I was making a circuit with 1 LED and 1 transistor. pretty primarily i was trying to use the transistor as a normal switch. Here is the diagram below.
So, this is an NPN transistor and i connected the -ve part of the D1 (transistor) with the -ve part of the LED and the other -ve leg of D1 to ground. I connected the +ve part of D1 with the positive end of 9V dc power supply. The LED should light up and if I disconnect the wire connecting the +ve part of D1 , the LED should turn off. But the problem that I am facing is that the LED isn't lighting up when I connect all the points and follow the schematic.
Also another problem that I am facing is that when I connect a wire with the +ve leg of D1 and don't connect the other end of the wire with the battery and just hold the wire by my fingers, the LED turns on but emits very little light. Is there any mistake in my schematic?
Best Answer
The base-emitter junction of a bipolar junction transistor is a diode:
simulate this circuit – Schematic created using CircuitLab
You have another diode in your circuit, the LED. And you obviously understand that you need a current-limiting resistor if you want to connect this diode to a 9V battery, because a diode has an approximately constant voltage drop that is less than your battery voltage.
The problem is the same with the transistor in your circuit. You have the base connected to the positive side of your battery, and the emitter connected to the negative side. You've done this, essentially:
simulate this circuit
What will happen here? A whole lot of current will flow until the battery can supply no more, or something melts. If you are using a 9V battery, it can't supply a whole lot of current, and I bet if you measure the battery voltage in your circuit, it's about 0.65V, unless the battery is dead by now.
So what you need is a current-limiting resistor on the base of the transistor, like this:
simulate this circuit
After you understand that, you might read Why would one drive LEDs with a common emitter?