Thomas:
If you are talking about the transistor for backlight control as shown in this schematic:



...It really depends a lot on the amount of current required to drive the backlight. In the schematic above the voltage supply is shown as the +5VDC source. With a supply like this you must first make sure that the source can supply sufficient current for the backlight. Then you need to make sure that the transistor has a rating to be able to conduct this amount of current. Finally the base bias resistors must be able to permit enough base current to flow so as to fully saturate the transistor in the ON state when the required backlight current draw is required.

Lets take an example. Lets say that through some experimentation you determine that it takes 50 mA to light the LCD module backlight to a suitable level. So then make sure the 5 volt supply is capable of sourcing 50 mA in addition to the rest of the 5V needs on the board. If it is capable then select a transistor good for at least 50 mA rating. A common PNP transistor is a 2N3906. This part, in a TO-92 package, is rated at 200 mA so it is a suitable part. (See its data sheet here - ON Semiconductor 2N3906 Data Sheet).

This transisitor shows a minimum current gain of 60 at a collector current of 50 mA. This means that a base current of about 1 mA will be required to saturate the transistor (ie 50mA/60 = ~ 1mA). The series base resistor on the PNP as shown above in the schematic would require a size such that it can let the required base current pass when the microcontroller output goes low. The voltage drop across this resistor is 5V-Vbe-Vol. (Vbe of the 2N3906 and Vol of the microcontroller). Some quick math would show that this voltage drop is approximately 4 volts. So 4V/1mA = 4K ohms. Thus the base resistor should be 4K or less. A good design choice could be to choose a 3.3K resistor as the next lower standard size.

Keep in mind that the schematic also shows a resistor from the collector of the transistor to the Anode of the backlight LED array. Most LCD modules do not provide current limiting from the A to the K terminals of the module. This is so the end user can select the brightness as needed. Thus it is necessary to select this series resistor to a suitable value to limit the current to 50 mA (using our example above). The resistor value is equal to (5V-Vsat-Vled)/50mA where Vsat is from the 2N3904 transistor and Vled is the drop across the display from A to K at 50 mA. If Vsat was 1V and the Vled value was 2.7 volts then the resistor value is (5-1-2.7)/.050 = 26 ohms. Choose the standard value of 27 ohms for this resistor.

Hope that helps.
Michael Karas