As you can see, I am trying to switch on/off a LED diode using NPN transistor. During the normal operation, 1V8 pin output is controlled by MCU - this is done automatically.
As the LED should be used for debugs only, I do not want the LED circuit to run all the time, that is why there is a jumper in collector, which should be used to completly disconnect the load from the circutit.
What I want to know, if such circuit is well designed, or the jumper should be placed somewhere else? For example to base or emitter? When the jumper is off, I just want the lowest current possible to be flowing in the circuit, as it is a battery based application.
Yes, that is what I am aware off. Both resistors in the transistor are 10k.
Is there any better idea how to design the circuit? For example putting the jumper to base or emitter? What will be the current flowing via the circuit then?
Since there is already a 2.2k resistor in the collector circuit you may even move the jumper across the LED - connecting the jumper shorts out the LED and vice versa (opposite to what you have now). The only other difference is that whether or not there is a jumper, collector current will still flow when the transistor is turned on, so for actual use (like when connected to the next stage) you can have the jumper in place and you can quickly check if all is well by disconnecting the jumper.
In that case, it makes more sense to do it like this:
Anyway, is not there some little current flowing from collector to ground, even when the base is off? Some nA or uA? I am asking just for confirmation.
This solution is much better than the first one, because in the first one, there would be some mA current flowing even with the jumper in collector off.
Yes, maybe 1 to 100 nA. The manufacturer's data sheet, if complete, will state the open base leakage current as a function of collector-emitter voltage and temperature. But then, fingerprints on the PCB are conductive, too.
Right, but you can move the jumper between emitter and Gnd.
The resistor network is not properly designed. R2 should go to the input, not to the base. As is it forms a voltage divider that may prevent proper operation on 1.8V signal level.
