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Why Opamp with potentiometer?

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

  1. alec_t

    alec_t Well-Known Member Most Helpful Member

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    A single opamp will easily do that. Why are you intent on using such a complicated circuit? :)
     
  2. sagh

    sagh Member

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    I mean I want gain of 10 from this circuit. I have to buy matched transistor arrays to get this circuit working. In fact without matched transistor arrays when I change resisitor values in the current mirror sections and before that every thing is messed up. i can't afford to get transistor arrays on the other hand this circuit took me so much time to start working at least with specific resistor values. changing resistors at the last stage is a way to get away from all problems that I have with current mirrors.
     
  3. sagh

    sagh Member

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    It's really interesting there isn't any kind of quad transistor array even in Altium software. :-( I need it for my PCB. how can i create a model in altium?
     
  4. dave

    Dave New Member

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  5. alec_t

    alec_t Well-Known Member Most Helpful Member

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    I suggest you start another thread for that question. There may be Altium gurus out there who won't be following the present thread.

    Edit:
    Ah, I see you have another thread :)
     
    Last edited: Jun 22, 2014
  6. sagh

    sagh Member

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    Yeah but I'm not sure The post is in a right directory
     
  7. sagh

    sagh Member

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    To MrAl: In post # 134 you said if I change 10k resistors at the output of first stage to 1k, gain would be increased. I did more change in the schematic of circuit (below picture) to get gain of 10 and prepare it for PCB. I would be grateful if you tell me these changes are right or not, at least in theory.

    I decreased resistors value at the output of current mirrors to 300 ohm. In practice when I change resistors value at the output of current mirrors to 10k I get unbelievable and bad results because I don't have any matched transistor array so I decided to decrease it. I also changed 1k resistors at the last stage to 300 ohm because I want overall gain of 10 from this circuit. I didn't want to increase 10k resistors at the last stage so I did change 10k resistors at the output of first buffers to 1k to get a gain of 10 in overall. but I don't know can these changes affect the operation of transistors at the current mirrors and make them worse.
     

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  8. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    Before we can make changes that drastic we have to know what this circuit will be used for.

    We can make changes here and there, but if they get extreme like that then we have to know what we are to use this for. Changing the output stage resistors to 300 ohms is such a huge change it may not even work at all.
     
  9. sagh

    sagh Member

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    I have to prepare this circuit to get gain of 10 for less than 1v input voltage I mean when I input 0.2v I get 2v at the output. In fact I was told to get gain of 10 from this circuit in practice by only changing resistors no matter what happens to gain bandwidth of this circuit or other stuffs. At this time I just want this circuit to work at low frequency like 100hz or less than 100hz. This circuit may use in biomedical applications and it's not necessary to have a wide band frequency.
    Right now application of this circuit isn't really important. I should report a correct result by changing resistor values. I can put 1k resistor at the output of current mirrors also 100k and 1k at the last stage but I'm not sure last stage can work with 100k.
     
    Last edited: Jun 23, 2014
  10. alec_t

    alec_t Well-Known Member Most Helpful Member

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    For a serious purpose like that I would seriously consider simply using a decent instrumentation amp IC rather than this complex arrangement :).
     
  11. sagh

    sagh Member

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    I said application of circuit isn't really important right now. I just want to know why can't I get gain of almost 10.
     
  12. alec_t

    alec_t Well-Known Member Most Helpful Member

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    I don't see why you can't either. In theory it should be just a matter of selecting suitable resistor values.
     
  13. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Ok well i had to ask for the application and any other information that may be important because when you change the gain of the output stage, as said before, the gain bandwidth of the whole circuit changes. You wanted to use 300 ohm resistors and with 30k feedback resistor (and 30k resistor to ground) the gain would be 100, and if the output op amp GBW spec was 1000kHz, that means it would limit the useful gain bandwidth to only 1kHz. (1000/100=1). Since you have specified 100Hz, that relaxes the gain of the output stage a little.

    The choice of transistors was not a good one for this circuit. Better operation would occur i think if you used a transistor made for lower current operation. The type of transistor being used (NPN) is similar to the 2N2222, which was made for higher current operation like 150ma. If you used a transistor made for lower current it would probably function better with higher value output resistors on the current mirrors.

    300 ohms is too low however. That puts too much of a load on the previous stage. At 10 volts that would draw 33ma. You could check to see if the op amp can function with that much output current. Going to 1k would be better however. If i remember right, you need a gain of 10 with 1k resistors on the current mirrors and for the output stage, so the feedback and ground resistor would have to be 10k. That should work in real life although introduce a little offset. With 2mv input offset and a gain of 10, we could see 0.02v output offset. This is true of any pairs of resistor values that result in a gain of 10 however.

    There's also the possibility of using 1k on the output of the op amp input buffers and 1k on the output of the current mirrors. Did we try that yet?
     
    Last edited: Jun 24, 2014
  14. sagh

    sagh Member

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    Is 150ma high for transistor currents? You mean transistors with lower current provide lower current at the output of current mirrors so I could use
    high value resistors at the output of current mirrors.

    Yeah I had problem with this, too. because of all problems I decided to change pairs of 10k resistors at the last stage to 100k and didn't change resistors of first stage because they work ok with 1k at the output of current mirrors and 10k at the output of first input buffers. It's very strange to me why with 1k at the output of first buffers everything is messed up. Do these resistors affect current mirrors operation in theory?
     
  15. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Well the current mirrors throw a lot of things into question that the op amps dont. Questions like what is the best operating current, are they matched and how bad is the offset if they are not matched.

    With transistors that normally work at lower currents i would expect the gain to be better at the normal operating current. With higher current transistors the gain is better at higher currents like 150ma, which you do not want to use.
     
  16. sagh

    sagh Member

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    I don't understand you. You mean if I place resistors more than 1k at the output of current mirrors then the gain would be better. I did that but the current mirrors couldn't handle these changes. It's strange if operating current is high (150ma) then with higher resistors why doesn't work it correctly.
     
  17. sagh

    sagh Member

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    for gain of 10 I have to change 10k resistors at the last stage to 100k resistors. but I have no idea this change damage operation of circuit. In fact due to my problems with current mirrors I don't want to change resistors of first stage. with 1k and 10k at first stage and 1k and 10k at last stage the gain is almost 1.1
    I only want to change 10k resistor pairs at the last stage to 100k to get the gain of 10.
     
  18. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hello,

    For a gain of 10 the resistors would be 1k and 10k, that's it.
    For an overall circuit gain of 10 when the input resistors are 10k and current mirror output resistors are 1k, the output stage has to have a gain of 100 which means 1k and 100k.
    If the current mirror output resistor is 300, then that's at least 3.33 times less than 1k, so we need a gain of 333 at the output stage. With 1k and 333k we get this. If you dont like that, then try two stages each with a gain of sqrt(333) each.

    You dont seem to understand that the gain is related to the collector current and the part number of the transistor. That is, a transistor made for 150ma operation will have better gain at 150ma and less gain at other currents. That's why i suggested trying a lower current transistor so that there would be more gain available in the transistor. Look at the data sheet for a 2N2222 transistor.

    I am failing to see how you would want to use this for a medical oriented application circuit. The current mirrors dont seem to offer any advantage except as noted much earlier about the improved slew rate and therefore the usable bandwidth, but you dont seem to need any of that so i fail to see the reason for this circuit except as a purely academic endeavor of some sort.
     
    Last edited: Jun 24, 2014
  19. alec_t

    alec_t Well-Known Member Most Helpful Member

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    I totally agree. If, on the one hand, you have been told to investigate the circuit properties as an experiment then that's ok; you've demonstrated that it has problems. If, on the other hand, it has been suggested as a circuit for a medical application you really should ask whoever made the suggestion to justify its use and explain what supposed advantages it has over a simple opamp in that application. For most medical applications you would want a reliable circuit. Because of its complexity this circuit will have a much higher failure probability than a single opamp.
     
  20. MrAl

    MrAl Well-Known Member Most Helpful Member

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    Hi,

    Yeah and that reminds me, for a medical application you'd want as few connections as possible. You'd also want to read the fine print disclaimer on the data sheets for the op amps.

    Maybe someone stated this problem as an academic problem where it has to have the current mirrors and it also still has to work in a medical application :)

    It is an interesting circuit though you have to admit. If it were me though i would try to get some advantage out of that slew rate increase. That's significant.
     
  21. alec_t

    alec_t Well-Known Member Most Helpful Member

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    But isn't the overall slew rate of the complete circuit going to be limited by the choice of opamp, regardless of what the current mirrors themselves can achieve?
     

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