Non-Ideal Opamps and expressions

Discussion in 'Homework Help' started by jason_bourne, Oct 24, 2011.

  1. jason_bourne

    Thread Starter New Member

    May 4, 2010

    I have a test coming up soon and i would like to know how to solve/account for non-ideal characteristics in op amps. I know what they are and their definitions but i am unsure how it actually affects the output and operation.

    In addition i have difficulty deriving some of the expressions, I usually start simple by applying a positive/negative signal to input and see how the current flows however i can't seem to get expressions for the understanding that i have of unknown circuits. I have attached some of the questions in this post.

    a)inverting with zener (4.7) - (01_1) FILENAME
    Only need to do ideal analysis, when Vin is +
    I1 flows through R1, e+ =0 thus e- = 0, as it is ideal current can't flow into the terminal thus it has to go split between R2 and the zener. So do we just ignore the zener, will it act like a diode? when Vdrop is 0.7, if not then is it just a short circuit and Vo = -Vi?
    It's inverting so Vo will be -, but not sure if there is a gain or how the diode acts.
    When Vin is - assuming reverse current is reached the diode conducts and the combination of zener and R2 sees Vo = 4.7, so when its - Vo = 4.7?

    thus what is the Vo/Vi expression?

    b)The diode is 2.5V at 1mA (01_2)
    Need to find Vo, and then find all Resistor values so that Vo = 3. Finally how can we make this circuit source more current?

    Seems to be no input going in, so lets assume Vo = 15
    then e+ = 7.5 (assuming R1 = R2). So e- = 7.5
    voltage drop across R = 7.5
    Thus Vo = 2.5 + 7.5 = 10, ie Vo = kVo + Vz?
    So it is decreasing? eventually Vd = 0, then it will switch to 0
    then e+ = e- = 0, then Vo = Vz?

    Looks like a monostable timer but its using a resistor instead.

    c) This is a relaxation oscillator. I can derive the expression for frequency
    1/RCln(1+k/1-k) where k = R1/R1+R2.
    Anyway the question is how is this affected with non-ideal characteristics,
    Slew Rate of 13V/s, Input bias current of 50pA, output short circuit current of 20mA, input impedance of 1k (not good i reckon). So how does the expression and output change? In this case ideally it wants to charge up to 15 V (Vcc = 15), instead if goes to +-7.5 (due to e+ = 7.5).

    Also this is 50% duty cycle, how is it possible to make it 25% duty cycle with addition of a diode and resistor. I understand we now want k to be 0.25 instead of 0.5. We used voltage divider before, now what? Also how can we calculate the values when it is 25%?

    d)This is a capacitance multiplier. Need to find expression for Va/Vi
    This is what i did:
    Opamp1 is voltage buffer, therefore Vo1 = Vi
    The e- terminal for Opamp 2 is 0 therefore (no spec on input bias) not sure how to account for this so i just said current flows straight through to Va
    thus Va/Vi = -R2/R1

    Now it says that prove Vi/Ii = (1+R2/R1)C (this is Cequivalent)

    Ii = (Vi - Va)sC
    this gives 1/(1+R2/R1)sC
    which is -j/(1+R2/R1)wC not sure what i am doing wrong here

    Thanks a lot, help with any of the questions would be greatly appreciated