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The use of a reversed biased zener diode in voltage regulator

  1. Feb 26, 2012 #1

    Hello. I asked this question because I am confused of the use of a reversed Zener Diode in voltage regulator even after I built the circuit using MultiSim.

    I don't understand why we should put a zener diode right there and how it works. I looked at the I-V plot of the Zener diode and I know when it is connected in reversed biased, the zener diode will conduct at the breakdown voltage and current will start flowing. But after a lot of thinking, I will don't know how this characteristic is needed in a voltage regulator.

    This is not homework but just a question to myself. So I want to make sure myself understand the concept.

    Thank you
  2. jcsd
  3. Feb 26, 2012 #2
    It is a 6.2V zener, so anything over 6.2V and it will conduct. You have 1K from 17 volt so there is 11.3V drop across the resistor. Current drawn is 11.3/1000=11.3mA if there is no other component. Now if you put a 10K in parallel with the zener. it draw 0.62mA. BUT there are total of 11.3mA draw from the 1K, so 11.3-0.62=10.68mA still need to draw by the zener diode. So the zener still in reverse conduction and still hold 6.2V across. So the zener still regulate the voltage at the anode.

    You can lower the 10K to say 1K. At 1K, it draw 6.2mA away from the junction of the anode. So the zener still conducting 11.3-6.2=5.1mA. So the zener diode still regulate the voltage.

    But if the 10K is lower to 558Ω. At 6.2V, it draw 6.2/558=11.11mA. At this, the zener only draw 11.3-11.11=0.189mA. Almost to zero. The zener is at the edge of turning off. Any lower value than 558Ω, there will be no current into the zener and the zener turn off and the voltage will drop below 6.2V and you lost regulation.

    This is called shunt regulation. The shunt regulator will draw the excess current that the load ( in your case the 10K) cannot draw. It will regulate the voltage until the load draw more than the current that the zener originally programmed to draw ( by the 1K). to increase the capability of regulation, you need to lower the 1K so at default, the zener draw more current. Then you will have more room for regulation.

    For example if you reduce the 1K to 500Ω then the zener will draw 22.6mA to start. Then any load that draw less than 22.6mA will let the zener in regulation.
  4. Feb 28, 2012 #3
    Thanks for the reply. I understood more now.
    After listened to your explanation, I built the power supply with op-amp regulator in MultiSim. But I have some confusion:

    How do the output of the Op Amp work with the pass transistor? I know the voltage of V(+) and V(-) of an op amp will be the same. But like in the very beginning, when V(+) and V(-) are not the same, what do the V(output) of the op amp do to make V(+) and V(-) the same?

    I have searched online, it says when the current Iout which goes into the pass transistor increases, then the voltage coming out from the pass transistor will decreases. Is that the way it works?

    Here is the circuit that I built in MultiSim:


    Thank you.
  5. Feb 28, 2012 #4


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    Staff: Mentor

    Instead of feeding back a portion of U1's output, R4/R5 feed back a sample of the Probe6 signal, so OP-AMP operation causes the Probe6 signal to be proportional to the Probe3 signal. The transistor's emitter current is beta times the output current of the OP-AMP. Connect your external load between emitter and ground.
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