AboutDave Nyce Expertise I have been an electronics engineer for 25 years. I can answer questions on analog and digital circuits and my specialty is sensors.
Experience I am the inventor on 23 US patents, and also some foreign ones. Developed sensors for over 25 years. Licensed private pilot (airplane and rotorcraft), have HAM radio license. I'm not an expert in computer networking.
Expert: Dave Nyce Date: 10/2/2002 Subject: How Hard to Drive PNP?
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
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.
