0
\$\begingroup\$

i want to make reverse polarity protection without voltage drop ,i was searching and i found this

https://www.edn.com/design/analog/4368527/Simple-reverse-polarity-protection-circuit-has-no-voltage-drop

enter image description here

now , i want to size this relay .

load ( 30 A , 48 V ) 1440 Watt

Does load ( normally closed ) matters with relay?

Where load is normally closed so relay is not operated when power supply is connected correctly.

so if that right then

Does relay ( 12 V , 5 A ) 60 Watt can do the work ?

if all above right , then the load will taken into concern is when relay energized so the led ( in photo above ) and resistor here will be loads .

Thanks you .

\$\endgroup\$
  • 2
    \$\begingroup\$ Probably about the most ineffective reverse polarity protection in the entire world. What voltage does the relay need to operate? Say 9 volts - this means it won't protect until the reverse voltage is greater than 9 volts. \$\endgroup\$ – Andy aka Dec 10 '19 at 13:41
  • 1
    \$\begingroup\$ @Andyaka thanks for your reply , yea you are right , here i have constant but remote power supply so it can be connected reverse but the value is 48 voltage , so it cant be less than 9 volts , actually i think it will be good if it doesnt drop any volts . \$\endgroup\$ – Ahmed elmenshawie Dec 10 '19 at 13:44
  • \$\begingroup\$ @Ahmedelmenshawie So wait, you're supplying a 12V relay with 48V? That's not going to protect from reverse voltage at all, reverse voltage will just first destroy the relay and then destroy whatever it was supposed to be protecting. \$\endgroup\$ – Hearth Dec 10 '19 at 13:48
  • \$\begingroup\$ @Hearth oh surely you are right , i was totally mistaken here , so if relay is 48 volt same as input here , then how about the load current if its more than the relay can endure \$\endgroup\$ – Ahmed elmenshawie Dec 10 '19 at 13:50
  • 2
    \$\begingroup\$ If the relay is 48 volts then it might not reverse protect until maybe -36 volts. Same as before, it's plain bad. \$\endgroup\$ – Andy aka Dec 10 '19 at 13:52
0
\$\begingroup\$

In a word, no.

You say your load requires 30 A, but your relay is only rated for 5 A. Relays are resilient and can handle a bit of pulsed overcurrent, but you're trying to use this relay for six times what it's rated for--it won't survive.

There are better reverse-polarity circuits out there, by the way. Relays are expensive, but a MOSFET is much less so.

\$\endgroup\$
  • \$\begingroup\$ thanks for your reply , but i have a question , how the relay will not consume power at correctly connected and in same time the current of load will burn it ? \$\endgroup\$ – Ahmed elmenshawie Dec 10 '19 at 13:40
  • \$\begingroup\$ @Ahmedelmenshawie I'm not sure what you're asking, sorry. \$\endgroup\$ – Hearth Dec 10 '19 at 13:43
  • \$\begingroup\$ im not sure if i can but external links here , can u check this ? \$\endgroup\$ – Ahmed elmenshawie Dec 10 '19 at 13:45
  • \$\begingroup\$ edn.com/design/analog/4368527/… \$\endgroup\$ – Ahmed elmenshawie Dec 10 '19 at 13:45
  • 1
    \$\begingroup\$ @Ahmedelmenshawie As Andyaka points out, that won't protect from reverse voltages of less than 12V at all. (or at least, from reverse voltages less than what the relay's turn-on voltage is) \$\endgroup\$ – Hearth Dec 10 '19 at 13:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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