Simple op-amp analysis.

Discussion in 'Homework Help' started by tAllann, Oct 26, 2013.

  1. tAllann

    Thread Starter New Member

    Oct 26, 2013

    The book says that it is obvious that V0 is the negative of the ratio of the "feedback" resistor (8k ohms) to the input resistor (2k ohms) multiplied by the voltage the input terminal, thus it is -8 volts.

    It may be that my brain is currently fired, but the ratio of the resistors part through me for a loop. I was looking back trying to find the rule that allows that analysis to be done.

    I understand that the voltage at the node adjoining the 8 and 2 ohm resistors is 0. Before the 2 ohm resistor, the voltage is obviously 2 volts, and so the current flowing through both resistors is 1 Amp, so 1 Amp, and amp across the 8 ohm resistor yields a voltage drop of 8 volts. I just didn't understand the rule about the "ratio of the resistors".

    And you can't use voltage division right. Because 2 * 2 / 10 does not equal 2 ... But why can't you use voltage division?

    EDIT: My brain must be fried, I posted this in the wrong forum. Sorry about this. If you can move the thread that would be helpful. Thank you.
  2. LvW

    Active Member

    Jun 13, 2013
    tAllan, at first you mix ohms with kohms.
    At second, there are different ways to compute the gain.
    You have started already with the assumption that the voltage at the inverting opamp terminal is zero (in fact it is some mikrovolts, however it is common usage to neglect this small voltage).
    Now you can apply the superposition rule for this zero voltage that is caused by both the input signal and the output signal. This leads to the ratio Vout/Vin.
    Other methods use the current through the resistors or the classical feedback model with the feedback factor R1/(R1+R2).

    EDIT: "And you can't use voltage division"
    Of course, you can - no, you must if using the superposition principle(you have two voltage sources!).
    Last edited: Oct 26, 2013
    tAllann likes this.
  3. WBahn


    Mar 31, 2012
    Be sure that you understand that the voltage at the inverting input of the opamp is 0V only because two things are assumed to be true: First, that the opamp is operating in its linear (i.e., active) region; and second, that the gain of the opamp is sufficiently high so that there is very little voltage difference between the two inputs of the opamp. Many people make the mistake of just assuming that the voltage at the opamp inputs are 0V. In general, that isn't true. In this case, the voltage at the inverting input is 0V, even given these assumptions, only because the voltage at the non-inverting input is rigidly ties to 0V.

    Next, the analysis you did was just fine and will reveal the answer, but it is best seen by using symbolic values instead of specific resistor values and voltages.

    So let's say that the input voltage if Vin and the output voltage is Vout. Let's further say that the input resistor (the 2kΩ resistor) resistor Rin and the feedback resistor (the 8kΩ resistor) Rf.

    Answer the following with the symbolic quantities:

    Q1) What is the current (left to right) through R1?

    Q2) What is the voltage drop across R2 (positive at left side of R2)?

    Q3) What is the output voltage in terms of the input voltage?

    Q4) Do you see where the ratio of resistors comes from?
    tAllann likes this.
  4. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    Your analysis is spot on, now let's work on that rule about the "ratio of the resistors".

    Oh, I hate that term lol. But it works well enough here.

    The current thru the 2K is Vin/2K. Call this I = Vin/R1.

    The voltage across the 8K is 8K*I. Call this Vo=-R2*I

    (Bang back if you don't see where the "minus one" comes from.)

    Now put eq2 into eq1:

    Vo = -R2* Vin/R1

    Divide both sides by Vin:

    Vo/Vin = -R2/R1 = Gain

    It just follows from the math for the general case.
    tAllann likes this.
  5. tAllann

    Thread Starter New Member

    Oct 26, 2013
    Thanks. I completely understand now.