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Step of 5 12.001E Refer to Figure1 11.2 in the The two transistors have same characteristic, the collector-emitter saturation voltages are, = = 0.2V The positive limit of the linear region is determined by the saturation of thus the maximum output is, Substitute 15 V for and 0.2V for =15-0.2 =14.8V In the negative direction, depending on the values of I and the limit of the linear region is determined by turning off. The minimum output is, = Substitute 15 V for and 0.2V for -15+0.2 =-14.8V Thus, the output voltage swing is from Step of 5 The absolutely lowest (most negative) output voltage is that given by minimum output equation and is achieved provided the bias current I is greater than the magnitude of the corresponding load current, I I Substitute 15 V for for and 0.2V for I 1000 14.3 = 1000 14.3 mA Step of 5 Calculate the value of base voltage of the transistor, Substitute 15 V for and for Calculate the value of the resistor, I Substitute -14.3V for and 14.3 mA R |-14.3| 14.8x10⁻³ Thus, the value of the resistance, R is 0.97 Step of 5 Calculate the value of maximum load current, R2 Substitute for Vomax and kΩ for 14.8 1000 =14.8mA Calculate the value of minimum load current, Substitute -14.8V for Vomin and kΩ for -14.8 1000 =-14.8mA Step of 5 Calculate the value of minimum emitter current of the transistor, Substitute -14.8mA for and 14.3 mA for I = 14.8mA Thus, the value of minimum emitter current of transistor, Calculate the value of maximum emitter current of transistor, Substitute 14.8mA for and 14.3 mA for I =14.8mA + 14.8mA Thus, the value of maximum emitter current of transistor, is 29.6mA