Components Needed Part 1: The BJT Characteristie Curve Resistors: one 100 resist

Question

Components Needed Part 1: The BJT Characteristie Curve Resistors: one 100 resistor, one 3342 resistor One 2N3904 npn transistor (or equivalent) One rectifier diode (IN4001 or equivalent) One 12.6 V ac center-tapped transformer with fused line cord Part 2: BJT Switching Circuits Resistors: one 330 , one 1.0 ka. two 10 k One 10 k potentiometer Two small signal npn transistors (2N3904 or equivalent) One LED Part 1: The BJT Characteristic Curve 1. Measure and record the resistance of the resistors listed in Table 4-1. 2N3904 Table 4-1 Resistor Listed Measured Value Value 33 k Ri 2.L Enitser R, 1100 Collecse 0. 2N3904 Figure 4-1 Connect the common-emitter configuration illustrated in Figure 4-1. Start with both power supplies set to 0 V. The purpose of Ri is to limit base current to a safe level and to allow indirect determination of the base current. Slowly increase Vaa 2. until Yei is 1.6s V. This sets up a base current of s0 HA, which can be shown by applying Ohm's law to R 3. Without disturbing the setting of as, slowly increase Voc until +2.0 vis measured between the transistor's collector and emitter. This voltage is VCE Mesure and recornd I' for this setting, Record Fe in Table 4+2 in the column e8ov labeled 32

Explanation / Answer

1) The experimental data indicates Bdc is not exactly constant at all points on Ic vs Vce characteristic graph.

Yes. this Bdc chages the state of transistor from cutoff state to saturation state and followed by linear active region.

2) Higher value of Bdc makes the transistor driven in to saturation mode . Ic < ( Bdc * Ib)

3)The maximum power is dissipated at Ib = 150 micro amps, Vce = 8v and Ic = 0.0278 amps is Vce * Ic = 0.2224 W

  

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