# Replacing a variable resistor (rheostat) with MOSFET

#### Denis De Bremme

Joined Dec 29, 2015
10
Replacing a variable resistor (rheostat) with MOSFET

I have a separately excited DC electric motor. I control the speed by varying the field with a rheostat. I would like to substitute 2 n-channel power mosfet to this old and cumbersome rheostat taken from the lab. Once the motor has reached 1200 rpm, the bypass resistor is cut off and the armature is fed directly from the battery until it reaches 2000 rpm. Then the required variation of speed is between 2000 and 3000 rpm only. Of course, a limited variation of speed is possible by field weakening but this is enough for our application.

The motor is a ES-84-12 from D&D Motor Systems. At start, current is shunted around through a large 1 ohm (10 kw) protection resistor. The field is set at 6.25A for 14.8 ohms (13.6 from the rheostat + 1.2 from the field winding). Progressively the resistance is brought up to 19.2 ohms (18 + 1.2) to increase the current in the armature. At this point (1200 rpm) the motor reaches enough BackEMF to plug it directly to the pack.

So, your advice on substituting 2 (or 3 for a smoother start) SK1019 n-channel power mosfet would be appreciated. I include a sketch at: http://carrier-denis.net/11-39A_MOSFET_as_var_res.jpg

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#### wayneh

Joined Sep 9, 2010
17,498
So you want your device to control within the 2000-3000rpm range, once the motor has started up?
What's the running current draw of the unloaded motor? Or even better, do you know what typical current draw is under load? I'm wondering how much current the MOSFET needs to handle.

#### Denis De Bremme

Joined Dec 29, 2015
10
I'm wondering how much current the MOSFET needs to handle.
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Data for my electric motor (ES-84-12):

Field resistance: 1.18 ohms at 25°C

Field amps (full field max.) is If = 50 A

Field amps minimum is 5 A

Maximum armature current is 500 amps (for 20 seconds)

My post is about controlling (weakening) the field only (not controlling the armature). At start, when the current in the field must be at max, the amperage is 6.5A (this is the max. my vintage rheostat can handle). With no load, the start is quite smooth with 6.5 amps and there will never be any load applied at start. Load will be applied later on. So, 6.5A x 96v = 624 w. Each SK1019 MOSFET can handle 300w (from data sheet), two of them should be ok. This max last only a few seconds. It will diminish to 5A when the motor reaches 1200 rpm. Starting at a higher amperage would require a third MOSFET but with a smoother behavior.

#### ronv

Joined Nov 12, 2008
3,770
___________________________________________________________

Data for my electric motor (ES-84-12):

Field resistance: 1.18 ohms at 25°C

Field amps (full field max.) is If = 50 A

Field amps minimum is 5 A

Maximum armature current is 500 amps (for 20 seconds)

My post is about controlling (weakening) the field only (not controlling the armature). At start, when the current in the field must be at max, the amperage is 6.5A (this is the max. my vintage rheostat can handle). With no load, the start is quite smooth with 6.5 amps and there will never be any load applied at start. Load will be applied later on. So, 6.5A x 96v = 624 w. Each SK1019 MOSFET can handle 300w (from data sheet), two of them should be ok. This max last only a few seconds. It will diminish to 5A when the motor reaches 1200 rpm. Starting at a higher amperage would require a third MOSFET but with a smoother behavior.
It might be easier to use PWM to control the current. Kind of like an automotive regulator except for current.

#### wayneh

Joined Sep 9, 2010
17,498
OK, thanks, that (post #3) clears up a lot.

If the field is drawing 5A, the voltage across the field will be ~6V, and the MOSFET has to drop 90V and will dissipate 90V•5A=450W. Split in two, you should be OK in theory. But I'd take the data sheet rating with a grain of salt. I think they assume some very generous conditions. Also, it would be critical that the MOSFETs share the load in your application but the circuit you posted does not level the loading. One MOSFET will conduct more, get hotter, conduct still more, and so on until it pops.

As ronv noted, a PWM setup would allow you to use a single MOSFET and you probably wouldn't even need a heat sink. You'll need a few components to generate the control signal, or you could simply buy a ready-made PWM module off e-bay. They're only a few dollars.

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#### Denis De Bremme

Joined Dec 29, 2015
10
It might be easier to use PWM to control the current. Kind of like an automotive regulator except for current.

#### ronv

Joined Nov 12, 2008
3,770
@Denis De Bremme
Couple of questions:
What controls the change of the field?
Any idea what the field inductance is?
The power rating of the FETs is at a case temperature of 25C. This would be very difficult to hold no matter how big the heat sink is.

#### Denis De Bremme

Joined Dec 29, 2015
10
OK, thanks, that (post #3) clears up a lot.

.... it would be critical that the MOSFETs share the load in your application but the circuit you posted does not level the loading. One MOSFET will conduct more, get hotter, conduct still more, and so on until it pops.
.
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Thank you for taking some of your time to answer me. I appreciate.

About balancing parallel MOSFET’s I thought I could count on the fact that the FET "on" resistance (Rds) will increase as it's temperature rises so parallel FET will "self balance". So as an example, if you have two FETs in parallel, the one that conducts the most current will also be the one that runs the hottest. As a result, it's Rds(on) resistance will rise and the other FETs will conduct more current and thus balance the current load.

By the way, I will consider using a ready-made Pulse Width Modulation unit. Thank you for your suggestion

Happy New Year

#### Denis De Bremme

Joined Dec 29, 2015
10
@Denis De Bremme
Couple of questions:
What controls the change of the field?
Any idea what the field inductance is?
The power rating of the FETs is at a case temperature of 25C. This would be very difficult to hold no matter how big the heat sink is.
____________________________________________

Q. What controls the change of the field?

A. This is the subject of this post. My schematic shown as an attachment in my first post, I hope could be the answer after taking your advices into account.

Q. The field inductance.

A. I do not have this information.

Thank you for your caution about keeping the 2 MOSFET’s at low temperature. I know how important it is. But they are ON only for a short duration (until 1200 rpm is reached) after which, the bypass (protection) resistor is cut off and the motor is fed directly from the pack.

Thanks again for your help. Happy New Year.

#### ronv

Joined Nov 12, 2008
3,770
__________________________________________________

Thank you for taking some of your time to answer me. I appreciate.

About balancing parallel MOSFET’s I thought I could count on the fact that the FET "on" resistance (Rds) will increase as it's temperature rises so parallel FET will "self balance". So as an example, if you have two FETs in parallel, the one that conducts the most current will also be the one that runs the hottest. As a result, it's Rds(on) resistance will rise and the other FETs will conduct more current and thus balance the current load.

By the way, I will consider using a ready-made Pulse Width Modulation unit. Thank you for your suggestion

Happy New Year
FETs are easy to parallel when they are used as switches. But when they are linear they need to be controlled independently because of differences in threshold voltages mostly.
This can be done pretty easily with some sense resistors in the source and an op amp.
So you turn the pot by hand as the rpm changes?
I'm not sure I understand where the 1 ohm resistor is in your circuit.

#### Denis De Bremme

Joined Dec 29, 2015
10
So you turn the pot by hand as the rpm changes?
I'm not sure I understand where the 1 ohm resistor is in your circuit.
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So you turn the pot by hand as the rpm changes? Yes (at first)

I'm not sure I understand where the 1 ohm resistor is in your circuit

Please see attached sketch. At start, the armature is protected from too much current by the 1 ohm large resistor. When the motor reaches 1200 rpm, the back emf is enough so it can be fed directly from the 96v source.

#### ronv

Joined Nov 12, 2008
3,770
______________________

So you turn the pot by hand as the rpm changes? Yes (at first)

I'm not sure I understand where the 1 ohm resistor is in your circuit

Please see attached sketch. At start, the armature is protected from too much current by the 1 ohm large resistor. When the motor reaches 1200 rpm, the back emf is enough so it can be fed directly from the 96v source.View attachment 97725
Thanks.
Here are the 2 possibilities. Both need more study before you start, but you can get the idea. The PWM circuit requires just a small heatsink and smaller FET.
I'm thinking 3 in parallel for the analog version.

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#### dannyf

Joined Sep 13, 2015
2,197
I control the speed by varying the field with a rheostat.
That's some rheostat to do that job.

#### wayneh

Joined Sep 9, 2010
17,498
Denis, one thing you should be aware of is that a rheostat that can handle that kind of power is not cheap. If you make hobbyist electronics projects, one thing you soon learn is that a variable resistor will often be the most expensive component in your project. One that can handle power, instead of just small signals, gets to be much more expensive. That's why big, power-handling variable resistors get replaced with a tiny variable resistor controlling a PWM circuit. The entire PWM circuit board, ready-made, costs far less than a big rheostat. And PWM is actually a lot more efficient as well, since you're not burning off excess energy in the rheostat.

#### Denis De Bremme

Joined Dec 29, 2015
10
Denis, one thing you should be aware of is that a rheostat that can handle that kind of power is not cheap. If you make hobbyist electronics projects, one thing you soon learn is that a variable resistor will often be the most expensive component in your project. One that can handle power, instead of just small signals, gets to be much more expensive. That's why big, power-handling variable resistors get replaced with a tiny variable resistor controlling a PWM circuit. The entire PWM circuit board, ready-made, costs far less than a big rheostat. And PWM is actually a lot more efficient as well, since you're not burning off excess energy in the rheostat.
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Thank you for your advice. I am trying to get more familiar with DCP,s (Digitally Controled Potentiometers)

Denis

#### wayneh

Joined Sep 9, 2010
17,498
I would guess those are even more expensive. PWM is the cheapest solution.

#### dannyf

Joined Sep 13, 2015
2,197
I am trying to get more familiar with DCP,s (Digitally Controled Potentiometers)
Good luck finding that for your application.

#### Alec_t

Joined Sep 17, 2013
14,317
What would the DCP be used for? As far as I know they can handle only a few tens of mA at most, and supply voltages of 5V or so.

#### Denis De Bremme

Joined Dec 29, 2015
10
What would the DCP be used for? As far as I know they can handle only a few tens of mA at most, and supply voltages of 5V or so.
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The DCP would be used for controling the gate of the Mosfet. This requires only 250 uA at 5V as far as I know (from Data sheet). I intend to use an AD8400 from Analog Devices Inc. The mosfet could be a FDL100N50 which can handle up to 100A and up to 2500W

#### ronv

Joined Nov 12, 2008
3,770
________________
The DCP would be used for controling the gate of the Mosfet. This requires only 250 uA at 5V as far as I know (from Data sheet). I intend to use an AD8400 from Analog Devices Inc. The mosfet could be a FDL100N50 which can handle up to 100A and up to 2500W
You need to close loop control it like I showed above because as the FET heats up everything changes. For example threshold voltage.

Here is a link I found about short duty cycle heat calculations. I've been looking for this myself.
http://www.irf.com/technical-info/appnotes/an-949.pdf