Differential amplifier problem

Discussion in 'General Electronics Chat' started by rp_1, Aug 22, 2008.

  1. rp_1

    rp_1 Thread Starter New Member

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    Hello,
    I'm using an opamp as a differential amplifier. I have trouble figuring out the problem. So i need your help.I have attached the circuit- Fig 1. The gain is 100. V1 is supplied from a potentiometer which is 0V to 10V. I want the V2 to detect the voltage output from a transistor source pin.

    I have kept the gate voltage at 0V. So the transistor is OFF and no voltage should be present at the SOURCE pin.

    But when i change the V1 voltage (non inverting input) from 0V to 10V, i see the same voltage 0V to 10V at the source pin eventhough the transistor is OFF. I checked to see whether the transistor is bad but it is fine. I verified it with a diode (Fig 2) between SOURCE pin of the transistor and V2. Now i tested it again with 0V to 10V at V1. I did not see any voltage at source pin of the transistor. But i do see same voltage (0V to 10V) at V2.

    I dont understand why there is 0V to 10V at V2 when it should read only 0V from Source pin of the transistor. Could someone please help me.

    Thanks,
    RP

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  2. mik3

    mik3 Senior Member

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    Is it a p or n channel FET?
  3. rp_1

    rp_1 Thread Starter New Member

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    I'm sorry tht i didnt mention it. Its an NFET.
  4. beenthere

    beenthere Retired Moderator

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    Your op amp can't see a difference voltage. The output voltage is fed back to the summing junction through either the 100K resistor, or the FET and 1K resistor. The gain will always be 1.

    To make the op amp into a follower with gain, connect the source lead of the FET to ground and the drain to the end of the 1K resistor. That will give you your gain of 100 when it is turned on. With lower conduction, the gain will reduce.
  5. rp_1

    rp_1 Thread Starter New Member

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    Hi,
    Now i understand the voltage at V2. Thank you.
    The circuit in the attachment is only a part of the whole design. I chopped the gate part off from the original circuit. The transistor source is the system output for the entire design. So the source cannot be grounded. Are there any other possibility?

    Thanks,
    RP
  6. Audioguru

    Audioguru New Member

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    Negative feedback causes the voltages at both inputs of an opamp to be the same unless its output is clipping or its input voltage is too high or too low.
  7. rp_1

    rp_1 Thread Starter New Member

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    If i dont use a negative feedback, then the circuit is similar to using a comparator but there will be only 2 output states either High(+VCC) or Low(Gnd) which means no linear region of operation and there is no gain limitation.
  8. beenthere

    beenthere Retired Moderator

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    If you want comparator action, then use a comparator like an LM311. If you are running the op amp on a positive voltage and ground, then the op amp will not be able to swing the output all the way to ground, unless you are using a rail-to-rail device.

    All devices have a gain limitation. There is always a measurable transition between one state and the next.
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