The Transistor as a Switch

When a bipolar junction transistor (BJT) is used in any circuit, its function is determined by the devices characteristic curves.

For a BJT there are input, output and transfer characteristic curves, the most useful being the output characteristic curve. The output curves dictate the range of collector-emitter voltage Vce for variations in collector current, Ic. For use as an amplifier, biasing is arranged so that the linear part of the output curves (the almost horizontal sections) are used. If the circuit uses the transistor as a switch, then biasing is arranged to operate on regions of the output curves known as saturation and cut-off. See the diagram below:


The yellow shaded area represents the "cut-off" region. Here the operating conditions of the transistor are zero input base current, zero output collector current and maximum (supply rail) collector voltage. In "saturation", as depicted by the red shaded area, the BJT will be biased so that the maximum amount of base current is applied, resulting in maximum collector current flow and minimum collector emitter voltage. In both cut-off and saturation, minimum power is dissipated in the transistor.

Load Line
Knowing the circuit load current and operating voltage, the load line can be constructed. Imagine a transistor designed to switch a 20mA load, the supply voltage is 5 volts DC. When the transistor is off, the Ic will be zero and Vce will rise to supply voltage (5V). This is point A on the diagram above. When the transistor is on, Ic will be 20mA and Vce will be small (close to zero). This is point B on the diagram. This line is called the load line and shows that the transistor can be operated anywhere on this line (by choice of appropriate bias current). However, for use as a switch the device must work in the saturation and cutoff regions of the output curve. The bias circuit should be designed to work with the minimum value of hfe for the given transistor.

BJT Switch Calculations
Suppose a BJT with a 5V supply is designed to switch a 5V 20mA lamp on and off. The transistor chosen is from a batch with variations in hfe from 100-500. The switching configuration is for common emitter, the bias circuit is shown below. Find a value for Rb to work with any transistor in the same gain group.



As the transistor chosen may have any hfe between 100 and 500 then the minimum current gain is chosen (100). The collector current is 20mA, the required base current is therefore:

hfe = Ic / Ib

ib = Ic / hfe(min) = 20/100= 0.2mA

The value of Rb can now be found. As the switching input, Vin is 5V and the base emitter voltage of the transistor, Vbe is 0.6V then 4.4V is developed across Rb. As a base current of 0.2mA is required, then :

Rb = 4.4 / 0.2 = 22K
A transistor with a gain higher equal to or higher than 100 will easily work and light the lamp. The collector emitter voltage of the transistor will be very low (around 0.1 V) the power dissipated in the transistor is also low Ic * Vce = 2mW, and almost full power is developed in the load.

To Summarise:
You will always know the load current your circuit needs. Therefore use the minimum value of hfe for the transistor in your patch (found from the manufacturers data sheet) or catalogue. Calculate the bias current to reach this minimum value.
In pactice and for heavy load currents and power transistors you may allow 5 times more base current than actually required. This will ensure any transistor within a given gain group reaching saturation.

Points to Note:
The current gain hfe is lower in some power transistors at very high load currents. Therefore it may be wise to calculate the bias current Ib and allow the actual value to be two or three times higher. The base emitter voltage Vbe which varies between individual devices should be taken as the highest value. This is generally 0.6 or 0.7V with small signal transistors, but can be as high as 0.8V on some power transistors. In saturation, a heatsink is rarely required as little power is developed in the transistor. However in a power supply or other circuit where a transistor may be required to control large variations in current and voltage then significant power may be developed. If the power dissipation of the device is exceeded then it will be destroyed. In practise allow for the worst combination of currents and voltages and calculate accordingly.

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