Current amplifier with OpAmp

Discussion in 'Homework Help' started by PsySc0rpi0n, Nov 3, 2015.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    The circuit I have is this:
    2.2.png

    I' not sure if I'm supposed to add that current source there or if that symbol is just for information purposes!

    I also want to ask if the 2 thumbs-up rules also applies in this case! The rules are:
    1 - no current flows into any of the input terminals
    2 - the OpAmp tries by any means to keep the voltage drop as 0V between the 2 input terminals!
     
  2. MikeML

    AAC Fanatic!

    Oct 2, 2009
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    Add the word "almost" in front of #1

    If in doubt, look up the data sheet for your opamp, and check the parameter "input bias current".

    Similarly, check the parameter "input offset voltage".

    in.gif
     
    Last edited: Nov 3, 2015
  3. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Yes you need to add a current source

    Yes, this two rules hold for this circuit as long as opamp is not in saturation.
     
  4. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    And what about the voltage source Vcc in series with the current source? Do I have to add it too? If so, what is the purpose of having a voltage source and a current source on the same branch?
     
  5. MikeML

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    136.gif
     
  6. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I think the Vcc is not needed as LTSpice current source is able to work without a voltage source to "activate it".

    So I have the attached circuit and I had made sure that I_Load = I_R1 + I_R2

    Now I need to come up with an expression that gives me I_load in function of I_in. The only thing I can remember of by now is that I_Load = I_R1 + I_R2. But I think this is not what I'm expected to find!

    Any tips?
     
  7. MikeML

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    Not just in LTSpice... By definition, a current source will have whatever voltage between its terminals that is required to force the specified current. It makes no difference to the current-source what external voltages are around....

    I(Rload) =I(R2) + I(R1); KCL
    V(noninv pin) = V(invert pin) = V(output pin); per ideal opamp.
    should be enough for you to get started to write an expression for I(Rload) in terms of I(in), R1 and R2...
     
  8. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Ok, I'm aware of that about voltages and input terminals and Vout!

    Is it possible to come up with an equivalent circuit without the OpAmp?

    I've been working with some equations to see if I'm going anywhere with it, but I'm not going anywhere!
    I've tried to convert the OpAmp circuit into another circuit taking into account those 2 rules but probably I can't do it!
     
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  9. MikeML

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    The first two purple equations in aac.png are a good start. You shouldn't care what Vx is, so manipulate the second equation to get rid of Vx. What is left?
     
  10. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    You mean to get rid of vx by simplification or by substitution for something else? I'm not getting anywhere by trying to simplify the equation!
     
  11. MikeML

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    1. Iin*R2 = (Vo-Vx)

    2. Ir1*R1 = (Vo-Vx)

    3. therefore Iin*R2 = Ir1*R1

    4. solve 3 for Ir1:
    Ir1= Iin*R2/R1

    5. Iload = Iin + Ir1 = Iin + Iin*R2/R1 = Iin*(1+R2/R1)

    6. Iload/Iin = (1+R2/R1)

    Try your sim for R2= 9*R1

    2.2.gif

    As load goes to 10KΩ, what power supply would the opamp require?
     
    Last edited: Nov 4, 2015
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  12. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
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    Hum, ok... I wasn't close to that solution to the current relationship...

    About the OpAmp requirement it would require a 55V supply or it would saturate and those 2 rules wouldn't hold anymore! That's what I think...
     
  13. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    For type of a circuit (if we interest in current) you should write your equation in terms of a current.
    For example
    IL = Iin + IR1

    IR1 = (Vo - Vx)/R1 = [ (Iin*R2 + IL*RL) - IL*RL]/R1 = Iin*R2/R1

    IL = Iin + Iin*R2/R1 = Iin * (1 + R2/R1)

    Yes you right, but do not forget that the real op amp will have some saturation voltage. And this is why Vsup must be larger than this 55V.
    http://e2e.ti.com/blogs_/archives/b...nges-input-and-output-clearing-some-confusion
     
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