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Zener Diode Regulator

  1. Jul 31, 2010 #1
    My teacher was talking about power supply voltage regulators.

    I cannot get why does a zener diode regulator's output voltage drops after the load current through it exceeds some critical current?

    thanks in advance :smile:
     
  2. jcsd
  3. Jul 31, 2010 #2

    K^2

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    Here is the Voltage-Current characteristic of a Zener diode.

    240px-V-a_characteristic_Zener_diode.svg.png

    Do you see how at very, very low currents, the voltage actually drops bellow threshold?

    So let us see what happens in the simplest voltage regulator. Lets say some applied voltage Va is applied across the Zener diode with threshold voltage VZ and a resistor R1 connected in series. In parallel with Zener diode, you connect your load resistance, RL at regulated voltage VZ.

    So lets look at the currents. The load current is obviously given by this.

    [tex]I_L = \frac{V_Z}{R_L}[/tex]

    And the current across R1 is this.

    [tex]I_1 = \frac{V_a - V_Z}{R_1}[/tex]

    And of course, the current flowing through R1 then splits to go through RL and the diode. So we can easily find the diode current, IZ.

    [tex]I_Z = I_1 - I_L = \frac{V_a - V_Z}{R_1} - \frac{V_Z}{R_L}[/tex]

    If VZ remained constant despite absolutely any changes, this value would become zero whenever

    [tex]R_L = \frac{V_Z}{V_a - V_Z}R_1[/tex]

    Clearly, load resistance can be that low, or even lower. If it is lower, the value of IZ would become negative, which would mean the current is flowing against the potential, and that's nonsense.

    In reality, of course, as seen in the diagram, VZ begins to drop before you quite get to that point. In the limit where RL goes to zero, VZ also goes to zero.

    If the regulator you are thinking of is more complex, you might have to do a bit more work to get this result, but the idea is the same. Find the current that flows through the Zener diode as a function of RL, and you'll see when the regulator is not going to be very useful anymore.
     
  4. Aug 2, 2010 #3
    Thanks very much for the clear explanation, K^2!

    I can see the where the formulas come from since you explanation fits my regulator well.

    However, there is one thing I need to clarify: is the threshold voltage 17V in the diagram?
    And you're saying that at low current e.g. 1mA in the diagram, voltage drops to 0.65V?
     
  5. Aug 2, 2010 #4

    Born2bwire

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    The threshold voltage of interest here is the reverse breakdown voltage at around -17V. The idea being that once you setup a bias current that forces the zener diode into reverse breakdown voltage, then the voltage output is going to be fairly constant (-17 V) despite small variations in the bias current. So a simple AC-DC power supply can be made using a transformer to stepdown the AC voltage, rectifying the AC voltage to keep it always positive, passing it through a low pass filter to smooth out the rectified voltage a bit, and then using the voltage to create a bias current through a zener diode. The voltage output is taken off of the zener diode. There will still be a small amount of ripple voltage across the diode due to the variations in the bias current but it can be controlled via design to be below a certain threshold.
     
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