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I'm trying to use the GPIO pins in my raspberry pi to drive a relay. I have a 12 volt rechargeable battery (12.0 AH) and a 12 volt relay, along with a 2N2222 NPN transistor. I built a simple transistor switch circuit by using the 12V battery positive to a resistor (R1) followed by the relay coil and a diode at the collector, and connected the grounds and the emitter of the transistor. I've discovered that without anything connected to the base, the relay still trips if the battery is fully charged. Once the battery discharges a bit the circuit seems to work (GPIO plus R2 at the base turns on the relay). I've attempted to vary the value for R1 but 10K is too much (relay won't trip at all) and 5K is too little (relay always tripped regardless of base). And the charge level in the battery seems to affect this circuit. The relay is 12V. Any suggestions for this? Thanks so much! Here's the circuit I'm using (to the best of my abilities with ASCII): +12V from battery ----- Resistor R1 -- Relay Coil ----- Collector of NPN transistor Diode -|<--- tied to relay coil terminals +3.3V GPIO ----- Resistor R2 ----------------------------- Base of NPN transistor Ground of 12V tied with Ground of Raspberry Pi ---- Emitter of NPN transistor Relay I'm using: https://www.jameco.com/Jameco/Products/ProdDS/2167453.pdf Jameco part number: 2167453 Hopefully the formatting of this post don't get messed up when I submit. Thanks so much for the help.
And got the following answer:
Provide some kind of scrawled schematic if you can. Useful responses may result. I'm just not sure how to read what you wrote in terms of a schematic. (But the main idea that comes across to me wouldn't even work -- it sounds like you have two resistors with one from the I/O control line to the base and another one in series with the relay?) Also, name the relay... I'd like to know the pull-in current as well as the voltage required. More details would help a lot. EDIT: So, this? http://www.infinitefactors.org/schematics/cody_relay.png That won't work, if so. And did you wire up the diode in the direction I show? Or the OTHER direction I don't show? Either way, doesn't matter. Won't work. It appears that the GPIO can source or sink from 2mA to 16mA, in 2mA increments. With about 85mA required by the relay and a β=20 estimate for driving the 2n2222, I'd say you need MORE than 2mA. Probably should set it to the maximum. You won't need it, but it lowers the drop at the GPIO pin. May as well. Just use an appropriate value for R2, which you don't mention. The Vbe for a 2n2222 is about 0.7V at 2mA collector current and goes up by 60mV for every factor of 10. So, worst case would be about 0.82V. They guarantee only 2.4V on the output. So at 16mA setting, that works out to about 56Ω. So with a base drive of say 5mA you will see about 300mV drop and at 10mA that will be about 600mV drop at the pin. Let's assume 5mA is okay, for now. So you will have 3V at the IO pin and lose about 820mV at the 2n2222. This means you want a 390Ω or 450Ω resistor for R2. Go with the 390Ω and get a little more current there. If my diagram guess about your circuit is right, lose R1 by shorting it out. You don't want it. Make DARNED SURE you set up your GPIO for 16mA drive, too. Don't skimp on that. You can lighten the load on your GPIO, but that will require another BJT.
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