Help with bizarre current mirror

Discussion in 'The Projects Forum' started by zxdesign, Jul 14, 2009.

  1. zxdesign

    Thread Starter New Member

    Jul 14, 2009
    4
    0
    Hi all.

    I've spent two long days trying to work out what this circuit is doing:

    mirror.png

    I'm need to understand its operation so I can write a detailed description with equations, and be able to explain how to modify it to tailor its operation.

    If this consisted of a straight forward current-mirror, I'd find this easier. However, with emitter resistors R3 and R5 I'm struggling to find/formulate the appropriate equations.

    I have no idea what Q1 is really for. Calculating Ib of Q1 is easy of course, however Vce (or equivalent resistance) of Q1 is another matter... perhaps this is there for temperature stability?

    So my fundamental questions are:

    1. What is the purpose of Q1, and how do you calculate the voltages and currents?
    2. What are the voltages/currents associated with Q3?
    3. What effect does changing R5 have on the output of Q3?
    4. Pulling Raise to Vs (0.95) effects Q2 how? How does the output of Q3 alter?

    This circuit is implemented within an IC, so the transistors are matched and operate at the same temperature. I don't have values for hfe etc though, so I've been assuming sensible defaults.

    I'd be really greatfull for any help you can provide!

    Cheers,
    Chris
     
  2. millwood

    Guest

    this is NOT a current mirror. it is a bandgap reference.

    read on Widlar bandgap and you will understand it. Bob Pease used to write a lot about it.
     
  3. zxdesign

    Thread Starter New Member

    Jul 14, 2009
    4
    0
    Really?
    I drew the schematic from the poly layer of a chip im reverse engineering, so the layout is down to me. The two NPN's in what looked like a current mirror threw me off track, I guess.

    Been 15 years since I last did any bipolar design.... Clearly more than a little rusty! :-(

    Thanks for the pointer.
    Chris
     
  4. zxdesign

    Thread Starter New Member

    Jul 14, 2009
    4
    0
    Having looked at the bandgap references, I can see the striking similarity.

    My circuit is a D-to-A converter, and I've updated the schematic to show this (I tried to keep it simple last time). There is a 1K resistor, which is off chip, in the emitter of the output - I've added that to the schematic too.

    The Widlar bandgap does look like the closest approximation, but I'm puzzled by vout in my circuit being taken from the emitter follower, and completely at a loss as to what Q1 is doing.

    I'd really like to understand this circuit as it's the only bit of a 2200 transistor circuit (99% digital) I've back-annotated from silicon that I can't get to grips with.

    http://www.zxdesign.info/download/mirror.png

    Cheers
    Chris
     
    Last edited: Jul 15, 2009
  5. millwood

    Guest

    the emitter follower is to isolate the bandgap from the rest of the circuit; check out lm113 and see how it works.

    Q1 and other transistors there is to control the current density.
     
  6. zxdesign

    Thread Starter New Member

    Jul 14, 2009
    4
    0
    Okay millwood, I think see what you're getting at.

    Let me check if I've got this:

    Ignoring Q1 for the moment, due to the different emitter current densities of Q2 and Q3, the circuit produces the delta of the two Vbe's across R5 (assume switched in).

    I've measured the silicon dimensions of the transistors involved, and they are identical, so I'm left with:

    Delta Vbe = VT.ln( IC2 / IC3 )

    Which brings me to where I'm getting stuck - definitely missing something fundamental, but hopefully not obvious :rolleyes:

    Assuming IB is zero so IC=IE, I was about to wade into ohms law to calculate IC2 and 3, but I knew that that would assume no effective resistance across Q2 or Q3, and thus no voltage drop. Plus, that would give a current densities of J2 < J3, which gives a negative delta Vbe. Ouch!

    Also delta Vbe has a positive temp coefficient, so will Q4 introduce the classic negative temp coefficient when configured as an emitter-follower to give, a stable Vout?

    Incidentally, on reflection, should I have posted this to the Homework forum?

    Many thanks.
     
    Last edited: Jul 16, 2009
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