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# Assume transistor parameters of β 80 v be on 0 7 v

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= 12 mA to produce the specified output light. Assume transistor parameters of β = 80, V BE ( on ) = 0 . 7 V, and V CE ( sat ) = 0 . 2 V, and assume the diode cut-in voltage is V γ = 1 . 5 V. [Note: LEDs are fabricated with compound semiconductor materials and have a larger cut-in voltage compared to silicon diodes.] Specifications (Figure 2.20(b)): The inverter circuit in Figure 2.20(b) uses a pnp transistor. In this case, one side of the load (for example a motor) can be connected to ground potential. The required load current is I C 2 = 5 A. Assume transistor parameters of β = 40, V EB ( on ) = 0 . 7 V, and V EC ( sat ) = 0 . 2 V. Solution (Figure 2.20(a)): For v I 1 = 0, transistor Q 1 is cut off so that I B 1 = I C 2 = 0 and the LED is also off.

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Solution (Figure 2.20(b)): For v I 2 = 12 V, transistor Q 2 is cut off so that I B 2 = I C 2 = 0 and the voltage across the load is zero. Comment: As with most electronic circuit designs, there are some assumptions that need to be made. The assumption to let I C /I B = ( 1 / 2 in each case ensures that each transistor will be driven into saturation even if variations in circuit parameters occur. At the same time, base currents are limited to reasonable values. We may note that for the circuit in Figure 2.20(a), a base current of only 0.3 mA induces a load current of 12mA. For the circuit in Figure 2.20(b), a base current of only 0.25A induces a load current of 5A. The advantage of transistor switches is that large load currents can be switched with relatively small base currents. EXERCISE PROBLEM (a) Redesign the LED circuit in Figure 2.20(a) such that I C 1 = 15 mA and I C 1 /I B 1 = 50 for v I = 5V. Use the same Q 1 transistor parameters given in Example problem given above. (b) Redesign the circuit in Figure 2.20(b) such that I C 2 = 2A and I C 2 /I B 2 = 25 for v I = 0. Use the same Q 2 transistor parameters as given in Example. (Ans. (a) R 1 = 220 , R B 1 = 14 . 3 k ; (b) R B 2 = 141 ) When a transistor is biased in saturation, the relationship between the collector and base currents is no longer linear. Consequently, this mode of operation cannot be used for linear amplifiers. On the other hand, switching a transistor between cutoff and saturation produces the greatest change in output voltage, which is especially useful in digital logic circuits, as we will see in the next section.
Digital Logic Figure 2.21 A bipolar (a) inverter circuit and (b) NOR logic gate Consider the simple transistor inverter circuit shown in Figure 2.21(a). If the input V I is approximately zero volts, the transistor is cut off and the output V O is high and equal to V CC . If, on the other hand, the input is high and equal to V CC , the transistor can be driven into saturation, in which case the output is low and equal to V CE ( sat ) . Now consider the case when a second transistor is connected in parallel, as shown in Figure 2.21(b). When the two inputs are zero, both transistors Q 1 and Q 2 are in cutoff, and V O =5V. When V 1 =5V and V 2 =0, transistor Q 1 can be driven into saturation, and Q 2 remains in cutoff. With Q 1 in saturation, the output voltage is V O = V CE (sat) 0.2 V. If we reverse the input voltages so that V 1 =0 and V 2 = 5V, then Q 1 is in cutoff, Q 2 can be driven into saturation, and V O = V CE (sat) 0.2V. If both inputs are high, meaning V 1 = V 2 = 5V,then both transistors can be driven into saturation, and V O = V CE (sat) 0.2 V.

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