I have a simple setup with a BC547 C NPN transistor with nothing connected to the base as shown:
I'm using a USB cable plugged into the wall.
Why is it that when I touch the wire connected to the base, the LED brightly lights up?
Electronic – Why does the transistor saturate when I touch the base
transistors
Related Solutions
Use 10kΩ for R3.
I didn't look up a BC547, but if that's a typical small signal transistor as it should be, figure a gain of at least 50. 6V supply minus 700mV B-E drip leave 5.3V accross the resistor. 5.3V / 10kΩ = 530µA. Times the gain of 50, and that can support over 25mA of collector current. That's well more than any reasonable digital input will require, even if it has a pullup connected.
As others have said, you may need a pullup if whatever you are driving doesn't passively make the line go high. However, make sure the pullup goes to same voltage as the power supply of whatever will receive the digital signal. Your supply is 6V, which is probably higher than the receiver's supply. That could cause the receiver to not work or even to damage it if its power voltage is significantly lower.
The base-emitter junction (BE) of a NPN transistor behaves like a diode:
So you basically have this equivalent circuit:
As you can see, your LED is in series with the resistor and the transistor's diode (the BE junction). Some current (depending on the value of your resistor) is passing through this circuit, making the LED light.
One way to avoid this is, as The Photon said, using a MOSFET. Another way is using the NPN transistor in the common-emitter configuration:
From this comment of yours:
Actually, I read the specifications on the transistor and the maximum base input is 5V
I can understand why you think you should drive the transistor's base from 4.5V (< 5V). But you can drive the base with a higher voltage as long as you have a resistor in series. The BE junction, acting as a diode, will have a fixed voltage drop (typically 0.7V) across it. The rest of the voltage will go across the resistor which also limits the current through the BE junction.
Best Answer
I expect the main cause of the symptom is that you run it from a 5V charger plugged into the wall. Try the same thing again, but this time with a power bank, I guess the LED will not light up that bright as it does with the wall charger.
The reason is common mode noise emitted by the charger. "Common mode noise" means that while the "+5V" and "GND" keep having a potential difference of 5V all the time, the voltage between GND and actual earth potential can be oscillating quite strongly. As long as you do not connect anything earth referenced to your circuit, the common mode noise does not affect the operation of your circuit. But if you touch the base of the transistor, you provide a path from "actual earth" to the transistor. Your path to earth is partly leakage current, and partly capacitive coupling to earth.
The GND output of the supply oscillates between being above earth potential and being below earth potential. If you touch the transistor base, the LED lights up only during the time GND is below earth potential. As the switching frequency of these supplies is around 100kHz, you will not see any flickering of the LED, though.