I have looked up some existing questions and did my own calculations for my own problem.

I am switching a relay with a BJT transistor and would like to protect the transistor from voltage spike on relay coil, after the transistor stops conducting.

Circuit: enter image description here

I decided to connect two diodes in parallel with my relay. A schottky and a zener diode.

First I determined/checked up the relay datasheet. My relay has nominal coil voltage of 5V (this is what I will be using for switching my relay minus the Vce voltage drop) and 35mA coil current (I used 40mA in my calculations).

I then searched for a BJT. I found one with Vce_max equal to 40V and Vcb_max equal to 60V. Maximum power dissipation is 200mW.

Next I picked a schottky diode with Vf_max equal to 0.3V, repetitive reverse peak voltage 30V and Iforward 1A and a zener diode with 2V Vz @ 5mA Iz. I looked at the graph in the datasheet which shows current/voltage curve and at the coil current which is 40mA voltage drop across zener diode is 2.7V.

In my opinion these chosen elements are okay, can anyone confirm this?

The circuit is not meant to switch relay with high frequency. It's meant for opening/closing garage doors.

From the diodes I have chosen is my assumption correct that voltage drop across coil will be around 3V (sum of both voltages of zener and schottky) after the transistor stops conducting and coil starts to discharge through diodes?

Thanks in advance.

  • 3
    \$\begingroup\$ Add schematic!! \$\endgroup\$
    – winny
    Commented Dec 22, 2016 at 11:20
  • \$\begingroup\$ Why you have chosen a zener and a schottky diode completely mystifies me. Sounds like an XY problem to me. \$\endgroup\$
    – Andy aka
    Commented Dec 22, 2016 at 11:49
  • \$\begingroup\$ Andy, zener diode is added so that relay's coil dissipates energy faster. \$\endgroup\$ Commented Dec 22, 2016 at 13:33
  • \$\begingroup\$ I did add a schematic \$\endgroup\$ Commented Dec 22, 2016 at 13:36

2 Answers 2


There is no reason to use a Schottky in this circuit- the extra few hundred mV will actually help, and if you use a Zener you want it to result in the maximum voltage the transistor can safely handle.

Below are three possible circuits, of which the first is adequate for your purposes. D1 can be a 1N400x or 1N4148, it only needs to be rated for the supply voltage (+5) and to handle the relay coil current (tens of mA) for a short time. The zener (+ series diode drop, if there is one) has to be less than the voltage rating of the transistor, but it might have to be even less depending on the SOA (safe operating area) of the transistor. The first circuit is not as hard on the transistor and any modern jellybean part (don't try an RF transistor even if the ratings look like they might be okay) will work fine.


simulate this circuit – Schematic created using CircuitLab

The two circuits to the right will give you a bit more life from the relay because the contacts will open more quickly (less arcing) but probably not necessary for your application. D2 is necessary because the Zener diode will conduct happily in the forward direction- it breaks down in the reverse direction. The circuit with D7 involves one less part, and is similar if there is a bypass cap nearby.

  • \$\begingroup\$ I was talking about using the circuit number 2 on your schematic. I added a picture not so it is more clear. And I want to use zener diode because I want my relay to live longer like you mentioned it would if one is added. \$\endgroup\$ Commented Dec 22, 2016 at 13:41
  • \$\begingroup\$ Your zener voltage should be much higher, as high as 25-30V for a 40V transistor. The transistor briefly sees +5+Vz+0.6V (assuming a regular silicon diode). The higher the zener voltage + diode forward voltage, the faster the relay switches. \$\endgroup\$ Commented Dec 22, 2016 at 13:55
  • \$\begingroup\$ Basically you are saying that I shoul implement such a zener diode that reverse voltage would reach about 25-30V before zener diode starts to conduct? The power dissipated this way would be many time higher and thus relay would close faster. \$\endgroup\$ Commented Dec 22, 2016 at 16:49
  • \$\begingroup\$ @GalErženPajič Exactly correct, except that it won't affect the relay closing at all, only opening. To speed closing you'd have to do something different. \$\endgroup\$ Commented Dec 22, 2016 at 18:00
  • \$\begingroup\$ Sorry, I meant opening :) Thanks, what is your input about my method of determining parameters of the elements I mentioned apart from voltage of zener diode being too low? Do you think I am goin in the right direction/taking into account things I should? \$\endgroup\$ Commented Dec 22, 2016 at 20:39

You are way over-thinking this. A single ordinary silicon diode (e.g., 1N4001) is more than sufficient for this application.

Adding the zener might reduce the dropout time by a small amount, but that is completely irrelevant for your application.

  • 3
    \$\begingroup\$ Those interested in the suitability of 1N400X diodes as flyback diodes should look at cliftonlaboratories.com/diode_turn-on_time.htm. I have been using them in this application for 40 years. \$\endgroup\$
    – RoyC
    Commented Dec 22, 2016 at 13:28
  • \$\begingroup\$ Won't adding a zener diode extend the life of my relay? \$\endgroup\$ Commented Dec 22, 2016 at 13:32
  • \$\begingroup\$ @RoyC Thanks for the link. It's nice to see myself credited there. ;-) \$\endgroup\$ Commented Dec 22, 2016 at 13:49

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