BJT bias - VDB open goes to emitter-feedback

Discussion in 'Homework Help' started by sbixby, Dec 1, 2011.

  1. sbixby

    Thread Starter Active Member

    May 8, 2010
    I'm working on a problem in Chapter 8 of Malvino/Bates "Electronics Principles", and one in particular is driving me nuts.

    In this problem, we start with a VDB (voltage-divider bias) and open the R2 resistor; the result is essentially an emitter-feedback, with the BJT saturated.

    There is an example that specifies expected voltages at the transistor's connections for this situation, as well as a problem which does the same thing with another example, and I can't seem to calculate voltages for either example that makes sense. I've tried it a number of ways using the equations in the chapter.

    Attached are two screen-caps from Paul Falstad's applet that I was using to visualize things. For the "normal" one, I have no problem getting matching numbers, but for the R2-open one, I'm stuck - I'm not clear on how the saturated transistor gives the numbers shown.

    Could someone give me an example of getting the voltages at CBE in the "r2open" attachment?

    Note that the applet's model is using a Vce of about 0.6v rather than the "standard analysis assumed value" of 0.7v, but that still doesn't give me reasonably close numbers using the "emitter-feedback bias" equations from the book.
  2. Audioguru


    Dec 20, 2007
    When the transistor is saturated then it is simply a diode with no current gain. Therefore the base current adds to the emitter current which raises the emitter voltage. When the emitter voltage rises and since the transistor is saturated then the collector voltage also rises.

    All the current numbers add correctly.
  3. t_n_k

    AAC Fanatic!

    Mar 6, 2009
    So putting some numbers to your example.

    One has to assume a VCEsat value.

    To make it simple let's say VCEsat≈0, VBE=0.6 (as you say)

    Then VE=(Ib+Ic)*1k







    VE=[10-VE]/3.6 + [10-0.6-VE]/10

    Which solves to give


    From there you can find the other values.

    If you assume VCEsat is ~50mV you would get a slightly different answer.

    Also keep in mind that the value of the stated current gain HFE = 100 doesn't have much meaning in saturation.
  4. sbixby

    Thread Starter Active Member

    May 8, 2010
    From simulations, I have no doubt the numbers add up, but I'm not sure how we get to those numbers in the first place.

    IOW: How did you establish the currents? I know that Ie=Ib+Ic, and those numbers jive.

    How did you get Ie (or Ib or Ic) to start with?
  5. sbixby

    Thread Starter Active Member

    May 8, 2010
    t_n_k replied just before I posted my follow-up.

    Thanks, guys! Putting numbers into equations pointed out my misunderstanding - that we use 9.4v across R1 along with 10v across Rc to get the total current. After that, as you say, the rest of the numbers fall out.