bjt-3 - Copy - Solution If we assume an empirical constant...

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Solution: If we assume an empirical constant of n = 3, we have Comment: The breakdown voltage of the open-base configuration is substantially less than that of the C B junction. This represents a worst-case condition, which must be considered in any circuit design. Design Pointer: The designer must be aware of the breakdown voltage of the specific transistors used in a circuit, since this will be a limiting factor in the size of the dc bias voltages that can be used. EXERCISE PROBLEMS 2.2: The open-emitter breakdown voltage is BVCBO = 200 V, the current gain is β = 120, and the empirical constant is n = 3. Determine BVCEO . (Ans. 40.5 V). Breakdown may also occur in the B E junction if a reverse-bias voltage is applied to that junction. The junction breakdown voltage decreases as the doping concentrations increase. Since the emitter doping concentration is usually substantially larger than the doping concentration in the collector, the B E junction breakdown voltage is normally much smaller than that of the B C junction. Typical B E junction breakdown voltage values are in the range of 6 to 8 V. 2.3: A particular transistor circuit requires a minimum open-base breakdown voltage of BV CEO = 30 V. If β = 100 and n = 3, determine the minimum required value of BV CBO . (Ans. 139 V).
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2.1.7. Basic Transistor Applications Transistors can be used to: switch currents, voltages, and power; perform digital logic functions; and amplify time-varying signals. In this section, we consider the switching properties of the bipolar transistor, analyze a simple transistor digital logic circuit, and then show how the bipolar transistor is used to amplify time-varying signals. Switch Figure 2.19 An npn bipolar inverter circuit used as a switch If we let v I = V CC and if the ratio of R B to R C , where R C is the effective resistance of the load, is less than β , then the transistor is usually driven into saturation, which means that ---- (2.23) ---- (2.24) and ---- (2.25) In this case, a collector current is induced that would turn on the motor or the LED, depending on the type of load. EXAMPLE Calculate the appropriate resistance values and transistor power dissipation for the two inverter switching configurations shown in Figure 2.20 below. Figure 2.19 shows a bipolar circuit called an inverter, in which the transistor in the circuit is switched between cutoff and saturation. The load, for example, could be a motor, a light- emitting diode or some other electrical device. If v I < V BE (on), then i B = i C = 0 and the transistor is cut off. Since i C = 0, the voltage drop across the load is zero, so the output voltage is v O = V CC . Also, since the currents in the transistor are zero, the power dissipation in the transistor is zero. If the load were a motor, the motor would be off with zero current. Likewise, if the load were a light-emitting diode, the light output would be zero with zero current.
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Specifications (Figure 2.20(a)): The transistor in the inverter circuit in Figure 2.20(a) is used to turn the light-emitting diode (LED) on and off. The required LED current is I C 1 = 12 mA to produce the specified output light. Assume
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