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I am currently diving in in embedded development using ATMega Microprocessesors. I have now a question regarding Z-Diodes. My circuit is powered by a 9V battery that gets through an IC7805, thus getting a constant 5V power supply. Now I have a part that requires to be used at 3V. I made my thoughts how to achieve that and two things came across my mind: 1. Use an LM1086, which I felt is a but overpowered for just one part needing 3V or use a Z-Diode.

I build a little prototype on my breadboard using the IC7805 as supply and using a 3V Zener Diode to go down to 3V and power a LED with that. It so far works fine, but when I measure the power at the LED, I always only get 2,3V and if i measure the diode, i get 2,7V.

Here is the circuit plan I use:

enter image description here

(I just saw I forgot to add a value to R1, which is 100 Ohms)

Am I really getting just 3V at the diode? If yes, why cant i measure it with my multimeter.

Thanks in advance!

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  • \$\begingroup\$ The LED and zener are in parallel. If you measure 2.3 volts on the LED you MUST also see 2.3V on the zener. Measurement error or construction error or meter battery failing or human error. A data sheet link to the LED would be useful. \$\endgroup\$
    – Andy aka
    May 24, 2015 at 14:24

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Zener can be modeled as a resistor with a voltage source when current is flown through it.. thus output voltage varies a little...

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The LED is probably drawing an excessive amount of current and is causing a voltage drop on the wires with which it is connected to the Zener. This is (probably) why you measure different voltages across the LED and the Zener.

You have to put a resistor in series with the LED and limit its current to about 5mA-10mA (depending on the model, consult datasheet to be sure, unless it is a jellybean 5mm LED).

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  • \$\begingroup\$ The voltage across an LED is determined primarily by the LED's colour and chemistry, and varies onlly slightly with current. \$\endgroup\$
    – Peter Bennett
    May 24, 2015 at 15:29
  • \$\begingroup\$ @PeterBennett Please, reread my post. The voltage drop I referred to is across the wires. If the OP connected the LED to the Zener with tiny wires the drop on those wires could account for those ~400mV. The LED threshold voltage appears to be around 2.7V, so if you try to power it with a voltage source (the Zener) without a limiting resistor the excess voltage will make the LED draw "lots of amps", usually destroying the LED. In this case the excess voltage is tiny, so perhaps the wires are acting as unintended limiting resistor, saving the LED from its death. \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    May 24, 2015 at 17:42
  • \$\begingroup\$ @PeterBennett of course I meant "The LED threshold voltage appears to be around 2.3V". Darned copy/paste! :-) \$\endgroup\$
    – LorenzoDonati4Ukraine-OnStrike
    May 24, 2015 at 17:54
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You don't need or want a Zener diode (Z-diode is not a standard term) in this situation.

The LED will regulate its own voltage with the resistor so a zener is not required.

At best, the Zener diode will steal a poorly defined amount of current away from the LED, for no good reason.

Also, as Andy said in a comment, you must have the same voltage across the LED and Zener modulo bad connections and measurement error (resistance of the connections should be negligible in this situation).

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  • \$\begingroup\$ I just added The led to have a consumer there. Later i want to add a dcf77 receiver. \$\endgroup\$
    – Nidhoegger
    May 24, 2015 at 14:31
  • \$\begingroup\$ Don't use a zener for 3V- use a series regulator. If you insist on using a shunt regulation scheme- use an IC regulator such as a TL431. The LED will take current away from a shunt regulator (and will get too much current from a series regulator)- if you want to use it as a load (a resistor alone would be better) put a resistor in series of a few hundred ohms. \$\endgroup\$
    – Spehro Pefhany
    May 24, 2015 at 14:31

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