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uln2003 base resistor

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raman00084

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in my circuit the input and output of uln2003 is 24v dc what value of base resistor can i use in order to drive a 24v 400ma relay.
if my input is 5v then i can connect directely but for 24v what value of series resistor can i use. is 10k ok.
 

std_match

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10k is a good value.

How to calculate the approximate value that gives 5V at the ULN2803 input:
Input current at 5V = (5V - 1.4V) / 2700 Ohms = 1.3 mA (ignoring the 100 uA in the internal base-emitter resistor)
Wanted drop in the external resistor = 24V - 5V = 19V = R * 1.3 mA
R = 19V / 1.3 mA = 14.6k

10k will give a little more than 5V at the input, but they are specified to survive up to 30V.
I would use 10k.
 

betwixt

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The input current is specified at 1.35mA with 3.85V applied.
That means to drop the additional (24 - 3.85) = 20.15V at the same current, Ohms Law says R=V/I = 20.15/0.00135 = 14.925K.

So 12K to 15K would be a better choice but 10K would still be safe.

Brian.
 

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