Re: Need help with a MOSFET electronic load

On Tue, 4 Nov 2008 11:07:29 -0800 (PST), "jalbers@xxxxxxx"
<jalbers@xxxxxxx> wrote:

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.

https://ilocker.bsu.edu/users/jalbers/WORLD_SHARED/ElectronicLoad.PDF

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

---
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
.

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