Expert: Dave Nyce - 10/2/2002

Question
Dear Dave,

My apologies for asking a time consuming open ended question the other week.  I got carried away, but won't let it happen again.  I'll keep my questions concise from now on.  The information you've given has helped me tremendously.  I appreciate you are doing this on a volunteer basis and your time is precious.

If I may, I have a simple question on how hard to drive a transistor given that the load is 100mA .  This amounts to choosing the value of R in the circuit.   Studying your earlier answers I have some idea, but am not sure.  The question (short) is at

http://www.geocities.com/electronics_help/value_of_R_for_PNP.htm


Thanks again and apologies for any inconvenience caused earlier.  It won't happen again.

Regards,
Usuff


Answer
The gain of a transistor is usually specified to range over a 3 to 1 range at room temperature, and it also changes with temperature. The gain is specified for operation in the linear range, but you want a low Vce drop in a switching application. For these reasons, switching circuits are usually designed to overdrive the transistor to make sure that the load is always adequately driven.

So, in your circuit, place a 35 ohm resistor in series with the LED to limit the current to 100ma when the Vce is about 0.2 volts. 5.4 volts -0.2 Vce -1.7 VLED = 3.5 volts. 3.5volts/0.1A = 35 ohms.

I think the 3906 gain is rated at 50 to 150. We'll assume 25 for a switch application. 0.1A/25 = 4 mA. 5V/.004 = 1250 ohms.

So, put a 35 ohm resistor in series with the LED, and the base resistor is 1250 ohms. If you are using 2 LEDs at 50 mA each, place a 70 ohm resistor in series with each. (or 140 ohms each if using 4 LEDs at 25 mA each).

Note: If the transistor gets hot, that means it is not being turned on hard enough and has a larger Vce drop. I don't remember how much current your CMOS gate can pull down.

Hope this helps!

Dave