[lacrossanator1]

Where to start. Transistor that covers a lot of area. I will hit some high points.

I assume that you know what this means and can use it: Volts = Amp times Ohm. What a diode is and can use it.

Let start with the old PNP and NPN transistor and that is as far as I am going in the area of transistor.

1: PNP = Arrow Points N (in). That is our Diode. I will not explain the PNP but it is the same but the opposite.
NPN = Arrow Not Pointing N (in).

2: There are 3 leads. Named B= Base, C=Collector, and E= Emitter.

3: Think of the Arrow in the transistor symbol as an Diode. Put an line at the point of the arrow so it looks like a diode.

4: When a Diode is forward bias (Turn On) it will read a voltage of 0.2 to 1 Volt (V), most will say 0.7 and that is what I will use and also for the diode in the transistor.

5: I will talk about Voltage when talking about the Base Emitter junction, and Current (I) when talking Collector to Emitter.
We get a voltage drop from Base to Emitter and we have current flowing in Emitter to Collector.
OK let see if we can make the NPN transistor work.

1: Circuit We will input (In) to a 100K resistor to B, Tie E to Ground (G) and C to a 1K to +5V.
Draw it out and label.
0 to 0.6 V at In: Measure; 0 to 0.6V at B, 0V at E (it will always be 0V in this example, last time for it), C at 5V. ) 0I C to E.
The base to emitter is said to be off and the transistor is said to be off.
0.7V at In: With E tie to ground the transistor is switching and It is anybody guess what the V and I are.
0.8 to 5V at In: Measure; 0.7V at B, C at 0.2V ( less then 1V). The base to emitter is said to be on and the transistor is said to be on.
Remember we said 0.7V is the B to E voltage, if we apply 1V directly to the B it will blow up, just like a diode will. There is just enough Current flowing in the 100k resistor to drop the voltage to the B so it does not blow up. Remember Volts = Amp times Ohm. I use it a lot with transistor. Also the current is limited by the 5K resistor so transistor does not burn up.
Now if you place a LED and a resistor sized to the LED current in the Collector side circuit you could turn if on and off.



2: Circuit We will input (In) to a 100K to B, Tie E to a 100ohm resistor to G and C to a 1K to +5V.
Draw it out and label. This Circuit is said to have a gain of 10, 1K /100 Ohm=10
0 to 0.6 V at In: Measure; 0 to 0.6V at B, 0V at E , C at 5V. The base to emitter is off and the transistor is off. No current in E to C
Now this is the part you can use.
Remember we said this circuit has a gain of 10 so the useful input voltage is a 0 to 0.5voltage change since the supply voltage is 5V. 0.5V time 10 = 5V. Called hitting a voltage rail. Or from 0.7 to 1.3 input voltage.
So 0.8 In Volt at the B give us .1 volt at E (0.8-0.7) across the 100 ohm resistor and that is how much current? 0.001. That current is the current that is also flowing in the EC part of the transistor and it will drop how much volt across the 1K resistor, now minus that from the 5 V supply and you get 4V on the C. See you have just got a gain of 10. 0.1 volt change in, 1V change out.
Now try it for 0.9 in. You should get 3 volt out. If I have explain it correctly.
Remember at the start I said the useful range was 0.7 to 1.3 let see if that is true. We saw that .7 would give use 0 v across the 100 ohm so we would have 5 V at the collector now lets try 1.3v in we get 0.5V across the 100 ohm that would give us 0.005 Amp. Now time it against 1K = 5V. Now that from supply 5V =0V but wait a minutes, I can not have 0.5V on the 100 and 5V on the 1K that would be 5.5V and that is what is meant running into the rail. So the max input useable voltage is between 1.2 and 1.3.
If you run your current thru a resistor to give you a voltage and now with the 2 circuits, you should be able to work something out.

Hope this helps.

Thanks a lot for this help; I am starting to gather an understanding as to how I can use these! Where you say in input of for example 1v to the base, what would be the current draw of the transistors base so I can use ohms law to find what resistor I must put in series with the base in order for the led to be switched on? Using the equation r=V/I

Also, what resistor would be suitable for the application I talk of, I was thinking of ordering some bc548 to play with but I wasn't sure if they would be appropriate?
Thanks again, Thomas