Please help with 3-transistor current source question.

Discussion in 'Homework Help' started by HunterDX77M, Apr 2, 2012.

  1. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
    Please refer to the attached image for the circuit and problem in question.

    Hey guys,

    I'm stuck on this current source question and was hoping someone could help. At first I made the assumption that the transistors all had the same turn-on voltage of VBE,on = 0.7 V but I quickly found out that doesn't make any sense, because the problem would be a simple middle-school plug and chug other wise.

    I decided to go back to the basics and make some nodal equations, but I had more variables than equations so I scrapped that.

    Any help please?

    (The complete question and circuit is attached.)
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
    Normally, it is impossible to find analytical solution for this type of a circuit.
    We need to use numerical method or use iteration method.
    See the example problems (post 8) (page 4)

    But in point B they want you to find R3 resistor.
    And this can help you solve this circuit much faster.

    Simply use β and Ic = β*Ib , Ie = Ib + Ic = (β +1)*Ib to find Ic5 and Ic6.

    Ic5 = Ic6

    Ib6 = Ic6/β

    Ie7 = ( 2 * Ic6/β ) + IR3

    Ib7 = (( 2 * Ic6/β ) + IR3 )/ ( β + 1)

    And Iref = Ic6 + Ib7

    So assume IR3 and solve for Ic6

    Or you can try some different approach.

    Assume for example Ic5 = Ic6 = 9μA then use shockley equation to find Vbe and select R3 that meets this requirements.
    Last edited: Apr 2, 2012
  3. HunterDX77M

    Thread Starter Active Member

    Sep 28, 2011
    If anyone cares to know, I asked my professor and he said that the three currents "I" in the diagram are all the same. It made the whole thing very simple indeed. :)
  4. panic mode

    Senior Member

    Oct 10, 2011
    well that is what question suggested - all three are marked I.

    btw circuit is not correct, 1k resistors are not connected to -Vee. only when they are connected this becomes possible (need junction placed at Vee, connecting all three resistors to negative supply).

    btw, you are not supposed to assume value for any current, you should calculate it.
  5. panic mode

    Senior Member

    Oct 10, 2011
    read the question (and understand it) and you may come up with some helpful conclusions. for example:

    since all transistors are same, have same gain beta=200, and all collector currents are same, and all transistors are in same mode, they all must have same Q points. this means:
    Ic=I same collector current for all of them
    Ib = same base current for all of them
    Ie=Iref same emitter current for all of them

    using KCL we can write:

    Iref=Ib7 + I
    Ie7=Ie5=Ie6=Ib*(beta + 1)

    or simply


    since we know that
    we can find base current directly:
    Ib=9.4uA/201 = 46.766nA

    and using diode model we can find Vbe for all transistors:

    Vbe=Vt*ln(Ic/Is + 1)

    where Vt is known (thermal voltage is assumed constant at roomtemp: Vt=0.0258V)

    this means


    R3=Vb5/Ir3=58403 Ohm

    Edit: ok Empire may have had a chance :)
    Last edited: Apr 9, 2012
  6. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    Hi panic mode,

    I would probably have used either the emitter current or the collector current to find VBE via the Shockley equation - rather than using the base current.

    What do you think?
    panic mode likes this.
  7. panic mode

    Senior Member

    Oct 10, 2011
    you are right, btw. the equation used is Schockley equation just expressed in terms of Vbe.