Load current in Zener voltage regulator

Discussion in 'Homework Help' started by Javri, Feb 17, 2016.

  1. Javri

    Thread Starter New Member

    Jan 14, 2016
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    Hi! I'm having some difficulty understanding how the question wants me to solve its problem.

    A voltage supply varying between 8 V and 12 V is in series with a 5-ohm resistor and a parallel combination of a zener diode (Rz = 0) and a load resistor. The diode has a zener voltage of 5 V and can regulate the load voltage when the series current (through the 5-ohm resistor) is between 50 mA and 1 A. Find the range of the load current for regulation.

    Below is a quick diagram of the circuit from falstad. Note the values don't match (particularly, load resistance is not given).
    Circ.png

    From my understanding, the lowest series current in this set up is [ V(ss, min) - V(z) ] / R(s) = [ 8 - 5 ] / 5 = 0.6 A, which is a lot higher than the 50-mA lower end for regulation provided by the problem. On the other end, [ V(ss, max) - V(z) ] / R(s) = [ 12 - 5 ] / 5 = 1.4 A, which exceeds the upper limit for regulation, 1 A. So I'm not entirely sure what's going on; am I supposed to just ignore the values for V(ss) > 10 V and only consider 8 V < V(ss) < 10 V for regulation?

    I also do not know how to find the load current. I know V(load) = V(z) = 5 V in regulation, but the load resistance is left unspecified. The only other way I can think of to find load current is to subtract the diode current from the series current, but I can't figure out a way to figure out the diode current separately.

    Any help is appreciated.
     
  2. MrAl

    Well-Known Member

    Jun 17, 2014
    2,423
    490
    Hello,

    Well, it looks like you have to think about this a little.

    First, think about what happens when the output current changes from 0 to max 1.4 amps.
    The point where it goes out of regulation is when the output voltage equals the zener voltage (if the zener was taken out of the circuit temporarily), or just very very slightly below that point (probably ok to figure it as going out at the equality). Alternately, when the output current equals the series resistor current.

    For an example, say the current through the 5 ohm resistor was 100ma, and the output is regulating at 5v with no load yet. If we then connect a load, what load would draw 100ma at 5 volts? That is when there is no more zener current and thus the zener will not be able to regulate when the output current increases just a tiny tiny bit more.

    Since there is a range for the input voltage that means you'll have to find more than one solution, or a formula that expresses this relationship depending on what your instructor is looking for or what you have done in the past.
     
    Last edited: Feb 17, 2016
  3. WBahn

    Moderator

    Mar 31, 2012
    17,720
    4,788
    What you are looking for is the range of load currents that will ensure that the zener current is within limits regardless of what the source voltage happens to be (within its limits).

    If the source voltage is 8 V, what is the range of load currents that will keep the zener within limits?

    If the source voltage is 12 V, what is the range of load currents that will keep the zener within limits?

    What is the overlap of these two ranges.
     
  4. Javri

    Thread Starter New Member

    Jan 14, 2016
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    0
    Ah, I may have been overthinking this problem. Let me go through this problem to see if I actually understand it.

    For regulation, we expect the load voltage V(L) to take the same value as the zener voltage at breakdown, V(z) = 5 V. Because of the problem-set range for the supply V(ss) and the series current I(s), the maximum load current I(L) is its value as it approaches I(s), so that in the circuit, the diode current I(z) approaches 0.

    Specifically, at V(ss, min) = 8 V, assuming zener breakdown regulation, I(s) = [ V(ss) - V(z) ] / R(s) = [ 8 - 5 ] / 5 = 0.6 A. Therefore, I(L) takes a max value of 0.6 A with V(ss) = 8 V; this is because I(z) is 0 and any I(L) that is higher breaks regulation.

    But at V(ss, max) = 12 V, we find I(s) = [ 12 - 5 ] / 5 = 1.4 A, which is greater than what the problem allows for I(s); we then take this upper limit for I(s) = 1 A to occur at V(ss) = I(s)*R(s) + V(z) = 1*5 + 5 = 10 V. Basically, for regulation, we are only concerned with:
    • V(ss): 8 V < V(ss) < 10 V
    • I(s): I(s) < 0.6 A for V(ss) = 8 V, I(s) < 1.0 A for V(ss) = 10 V
    Incidentally, we can use these values to calculate the minimum load resistance to satisfy the max current, i.e. R(L) = 5 V / I(s), but I do not think we actually need a load resistance value to solve the problem.

    On the other end, the minimum of I(L) is realized when I(z) approaches its maximum, i.e. I(z) approaches I(s). Just from inspection, I(L) must be greater than 0 A for V(L) to equal V(z) = 5 V with any realistic resistor.

    So in conclusion, at V(ss, min) = 8 V, our load current must be I(L): 0 A < I(L) < 0.6 A. And at V(ss, max) = 10 V, our load current must be I(L): 0 A < I(L) < 1.0 A. Or, at any V(ss) between 8 and 10 V, I(L) must be less than [ V(ss) - 5 ] / 5. At any other point, regulation is broken.

    Is that all there is to this problem?
     
  5. WBahn

    Moderator

    Mar 31, 2012
    17,720
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    Reread your next to last paragraph. What if Vs = 12 V, which the problem allows for? Why are you only working with Vs between 8 V and 10 V?

    Also, you are neglecting the spec that the zener won't regulate properly unless the current through it is at least 50 mA.

    Other than that you are getting the right idea.
     
  6. Javri

    Thread Starter New Member

    Jan 14, 2016
    5
    0
    Sorry, I worded the problem poorly. The range of current provided by the problem is for the current through through the series resistor, i.e. the one in series with the voltage supply before the circuit branches off in to zener/load parallel combination. In the diagram from my first post, I(s), 50 mA < I(s) < 1 A is the current flowing through the top left resistor, in between the supply and the diode.

    Knowing that, if V(ss) was 12 V (or above 10 V), for the zener diode to maintain a 5V drop across the load, I(s) would then be [ 12 - 5 ] / 5 = 1.4 A, which is outside the set limit for I(s). Similarly, if I(s), not the diode current, is 50 mA (or below 0.6 A), then the voltage drop across the resistor is V(s) = I(s)*R(s) = 0.25 V. For the zener diode to maintain a 5V drop across the load, the voltage supply must then be 5.25 V, which is outside the limit provided for the V(ss) = 8 V.
     
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