Ee362l Tests Summer05

VERSION 2 (V2) 75 minute test. Ten multiple choice questions. One 8.5x11” note sheet permitted (both sides). No laptop computers. No credit for guessing - you must show sufficient reasoning, equations, and steps to justify your answer. Circle the correct answer. I intended for the correct answer to be one of the choices, but if not, then choose “other” (if provided) and write in your answer. Do not unstaple. Synchronized starting. Seating assignments. No partial credit. If you finish more than 15 minutes early, quietly turn in your test. Else, wait until the end of class to avoid disturbing others. Problem 1. A MOSFET is operating with 5W of average power loss. The MOSFET is mounted on a heat sink that has a thermal coefficient of 2.0°C rise/W. The ambient temperature of the room is 30°C. The heat sink temperature is monitored by a thermistor (10kΩ @ 25°C) whose resistance variation is shown in the table below. Compute the resistance of the thermistor at the operating condition described. Temp T - ºC 0 5 10 15 20 25 30 35 40 45 50 55

a. 6530Ω

RT/R(T=25ºC) 3.28 2.55 1.993 1.573 1.250 1.000 0.806 0.653 0.532 0.437 0.360 0.298

b. 4370Ω

c. 3600Ω

d. 5320Ω

e. Other _________________

Problem 2. A 2,500µF capacitor is charged to 32V and then connected across a load resistor R. The response is shown on the oscilloscope snapshot below. Compute the resistance of load resistor R. 0.1sec/div

10V/div

0V

Values in Ω a. R ≤ 25

b. 25 < R ≤ 50

c. 50 < R ≤ 75

d. 75 < R ≤ 100

e. R > 100

VERSION 2 (V2) Problem 3. A diode bridge rectifier has a 3,300µF capacitor and is powering a 10Ω load resistor. The peak voltage across the load resistor is 40V. The AC input current waveshape to the DBR is shown below. When the input current is not flowing, the capacitor is discharging into the load resistor, so that the load voltage falls a bit before the AC current starts to flow again. Use the RC time constant of the DBR and load to compute the minimum voltage VL that appears across the load resistor. 2msec/div

Values in volts a. VL ≤ 30

c. 30 < VL ≤ 32

c. 32 < VL ≤ 34

d. 34 < VL ≤ 37

e. 37 < VL ≤ 40

Problem 4. The inductor in the circuit below is in periodic steady-state and has the current shown in the graph. The circuit is delivering ripple-free current to a load. Determine the peak-to-peak value (i.e., max minus min) of the capacitor current. iL

iin

L

Vin

id

3A

Iout C

iC

+ Vout –

2A 1A 0A

a. 1.0A

b. 2.0A

c. 5.0A

d. 2.5A

e. Other _________________

VERSION 2 (V2) Problem 5. In the circuit shown below, Vin = 40V, and L = 160μH. The inductor current is initially zero. Then, at t = 0, the switch closes. Compute the time required for the inductor current to reach 10A.

iL

Vin

a. 40μs

– v +

L

b. 16μs

c. 10μs

d. 25μs

e. Other _________________

Problem 6. Consider the circuit shown above in Problem 5. Once the inductor current reaches 10A, the switch opens, and the inductor current continues to flow for a while through the freewheeling diode. The diode is not lossless. Its i-v characteristics are shown below. When forward biased, the diode has a 1V drop. Compute the time needed for the inductor current to drop to 5A. i 0A v 0,0

a. 1.6ms

1V

b. 1.0ms

c. 800μs

d. 500μs

e. Other _________________

VERSION 2 (V2) Problem 7. Two current sources combine in parallel to provide power to resistor R. Source i(t) is periodic, and source Ix is constant. Compute the Ix that causes the average resistor power to be 80W. i(t)

3A 2A i(t)

Ix

R = 10Ω 1A

2 = I 2 + 1 (ΔI )2 [Hint – for a sawtooth waveform, I rms DC 12

a. 1.11A

b. 3.11A

c. 0.769A

0A

]

d. 2.77A

e. Other _________________

Problem 8. Consider the circuit shown above in Problem 7. Assume Ix = 3A. Compute the peak instantaneous power absorbed by resistor R.

a. 93.3W

b. 160W

c. 360W

d. 253W

e. Other _________________

VERSION 2 (V2) Problem 9. Assume that a solar panel is mounted horizontally atop ENS (i.e., is zero tilt angle). Your kWh(longest day of year ) task is to estimate the panel’s ratio. Assume kWh( shortest day of year ) • •

that both days are crystal-clear, with equally-intense direct normal W/m2, and practically no diffuse-horizontal W/m2, so that the amount of solar power hitting the panel surface is proportional to the cosine of the solar zenith angle (which is shown below for the longest and shortest days).

Neglect any temperature-related variation in panel efficiency. (Hint – the ratio of kWhs is the same as the ratio of areas under the cosine curves) Cosine of the Solar Zenith Angle for Jun. 22 and Dec. 22 in Austin 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

Hour of Day (Solar Time)

a. ratio ≤ 0.6

b. 0.6 < ratio ≤ 1.0

c. 1.0 < ratio ≤ 1.5

d. 1.5 < ratio ≤ 1.9

e. ratio > 1.9

VERSION 2 (V2) Problem 10. An 80Vrms (i.e., 113V peak), sinusoidal, 60Hz voltage is applied to the triac light dimmer circuit shown. Initially, the potentiometer resistance Rpot = 250kΩ, and there is no “firing.” With no ⎡ ⎤ 1 firing, the voltage across the diac is a phasor with expression Van ⎢ ⎥, ⎣1 + jωRC ⎦ where R = Rpot + 3.3kΩ. Gradually, Rpot is lowered until the peak magnitude of the diac voltage reaches 35V, whereupon firing begins. Compute the R at which firing begins. (Hint – use the squared magnitude of the phasor expression to find R)

a

100W light bulb b 3.3kΩ

+ 80Vrms, 60Hz –

MT2

250kΩ linear pot 0.1µF

c

G

35V bilateral trigger diac

Triac MT1

n

Values in kΩ a. R ≤ 60

b. 60 < R ≤ 100

c. 100 < R ≤ 140

d. 140 < R ≤ 180

e. R > 180