# The Bipolar Transistor as a Switch

When a transistor is used as a switch it is either biased to be non­conducting or OFF, or it is biased to conduct the maximum possible collector current or be ON. If the base current of a transistor with a collector resistor rl is gradually increased from zero the collector current will increase as well, since IC = hFEIB. The collector-emitter voltage VCE will fall since, from Fig. 20, VCE = Vcc - IcRL. Eventually, the point will be reached at which VCE(SAT) has fallen to its minimum possible value. This minimum voltage is known as the collector saturation voltage VCE(SAT), The transistor is then said to be saturated; the collector current at saturation is labelled as Ic(SAT) and the base-emitter voltage that just produces saturation is labelled as Vbe(saT) Typically, VCE(SAT) is 0,7 V and VCE(SAT) is 0.2 V.

voltages in a simple bipolar transistor circuit

Typical variations of VCE(SAT) and VBE(SAT) with change in collector current are shown by Figs. 21 (a) and (b) respectively.

Fig. 22 shows a typical set of output characteristics for a transistor. A d.c. load line has been drawn on the characteristics between the points Vce = Vcc = 6 V, Ic = 0 and Ic = VCC/RL = 6/1000 = 6 mA, VCE = 0. When a transistor is used as an amplifying device, its operation is restricted to the linear part of its characteristics in order to minimize distortion of the applied signal. When used as a switch, a transistor is rapidly switched between two states. When the base current is zero the transistor is held in its OFF condition. When the transistor is OFF it only conducts the very small collector leakage current. The voltage then dropped across the collector resistor is negligibly small and so the small voltage across the transistor in the OFF state is equal to the collector supply voltage Vcc, When the transistor is driven into saturation it is then in its ON state. The voltage across the transistor is now its collector saturation voltage VCE(SAT).

The transistor can be switched rapidly between its ON and OFF states by the application of a rectangular voltage waveform to its base-emitter terminal (Fig. 23). When the input waveform is at zero potential with respect to earth, the transistor will be switched into its OFF state; the voltage which appears at the output terminals is then equal to the collector supply voltage since there will be zero voltage drop across R^. When the transistor is switched into its ON state by the rectangular base signal voltage, the transistor will conduct heavily and a large voltage, approximately equal to Vcc, is dropped across RL. The output voltage of the circuit is then equal to the saturation voltage Vce(sat) of the transistor. The voltage at the output terminals the circuit switches rapidly between Vce(sat) and VCC volts. It should be noted that when the input signal voltage is positive the output voltage is VCE(SAT), and when the input signal voltage is zero the output voltage is Vcc volts (See Fig. 24). This means that the input waveform has been inverted; and the circuit has performed the logical function NOT.

The more positive (less negative) voltage level can be regarded as representing logical 1 and the less positive level as giving logical 0. This convention is known as positive logic. Conversely, logical 1 can be taken as being the less positive, or more negative, voltage level with the more positive voltage being labelled as logical 0. This second convention is known as negative logic. Clearly, the convention employed in a particular case must be clearly stated, or understood. Positive logic is the more commonly employed and is employed in the rest of this book.

### Example 9

The transistor used in the circuit of Fig.23 has Vce(SAT) = 0.15 V, RL = 1200 Ω and Vcc = 5 V. Calculate (a) the saturated collector current IC(SAT) and (b) the power dissipated in RL when the transistor is (i) ON and (ii) OFF.

### Solution

(a) Ic(SAT) = (5 - 0.15)/l200 = 4.042 mA (Ans.)

(b) (i) P = (4.042 X 10-3)2 X 1200 = 19.6 mW (Ans.)

(ii) Ic= 0 so P = 0 (Ana.)

### Example 10

The circuit of Fig. 22 has Vcc = 12 V, RL = 1 kΩ, VCE(SAT) = 0.2 V , Vbe(sat) = 0.7 V and RB = 30 kΩ. If a 5 V voltage is applied to the input terminals determine the minimum value of hFE which will produce saturation.

### Solution

IB = (5 - 0.7)/(30 X 103) = 143 ΩA

IC(SAT) = (12 - 0.2)/1000 = 11.8mA

Therefore

hFE(MIN) = (11.8 X 10-3)/(143 X 10-6) = 82.5 (Ans.)

### Speed of Switching

Rapid switching between the ON and the OFF states of a transistor is desirable in order to minimize the power dissipation within the device .the power dissipated when the transistor is OFF is POFF = VCCICEO ≈ 0, and when it is ON the dissipation is PON = VCE(SAT)I C(SAT). is only a fraction of a volt the power dissipation in either state is very small. Most of the power dissipation occurs while the transistor is actually changing state and so the time taken to achieve this should be as small as possible .

The switching speed of a transistor is the time that elapses between the application of a voltage to the base-emitter terminals and any resulting change in the output voltage. The concept is illustrated by Fig. 25. The base-emitter .voltage pulse. Fig. 25(a), does not cause die collector-emitter voltage,. Fig. 25(b), to change is state instantaneously. Instead, it remains at the supply voltage Vcc volts for a short time and has only fallen to 0.9 Vcc after a time period td .This is the time needed to fully charge die capacitance of the base-emitter p-n junction. After , td seconds the collector-emitter voltage VCC falls at rate that limited by the need to charge the collector-base capacitance , but eventually it reaches its saturation value VC(SAT) . the fall-time tf is the time taken for VCE to fall from 0.9 VCC to 0.1 VCC . the turn-on time tON is the sum of the delay time and the fall-time, i.e. tON = td + tf

Once the collector-emitter voltage has fallen to its saturation value any further increase in the base current will not give a corresponding increase in the collector current. The excess chargers stored in the base region of the transistor. The base storage effect ensures that when the base-emitter voltage is reduced to zero the transistor does not cease conduction instantaneously. there is an initial storage delay ts during which the excess base charge is removed from the transistor. At the end of this period the collector current falls, and, so the collector-emitter voltage rises towards Vcc volts. The rate at which the collector-emitter voltage is able to increase is limited by the need for the junction and load capacitances to be discharged . The risetime tr is the time taken for the collector-emitter voltage to increase from 0.1 Vcc to 09 VCC volts.

The turn-off time tOff is equal to the sum of the storage time and the risetime , i.e. toff = ts + tr

Typical figures for the turn-on and the turn-off times of a transistor are tOFF = 360 ns and ton = 55 ns for a general-purpose transistor and toff = 125 ns and ton = 27 ns for a switching transistor.

### Switching Circuits

A simple switching circuit can be designed using a single transistor with the load in the collector circuit and a current-limiting resistor in the base circuit. A basic lamp-control circuit is shown in Figs .26(a) and (b).

When the base current is" zero there will be zero collector current and the lamp will not light. When a voltage pulse is applied, Fig, .26(a), or the switch is closed. Fig. 26(b), a base current, and hence a collector current, flows and the lamp lights. The transistor that is employed should have adequate current gain, a low collector saturation voltage, and, if high-speed operation is desired, tow values of turn-on and turn-off times. The value of the base resistor RB should be chosen to ensure that the transistor saturates. The base current required will need to be some what greater than IC(SAT) because the value of hFE will fall as the collector-emitter voltage approaches its saturation value. The minimum value of hfE obtained from the data sheet should be used.

Applying Kirchoff’s law to the base circuit of Fig. 25(a) gives V = IBRB + VBE,

Or

Rb = (V-VBE)/IB (3.21)

If 1B is to be the value which just produces the saturation collector current Ic(sat) then VBE should be the saturation value VBE(SAT) obtained from the data sheet.

### Example 11

In the circuit given in Fig. 26(a) the collector supply voltage Vcc is 5 V and the resistance of the lamp when lit is 50 Ω. The transistor parameters are hFE = 100, VBE(SAT) = 0.7 V and VBE(SAT) = 0.2 V. Calculate the necessary value of the base resistor if the applied voltage is 5 V.

### Solution

The saturated collector current is

Ic(SAT) = (5 - 0.2)/50 = 96 mA.

The minimum base current for saturation is

IC(SAT)/hFE = (96 x 10-3)/100 = 960 μA

From equation (3.20)

RB = (5 - 0.7)/ 960 X 10-6) = 4479 Ω

A lower value should be selected to ensure the transistor Saturates. The nearest preferred value is 4300 μ (Ans.)

These circuits power the load whenever the transistor has been turned ON by a HIGH input. The circuit shown in Fig. 27 passes current through the load when the input is LOW; then T1 is OFF and the HIGH voltage at its collector turns T2 ON.