Article : Andy Collinson
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Overview
A bipolar junction transistor (BJT) can be used in many circuit configurations; as an amplifier, oscillator, filter rectifier or just used as a switch. If the transistor is biased into the linear region, it will be used as an amplifier or other linear circuit, if biased into saturation or cut-off, then it will be used as a switch, controlling other circuits. This article describes the BJT when used as a switch.

BJT Characteristic Curves
For a transistor there are input, output and transfer characteristic curves. The output characteristic curve is especially useful as it shows the variations in collector current, Ic for a given base current, Ib over a range of collector-emitter voltage, Vce. A sample plot for a 2N3904 is shown below. The centre, cyan curve shows that for a base current of 42uA the collector current is about 8.5mA. This gives a static dc current gain, (also called beta or HFE ) of 8.5/0.042 = 202.

BJT Output Curves

When used as an amplifier, the biasing is arranged so that the transistor operates in the linear region ( shown above as almost horizontal sections). The linear region is for Vce > 0.5V and extends to the full supply voltage. An amplifier will usually be biased to about half the supply voltage to allow for maximum output swing.

For use as a switch, the BJT operates in the regions of the output curves called saturation and cut-off. See the diagram above.

The yellow shaded area represents the "cut-off" region. In cut-off the BJT is fully off and its operating conditions are zero base current, zero collector current, and collector-emitter voltage Vce will be high.

In "saturation" as depicted by the red shaded area, the transistor is fully on, and conditions are maximum base current, maximum collector current, and minimum collector-emitter voltage Vce. In both cut-off and saturation, minimum power is dissipated in the transistor.



Current Gain in Saturation
In the linear region, the dc current gain, HFE is fairly constant over a wide range of collector-emitter voltages. However, in saturation, an important change takes place. A zoomed view of the output curves from 0 to 0.4 Volts Vce is shown below:

BJT in Saturation

The change in gradient means that a change HFE has taken place. The purple trace shows a base current, Ib of 62uA. At Vce = 0.1 Volts the collector current has now fallen sharply to 2.5mA. The dc beta is now HFE = 2.5/0.062 = 40. Current gain in saturation will also vary with the amount of collector-emitter voltage (i.e. the amount of saturation) and transistor type. Because of this the bias circuit should be designed to work with the minimum value of HFE for any given transistor, and a rule of thumb is to assume of current gain of only 20x.

For higher current loads, it is important that the transistor remains in the saturation region. The graph below, shows collector current on a logarithmic scale, plotted against collector-emitter voltage.

Ic versus Vce in Saturation

For small base currents below 1mA the current gain is 20. However at 10mA base current the current gain drops to just 5. One way to design for a power transistor is allow 5x more base current than you actually need, to make sure the device remains saturated.


BJT Switch Example
Suppose a BJT with a 5V supply is designed to switch a 5V, 20mA lamp on and off. The transistor data sheet shows 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 batch.

As the transistor will be used in saturation then assume that current gain will be 20. As collector current is 20mA, the required base current is therefore:

ib = Ic / hFE(min) = 20/20= 1mA

The value of Rb can now be found. Assume the switching voltage, Vin is 5V and the base emitter voltage of the transistor, Vbe is 0.6V. Large power transistors can often have base emitter voltages as high as 0.8 or 0.9V in saturation. Small signal transistors the value is somewhere between 0.6 and 0.7 Volts. For this example, assuming a Vbe of 0.7V and an input signal of 5V, then 4.3V is developed across Rb. As a base current of 1mA is required, then :

Rb = 4.3 / 1m = 4.3k (use nearest preferred value)

All transistors in the same batch should easily work and light the lamp. The collector emitter voltage of the transistor will be very low (around 0.1 V) and power dissipated in the transistor is also low Ic * Vce = 2mW, and almost full power is developed in the load.

Points to Note:
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 much as 0.8V or 0.9V on some power transistors.

In saturation, a heat-sink 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 practice allow for the worst combination of currents and voltages and calculate accordingly.


Summary
It is difficult to predict when a transistor comes out of saturation. The load current must always be known and assume a minimum value for hfe in saturation. Calculate bias currents using the minimum current gain of 20 or 10 for power transistors. 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. Vce should be a minimum value of 0.2 Volts or less in the saturated region.