Need help with a MOSFET electronic load

If 20 ohms is your low value, how about paralleling a couple of 50 ohm 100 watt resistors, then using a MOSFET to adjust from 25 ohms to lower values. If you use the resistors then your MOSFET will only dissipate 50 watts instead of 200 watts. You can probably find 50 ohm resistors at surplus prices. Mike

The '>>' indicates that R1 should be 'much greater than' Rs. Your equation for the effective resistance of the circuit only considers the current through Rs. You have neglected the current flow through the series combination of R1 and R2. The actual effective resistance of the total circuit is the parallel combination of your R and R1+R2. For large values of R1+R2 (i.e. R1+R2 much larger that R) it is reasonable to neglect the current flow through R1+R2.

A low value for Rs will reduce the power dissipation in Rs. However you have to change R2/R1 to compensate to give the resistance that you are trying to mimic. For a desired mimic resistance (Rm), the voltage across the mosfet and Rs is I * Rm. The power being dissipated in the pair (Rs and the mosfet) is I^2*Rm. This power is a function of only the desired mimic resistance and the current. If you reduce the voltage across Rs, you will need to increase the voltage across the mosfet to maintain the same Rm. Thus reducing Rs will reduce the power dissipation in Rs but it will increase the power dissipation in the mosfet.

Yes, Rs needs to be less than the resistance that you are trying to mimic. You can see from your equation R = Rs(1+R2/R1) that it is not possible to to get R2/R1 to be less than zero (unless you have negative resistors in your pocket).

See below.

snip ...

Yes, you have already answered your own question. (Since you already knew the answer, why did you bother to ask?)

Dan

-- snip --

• Get a handful of resistors and make a power resistance decade box.

• Get some toaster ovens (or toasters) from Goodwill (or a dumpster) and use their heating elements for cheap power resistors.

• What in heck is the matter with light bulbs?

--

Tim Wescott
Wescott Design Services
http://www.wescottdesign.com

Do you need to implement control loops in software?
"Applied Control Theory for Embedded Systems" gives you just what it says.
See details at http://www.wescottdesign.com/actfes/actfes.html

Look at gain versus temperature and you'll see the light. Self heating changes gain and in turn, changes current.

--


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--- Here's another way to look at it: (View in Courier)

First, let's look at the voltage across and current through the MOSFET which'll make it look like a 20 ohm, 200 watt resistor.

Since:

P = I²R,

We can rearrange and solve for the current like this:

P 200W I = sqrt --- = sqrt ----- ~ 3.16 amperes R 20V Then, since:

P = I E

we rearrange and solve:

P 200W E --- = ------- ~ 63.3 volts I 3.16A

Next, here's your circuit in ASCII:

+V---+----------------+--E1 | | [R1] | | | E2--+-------|+\\ D | | >---G [R3] | +-|-/ S | | | [R2] +----------+--E3 | | | [R4] | | GND>-+----------------+

with the reference designations changed, the MOSFET labeled "R3", and some voltage nodes added for later.

What happens with opamps is that when the voltage on the + input goes more positive than the voltage on the - input, their output voltage will rise toward the positive rail (+V) and when the voltage on the - input goes more positive than the voltage on the + input, the output voltage will fall toward ground (GND).

Now, looking at your circuit and assuming the MOSFET is off, the - input will be looking at 0V (GND) through R1. Then, since the + input is connected to the voltage divider R1R2, it'll be more positive than the

-input, and the opamp's output will go positive and start to turn on the MOSFET.

As the MOSFET turns on more and more, the voltage on the - input of the opamp will rise more and more until it equals the voltage on the + input and the MOSFET will then be the resistance you want it to be! :-)

But, how to pick that point?

Well, it depends mainly on how much power you're willing to waste in R4 and how close to ground the opamp will allow the input signals to be and still work. 1 volt is easily doable, so let's say that with the target current, 3.16A through R4 that's the voltage we want.

Then, from Ohm's law, we can say:

E 1.0V R = --- = ------- = 0.316 ohm I 3.16A

So R4 will have a resistance of 0.316 ohm and the power it'll dissipate will be:

P = IE = 3.16A * 1.0V = 3.16 watts.

Not bad, and all we have to do now is set the reference voltage on the opamp's + input to be equal to 1V and we'll be done.

Well, almost...

We do have to consider the opamp's supply voltage, the voltage divider should have a fairly stiff source to work off of, and we need to make sure that there's enough headroom in the HV supply so that the input to the MOSFET never falls below 64.6V. (63.6V for the MOSFET and 1.0V for the 0.316 ohm shunt)

So, how about some more details?

Like, where does the EDM get connected into this scheme, what kind of a power supply (voltage, current) are you using, etc, etc, etc.?

JF

^ R

JF

** I hope you realise that all power MOSFETs have inbuilt diodes in parallel.

So can never behave like a physical resistor.

.... Phil