AnalogNewb
Junior Member level 1
Hello everyone,
I've got to put a 12V relay on my new board, driven from a 3.3V logic output. Could you please take a look a my circuit and confirm that I've done it right?
Q1 is a 2N3904, used because I'm slightly familiar with it. I know that Hfe is an unreliable number, but it seems to have a minimum Hfe of 40.
K1A is the relay coil, 1028 Ohms (+/- 10%). Max pick-up voltage is 9V. Min drop-out volate is 1.2V.
So, figuring on an Hfe of 20 (for margin), and knowing that I want at least 11.6 ma (12V/1028Ohms) to flow through the coil, I need a base current of 11.6/20 or 0.58ma.
Subtracting the voltage drop from my logic voltage of 3.3V gives me 2.6 volts, which means I need at most a 4444Ohm base resistor. I've halved that for good measure, so R83 is a 2.21K resistor.
That should cause the little transistor man inside the transistor to do whatever he can to make 93 ma flow from collector to emitter. Of course that will never happen because the relay is has a 1028 Ohm resistance, but it will ensure that the transistor is plenty "turned on" to activate the relay.
Throughout all this it looks like my transistor will dissipate:
(2.6V * .00117A) + (11.3V * .0117A) = 0.135 watts
Which is well under it's maximum power dissipation of 625mW.
As far as turning the relay off goes, R84 is a 10.0K resistor from base to ground. I don't know *exactly* why this is needed if my logic output goes to zero volts, but I've heard (horowitz and hill) that it's a good idea.
Of course the Diode is for back-EMF protection.
THANK YOU for helping me with this. This is going to be a very expensive board to manufacture, and I am doing everything I can to try to get it right the first time.
I've got to put a 12V relay on my new board, driven from a 3.3V logic output. Could you please take a look a my circuit and confirm that I've done it right?
Q1 is a 2N3904, used because I'm slightly familiar with it. I know that Hfe is an unreliable number, but it seems to have a minimum Hfe of 40.
K1A is the relay coil, 1028 Ohms (+/- 10%). Max pick-up voltage is 9V. Min drop-out volate is 1.2V.
So, figuring on an Hfe of 20 (for margin), and knowing that I want at least 11.6 ma (12V/1028Ohms) to flow through the coil, I need a base current of 11.6/20 or 0.58ma.
Subtracting the voltage drop from my logic voltage of 3.3V gives me 2.6 volts, which means I need at most a 4444Ohm base resistor. I've halved that for good measure, so R83 is a 2.21K resistor.
That should cause the little transistor man inside the transistor to do whatever he can to make 93 ma flow from collector to emitter. Of course that will never happen because the relay is has a 1028 Ohm resistance, but it will ensure that the transistor is plenty "turned on" to activate the relay.
Throughout all this it looks like my transistor will dissipate:
(2.6V * .00117A) + (11.3V * .0117A) = 0.135 watts
Which is well under it's maximum power dissipation of 625mW.
As far as turning the relay off goes, R84 is a 10.0K resistor from base to ground. I don't know *exactly* why this is needed if my logic output goes to zero volts, but I've heard (horowitz and hill) that it's a good idea.
Of course the Diode is for back-EMF protection.
THANK YOU for helping me with this. This is going to be a very expensive board to manufacture, and I am doing everything I can to try to get it right the first time.
