Inverting Op-amp Questions

Discussion in 'Homework Help' started by cheney23, Aug 12, 2013.

  1. cheney23

    Thread Starter New Member

    Aug 10, 2013
    1
    0
    Hi guys, just hoping you could help clarify a few things for me.

    The questions are based off this data sheet: http://www.utm.edu/staff/leeb/LM301.pdf

    and this circuit:

    [​IMG]

    1. Estimate A_{o}, \sigma, and GBP for the LM301A.
    A_{o} = \frac{v_{o}}{v_{+}-v_{-}}

    The data sheet specifies the typical input offset voltage as 2.0mV and maximum of 7.5mV. So in the most extreme case, v_{+}-v_{-} = 7.5mV
    The power supply is 15V, so the highest we could expect v_{0} to be is 15V. (Not sure if I'm going about that the right way).

    As for σ (the open-loop bandwidth), with the supply voltage of 15V, the voltage gain is guaranteed to be 88dB at minimum. Then reading off the Open Loop Frequency Response graph, the frequency is approximately 10Hz with a capacitor value of 30pF, which I believe we will be using.

    GBP = A_{o}\sigma
    2. What is the approximate input impedance of the amplifier circuit (as seen by the source V_{i})

    Should I assume an ideal op-amp, and that v_{-} is a virtual ground, and therefore the input impedance as seen from V_{i} is simply R_{i}?

    Any input would be much appreciated, it's been a few years since I last did any work involving circuits. Thanks!
     
  2. #12

    Expert

    Nov 30, 2010
    16,337
    6,820
    The highest output depends on the load current, but it never gets to Vcc. See graphs.

    You seem to be wrong or incomplete in your statement about 10Hz with a 30pf capacitor. The gain bandwidth product is still a trade off between gain and frequency with the capacitor added. You might properly say the nominal gain will be 98 db at 10 Hz with a 30 pf capacitor, according to the graph, but it could be lower if your particular chip is in the low range of allowable gain.

    The input bias current could be as high as 250na, according to the datasheet, and that is where you start calculating input impedance. From Page 10 you can see that (up to) 250 na of electrons will flow out of the input pins. Compare that to your intended gain resistors to find the error it will induce.

    Know this: The LM301A is an antique chip. I was working with it in 1973.
    Ask again if I missed something.
     
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