I have below design and want to know it is correct or need to change anything. Details are as below Input voltage is from 18V to 40V and can draw only 2mA current from source voltage. so I have current limiter using PNP. But ratings of coil are 8.4V, 10.5mA. Can attached circuit will work? Please advise, let me know your comments Thank you.

enter image description here

  • \$\begingroup\$ Looks good. The current will become even smaller, as temperature rises. \$\endgroup\$ Dec 7, 2018 at 13:02
  • \$\begingroup\$ So where are you going to get the extra 8.5mA from to make your coil work? \$\endgroup\$
    – Finbarr
    Dec 7, 2018 at 13:12
  • \$\begingroup\$ I have used C13 for temp. voltage and it will used for extra current, thats where I also have doubts \$\endgroup\$
    – Micro
    Dec 7, 2018 at 13:24
  • \$\begingroup\$ Since relay used is latched type relay, monentary current for 10msec, will not discharge Cap C13, compaired to its charging rate. \$\endgroup\$
    – Micro
    Dec 7, 2018 at 13:34
  • \$\begingroup\$ Show us your working. What's the resistance of the relay coil? \$\endgroup\$
    – Finbarr
    Dec 7, 2018 at 13:39

1 Answer 1


The circuit looks good with two recommended additions: First, you probably need a diode across your relay coil. Otherwise, the voltage on the Q6 drain-to-source will exceed its 50-volt rating when Q6 is turned off, and Q6 will break down and have to absorb the coil energy. I would also place a diode across the collector to emitter of Q3. If you don't, and your power supply is suddenly shut down to ground, you will have a potential reverse voltage of 40 volts across Q3 collector-to-emitter from the charge on the 47-uf capacitor. Both diodes are reverse-biased in normal operation, so choose the reverse voltage rating accordingly. Good luck!


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.