A Practical Application:

to start off with we would need to select a transistor that has a high enough voltage and current rating for the job, so then we would need to know a few details about the load that the transistor would be powering, so lets assume that our load needs 10 volts at 1 amp, we would need to select a transistor with a Vce rating of at least 10v, and a collector current rating of at least 1 amp right? wrong! it is good design practice to pick a transistor rated at least 30% higher than we will ever need, and by doing so the transistor is likely to never fail, so a transistor with a Vce rating of 15v or higher, and a max collector current of 1.5 amps or higher, would do nicely,

ok now lets assume that our transistor has a current gain of 100, so to get started lets start with the collector current, we already know that our load is going to have 10 volts dropped across it, and a current of 1 amp flowing through it, so our collector current is obviously going to be 1 amp,

now lets calculate the load resistance using this formula:

V

_{cc} / I

_{c} = R

_{L}where:

V

_{cc} = collector supply voltage

I

_{c} = collector current

R

_{L} = load resistance

so the voltage dropped across the load divided by the current flowing through the load equals the load resistance, so:

10v V

_{cc} / 1A I

_{c} = 10Ω R

_{L}so R

_{L} resistance is 10 ohms,

now lets calculate the value for or base bias resistor, since our transistor has a current gain of 100 then we already know that our base current should be 10 milliamps, because 1 amp or, 1000 milliamps collector current divided by the current gain of 100 = 10, here is the formula:

I

_{c} / β = I

_{b}where:

I

_{c} = collector current

β = current gain

I

_{b} = base current

now lets assume that our base bias supply is also 10 volts, and remember that it takes approx 0.6 volts on the base to turn the transistor on, so that 0.6 volts will be subtracted from the base bias supply voltage, now we can find the value needed for our base bias resistor by using the formula:

(V

_{bb} - V

_{be}) / I

_{b} = R

_{b} where:

V

_{bb} = base bias supply voltage

V

_{be} = base emitter voltage

I

_{b} = base current

R

_{b} = base bias resistor value

so:

(10v V

_{bb} - 0.6v V

_{be}) / .01A I

_{b} = 940Ω R

_{b}so our base bias resistor R

_{b} comes out at 940 ohms,

ok i think that just about does it for this little project, you could add a switch in series with the base bias supply to turn it on and off if you want, and it doesn't have to be a big switch either, in fact it would only need to be able to conduct 10 milliamps of current, the transistor is taking care of the heavy 1 amp of current that the load is drawing, which was really the purpose of using the transistor in the first place, so that a very small amount of current flowing through the base can be used to control a much larger amount of current flowing through the collector,