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I have just started learning about Zener diodes and this is a quite a basic question I think.

In the following circuit, I'm supposed to calculate the current in the LED, the power supply has 6.5 V, so shouldn't the Zener diode act as a voltage regulator and be at 5 V, and then we can calculate the current which will be 20 mA.

But in the solution of my professor, it says that the answer will be 15 mA, I'm not sure how he got there as I only have the result. How does that work and what happens in the circuit?

enter image description here

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    \$\begingroup\$ Welcome! What’s the Vf of the LED? \$\endgroup\$
    – winny
    Jul 12 at 10:38
  • \$\begingroup\$ The circuit is bad designed because voltage across zener never reach 5V. So the zener will not act as regulator. You must lower the resistor connected to source. \$\endgroup\$
    – Michal Podmanický
    Jul 12 at 10:43

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Ignore the Zener for a moment.

Let's suppose the LED is a typical red one with Vf of 2 V.

Now you have 6.5 V - 2 V = 4.5 V dropping over 300 Ω, which is 15 mA.

The top of the Zener will be 2 V + (4.5 V / 2) = 4.25 V

As this is below the Zener's voltage, it won't conduct.

Try experimenting with the simulator here, changing V1 to say 10 V to get the Zener to clamp, and see what happens to the current.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ Thank you, but I don t get why we should ignore the zener in the begining and not the LED. \$\endgroup\$
    – Anas
    Jul 12 at 11:47
  • \$\begingroup\$ Because until the voltage gets above the operating voltage of the Zener, the diode doesn't conduct. ie, is basically open-circuit. Try removing it from the circuit in the simulator. \$\endgroup\$
    – jonathanjo
    Jul 12 at 11:49
  • \$\begingroup\$ @Anas If you want to, you can ignore the LED at first, but you'll soon find that doing so gives you results which don't make any sense. Experienced EEs have come to recognize that if they see a zener in parallel with some other stuff then the 1st thing to do is figure out if the zener would even be having any effect in the circuit. And the way to do that is to ignore the zener and calculate the effect the other stuff has to work out if the voltage across the zener ever reaches its rated level. \$\endgroup\$
    – brhans
    Jul 12 at 12:20
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    \$\begingroup\$ @brhans this appears to be the exact Zener lesson OP's professor had in mind. \$\endgroup\$
    – jonathanjo
    Jul 12 at 12:22
  • \$\begingroup\$ Okay, I think that makes sense, thank you all for your help. \$\endgroup\$
    – Anas
    Jul 12 at 12:27

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