1
\$\begingroup\$

I have a question regarding 12v dc relay circuit design.

About 350mA flows through the relay coil from the 12v power supply, so the pmosfet specification was selected as vds -30v, id -2A class.

The first concern is, is there any possibility of fet burnout if 12v power is suddenly applied and an inrush current occurs through the coil? Of course, it seems that the inductor component may block the AC component (inrush current), but it is a concern.

The second concern is that the diode is selected as 1N4007, but it is questionable whether the 1N4007, which has a specification of 1000V 1A, can withstand the emf when the relay voltage is off.

Is it correct to select Fly back diodes 1n4007 in my design?

enter image description here

\$\endgroup\$
2
  • \$\begingroup\$ The sizing of diode depends on the energy stored in the relay coil. For most applications the 1N4004… is more than adequate. It is not necessary to have a 1kV rating - the diode clamps any flyback voltage to one diode drop. You mention a p chan mosfet but your circuit suggests a n chan mosfet. I think you want a n chan. The 10k resistor as drawn does not make sense. \$\endgroup\$
    – Kartman
    Commented Sep 6, 2021 at 7:46
  • \$\begingroup\$ Thank you for answer. It says npn on/off, I couldn't draw the circuit below, but I designed an npn transistor on the pmosfet gate stage. I have configured the switch circuit with npn and pmosfet. \$\endgroup\$
    – 이동재
    Commented Sep 6, 2021 at 8:00

1 Answer 1

1
\$\begingroup\$

An inductor has NO DC inrush current.
Current in inductor current cannot change instantaneously, so if there is no current in the coil at turnon current will start at zero and increase.

Diode current is L_coil l at turnoff. 350 mA in this case. Any 1N400x diode rated higher in voltage than Vsupply will easily suffice.

\$\endgroup\$
2
  • \$\begingroup\$ Thank you for your kind reply. \$\endgroup\$
    – 이동재
    Commented Sep 6, 2021 at 8:01
  • \$\begingroup\$ Even a 100 mA Diode will do for this pulse. With steady Voltage there will be a inrush "ramp" up due to V/DCR=I, T=L/R [s] ratio unlike caps which have a decaying pulse current. \$\endgroup\$ Commented Sep 6, 2021 at 20:38

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.