Voltage Regulators - Problem 5

Discussion in 'Homework Help' started by PsySc0rpi0n, Jan 28, 2016.

  1. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Hi guys...

    I'm back again with Electronics problems.

    I'm trying to find the output voltage at R_Load in the attached circuit.

    I have written a closed net loop equation including the Zener, Q2 and R2 as:

    -Vz - Vbe2 + I1*R1 = 0
    I1 = (Vz + Vbe2) / R1 = 4.57 mA

    From here I found V_R1 by:
    V_R1 = I1*R1 = 4.57 V

    Can I say that Vo = 2*V_R1 ??? I think I can but I'm not sure...

    I also have another solution that gives me the same result but by the same reason, I'm not sure why we can do that, which is:

    Vo = (Vz + Vbe2)*(1 + R2/R1)

    Why can we say directly that Vo is a voltage divider at R1 and that the Q2 base branch has no influence on this calc?
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  2. sailorjoe


    Jun 4, 2013
    Theoretically, the current through R1 feeds both R2 and the base of Q2. But the amount of the base current is quite small compared to the current through R1 or R2, so while not technically correct, it can be ignored.
  3. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014

    I now want to demonstrate, by wording, that this circuit actually regulates the voltage.

    I've already started with the help of a friend and he was talking about a relationship between Vbe and Ic, something like Vbe = exp (Ic)... I'm not sure I'm understanding this relationship! Can anyone explain?
  4. RBR1317

    Active Member

    Nov 13, 2010
    There is a theoretical relationship where Vbe controls Ic in a transistor; however, the relation is non-linear and with a large dependence on junction temperature (see Ebers-Moll) which makes the "Vbe controls Ic" theory not very practical for design & analysis (unless you are performing a temperature sensitivity analysis.) Nevertheless, for small-signal applications it is often useful to treat the transistor as a transconductance amplifier when the variation in Vbe due to the input signal does not exceed 10 mV. In that case fairly good linearity can be maintained.

    For large-signal calculations such as transistor biasing or power supply regulation it is usually best to treat the transistor as a current amplifier with a constant Vbe and the base current is controlled by the external circuit.

    For word descriptions of circuit operation, it should be sufficient to describe what causes base current to flow, what effect a change in collector current has, and whether the collector current, in turn, changes the base current.
  5. WBahn


    Mar 31, 2012
    You haven't defined I1. If I assume (and you've been told repeatedly that engineering is not about assuming such things) that you mean the current flowing downward through R1, then you haven't formed a loop equation because R1 connects the base of Q2 to the top of the load. Are you sure you don't mean

    -Vz - Vbe2 + I2*R2 = 0

    where R2 is the current flowing downward through R2?

    Again, assuming that you mean I2 and R2.

    Don't go plugging numbers into your equations so soon -- work symbolically so that you can get general results.

    How are you coming up with 4.57 mA? It would sure be nice to see some work. Working backwards, you appear to be using Vbe = 0.67 V. What an odd choice, but it's certainly reasonable.

    Again, assuming that you mean I2 and R2.

    You've already determined V_R2 -- it is Vz + Vbe2 = 3.9 V + 0.67 V = 4.57 V.

    Again, assuming you mean V_R2.

    What makes you think you can say this? What if R1 and R2 were not the same valued resistors?

    Where did this come from?

    Because the circuit was designed that way. If we made R1 and R2 a lot larger, say 100 kΩ or 1 MΩ, then we couldn't say that the Q2 base current has no influence. Instead, we choose the R1 and R2 values such that the amount of current that is siphoned off by Q2 is a negligible fraction (a few percent, typically) of the current in R1 and R2. We can afford to do that because the tolerance on R1 and R2 are usually on that order.
  6. WBahn


    Mar 31, 2012
    The best way, probably, is to start with the assumption that the output voltage is what you've calculated and then show that this assumption is valid. You do this by walking through the changes that would happen if the output voltage were to drop slightly and show that those changes would serve to raise the output voltage back up, and then show that the opposite would be true if the output voltage were to rise slightly.
  7. PsySc0rpi0n

    Thread Starter Well-Known Member

    Mar 4, 2014
    Sorry to come back to this thread so late!

    I have already worked it out so I'm now moving to another problem! I'm going to have exam this next Monday so the time is short to post here all my calcs and explanations!

    Anyway, thanks to who helped...

    @WBahn I read your post and I have mentally answered to almost all your questions and suggestions! So, thanks once more!