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I have searched in over voltage protection (like MOSFET.) I wanted to use a Zener diode in parallel so it won't dissipate power unless in over voltage protection, but its power is low so I read that I need a resistor to control the current passing through it.

I found people put the resistor in this shape:

enter image description here

This resistor will dissipate power, so I thought if we can put resistor like this:

enter image description here

In this situation the resistor (R3) will only dissipate in case of over-voltage but in desired voltage current won't pass through zener so won't pass through the resistor.

  • Is the voltage going to the LED equal to source or to the Zener?
  • Is the new placement for the resistor right? If wrong, why?
  • Is there another placement for resistor if this doesn't work ?**
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The first circuit you show, with the resistor in-line with the load current, is what you require. That way, the Zener diode will limit the voltage, and the excess voltage will be dropped across the resistor.

In your suggested circuit, the excess voltage will also be dropped across the resistor, but since the resistor is in series with the diode, and the combination is in parallel with the load, the load will still see the full voltage - the load voltage will not be limited by the Zener.

The circuit shown here uses the Zener to regulate to 3.3V, and R2 limits the current through the LED at that voltage.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ thanks for your answer , i have more questions in the first picture ,does load current affects on resistance ? when choose resistance value i will take only the max current of load in my mind ? , and does resistor voltage drop affected by current drawn by load? \$\endgroup\$ Sep 23 '19 at 0:39
  • \$\begingroup\$ In the first circuit, the load current flows through the resistor, so affects the voltage drop across the resistor - all determined by Ohm's Law. \$\endgroup\$ Sep 23 '19 at 2:13
  • \$\begingroup\$ thanks sir @Peter Bennett \$\endgroup\$ Sep 27 '19 at 12:24

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