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Background

I have a 12V power supply that is connected to a switching regulator to get a 1.8V power signal. I have a microchip running at 1.8V and want to control a 12V relay via GPIO. The relay is used to open and close a lock that only functions on 12V.

Help in

1:Must-operate voltage

2:Must release voltage

3:Coil voltage

4:Switching voltage

5:Current rating

What I have

I think I need operating voltage at 1.8V because the chip sends 1.8V GPIO signals. A release at 12V because that's the power needed for the lock. A coil at 12V with switching at 12V because of the lock. And current at 2A because the lock runs at 2A.

References for the lock

enter image description here

I need help figuring out the right relay to use and how to hook up the relay to the microchip and lock.

vu2nan
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    Why a relay? Why can't you just actuate the lock with an amplified signal from the GPIO? – TimWescott Oct 22 '20 at 01:17
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    Go to Digikey and look for something called "Trench FET". Choose one that has appropriate ratings. Look at Gate voltage as well as maximum Drain voltage and desired maximum current. – Dwayne Reid Oct 22 '20 at 02:04
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    Ah, 1V8 is indeed very low. There are a couple of workarounds: (1) Use a general device to shift up 1V8 GPIO signals to 3V3 or 5V0, (2) Use a specific transistor BJT or MSOFET in open collector/drain to do the shift up. (3) Use a JD-Vcc relay, remove the cap for separate power, and power the opto-isolator power with 1V8 (might need to do minor modifications, such as adding a current limiting series resistor, ... Ref: https://electronics.stackexchange.com/questions/505318/how-to-properly-use-a-relay-module-with-jd-vcc-from-arduino-raspberry. / to continue, ... – tlfong01 Oct 22 '20 at 02:28
  • You can use power MOSFETS, such as IRL540N instead of the relay, but there are pros and cons: (a) relay is bidirectional, more noise free, ... (b) power MOSFET is unidirectional, the Vgs(th) is 1-2V, so you need to pick a very low Vgs(th) for your 1V8 signals.Good luck and cheers, – tlfong01 Oct 22 '20 at 02:28
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    What are the manufacturer and part number of the "microchip"? Depending on the answers, this might take just one transistor. – AnalogKid Oct 22 '20 at 03:13

1 Answers1

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You could use a BJT + MOSFET or a very low Vth MOSFET.

An example of the latter would be the SQA410, which will dissipate less than 1/4W when on.

schematic

simulate this circuit – Schematic created using CircuitLab

The 2A Schottky diode protects the MOSFET from the inductive kick when the coil is turned off. R1 is to ensure the MOSFET is off (and not partially on) if the MCU output is floating. R2 provides some measure of protection against ground bounce.

Spehro Pefhany
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