lednpnswitches
I'm very, very new to EE, so this might be a really basic question. I am trying to make a very basic LED-switch using a bipolar NPN-transistor. The circuit I use can be seen below.
simulate this circuit – Schematic created using CircuitLab
Now, when I prepare this circuit on the breadboard, the LED will dimly emit some light even when the switch is open. When the switch is closed, the LED turns fully on, as desired.
Does anyone know why the LED is on, even though the switch is open?
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:
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:
After you understand that, you might read Why would one drive LEDs with a common emitter?
Your second circuit is perfect, provided that the pullup on the input is enabled.
When the switch is open, the anode of D1 is at VDD, and the cathode of D1 is at 13V-Vf. The diode prevents the higher voltage from reaching the input pin and destroying the MCU.
When the switch is closed, the cathode of D1 is at 0V, pulling the anode down to 0.7V and registering a low at the input.
You don't need the resistor since the pullup will not source enough current to damage the diode, but you must have the diode in order to prevent the higher voltage from reaching the MCU.
Best Answer
The QED 123 is a infrared LED. Light is not visible. The 1k is to high. Calculate the correct value. The QED 123 = 1.7V 100mA says the datasheet. So 1.3V (3V-1.7V) over the resistor (when the transistor is fully open). The (If) 100mA 1.3/0.1= 13 ohm. 15ohm is ok. The switch circuit should work. But best to connect the base to ground with a resistor if the transitor opens slightly a wierd way when the switch is open.