Hi,
I use a simple BJT transistor switch to control the flow of current through the emitter. The base of the transistor is connected to a digital output pin that comes from a PC USB I/O board. The collector is connected to a 12V battery.
My problem is that even if there is no input signal (the output pin is low) the transistor is "on" (current flows out of the emitter). When I check the voltage between base-emitter its 0.7 V so this explains why the transistor is "on".
If I connect the base of the transistor to ground it switches "off" as expected
So my questions are:
1. How come the base-emitter voltage is 0.7 V even if I completely remove the connection to the base of the emitter? Is it related to a problem with "floating" input?
2. What can I do to solve this problem?
Obviously the I/O card does not connect its output to ground when output is low since this would turn off the switch instead it just lets the output "float" !?!?