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uA741 op amp has 85K of resistance?

Discussion in 'General Electronics Discussion' started by komalbarun, Sep 17, 2013.

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  1. komalbarun


    Nov 25, 2011
    I connected a uA741 with a single supply of 9V and a resistor of 5K connected to the positive supply leg and measured the voltages across the resistor and the op map. Negative supply leg connected to ground.

    Vresistor : 0.5V
    VopAmp : 8.5V

    doing the maths:
    [ 5000/(5000+R?) ] * 9 = 0.5
    45000 = 2500 + 0.5R
    RopAmp = 42500 / 0 .5
    RopAmp = 85000ohms

    I got the OpAmps resistance to be 85K??

    Is this right or maybe the -ve leg must be connected to a -ve voltage to get a proper value?
  2. GreenGiant


    Feb 9, 2012
    That "resistance" will probably vary a little bit with different voltages, and with different operations of the chip.

    Your math is correct, but Im not sure why you need to know this, in the spec sheet it should state the current draw of the device.

    Why did you want to find this information?
  3. komalbarun


    Nov 25, 2011
    well I read somewhere that for semiconductors' specs its not the current drawn that is shown, but that is the maximum current it can withstand before getting damaged. Maybe am confusing it with diodes xD.

    Anyways, how can I use 2 9V batteries to supply a +ve and -ve volt to the op amp?
  4. BobK


    Jan 5, 2010
    The Quiescent current spec tells how much current is used by the op amp when no current is sourced or sunk from the output.

    To use two 9V for a dual supply, hook the + of one to the - of the other. This connection is then ground. The other + is V+ and the other - is V-.

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