What is happening in this circuit?

Discussion in 'Homework Help' started by geminax, Dec 12, 2018.

  1. geminax

    Thread Starter New Member

    Dec 12, 2018
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    upload_2018-12-12_14-43-52.png
    Hi, I was given a question to express Vload in terms of Rload, R and Vth. However, I am a bit confused as to what exactly is going on at Vout_a.
    Assuming the op-amp is ideal, I first made statements regarding what voltages were equivalent.

    Vload = Vout
    Vout = Vn (negative terminal of op amp)
    Vn = Vp (positive terminal of op amp)
    Vp = Vout_a

    The problem that I run into is figuring out what Vout_a is. I tried doing KCL at that node, but all i get is that Vout_a is equal to Vth since no current can pass through the input terminals of an op-amp. And that's where I'm stuck because a lot of things just don't seem to add up with my thought process
     
  2. geminax

    Thread Starter New Member

    Dec 12, 2018
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    Since the op amp is just a buffer, can I just act like it isn't there? Meaning Vload is just Vth - IR
     
  3. ebp

    Well-Known Member

    Feb 8, 2018
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    What would make Vout_a differ from Vth?

    Is the expectation that you analyze this on the assumption of an ideal op amp or a real op amp? It makes a big difference, not necessarily in the magnitudes but in the complexity of the equations required. It you can use an ideal amp, what do you know about the characteristics of an ideal amp?
     
  4. geminax

    Thread Starter New Member

    Dec 12, 2018
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    I believe we are analyzing this on the assumption of an ideal op amp. Since an ideal op amp has infinite input impedance, no current would travel. I'm also considering the characteristic of a buffer op-amp, Vout = Vin. Therefore, Vout would be equal to Vth? But how can that be true if there is no current traveling?

    And the resistor would mean a voltage drop if there was current traveling, no?
     
  5. WBahn

    Moderator

    Mar 31, 2012
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    That no current flows and, as a result, Vout_a is equal to Vth is precisely the point of using the opamp.
     
  6. WBahn

    Moderator

    Mar 31, 2012
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    Which 'I'? Iload?

    Is Iload the current flowing in R?

    If not, then you can't just through the nearest I and the nearest R at Ohm's Law.
     
  7. WBahn

    Moderator

    Mar 31, 2012
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    Current traveling where? You need to be specific.

    In the load? There IS current flowing in the load.

    In R? If there were current flowing in R, the that would require that Vout NOT be equal to Vin.

    And don't forget that an opamp is also connected to a power supply.

    Again, traveling where? Yes, there would be a voltage drop across a resistor IF current is flowing in THAT resistor.

    You are trying to assert that R and Rload are in series (meaning that whatever current flows in one must flow in the other). NOT the case!
     
  8. geminax

    Thread Starter New Member

    Dec 12, 2018
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    So what is the case then?

    What I think to be true:
    No current flowing through R.
    There is obviously current flowing through R_LOAD
    Vout = Vn = Vp = Vout_a

    My question is where does Vout_a get its value if there is no current flowing on that LEFT side of the circuit
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes.

    This is wrong. Because the opamp will "set" his output at Vin.

    From the external voltage source (Vcc , Vee) and internal circuit inside opamp ensures that Vin=Vout in this case.

    https://www.allaboutcircuits.com/te.../chpt-8/single-ended-differential-amplifiers/
    https://forum.allaboutcircuits.com/...ns-but-im-troubled-by-them.64696/#post-444315
     
  10. WBahn

    Moderator

    Mar 31, 2012
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    It get's its value from the voltage source.

    Image two drums filled with water and with a pipe one the ground connecting them at the bottom. The only way for there to be no flow in the pipe is for the fluid level in both drums to be the same. If you start out with fluid in one drum and not the other, then fluid will flow (and there will be a pressure drop across the pipe) but only until the fluid levels match. It's the same here. Initially a very small amount of current flows but after a fraction of a nanosecond (maybe a bit longer), things have equalized and the voltage at the opamp input is essentially identical to the supply voltage. Otherwise, current would flow but, with no where to go, it would result in the voltage at the opamp input moving even closer to the supply voltage.
     
  11. crutschow

    Expert

    Mar 14, 2008
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    A significant reason for using a op amp as a follower is so that the output current is provided by the op amp and is not seen by the source voltage, and thus the output voltage equals the input voltage independent of load and source resistance (within the capabilities of the op amp of course).
     
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