Hello all,

This is my first time actually asking a question, but long time lurker. Basically, I don't understand how the attached circuit works. Or more specifically, why it works like it does. I don't understand why it lights up when the circuit is open, but turns off when the circuit is closed. What are the actual calculations that drive the transistor to send enough current through to light the LED when the switch is thrown?

The transistor in the picture is a 2N2222.

Thanks for any help!

 

The caption under the photograph pretty much explains it! I think that what you are really looking for is an explanation of how transistors and semiconductors work. Google is your friend!

 

Odd to find that in a textbook. I can't see any good reason to put the LED below the transistor. It should be above it with the 680R resistor. Sure it will work at 12V, but it wouldn't work properly with low voltages.

 

Thanks for the replys. First off, as a reference, this circuit is supposed to be a simple version of some type of door alarm or window alarm. The book keeps building it more complicated, adding components.

I do believe I understand how transistors work. In this instance, you apply a small amount of current to the Base, and a large current comes out of the emitter.

I guess where I'm really stuck is, why does the current into the base high enough when the switch is open to light the LED but not when it is closed? How can I figure out what the values are?

 

With the switch closed the voltage at the base is around 1V which would normally be enough to fully turn on the transistor.
The base emitter junction acts as a diode so below about 0.7V it blocks all current.
The LED also has a voltage drop (unknown but maybe about 1.6V). This means that the voltage across the base emitter is low enough to turn it off.
With the switch closed the base gets about 1mA through the 10K resistor.

 

Ok, I feel like I'm making some progress here. I've found a similar example of this circuit http://www.wisc-online.com/objects/ViewObject.aspx?ID=SSE3803 and those calculations helped too. A question I have now is what is the VBase if the switch is open? Is it 0 because there is no voltage divider?

 

what is the VBase if the switch is open? Is it 0 because there is no voltage divider?

No.
In your original circuit the base voltage is the LED voltage (1.8V to 3.6V depends on its colour) plus the approx. 0.7V base-emitter voltage of the transistor when the switch is open.

The voltage divider action of the 1k resistor is not needed in the circuit and the switch can directly connect the base of the transistor to the negative power supply voltage.

 

Got it! I went home during lunch and tried out some scenarios with the multimeter and I actually understand whats going on. Thanks guys!