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Op Amp adder to control current flow?

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

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  1. 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?

    --
    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.
    But keep in mind you pay extra bandwidth for their bloated code
     
  2. 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. 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.
    But keep in mind you pay extra bandwidth for their bloated code
     
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