Instructor
Table Name: number: Solution
ECE 327: Electronic Devices and Circuits Laboratory I
Lab 3: Voltage Regulators Quiz A (10 points)
Description. This quiz tests your comprehension of what youve read on voltage regulators. Complete this quiz with closed book and closed notes.
Problem Q3-1: Zener diode regulators
1. On the axes below, plot an approximate iZ,forward -vs-vZ,forward for the Zener diode shown beside it. Be sure to label the VZ that we use in this laboratory. (4 points) iZ,forward vZ,forward iZ,forward VZ vZ,forward
Compare the shape of the curve with the Z -shaped hat of the Zener diodes symbol. (exponential; curved)
0V
(linear; straight)
2. The line regulation of a Zener diode regulator can be improved by reducing the Zener diodes dynamic resistance. Why? (3 points) Solution: Line regulation measures the change in a voltage regulators output when there is a change in the voltage regulators input. Ideally, a voltage regulators output would be independent of input changes, and so it would have a low line regulation. The dynamic resistance rZ of a Zener diode models how closely the diode behaves as an ideal voltage source when adequately reverse biased. In particular, 1/rZ is the slope of the reversebiased region of the Zener diodes iv curve shown in question Q3-1.1. As rZ is decreased, the curve becomes more vertical, and so the voltage dropped across the Zener holds more closely to a constant value over a wider range of currents. Hence, a low dynamic resistance implies a low (i.e., good) line regulation.
Problem Q3-2: Load Regulation
Qualitatively dene load regulation. That is, what does load regulation measure? (3 points)
Solution
Load regulation measures the changes in a voltage regulators output when the load on the output changes. Ideally, a voltage regulators output would be constant regardless of load, and so it would have a low load regulation.
Copyright c 20072009 by Theodore P. Pavlic All rights reserved.
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ECE 327 [Lab 3: Voltage Regulators Quiz A (10 points)]
Problem Q3-3: Make Some Adjustments (BONUS)
Assume that the nominal baseemitter drop for the ideal transistor in Figure Q3-1 is 0.7 V. Also, assume that the circuit parameters are chosen so that the bipolar transistor is in active mode.
5.5 250 V 1 k vout
A
3.3 VDC
4 k
L
0V
Figure Q3-1: A discrete transistor voltage regulator with load L. For the following two bonus questions, assume that the switch is open. 1. What is the voltage at node A ? (3 bonus points) Solution: Node A is the base of the npn transistor. Because there is 3.3 V at the transistors emitter and the nominal baseemitter drop is 0.7 V, the voltage at node A is (3.3 + 0.7) V, or 4 V. 2. What is the voltage vout ? (3 bonus points) Solution: As discussed, there is 4 V at node A . We assume that there is negligible current going into the base of the transistor, and so the current through the 4 k resistor matches the current through the 1 k resistor. The 4 k resistor carries (4 V)/(4 k), or 1 mA. That 1 mA across the 1 k resistor causes a 1 V drop. The voltage vout is the sum; that is, vout = 4 V + 1 V, or 5 V. For the following bonus question, assume that the switch is closed. Also, assume that the load (located below the switch) is a simple resistor with resistance L. 3. In this circuit, the output vout is constant over a wide range of loads. What is the smallest value of load L for which vout is constant? Explain. (4 bonus points) Solution: As the load L is decreased, it draws more current through the 250 resistor and tends to reduce vout . As vout is depressed, so is the voltage at node A , and so the trans...