# Need help with a MOSFET electronic load

• posted

I have been looking at some electronic load circuits. I would like to have a circuit that uses a MOSFET that pretends it is a high wattage fixed resistor. My application is a homemade RC relaxation oscillator EDM machine. Since I am experimenting, I don't know what resistance value I am ultimately going to need so I would like to make it adjustable. I also wanted to try my hand at coming up with something more elegant than using light bulbs as power resistors. I have a few questions about circuit 1 that I copied from an internet source.

Is the math correct for circuit 1? To behave like a fixed resistor I=V/R. V+ = V * R1/(R1+R2) and V- = I * Rs V+ = V- V * R1/(R1+R2) = I * Rs I = V / (Rs(R1+R2)/R1)) R = Rs(1+R2/R1)

Suppose that I want the circuit to behave like a fixed 20 ohm, 200W power resistor for example. What would be logical choices for Rs, R1, and R2? A source that I have been reading says to make R1>>Rs. What does the >> mean?

I believe that Rs should be a very low value to keep the I^2*R power dissipation down. Also it would have to be less than the total resistance that the circuit is trying to mimic in the first place. Is this true?

Why is the R1, R2 voltage necessary. Why wouldn't circuit 2 work?

Also there was some mention of adding an RC to the output of the op amp to prevent oscillation. Could someone provide more details on that?

Why wouldn't circuit 2 work?

Is the math correct for circuit 2? To behave like a fixed resistor I=V/R. V+ = V V- = I * Rs V+ = V- V = I * Rs I = V/Rs R = Rs

NEVER MIND! Rs would have to be the power resistor that I am trying to have the circuit mimic in the first place.

Any help would be greatly appreciated. Thanks

• posted

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

• posted

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 ...

Dan

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-- 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```
• posted

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

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• posted

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

First, let\'s look at the voltage across and current through the MOSFET```
• posted

^ R

JF

• posted

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

So can never behave like a physical resistor.

.... Phil

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