make that work and maybe
you don\'t want to reduce the output voltage.

But just shutting off could work fairly easily.

Shutting it off would worsen the problem - the other supplies would then
be likely to hit their limits.

Surely the TO-247 MOSFETs can take a high current? Or is that only when
turned fully on? I guess it would be complicated to make them pulse,
and probably upset the SMPS it\'s adjusting.

At the moment I\'m just going to sort out the biggest imbalance - two of
the supplies being 12.85V, two being 12.35V, and one adjustable. The
two high voltage ones I can put big TO-220 schottkys on to drop half a
volt.


Simplest method:
Open up the power supplies and find the resistive divider in the voltage
sensing circuit that sets the output voltage.

I thought of that, but they\'re tightly packed complex circuits, and I don\'t even have a pinout. The power output was obvious, the huge connectors. I had to find the power on pin by trial and error.

Put a trimpot with series
resistor across one of the resistors, to make it slightly adjustable,
but with hardly any more range than needed to reach the desired voltage.
Then adjust each of the SMPS to the same output voltage. Then, rather
than paralleling all of the SMPS with fat wire and running longer wire
from there to your load, instead run separate, fairly long (maybe longer
than physically necessary) (and equally long for each PSU) wires, no
thicker than necessary, from each SMPS separately to the load. This will
provide a small ballast resistor between each SMPS and the load. You can
still twist the positive and negative output wires of each SMPS, to
minimise the loop area between them and thereby minimise inductance. If
the inductance of the long wires is still a problem, you could add lots
more electrolytic capacitors at the load end.

Proper method:
Buy SMPS designed for parallel operation.

They are, but I don\'t have the specs sheet. They\'re from an HP blade server, and the end user is supposed to just plug them in and not know what the pins do.

At £14 for 2600W at 12V, I wasn\'t going to buy anything else!

 

[Jan Panteltje]

On a sunny day (Thu, 15 Jun 2023 13:45:13 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16ky1naomvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will
rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the
source-drain path.

It \'reacts\' to the difference in voltage between gate and source.
Look in the dataheet, anything from 1 V to maybe 10 or more volts (depending on the model) is needed between gate and source to get that low conductance.
So the source is _always_ lower than the gate in a source follower.
+12 drain source ---- load
less than +12
gate
|
+12

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

Yes, those exist, note the gate to source voltage \'Vgs_on\' needed,
datasheets usually provide a graph for Vgs versus Id

 

[John Larkin]

On Thu, 15 Jun 2023 13:45:13 +0100, \"Commander Kinsey\"
<CK1@nospam.com> wrote:

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the source-drain path.

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

it will not get very hot in such a case at 50 A (but still calculate it
and use a heatsink, for example 50 mOhm Rds_on at 50 A, gives P = i^2 x R so 50 * 50 * .05 still makes 125 Watt!

The 1 milliohm one at 50A will produce only 2.5W when fully on, and 25W when dropping 0.5V.

If you just want to limit to 50 A as in blowing a fuse
then you need a sensing shunt and a trigger circuit to quickly switch the MOSFET off.

As a switch you could use the MOSFET upside down like I do here for example to get enough gate drive:
https://panteltje.nl/panteltje/cb/tx_power_switch2.jpg
but even that is 30 A peak, not continous...

So maybe simpler to just use a fuse?

No, I definitely don\'t want it cutting out. Then the other supplies would be even more overloaded, they would cut out, and I\'d have no power.

There is more to it.
Else just get a few power MOSFETs and play with those in some test circuit to get the hang of it.

I have ordered some and will play around with them and some lightbulbs as a load and see what voltages I can get.

Why not use resistors or, as someone has suggested, wire?

 

[Tabby]

On Thursday, 15 June 2023 at 07:37:17 UTC+1, Commander Kinsey wrote:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <al...@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <C...@nospam.com> wrote in <op.16jf5...@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
I\'d be dropping maybe 0.1V. 5W isn\'t a problem with a heatsink.

Yes, go for it. Will work fine. At least until it limits, then you\'ll have a hole where the fet was. Stop wasting our time, go do it & get back to us when you\'ve realised you don\'t know what you\'re doing. Meanwhile it\'s time stop feeding the dumb troll.

 

[Commander Kinsey]

On Thu, 15 Jun 2023 17:17:42 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:

On Thu, 15 Jun 2023 13:45:13 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the source-drain path.

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

it will not get very hot in such a case at 50 A (but still calculate it
and use a heatsink, for example 50 mOhm Rds_on at 50 A, gives P = i^2 x R so 50 * 50 * .05 still makes 125 Watt!

The 1 milliohm one at 50A will produce only 2.5W when fully on, and 25W when dropping 0.5V.

If you just want to limit to 50 A as in blowing a fuse
then you need a sensing shunt and a trigger circuit to quickly switch the MOSFET off.

As a switch you could use the MOSFET upside down like I do here for example to get enough gate drive:
https://panteltje.nl/panteltje/cb/tx_power_switch2.jpg
but even that is 30 A peak, not continous...

So maybe simpler to just use a fuse?

No, I definitely don\'t want it cutting out. Then the other supplies would be even more overloaded, they would cut out, and I\'d have no power.

There is more to it.
Else just get a few power MOSFETs and play with those in some test circuit to get the hang of it.

I have ordered some and will play around with them and some lightbulbs as a load and see what voltages I can get.

Why not use resistors or, as someone has suggested, wire?

Because I want to be able to tweak it. A variable resistor of that power output, and at fractions of an ohm, is impossible to obtain.

If the mosfet works, I can just turn a dial until my amp clamp says they\'re leveled.

I\'ve found a 200A 1 milliohm mosfet, that oughta do the trick. FFS I had to order it from America, couldn\'t even find stock in China.

 

[Commander Kinsey]

On Thu, 15 Jun 2023 16:24:53 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 13:45:13 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16ky1naomvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will
rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the
source-drain path.

It \'reacts\' to the difference in voltage between gate and source.
Look in the dataheet, anything from 1 V to maybe 10 or more volts (depending on the model) is needed between gate and source to get that low conductance.
So the source is _always_ lower than the gate in a source follower.

Yes, I have read the datasheet now, looks like 2 to 4 volts is the sweetspot for the one I chose (SUM40014M). So I could stick a pot and a resistor either side in series, ground to VCC. Then vary 2-4V by turning the pot.

+12 drain source ---- load
less than +12
gate
|
+12

That doesn\'t line up, even pasted into notepad using fixed width font.

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

Yes, those exist, note the gate to source voltage \'Vgs_on\' needed,
datasheets usually provide a graph for Vgs versus Id

Yes, the graphs from Vishay are very helpful.

 

[Phil Hobbs]

On 2023-06-15 01:23, Jan Panteltje wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?

Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

--
Dr Philip C D Hobbs
Principal Consultant
ElectroOptical Innovations LLC / Hobbs ElectroOptics
Optics, Electro-optics, Photonics, Analog Electronics
Briarcliff Manor NY 10510

http://electrooptical.net
http://hobbs-eo.com

 

[John Larkin]

On Thu, 15 Jun 2023 21:57:37 +0100, \"Commander Kinsey\"
<CK1@nospam.com> wrote:

On Thu, 15 Jun 2023 17:17:42 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:

On Thu, 15 Jun 2023 13:45:13 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the source-drain path.

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

it will not get very hot in such a case at 50 A (but still calculate it
and use a heatsink, for example 50 mOhm Rds_on at 50 A, gives P = i^2 x R so 50 * 50 * .05 still makes 125 Watt!

The 1 milliohm one at 50A will produce only 2.5W when fully on, and 25W when dropping 0.5V.

If you just want to limit to 50 A as in blowing a fuse
then you need a sensing shunt and a trigger circuit to quickly switch the MOSFET off.

As a switch you could use the MOSFET upside down like I do here for example to get enough gate drive:
https://panteltje.nl/panteltje/cb/tx_power_switch2.jpg
but even that is 30 A peak, not continous...

So maybe simpler to just use a fuse?

No, I definitely don\'t want it cutting out. Then the other supplies would be even more overloaded, they would cut out, and I\'d have no power.

There is more to it.
Else just get a few power MOSFETs and play with those in some test circuit to get the hang of it.

I have ordered some and will play around with them and some lightbulbs as a load and see what voltages I can get.

Why not use resistors or, as someone has suggested, wire?

Because I want to be able to tweak it. A variable resistor of that power output, and at fractions of an ohm, is impossible to obtain.

Measure the power suppies and pick the resistor or wire length.

If the mosfet works, I can just turn a dial until my amp clamp says they\'re leveled.

I\'ve found a 200A 1 milliohm mosfet, that oughta do the trick. FFS I had to order it from America, couldn\'t even find stock in China.

We stock IXFH400N075T2, a 1000 watt, 1000 amp mosfet. 2.3 mohms with
10 volts on the gate. You\'ll need 10 mohms, roughly.

Given the heat sinking issue, you could use two or three smaller fets
in parallel.

 

[boB]

On Wed, 14 Jun 2023 23:41:50 +0100, \"Commander Kinsey\"
<CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 23:18:04 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 19:37:16 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 18:31:09 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 18:07:05 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 17:59:32 +0100, Commander Kinsey <CK1@nospam.com> wrote:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

If it\'s simpler, a potentiometer to vary the voltage drop across the mosfet would do.

I\'m trying to balance a few power supplies all running in parallel, so they do the same work each. I don\'t mind tweaking a potentiometer on each one to make the current about equal on each.


You could make a circuit that shuts the FET OFF when you hit 51
amps. If you want to limit current by lowering the output voltage
using that FET, then you are asking for blown up FET I think.

You\'d need a huge amount of dissipation to make that work and maybe
you don\'t want to reduce the output voltage.

But just shutting off could work fairly easily.

Shutting it off would worsen the problem - the other supplies would then be likely to hit their limits.

Surely the TO-247 MOSFETs can take a high current? Or is that only when turned fully on? I guess it would be complicated to make them pulse, and probably upset the SMPS it\'s adjusting.

At the moment I\'m just going to sort out the biggest imbalance - two of the supplies being 12.85V, two being 12.35V, and one adjustable. The two high voltage ones I can put big TO-220 schottkys on to drop half a volt.

Well, if you have, say, a 1 milliOhm FET which are obtainable, then at
50 amps that is 2.5 watts which can be OK with a good heat sink.

If you were to say, drop 1 volt across the FET at 49 amps, now you are
talking a out 1 X 49 = 49 watts which is NOT going to work even with
a TO-247 package.

A TO-247 on a big heatsink and fan arrangement could easily dissipate 50W.

So, you turn off the FET and latch off if you want to limit current
and then it only dissipate a few watts at most while fully on.

But I want to LIMIT the current, not fuse it.

OK then.

Use an N-channel FET with a small Ohm resistor in series with its
source for the output. 15 milli-Ohms is about right for this one.

Drain goes to your source voltage. 12V here, right ?

Then, tack in a small NPN transistor with its emitter to the output
load, its base to the source of the FET and its collector to the gate
of the FET. Choose a series R that drops around 0.7 volts at 50 amps
and then the NON will turn on, reducing the gate-source voltage and
therefore reducing the output current. That\'s about the simplest
that can be done.

You will need to use a resistor driving the gate so that the NPN can
easily reduce that Vgs to limit the current.

boB

 

[John Larkin]

On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?

 

[Phil Hobbs]

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount?
Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?

Gas. Some of those jokey IR datasheets give new meaning to ‘vaporware’.

--
Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC /
Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics

 

[John Larkin]

On Fri, 16 Jun 2023 02:19:21 -0000 (UTC), Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:
On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount?
Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?



Gas. Some of those jokey IR datasheets give new meaning to ‘vaporware’.

You mean the dpaks that run at 200 amps?

When we were doing NMR gradient amps, we tested lots of mosfets to
destruction. We used shrapnel shields. Then we measured voltage and
current in real time and computed junction temperature for shutdown.
That about tripled how hard we dared to push the fets.

https://www.dropbox.com/s/4nxm7m2q3j3buvc/ExFets.jpg?dl=0

 

[Phil Hobbs]

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:

On Fri, 16 Jun 2023 02:19:21 -0000 (UTC), Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:
On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount?
Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?



Gas. Some of those jokey IR datasheets give new meaning to ‘vaporware’.

You mean the dpaks that run at 200 amps?

When we were doing NMR gradient amps, we tested lots of mosfets to
destruction. We used shrapnel shields. Then we measured voltage and
current in real time and computed junction temperature for shutdown.
That about tripled how hard we dared to push the fets.

https://www.dropbox.com/s/4nxm7m2q3j3buvc/ExFets.jpg?dl=0

Any reliability issues show up?

--
Dr Philip C D Hobbs Principal Consultant ElectroOptical Innovations LLC /
Hobbs ElectroOptics Optics, Electro-optics, Photonics, Analog Electronics

 

[John Larkin]

On Fri, 16 Jun 2023 03:32:30 -0000 (UTC), Phil Hobbs
<pcdhSpamMeSenseless@electrooptical.net> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:
On Fri, 16 Jun 2023 02:19:21 -0000 (UTC), Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:
On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount?
Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?



Gas. Some of those jokey IR datasheets give new meaning to ?vaporware?.

You mean the dpaks that run at 200 amps?

When we were doing NMR gradient amps, we tested lots of mosfets to
destruction. We used shrapnel shields. Then we measured voltage and
current in real time and computed junction temperature for shutdown.
That about tripled how hard we dared to push the fets.

https://www.dropbox.com/s/4nxm7m2q3j3buvc/ExFets.jpg?dl=0





Any reliability issues show up?

No, we rarely lost a fet in the field.

We digitized the voltage drop and current of the fets about once a
millisecond. We computed estimated junction temp with a simple 1st
order time constant, using base heat sink temp and calculated power.

The gradient drivers were constant-current sources, so the max
possible power dissipation was known, which helped. Nothing would blow
up in one millisecond.

The smallest unit could deliver about 50 watts peak into a gradient
coil, and the biggest was 20 KW. The big one was always sold three at
a time.

It was great business until Agilent acquired Varian and shut down the
NMR division.

https://www.dropbox.com/sh/ykupptx6maexp6l/AAD30JqJE-3OVEO-Ozjo5iOMa?dl=0

The big ones were for MRI-style imaging of small things, up to mice
and rabbits. The little ones did some sorts of quantum mechanical spin
flipping that helped resolve molecular structures, with complicated
recipes of RF excitation and gradient mag fields.

 

[Commander Kinsey]

On Thu, 15 Jun 2023 23:03:50 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:

On Thu, 15 Jun 2023 21:57:37 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Thu, 15 Jun 2023 17:17:42 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:

On Thu, 15 Jun 2023 13:45:13 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Thu, 15 Jun 2023 12:58:43 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Thu, 15 Jun 2023 08:58:58 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16klskj8mvhs6z@ryzen.home>:

On Thu, 15 Jun 2023 06:23:51 +0100, Jan Panteltje <alien@comet.invalid> wrote:

On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.
You could use a switcher with filter.
So define what you find \'simple\'
?

Can I just do this?

https://electronics.stackexchange.com/questions/377246/how-do-mosfets-and-potentiometers-work-together

What would the minimum voltage drop be?

That is a source follower,
Basicaly you put a voltage on the gate, the drain goes to the power supply output,
and the load to the source.
The problem is that at 50 A and when you drop 5 V over the MOSFET,
it will dissippate 5 x 50 = 250 W when the load is 50 A, less with a lower load current.
Basically the MOSFET will melt.
If it is on a heatsink and heatsink plus thermal resistance of the MOSFET is 1.5 degrees C per Watt then at 250 W it will rise
in temperature by 1.5 * 250 = 375 degrees C.
Add the 350 to say 20 degrees C ambitient temperature and the MOSFET internal will then be at 395 degrees C.
See the problem?

But I\'m not dropping 5V. I\'m dropping about 0.5V.

Plus you need a high enough voltage at the gate to go all the way up to 12 V, a souce follower does not provide that.

Doesn\'t it try to match the voltage? If I put 12.85V form the supply into the gate, it will fully open to 12.85V through the source-drain path.

The minimum voltage drop depends on the MOSFET\'s \'on\' resistance and you can find that in the dataheets as Rds_on,
it maybe very low, some milli-ohms,

I found one with 1 milliohm.

it will not get very hot in such a case at 50 A (but still calculate it
and use a heatsink, for example 50 mOhm Rds_on at 50 A, gives P = i^2 x R so 50 * 50 * .05 still makes 125 Watt!

The 1 milliohm one at 50A will produce only 2.5W when fully on, and 25W when dropping 0.5V.

If you just want to limit to 50 A as in blowing a fuse
then you need a sensing shunt and a trigger circuit to quickly switch the MOSFET off.

As a switch you could use the MOSFET upside down like I do here for example to get enough gate drive:
https://panteltje.nl/panteltje/cb/tx_power_switch2.jpg
but even that is 30 A peak, not continous...

So maybe simpler to just use a fuse?

No, I definitely don\'t want it cutting out. Then the other supplies would be even more overloaded, they would cut out, and I\'d have no power.

There is more to it.
Else just get a few power MOSFETs and play with those in some test circuit to get the hang of it.

I have ordered some and will play around with them and some lightbulbs as a load and see what voltages I can get.

Why not use resistors or, as someone has suggested, wire?

Because I want to be able to tweak it. A variable resistor of that power output, and at fractions of an ohm, is impossible to obtain.

Measure the power suppies and pick the resistor or wire length.

I\'ve never seen milliohm huge variable resistors.

If the mosfet works, I can just turn a dial until my amp clamp says they\'re leveled.

I\'ve found a 200A 1 milliohm mosfet, that oughta do the trick. FFS I had to order it from America, couldn\'t even find stock in China.

We stock IXFH400N075T2, a 1000 watt, 1000 amp mosfet. 2.3 mohms with
10 volts on the gate. You\'ll need 10 mohms, roughly.

Too late, already paid the ludicrous postage.

How much would two of yours have cost me to get shipped to Scotland if you\'re not in Scotland. Judging by \"highland\" I\'m guessing you are.

Given the heat sinking issue, you could use two or three smaller fets
in parallel.

I was going to until I found one which would do. They\'re going on 106A supplies which I will be keeping at 50% load max.

 

[Commander Kinsey]

On Fri, 16 Jun 2023 00:15:40 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 23:41:50 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 23:18:04 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 19:37:16 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 18:31:09 +0100, boB <boB@k7iq.com> wrote:

On Wed, 14 Jun 2023 18:07:05 +0100, \"Commander Kinsey\"
CK1@nospam.com> wrote:

On Wed, 14 Jun 2023 17:59:32 +0100, Commander Kinsey <CK1@nospam.com> wrote:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount? Or is it not that easy?

If it\'s simpler, a potentiometer to vary the voltage drop across the mosfet would do.

I\'m trying to balance a few power supplies all running in parallel, so they do the same work each. I don\'t mind tweaking a potentiometer on each one to make the current about equal on each.


You could make a circuit that shuts the FET OFF when you hit 51
amps. If you want to limit current by lowering the output voltage
using that FET, then you are asking for blown up FET I think.

You\'d need a huge amount of dissipation to make that work and maybe
you don\'t want to reduce the output voltage.

But just shutting off could work fairly easily.

Shutting it off would worsen the problem - the other supplies would then be likely to hit their limits.

Surely the TO-247 MOSFETs can take a high current? Or is that only when turned fully on? I guess it would be complicated to make them pulse, and probably upset the SMPS it\'s adjusting.

At the moment I\'m just going to sort out the biggest imbalance - two of the supplies being 12.85V, two being 12.35V, and one adjustable. The two high voltage ones I can put big TO-220 schottkys on to drop half a volt.

Well, if you have, say, a 1 milliOhm FET which are obtainable, then at
50 amps that is 2.5 watts which can be OK with a good heat sink.

If you were to say, drop 1 volt across the FET at 49 amps, now you are
talking a out 1 X 49 = 49 watts which is NOT going to work even with
a TO-247 package.

A TO-247 on a big heatsink and fan arrangement could easily dissipate 50W.

So, you turn off the FET and latch off if you want to limit current
and then it only dissipate a few watts at most while fully on.

But I want to LIMIT the current, not fuse it.


OK then.

Use an N-channel FET with a small Ohm resistor in series with its
source for the output. 15 milli-Ohms is about right for this one.

Drain goes to your source voltage. 12V here, right ?

Then, tack in a small NPN transistor with its emitter to the output
load, its base to the source of the FET and its collector to the gate
of the FET. Choose a series R that drops around 0.7 volts at 50 amps
and then the NON will turn on, reducing the gate-source voltage and
therefore reducing the output current. That\'s about the simplest
that can be done.

You will need to use a resistor driving the gate so that the NPN can
easily reduce that Vgs to limit the current.

That sounds feasible, apart from dropping 0.7 volts. I\'ve got supplies at 12.35V, and supplies at 12.85V. I want to drop the 12.85 to match the 12.35 ones. If they drop below 12.35 I\'d have to put something on the 12.35V ones too, and end up with less than 12V on the load, which works best nearer the max of 12.6V rather than the min of 11.4V.

 

[Commander Kinsey]

On Fri, 16 Jun 2023 04:16:02 +0100, John Larkin <jlarkin@highlandsnipmetechnology.com> wrote:

On Fri, 16 Jun 2023 02:19:21 -0000 (UTC), Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

John Larkin <jlarkin@highlandSNIPMEtechnology.com> wrote:
On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
pcdhSpamMeSenseless@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <CK1@nospam.com> wrote in <op.16jf5imhmvhs6z@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount?
Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?



Gas. Some of those jokey IR datasheets give new meaning to ‘vaporware’.

You mean the dpaks that run at 200 amps?

Uh oh....
I got these: https://www.vishay.com/docs/77646/sum40014m.pdf

 

[Fred Bloggs]

On Friday, June 16, 2023 at 12:58:15 PM UTC-4, Commander Kinsey wrote:

On Fri, 16 Jun 2023 04:16:02 +0100, John Larkin <jla...@highlandsnipmetechnology.com> wrote:

On Fri, 16 Jun 2023 02:19:21 -0000 (UTC), Phil Hobbs
pcdhSpamM...@electrooptical.net> wrote:

John Larkin <jla...@highlandSNIPMEtechnology.com> wrote:
On Thu, 15 Jun 2023 17:19:23 -0400, Phil Hobbs
pcdhSpamM...@electrooptical.net> wrote:

On 2023-06-15 01:23, Jan Panteltje wrote:
On a sunny day (Wed, 14 Jun 2023 17:59:32 +0100) it happened \"Commander
Kinsey\" <C...@nospam.com> wrote in <op.16jf5...@ryzen.home>:

Is it easy to make a simple circuit to limit a 12V DC current to 50A using a mosfet?

Can I just give it a variable voltage to turn it on a certain amount?
Or is it not that easy?

Dissipation is your issue, using a MOSFET in series
50 A with a voltage drop over it will bake it
volts multiplied by voltage dropped makes watts.

You could use a switcher with filter.
So define what you find \'simple\'
?


Not to mention that a good many modern FETs have safe operating areas
that are hard to distinguish from the Y axis.

Cheers

Phil Hobbs

On. Off. What else is there?



Gas. Some of those jokey IR datasheets give new meaning to ‘vaporware’.

You mean the dpaks that run at 200 amps?
Uh oh....
I got these: https://www.vishay.com/docs/77646/sum40014m.pdf

Wildly unstable at 300 Amps/Volt VGS which itself varies precipitously with TJ.