1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Voltage Regulator w/ Zener Diodes

  1. Feb 8, 2014 #1
    1. The problem statement, all variables and given/known data

    The diagram:
    http://images.fr1ckfr4ck.fastmail.fm/probB.jpg [Broken]

    These zener diodes;
    • Have a test current of 5mA when 6.8V is applied (V_z=6.8V, I_z=5mA)
    • Have an incremental resistance (r_z) of 20 ohms
    • Knee current (I_zk) is 0.2mA

    • Calculate the output voltage of the voltage regulator, assuming no load (R_L = infinite) and V+ is at nominal value.
    • Calculate the line regulation of the voltage regulator, assuming no load.
    • Calculate load regulation of the voltage regulator.

    2. Relevant equations

    see below

    3. The attempt at a solution

    To get the output voltage at nominal voltage and no load, I first went to solve for V_zo using V_z=V_zo+r_z*I_z and the test current I_z=5mA. That told me V_zo = 6.7V. Then I solved for I_z using I_z=(V+ - V_zo)/(R+r_z). I decided to go ahead and find I_z for one diode, then solve for half of V_o later. I_z turned out to be, for one diode, I_z = (15 - 6.7)/(500+20) = 16mA. So then to solve for output voltage V_o, I used
    I_z being not the test current but the one I just solved for. So;


    V_o=14.04 V

    It makes sense to me that the voltage should be ~1V less under a 500 ohm resistor, so I think that's a good answer.

    As for line regulation - I'm looking for the change in V_o resulting from the +/- 1 volt change in V+ (14 to 16 V).
    If there were only one zener diode it would be as simple as ΔV_o=ΔV+*(r_z/(R+r_z))
    But since there's two in series, I have to decide what to double and when.

    Since that equation is asking what the change in V_o is when you consider the change contributed by the zener diode, I decided that the equation should be done twice and involve both diodes. It looks like voltage division to me, so if there's three resistors it should involve all three.
    So, I changed r_z to 2*r_z:

    ΔV_o=ΔV+*(r_z/(R+2*r_z)) + ΔV+*(r_z/(R+2*r_z))

    Since ΔV+=2V, it all comes out to ΔV_o=2(20/(500+40))+2(20/(500+40))=8/54=0.148
    expressed in mV/V --> 148 mV/V
    and since it's a deviation they want I'd say the answer is (+/-) 74 mV/V.

    Load Regulation:
    Looking for the ratio of the difference in V_o vs. the difference in I_L, the current through the load.
    A load resistance that draws a current of 1mA will cause the current through the zener diodes branch to decrease by 1mA. Normally if there's one diode that change would be;
    But in this case there's two. Still, though - since the total current change will remain 1mA, it should be divided evenly between both diodes, and the end result stays the same. The load regulation should be proportional to r_z; -20 mV/mA

    I have a sinking feeling I messed this all up. Any tips on where I went wrong or what I did right, would help. Thanks in advance.
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 8, 2014 #2


    User Avatar

    Staff: Mentor

    There won't be anything like 16mA through that 0.5k resistor; that would need 8v across it, and the pair of zeners don't leave that resistor even 2v.

    Why only one zener considered here?
  4. Feb 9, 2014 #3

    ok. I guess I could answer that if I was seeing the big picture here, but I'm not. I thought there was one I_z to find that's true for both zeners but I guess not. I'm going back over a similar problem using one zener and see how to adapt it. meanwhile, any tips on a general approach would help, thanks.
  5. Feb 9, 2014 #4


    User Avatar

    Staff: Mentor

    Replace both zeners with an equivalent model consisting of a resistor, ideal diode, and voltage source (Initially assume that sufficient current will flow to cause zener action). Analyze the circuit: Use Kirchhoff's laws to determine the actual current and the output voltage.
  6. Feb 9, 2014 #5
    Right that's just what I'm doing and where I'm getting stuck is this; I have replaced the two zener diodes each with voltage sources of voltage V_Zo=6.7v, and resistors r_z=20 ohms. Now, I could say the total current going into the node above, which should equal the current through R, would be I_z=(V-V_R)/(R+r_z)=(15-6.7)/(500+20)=16mA
    but then if I just take that current and double it, one for each zener diode, I get 32mA, and that doesn't jive with the resistor R above because it would mean that V_R (voltage across that resistor) = (32mA)(500ohm)=16V.
    So, what, then it means that the current is the same through two zener diodes as it is through one? Then V_R would = 8V, and V_o would = 7V. doesn't seem right either.

    Big question for me is how to account for the current at the node above the two zener diodes, because I've only solved problems with one zener before and I don't know how to handle having two of them.
  7. Feb 9, 2014 #6
    Ok. So. The only thing that's making sense to me right now for that current is using I_z=(V-Vzo*2)/(R+r_z*2), because that makes I_z = 2.96mA, which in turn means V_R = 1.48V. It seems to make sense that under no load the voltage drop V_o would be around 13.52V, at least intuitively.
  8. Feb 9, 2014 #7


    User Avatar

    Staff: Mentor

    Just call it "the voltage V_o". The label "voltage drop" isn't approprate here.

    You could refer to the voltage across R as a voltage drop. Or ΔV_o as a voltage drop under load.
  9. Feb 9, 2014 #8
    Thanks for the pointers on terminology but my burning question is if I'm finding the potential there at V_o correctly.

    Does that seem correct, then? using I_z=(V-Vzo*2)/(R+r_z*2) for the current going into that node, and V_o being 13.52V?
  10. Feb 10, 2014 #9


    User Avatar

    Staff: Mentor

    That is correct.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted