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# Failure to activate relay

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#### Whimsy

##### New Member
I've breadboarded a circuit that is not working as intended. Hopefully readers can reference my PNG attachment for clarification. I have a bridge, the input to which is 23 VAC, and the output is approximately 16 VDC. I'm switching this voltage by turning on a NPN transistor which I was hoping to activate a 5 Volt relay. I thought that a 5 volt drop would be created by the 4.7K resistor, and that this drop would fire the relay. I'm assuming that this didn't work because the coil resistance of 124 ohms in parallel with the 4.7K messed up the drop and therefore lacked the necessary voltage to activate the relay. Where did I go wrong? The ultimate goal is to transform the 23 volts to the 5 volts needed by the relay. By the way, without the relay in place there is a 5 V drop across the 4.7K resistor. So I'm not a complete idiot.

I'm assuming that this didn't work because the coil resistance of 124 ohms in parallel with the 4.7K messed up the drop and therefore lacked the necessary voltage to activate the relay. Where did I go wrong?
You just stated where.
The coil is in parallel with the 4.7k which gives a equivalent resistance of about 121 ohms.
Thus virtually all the drop is occurring across the 10k ohm resistor.

The way to drop the voltage is just to eliminate the 4.7k resistor and then calculate the value for the resistor in series with the relay to give the voltage you want.

Make sense?

Note that for the 42mA coil current of the relay at 5V, the transistor base current should be about 1/10th of that for good turn-on saturation.

If the intention is purely for the relay to switch on when the 16V AC is applied, the transistor is pointless.
All you need is a series resistor for the relay.

Just remember the resistance of the relay coil itself forms part of the voltage divider.

Or you could use a 24V relay?

Depending on your relay characteristics, in the absence of a smoothing capacitor for its DC supply the relay may chatter.

Thanks to all who replied, especially rjenkinsgb. I was overthinking the problem with the addition of the transistor, and only needed a resistor in series with the relay coil to knock down the 16 volts to 5 volts.

This is a great forum. It's been beneficial to me over the several months that I've been signed up. Good work people.

Remember that the voltage across the resistor will be about 11 Volts. Use Ohms law to calculate it's power dissipation.

42 mA x 11 Volts will be about 1/2 watt. Use a 1 watt resistor for good safety margin.

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