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Why Opamp with potentiometer?

Discussion in 'General Electronics Chat' started by sagh, May 5, 2014.

  1. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    I'm sorry, it connects (or rather should connect) to a virtual voltage source (which is almost like a virtual ground), which is seen (or should be seen) at the inverting terminal.

    If the lower 1k went to ground then the non inverting terminal would get a voltage. Right now you effectively have a current source powering the lower 1k resistor in series with a 500k resistor. So even the smallest current could develop a very high voltage.

    If you look back at the original circuit (with the two current mirrors only) you'll see the 1k connecting to the inverting terminal which acts as a virtual ground so the current mirrors can create a voltage (or just a current through the 1k if you look at it that way). With the new circuit it goes to (or should go to) a virtual voltage source.

    To make a long story short, if the output of the lower two current mirrors went to a resistor of 1k then they would create a current that would show up as a voltage across the resistor and that would be driving the non inverting terminal. If you dont understand why, then maybe you can explain why you chose to use a 500k resistor on the non inverting terminal to ground.
     
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  2. sagh

    sagh Member

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    A virtual voltage source is like a virtual ground? why?

    yes, You are right but what's the problem of having high voltage at the non- inverting port of the second op amp? I simulated this circuit in pspice and I got zero volt when both of inputs were (1 Vac)! but I think one thing is wrong in lower current mirrors because base of top transistor has (4.35 Vdc) while the power supply of transistor is 15V!!! sometimes I really go crazy due to the strange result!
     
    Last edited: May 14, 2014
  3. atferrari

    atferrari Well-Known Member

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    The idea of sensing both +/- V currents, reminds me a design of an analog multiplier with two op amps using the "quarter square" concept.

    Anyway, IIRC, the main difference is that the second op amp was a differential one to sense difference between currents.

    But you know, is quite frequent that I am proved to be wrong.
     
  4. dave

    Dave New Member

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  5. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    A virtual ground is 0.000 volts constant, while a virtual voltage source is some other voltage which is also constant relative to the non inverting terminal. So if you set the non inverting terminal to say 2.500 volts then the inverting terminal will be close to that voltage, and so that makes the virtual voltage source equal to 2.500 volts or close to that.

    If you have something simple wrong with the circuit like a wrong connection or a bad part, you have to fix that first before anything is going to work at all. Then you can look into the lower 1k to ground issue.

    I'll also see if i can simulate this circuit myself later, and/or we can look at some equations for this circuit.
     
  6. atferrari

    atferrari Well-Known Member

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    Post edited twice. Sorry for previous unclear texts.

    This is what I got by simulating the circuit with 4 transistors. Am I off track?

    Clipboard01.png
     
    Last edited: May 14, 2014
  7. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Atferrari what are you trying to say there?
     
  8. atferrari

    atferrari Well-Known Member

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    I edited my post. Hopeffully the text makes sense now.
     
  9. sagh

    sagh Member

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    I checked all the parts and connections 5 times in Pspice. I really don't know what's wrong with this circuit. Unfortunately base of the top transistor in lower current mirror still has 4.35 Vdc. It's so weird!!!! :(
     
  10. sagh

    sagh Member

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    Yes you are! I'm talking about simulation of whole circuit not half of it.
     
  11. atferrari

    atferrari Well-Known Member

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    Thanks for the exclamation point.

    What half of the circuit is not included in the simulation? Vin and Vout are shown as an outcome of the simulation.
     
  12. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Atferrari:
    I believe it is the lower section of his "new" circuit with four current mirrors.
    He is a little hard to help though because he seems to be not willing to make changes as suggested even though he has these weird problems that he can not explain.
    Try simulating the lower section of his new circuit and see what you find.

    sagh:
    You have to be willing to make some changes as suggested or else you'll never get this to work.
     
  13. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Since your 4 transistors are not included in the negative feedback loop of the first opamp then when the level is higher the distortion might be awful.
    Here is an audio power amplifier that has the transistors connected properly:
     

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  14. sagh

    sagh Member

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    I attached the image of the lower current mirror. As you see the base' s voltage of top transistor is 4.35 v and the lower transistors have 5.37 v at their base. My main question is why these transistors have low voltages at their base while 15 v has been applied to them. Actually I don't know why base's voltage of top transistors aren't for example 14.3 or sth like that because (Vbe) is usually between 0.7 and 0.6. The whole circuit has been attached, too.
     

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    Last edited: May 16, 2014
  15. audioguru

    audioguru Well-Known Member Most Helpful Member

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    The base-emitter voltage of an ordinary transistor is about 0.7V.
    Since your simulation program shows 10V or 20V then IT IS COMPLETELY WRONG!

    But YOU are also wrong since the schematic with a single opamp shows TWO 15V batteries and THE WRONG battery is labelled -15V. The schematic with 3 opamps has the battery polarities labelled correctly.

    Pin 4 of a quad opamp is supposed to have a positive supply and pin 11 is supposed to have a negative supply. Look at the datasheet to see it.
     

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  16. sagh

    sagh Member

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    What part of my simulation program shows 10v or 20v?
     
  17. kubeek

    kubeek Well-Known Member

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    It would be interesting to see the current in that simulation, I bet the current is in the order of tens of amps, mainly because the opamp has the power supply swapped around.
    untitled1 shows -15V at the emitter and +5V at the base, that is 20V difference insted of normal 0.7-1V
     
  18. sagh

    sagh Member

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    That's my main problem, and I don't know why?!
     
  19. audioguru

    audioguru Well-Known Member Most Helpful Member

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    We told you that your batteries are connected BACKWARDS!

    I repeat: "Pin 4 of a quad opamp is supposed to have a positive supply and pin 11 is supposed to have a negative supply. Look at the datasheet to see it."
     
  20. sagh

    sagh Member

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    I hope this time the schematic is right. when I try to see DC voltages in pspice. At inverting input of right op amp I have -796.1 mv which is drawn in red circle. I don't know why voltage has been increased after R1. before R1 is -846mv and after that is -796.1 mv.
     

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  21. audioguru

    audioguru Well-Known Member Most Helpful Member

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    Why do you have R5 and R7? They do nothing so replace them with pieces of wire.

    The top opamp with current mirror has a 1k ohms load (R1) but the bottom opamp with current mirror has an 11k ohms load so they will have different amounts of attenuation because the current mirrors produce a high impedance signal source.
    WHY do you have those current mirrors?
     

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