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Table Name: number: Solution
ECE 327: Electronic Devices and Circuits Laboratory I
Lab 1: The Bipolar (Junction) Transistor Quiz A (10 points)
Description. This quiz tests your comprehension of what youve read on bipolar junction transistors (BJTs). Complete this quiz with closed book and closed notes. The following questions refer to Figure Q1-1, where (a) refers to Figure Q1-1(a) and (b) refers to Figure Q1-1(b).
Base
Emitter Collector
Base
Collector Emitter
(a)
(b)
Figure Q1-1: BJT Transistors
Problem Q1-1: Transistor Types
Fill in the blanks with either (a) or (b). (1 point) An npn transistor is depicted in A pnp transistor is depicted in (b) (a) . (emitter arrow is not pointing in) . (emitter arrow points in proudly)
Problem Q1-2: Transistor Labeling
2.1. Label the base, collector, and emitter terminals of Figure Q1-1(a). (3 points) 2.2. Label the base, collector, and emitter terminals of Figure Q1-1(b). (3 points) 2.3. The transistor symbols arrow points in the direction of current ow . (1 point)
Solution
The correct labels are shown in Figure Q1-1. The base is usually considered to be the input or control of the transistor. It is depicted as the solid dark line perpendicular to its terminal. The emitter is labeled with an arrow that points in the direction of active-mode positive current ow. A pnp transistor emits positive charge carriers, or holes, into its base. So the arrow on a pnp transistor points into the base. A npn transistor emits negative charge carriers, or electrons, into the base. Positive current ow is opposite direction the of electron ow, and so the arrow on an npn transistor points out of the base. The collector collects any charge carriers that dont make it into the base, and so it is opposite the emitter.
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ECE 327 [Lab 1: The Bipolar (Junction) Transistor Quiz A (10 points)]
Problem Q1-3: Transistor Gain
Assume the transistors above are in active mode with current gain . Express collector current iC in terms of and base current iB . (2 points)
Solution
By the EbersMoll model of a transistor, iC = iB in active mode. The collector current iC joins with the base current iB to form the emitter current iE , and so iC = iC + iB = ( + 1)iB (in active mode).
Problem Q1-4: Limiting Ratio of Currents (BONUS)
Assume the transistors above are in active mode with current gain . Call the collector current iC and the emitter current iE . The following questions refer to E...