Op amp Equivalent Circuit Question

Discussion in 'Homework Help' started by seal308, May 24, 2015.

  1. seal308

    Thread Starter New Member

    Jun 30, 2012
    11
    0
    Hello,

    I am asked to find the dependent voltage source of the following and choose the right polarity. (see attached file)
    I know the dependent source is A0Vin , so I have to do KVL for it. My current is clockwise for each loop
    From figure 3:
    KVL1: IA0Vin + IRL
    = 0
    and I know, I = V
    0/RL
    Therefore KVL1: (V0A0Vin)/RL + V0 = 0
    Then I did a KVL for the left circuit.

    KVL2: -V
    s + I2Ra + I2Rb = 0
    Also I
    2=Vx/Rb
    Therefore KVL2: -Vs + VxRa/Rb + Vx = 0

    And that's as far as I got so far.
    From the diagram it looks like the 2 loops in Figure 3 do not connect, also there is no feedback of the op amp in the diagram. So i'm confused how to relate.
    I'm wondering if in KVL1 if V
    in = Vs
    But I'm not sure.
     
  2. panic mode

    Senior Member

    Oct 10, 2011
    1,320
    304
    what is unit of gain A0?
    how does that affect your KVL1?
     
    Last edited: May 24, 2015
  3. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hello there,

    Figure 3 uses a voltage dependent voltage source which means it is a voltage source controlled by another voltage somewhere in the circuit. That control voltage for this circuit is Vx, and it is referenced to ground so the required gain for the source will be in the form:
    Vvcvs=Vout=Vx*Ax

    where Ax is the gain that you will calculate from the first circuit with the op amp in it.

    A simple way to analyze the op amp circuits is to use the fact that Vout is the internal gain times the voltage differential between the non inverting terminal (vp) and the inverting terminal (vn). This would be written as:
    Vout=(vp-vn)*Aol

    where Aol is the internal gain of the op amp. Once you calculate vp and vn you can then take the limit as Aol goes to infinity, and you will have a neat little equation where you can find whatever you need, like the gain Ax for the above.
     
  4. seal308

    Thread Starter New Member

    Jun 30, 2012
    11
    0
    Ok so you said that labeled as Vin was really Vx.
    Knowing that I updated my KVL equations.

    KVL1: (V0A0Vx)/RL + V0 = 0

    Because KVL1 and KVL2 both = 0, I set them equal to each other.
    (V0A0Vx)/RL + V0 = -Vs + VxRa/Rb + Vx

    This leaves me with A0Vx = [[-Vs + VxRa/Rb + Vx - V0] RL]/V0

    But I don't think this really tells me anything, or anything about the polarity.
    Also to be honest I didn't quite understand your post. I don't think I need to take the limit of anything, we never did anything that advance in this course yet.
    I think I'm supposed to use KVL, but I'm not sure if I'm doing this right.
     
  5. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    What i was saying was that the output of your second circuit will be in the form of:
    Vout=Vx*G

    where G is the gain you calculate, and of course Vx=Vs*Rb/(Ra+Rb).
    So the entire output expression of the second circuit will be:
    Vout=Vs*Rb*G/(Ra+Rb)

    where G is what you have to calculate. Once you have the right value or expression for G, you'll get the same Vout with either circuit for any input Vin.

    I will use a different method then to show how this works, and you can use whatever method you were taught, and we can go over that too if you like.

    So we start with the top circuit. What we dont know is what model you can use for the op amp, so we'll assume a simple model where the gain is infinite and so is the bandwidth.

    Since the circuit forms a non inverting op amp amplifier, the gain from Vx to output is:
    Vout=Vx*(R2/R1+1)

    so if we expect to get the same output in the bottom circuit we need a gain G=R2/R1+1.
    The voltage controlled voltage source is therefore non inverting, and has gain R2/R1+1.

    To check:
    Since Vx=Vs*Rb/(Ra+Rb) we get in total:
    Vout=Vs*Rb*(R2/R1+1)/(Ra+Rb)

    Using another method, i get:
    Vout2=(Rb*Vs*(R2+R1))/((Rb+Ra)*R1)

    and rearranging the previous equation for Vout i get the same thing:
    Vout=(Rb*Vs*(R2+R1))/((Rb+Ra)*R1)

    so it checks out ok.

    So what you are after is the gain G and once you have that you can label the voltage controlled voltage source with that gain, and also show the polarity, and since it is non inverting the plus goes on top and minus on bottom.

    Does this help a little more?

    If you need to incorporate a finite gain instead (Ao) then we would do:

    vn=Vout*R1/(R1+R2)
    vp=Vx

    Vout=(vp-vn)*Ao

    Vout=Ao*(Vx-(Vout*R1)/(R2+R1))

    solving for Vout explicitly we get:
    Vout=(Ao*Vx*(R2+R1))/(R2+Ao*R1+R1)

    and now pull Vx out by dividing both sides by Vx and we get:
    Vout/Vx=(Ao*(R2+R1))/(R2+Ao*R1+R1)

    so the gain is now the expression on the right side above.

    You should be able to reach this conclusion your own way too.

    In case you are up to it, if we take the limit as Ao goes to infinity, we get the same result as before using the 'perfect' op amp. You probably dont really need this right now though.
     
    Last edited: May 25, 2015
  6. seal308

    Thread Starter New Member

    Jun 30, 2012
    11
    0
    Yes this makes a lot more sense.
    The one thing I am confused about though is V0 = VxG.
    Does that mean that Vo is actually the voltage controlled voltage source. I was confused as to what Vo was from diagram so I assumed it was some node between the dependent source and the resistor.
     
  7. MrAl

    Well-Known Member

    Jun 17, 2014
    2,433
    490
    Hi,

    In this case Vout is the controlled voltages source because there is nothing else between Vout and the top node of the source and the bottom node of the source connects to ground (0v). So anything the source puts out is Vout.
    This isnt always the case, but here it is because of that connection.
    So if the source puts out 2v, then Vout=2v, and if the source puts out 3v, then Vout=3v, etc.

    A voltage controlled voltage source has four terminals:
    1. Input +
    2. Input -
    3. Output +
    4. Output -

    In the second circuit, the Output + is connected to Vout, the Output - is connected to ground (0v), the Input + is connected to Vx, and the Input - is connected to ground (0v). So when Vx varies the output varies according to Vx*G.

    Note that if the original circuit was an inverter, then we would have had the Output - go to Vout and Output + go to ground, or else reverse the input terminals (either way we get an inverter).
    This also means that a second way to connect it for a non inverting output would be to run the Output - to Vout and Output + to ground but then also reverse the input terminals. It is simpler just to do it the above way though.

    Sometimes the controlled source is drawn like a sort of IC chip, like a square with four terminals, two for input and two for output. Looking at it that way we can see that the entire circuit is connected together. The way your second circuit has it drawn though is also very common because it makes drawing the schematic easier (less crossing lines for more complex circuits). So you will often see a controlled source with just the 'output' part shown connected to something (like the second circuit where it connects to Vout) while the 'input' part is just a variable like that Vx. If we drew this all out we would see the input as part of the whole source and then we would see Vx also connecting to that 'block' that makes up the controlled source.
    (See attachment).
     
    Last edited: May 26, 2015
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