Need help with a MOSFET electronic load

[jalbers@bsu.edu]

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

 

[amdx]

<jalbers@bsu.edu> wrote in message
news:285bdab7-5093-4e2f-831b-9fe115c68a65@x1g2000prh.googlegroups.com...

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

Suppose that I want the circuit to behave like a fixed 20 ohm, 200W
power resistor for example.

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

 

[Dan Coby]

jalbers@bsu.edu 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?

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.

I believe that Rs should be a very low value to keep the I^2*R power
dissipation down.

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.

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?

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).

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

See below.

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?

snip ...

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

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


Dan

 

[Tim Wescott]

jalbers@bsu.edu 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.

-- snip --

Any help would be greatly appreciated. Thanks

* 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

 

[default]

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

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


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[John Fields]

On Tue, 4 Nov 2008 11:07:29 -0800 (PST), "jalbers@bsu.edu"
<jalbers@bsu.edu> 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 = I˛R,

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

 

[John Fields]

On Tue, 04 Nov 2008 17:51:47 -0600, John Fields
<jfields@austininstruments.com> wrote:

On Tue, 4 Nov 2008 11:07:29 -0800 (PST), "jalbers@bsu.edu"
jalbers@bsu.edu> 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 = I˛R,

We can rearrange and solve for the current like this:

P 200W
I = sqrt --- = sqrt ----- ~ 3.16 amperes
R 20V
^

R

JF

 

[Phil Allison]

<jalbers@bsu.edu>

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.

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

So can never behave like a physical resistor.




..... Phil