Basic Transistor Drivers

Discussion in 'General Electronics Chat' started by Wendy, Jul 8, 2012.

  1. Wendy

    Thread Starter Moderator

    Mar 24, 2008
    Transistors are used in digital modes frequently to turn loads on or off. They are not complex, but there is a small list of math simplifications used to make using transistors easier. Much of this material is covered in Chapter 10 of my LEDs, 555s, Flashers, and Light Chasers article, but this will be a bit more in depth. I'll uses LEDs and resistors as the loads because they provide a quick indicator what is happening.

    Common Emitter

    This is the most basic of transistor configurations for basic on/off. To work correctly, you must reach saturation. This is the state where the transistor can not turn on any harder, it is an conductive as it can get. However, when this condition is met some of the base current is "stolen", so using the gain of a transistor does not work well. So a simple rule of thumb was developed to KISS it (Keep It Simple Stupid).

    Ib = Ic / 10

    Figure 1.0 below shows how this might work.

    ..................................................Figure 1.0
    .....................................The Common Emitter BJT driver

    It does not matter what the load is, just what the currents are. In the case of a really large currents the B-E currents can also be large, which can be a problem. However, the voltage drop between the C-E is typically very low, less than 0.1 volts, so there is usually not much heat associated with this kind of switch.

    So lets try some real world values to these schematics. The LEDs will use 20ma each and wil drop 2.0 V (Vf). The power supply on both circuits is 12VDC (Vcc).

    The first circuit calculates R1 to be 500, which is close to a 470. The base current should be 2ma, since the collector current is 20ma. The base resistor calculates out to 5650, which is close to a standard 5.6K resistor.

    On the second circuit you have 100ma (20ma X 5) going through the collector. This means you want 10ma through the base, which calculates out to 113 (which rounds to 100 or 120). Incidentally R1 - R5 calculates to 300, which is already a standard resistor.

    If for whatever reason you need a smaller current to turn on the arrangement you can always go to a Darlington or S pair. The 1/10 rule still applies, but with each transistor contributing some gain you can change it to 1/100 Base current to Collector current (Gain 10 * Gain 10). Figure 1.1 shows how this might work.
    Last edited: Apr 8, 2014