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I want to select x as resistance for the protecting resistor (upper).

The zener diode can hold up to 4A

If I was guaranteed to have current through zener lower than 4A, I could consider its voltage to be 5.1V and the voltage through load to be 5.1 V.

But I'm not. So, I can't consider the voltage to be 5.1 V and I can't calculate a voltage drop for protecting resistor in terms of x, because I don't know the current in curcuit because I don't know the resistance of zener.

  • \$\begingroup\$ You would need to know load current in order to do this calculation as well as what the required Iz is in order to keep the zener in zener breakdown. \$\endgroup\$
    – efox29
    Dec 8, 2014 at 9:00
  • \$\begingroup\$ What are you doing? Do you really want to put 4A through the zener? 4 A at ~5 V is ~20 Watts of power? Can the Zener really handle that much power. (link to spec sheet or part number please.) (For what you want a ~2.5 ohm resistor would work.. but it would be dissipating ~40 Watts!) \$\endgroup\$ Dec 8, 2014 at 14:48

1 Answer 1


To find the limits of the resistor needed x, proceed as follows :

1) For zener to just turn on, Voltage across the load resistance ( 2400 ohms in your case) must be greater than or equal to the zener voltage.

Applying Voltage divider rule ;

[ 2400 / (2400 + x) ] * Vi > Vz

Vi = Input Voltage

Vz = Zener Voltage

From here, you get the maximum resistance x needed.

2) For Zener to be in its working limits, proceed as follows :

Current through the zener will be max ( Say Izmax). So

Izmax = [ Ix - IL ] 

Ix is current through x

IL is current through Load resistance, which equals [ Vz/ RL ].

From here, you compute the current through x.

Since Ix = [ Vi - Vz ] / x ; you get the second limit for your resistance.

  • \$\begingroup\$ -1 because of incomplete answer, since resistor wattage and need of capacitor in paralell to zener diode it is not included to the answer. I will remove the downvote if you comeback.. \$\endgroup\$
    – GR Tech
    Dec 8, 2014 at 12:17
  • \$\begingroup\$ @GRTech. I answered it with reference to the circuit given. \$\endgroup\$ Dec 8, 2014 at 12:19
  • \$\begingroup\$ Please edit your answer as a professional engineer \$\endgroup\$
    – GR Tech
    Dec 8, 2014 at 12:37
  • \$\begingroup\$ @GRTech . I'd love to give a more detailed info, only if I knew. I saw the question and immediately recognised it as one of the text book problems which I solve often. So I gave the answer. If you feel it is incomplete, please be kind enough to add the details, so even I can learn-- An aspiring engineer(still 2 years to go). \$\endgroup\$ Dec 8, 2014 at 12:41
  • \$\begingroup\$ Your answer it is perfect but not complete. I got lot of downvotes becausae of incomplete answers on this forum. So please dig more. It will be benefit for you and to the people they are posting questions here. Sincerely... \$\endgroup\$
    – GR Tech
    Dec 8, 2014 at 12:47

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