# Op Amp adder to control current flow?

Discussion in 'Electronic Basics' started by The little lost angel, Dec 15, 2003.

1. ### The little lost angelGuest

I'm trying (again, picking up from the last school vacation) to design
a better load bank with a fixed resistive part and a self adjusting
part. So that I don't have to use expensive matched parts.

/S = my symbol for a switch

+V1--+--R1--+----------+--/S --R2a--+
| | | |
| | +--/S --R2b--+
| | | |
| | +--/S --R2c--+
| | | |
+--\ | | |
VRef1 --/ | | |
| +------------Gnd
V2

Basically, R2A, 2B, 2C are fixed resistors drawing current in a binary
ladder amount when the switches are closed. They are going to be
valued at slightly higher than needed.

E.g. if V1=12V,
R2A= 13 Ohms / 0.92A
R2B= 6.2 Ohms / 1.94A
R2C= 3.3 Ohms / 3.6A

The slack is then taken up by the OpAmp/mosfet part where

R1 is a current sensing resistor. The ground of the OpAmp is not at
common ground so that the OpAmp will read the voltage drop across R1
and not the voltage remaning after R1.

Then a reference voltage VRef1 will allow the OpAmp to check the
current drawn, e.g. 92mV when R2A is closed against a VRef of 100mV,
the OpAmp should put a voltage to the mosfet such that the mosfet
resistance goes to 150 Ohms and the net effect is 1A is drawn.

Similarly if R2A and R2C are both closed, they draw 4.52A nominal,
while the target is 5A and VRef should go to 500mV.

Would something like an OpAmp adder (the web sources I read seem to
indicate it can be used to add multiple voltage inputs) be used? Or do
I need some kind of logic/binary component?

Of course, the other thing I haven't managed to figure out is, how to
prevent the OpAmp from dropping its output once the current drawn goes
up. Since when the mosfet starts drawing power, the voltage difference
between the two inputs of the OpAmp becomes zero, it will shut off
output. The FET will get no gate voltage and the net current drops
down. This causes a cyclical up and down I would like to avoid.

Are there better ways of doing this?

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2. ### HermanMunsterGuest

Yes.
something is wrong with your circuit or description. If V1 is an
uncontrolled source or fixed voltage referanced to ground, and I assume Vref
is fixed, then your opamp will saturate full on or full off. No regulation
will occur. This circuit will function as a programmable shunt, not a load
bank.

3. ### The little lost angelGuest

Hmm, wouldn't putting the ground/-ve of the OpAmp to the wire after R1
make the input to the OpAmp as the voltage dropped over R1?

E.g. V1=12V , R1=0.001 ohms, I=10A
The the voltage after R1 will be 11.99V

12V will go to the +ve input, while 11.99V will go to the ground of
the OpAmp, wouldn't the OpAmp then see the input voltage as 0.01V?

Thanks!

--
L.Angel: I'm looking for web design work.
If you need basic to med complexity webpages at affordable rates, email me
Standard HTML, SHTML, MySQL + PHP or ASP, Javascript.
If you really want, FrontPage & DreamWeaver too.
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