Hi everyone.
I'm still new to understanding/reading circuits, and am having some difficulty on deciding how to proceed with a solution to a problem I have.
What I would like to do is trigger a 12v relay with an NPN transistor using a 2v 20mA trigger source (3mm green LED).
I found a circuit online that's very close, but with a slightly different input trigger voltage. I modified it to reflect my components, however, I lack the experience to know if this will still work. The revised circuit is attached. Should this be ok, or will I need to use a different NPN and possibly a different value resistor? Any help would be greatly appreciated. Thanks.
Edit: The specsheet for the relay says the current draw is 12v 9.1mA
I'm still new to understanding/reading circuits, and am having some difficulty on deciding how to proceed with a solution to a problem I have.
What I would like to do is trigger a 12v relay with an NPN transistor using a 2v 20mA trigger source (3mm green LED).
I found a circuit online that's very close, but with a slightly different input trigger voltage. I modified it to reflect my components, however, I lack the experience to know if this will still work. The revised circuit is attached. Should this be ok, or will I need to use a different NPN and possibly a different value resistor? Any help would be greatly appreciated. Thanks.
Edit: The specsheet for the relay says the current draw is 12v 9.1mA
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depends on the current it takes to pick up the relay. Some are 10 ma, some are 500 ma. you should put 12 v on your relay, put the milliammeter of your DVM in series with it, and see what current it draws. This will tell you what your transistor has to sink.
2v source - .7v to turn on a transistor is 1.3v. So the base of the 2n3904 gets 1.3 ma drive current. 20 gain would saturate (go to 1 v drop) 26 ma relay current, 40 gain would saturate 52 ma. Both pretty light for driving a relay.
if you lower the 1k resistor to 200 ohms you would have more margin, but I don't know if your LED driver circuit could take that much current draw.
OTOH if you use a darlington transistor like a MPSA13 or something, gain of 500 to 1000 should saturate very well . The base drop on a darlington is about 1.4 v so use that to subtract from 2 v to calculate your base current.
Best of luck.
2v source - .7v to turn on a transistor is 1.3v. So the base of the 2n3904 gets 1.3 ma drive current. 20 gain would saturate (go to 1 v drop) 26 ma relay current, 40 gain would saturate 52 ma. Both pretty light for driving a relay.
if you lower the 1k resistor to 200 ohms you would have more margin, but I don't know if your LED driver circuit could take that much current draw.
OTOH if you use a darlington transistor like a MPSA13 or something, gain of 500 to 1000 should saturate very well . The base drop on a darlington is about 1.4 v so use that to subtract from 2 v to calculate your base current.
Best of luck.
The input is (2V-0.6V)/1k = 1.4mA base current, and the minimum DC beta of the 2N3904 is around 30,
so you will get at least 1.4mA x 30 = 42mA collector current. Does your relay coil need more than this?
If so, decrease the 1k base resistor accordingly. For example, if your relay needs 100mA, the maximum
value for the resistor would be (2V-0.6V)/(0.1A/30) = 420R.
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Thanks guys. Sorry, I had edited my post.
Yes, the relay's spec states it only requires 9.1 mA at 12v.
Yes, the relay's spec states it only requires 9.1 mA at 12v.
Then the 1k is fine. Make sure the input voltage remains at 2VDC when the transistor is on.
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Okay great! If I leave the LED connected, is there a chance I could end up drawing too much from the source circuit? Or best to remove it?
Unfortunately I don't have the circuit diagram for my decoder and am unable to find it online. Once I open the unit I'll trace the led circuit.
Thanks so much.
The 2N3904 base should draw (2V-0.6V)/1k = 1.4mA.
LEDs are often biased with current from 1mA to 10mA, so it may or may not work.
The maximum resistor for 2V input is (2V-0.6V)/(9.1mA/30) = 4.6k,
and the maximum base current would be (9.1mA/30) = 0.3mA, so a 4.7k should work.
Give it a try, and see if the relay triggers reliably.
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Oh wow, I believe I understand. So the 4.7k would be the bare minimum with the lowest draw on the led circuit in order to engage the relay. If it doesn't I could always drop the resister impedance value in small increments, which would help improve the reliability of the relay clicking, however requiring a higher draw from the led circuit.
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