High side current monitor. How does it work?

Discussion in 'General Electronics Chat' started by hp1729, Feb 2, 2016.

  1. hp1729

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    I would be at a loss to explain how this circuit works. The Emitter-Base of the transistor is used as the feedback circuit. Can somebody put into 100 words or less the details of why this works?
    I can see how the input voltage is developed across the sensing resistor. I recognize this as an inverting amplifier. The more positive voltage is applied to the inverting input so the output is going to go lower, increasing conduction in the transistor, thus giving a higher voltage out.
    ???? How do you calculate gain when the feedback is part of the transistor?

    When I try to explain an op amp operation to myself like this I try to explain that the voltage applied to the non-inverting input drives the output to the point where the inverting input equals the non-inverting input. In this case the more negative voltage applied to the non-inverting input drives the output low. But the E-B junction is not a fixed resistance. I get lost right there.

    This circuit isn't covered in the op amp tutorials here. :)
     
    Last edited: Feb 2, 2016
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  2. hp1729

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    In this application we should have 1 Volt out per 1 Amp of current through the load.
    Looking at currents I'm sure it isn't coincidence that output current, 0.1 mA (1 V / 10K ohm), is the same as the voltage across the sense resistor divided by R3 (0.2 V / 2K). But I can't work this into the story of operation.
     
  3. kubeek

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    I don´t think the circuit or the equation are correct. R3 should not have any influenece on the transfer function, since it is only connected to the noninverting input of the opamp, it probably should be R4.
    In essence this is a voltage to current converter, so whatever current is flowing through R4 should flow through Q1 and R1, the ground of R1 could be at a much lower potential.
    And since voltage at R4 should be equal to voltage at R2 you get the equation you wrote there.
     
  4. dannyf

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    The current flow through the sense resistor creates a voltage drop, thus current, through the resistor on the inverting input pin. That current goes through the emitter and then creates a voltage drop on the collector resistor.

    So those three resistors determine the gain of this amp.

    It also shows that in very low current applications, it is better to use a fet.

    The key here is to know that the inverting and non inverting terminals are at the same potential. That's true because for the combined amp (ie the opamp plus the pnp) the inverting and non inverting pins are swapped.
     
  5. Jony130

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    This diagram explains everything
    11.PNG

    Thank to op amp "action" we have Vn = Vp. And the voltage at output (voltage drop across RB) is equal to

    Vout = Iload * RS * (Rb/Ra) *β/(β+1) ≈ VRs *Rb/Ra

    In real life if op-amp is supply from the same voltage as a main circuit and the opamp Voh (positive saturation voltage) is large then 0.6V. We are force to add a Zener diode in series with the transistor base to lower the voltage the the op-amp output, also series resistor with the base is also needed.
    Or we can use a FET instead of a BJT
     
  6. crutschow

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    I don't see why. :confused:
     
  7. Jony130

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    Just in case some bad things happen.
    And I know and understand that Ra resistor together with transistor beta "sets" the base current.
     
  8. hp1729

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    R3 vs R4
    Your reasoning makes sense but since R3 = R4 it really doesn't change the result.

    Re: V to I converter
    Okay, that makes sense. So the non-inverting input drives the output until feedback current is equal and opposite to input current? Is that acceptable?
     
  9. kubeek

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    Just because they happen to be the same value doesn´t make them interchangable in the equations ;) R3 has the same value most likely beacuse the author thought that is the correct way to balance out the input bias current of the opamp, but I don´t think he did a very good job anyway since the input impedance seen on the negative terminal will be close to zero due to the diode drop right from the output terminal, if I am not mistaken.

    Yes the way you describe it is correct.
     
  10. hp1729

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    Ah! here is where you lose me. 'Diode drop" is the E-B junction? How does that make the input resistance close to zero? I do not grasp the effect of a diode as the only feedback component.
     
    Last edited: Feb 2, 2016
  11. kubeek

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    Now that I think about it it seems I has wrong with that statement, since the positive and negative inputs will be held at the same potential.
     
  12. hp1729

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    Help me make sense of this idea of a diode feedback.
    I tried to come up with a circuit to help me make sense of a diode as a feedback or the PNP use.
    With 0.100 in on the diode feedback example I get -0.470 V out ????
    The PNP example I got no output at all. The output of the op amp changed to the negative as input went positive, but at the collector I got 0 V.
    Any help here would be appreciated. Is there a better circuit I could use as an example to tinker with?
     
  13. crutschow

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    Normally the maximum op amp output current, even when saturated, would be limited to a value less than would damage the transistor.
    The only bad thing I see that could happen is if the op amp output shorts to ground (minus rail) which seems rather unlikely, and even then the current would be limited by the 1kΩ emitter resistor. ;)
     
  14. dannyf

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    No need to worry about the b-e junction in the feedback loop: the opamps output will be adjusted so that the same amount of current going through the resistor on the inverting input will go through the diode - the device here is essentially operating on current, not voltage.
     
  15. crutschow

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    In the first example the op amp generates an output voltage such that the current through the diode equals the input current through R1.
    This is to keep the minus input voltage equal to the plus input voltage (ground) as op amps do in a closed loop.
    This means the voltage at the diode must be a diode drop below ground (which is .47V in this low-current case).

    The second example doesn't work because there's no source of current for the emitter- collector through R4.
    (Where would it get that current)?
    To work, the emitter needs a resistor and a voltage source as the original circuit had.

    You need to think about how op amps and transistors work and the flow of the current in all the nodes.
    Ask the question: What does the op amp output have to do to keep both inputs at the same voltage?
    Random connection of devices gives random results. :rolleyes:
     
  16. dannyf

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    The 2nd example doesn't work because the collector is connected to ground via that resistor. As the base goes to below zero, the transistor is shut down.

    To make it work, you can connect the collector to a negative voltage and the voltage drop over that collector resistor should be 10k / 2k * 0.1v.
     
  17. dannyf

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    BTW the high side sense amplifier can also be built with npn transistors.
     
  18. hp1729

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    Thanks, that helps some. The math isn't so different from my earlier post. The relationship between the parts is not my problem. Using the E-B as the feedback is what is throwing me. How do I predict behavior if I were to try to design such a circuit?
    Yes, I have seen similar circuits using FETs. Let me stay confused on one circuit at a time, please. :)
    Re: beta
    It doesn't seem to matter what transistor I use. Most any PNP seems to work.
     
  19. hp1729

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    Re: "No need to worry about the b-e junction..."
    Maybe I'm trying to make this harder than it really is?
     
  20. hp1729

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    Ah, thank you.
     
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