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I am trying to switch a pair of 12v relays with approx 750r coils using a circuit and pre-fabricated pcb that I was given. The idea is that one relay is on when the other is off. The control comes from a signal from an arduino nano. Unfortunately it doesn't seem to like working. I have one relay that it works fine with but another that it doesn't (replacing one with the other in the same position of the circuit). I have tested the relays and they are fine. I'm guessing the transistor is not completely shutting off or something of that ilk (could it be due to the control signal from the arduino being 5v and the relays being 12v?) I have little knowledge of transistors. Can anyone suggest a solution, preferably one that I could try without having to get a fresh pcb made? Would using an npn and pnp mosfet solve my problem? Would I have to change any part of the circuit?

Many thanks!

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  • \$\begingroup\$ Which relay/LED is not turning off? Is it K2 & LED2? That would make sense because with a maximum of +5V out of the Arduino, the bottom transistor - the PNP - is not turning off. \$\endgroup\$
    – SteveSh
    Commented Feb 7, 2020 at 18:23
  • \$\begingroup\$ The leds work fine. With the arduino output high, led1 is lit. With the arduino output low, led2 is lit. I have two "identical" 12v dpdt relays. Assuming I'm concentrating on K1, one relay in that position works. If I swap it out for the other relay, it doesn't work. Yet, both relays definitely work when not used in this circuit. \$\endgroup\$
    – Paul B
    Commented Feb 7, 2020 at 19:11

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The problem its caused by the fact that \$Q_{18}\$ is always ON since even when the input level fro Arduino Nano+ is high, it does not reach the 12V needed to bring \$V_{BE_{Q_{18}}}\simeq 0\$. This is due to the fact that the Arduino output reach at most +5V, while the emitter of \$Q_{18}\$ is connected with various loads (the relay coil and the series of the red LED and \$1\mathrm{k}\Omega\$ resistor. Therefore if you want your circuit to work correctly you have two possible choices:

  1. Use two 5V coil relay and remove the 12V connect of the coils by connecting them to the 5V power supply as the two LEDs LD1 and LD2. This implies no circuit changes.
  2. If you cannot use 5V relays since your supply current is limited, you can use two NPN transistors connected as shown below

schematic

simulate this circuit – Schematic created using CircuitLab

This circuit can work perfectly with the 5V/12V supply arrangement, at the cost of changing the \$Q_{18}\$ transistor.

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    \$\begingroup\$ Hmmm...Gotta do a thorough analysis. With Q17 OFF, there may be enough current through R61 and the base-emitter of Q18 to turn on LED1 and energize relay K1. If that's the case, then K1 will never de-energize. Need to know the characteristics of LED1 and K1 (the minimum current at which K1 will energize). \$\endgroup\$
    – SteveSh
    Commented Feb 7, 2020 at 19:48

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