Transistor as a Switch
The Transistor as a Switch
– One major application of a transistor is as an amplifier.
– The other major application is switching applications.
– In this case, it is operated alternately in cutoff and saturation.
– Analyze the following graph.
- In (a), the device is in the cutoff region because the base-emitter junction is not forward biased.
– In this condition there is, ideally, an open between collector and emitter.
– In (b), the transistor is in the saturation region because the base-emitter junction and the base-collector junction are forward-biased and the base current is made large enough to reach its saturation point.
– In this condition there is, ideally, a short between collector and emitter.
– Actually, a drop of up to a few tenths of a volt normally occurs.
i) Conditions in cutoff
– A transistor is in cutoff region when the BE junction is NOT forward biased.
– Neglecting leakage current, all currents are zero and VCE = VCC.
ii) Conditions in saturation
– When the BE junction is forward biased and there is enough base current to produce a maximum collector current, transistor is saturated.
IC(sat) = (VCC – VCE(max))/RC
– Minimum value of base current needed to produce saturation is
IB(min) = IC(sat)/bDC
Example:
Consider the following circuit.
The LED requires 30 mA to emit a sufficient level of light. Therefore the collector current should be approximately 30 mA. For the following circuit values, determine the amplitude of the square wave input voltage necessary to make sure that the transistor saturates. Use double the minimum value of base current as a safety margin to ensure saturation. VCC = 9 V, VCE(sat) = 0.3 V, RC = 270 W, RB =3.3 kW, and bDC = 50.
Solution:
When the square wave is at 0 V, the transistor is in cutoff and, since there is no collector current, the LED does not emit light. When the square wave goes to its high level, the transistor saturates. This forward-biases the LED, and the resulting collector current through the LED causes is to emit light.
IC(sat) = (VCC – VCE(sat))/RC = (9 V – 0.3 V)/270W = 32.2 mA
IB(min) = IC(sat)/ bDC = 32.2 mA/50 = 644 mA
To ensure saturation, use twice the value of IB(min), that is, 1.29 mA. Then
IB = VR(B)/RB = (Vin – VBE)/RB = (Vin – 0.7)/3.3kW
Solving for the voltage amplitude of the square wave input, Vin, we get:
Vin – 0.7 = 2 IB(min)RB = (1.29 mA)(3.3 k kW)
Vin = (1.29 mA)(3.3 k kW) + 0.7 V = 4.96 V
