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Unformatted text preview: Scores:
DQS
2).;2 s 3)05
4013 Total / 9(9 \2 I pledge on my honor that I did not copy any of this exam, and that this work is entirely my own.
Furthermore, I did not use PSpice during this exam. Page 1 of 9 Problem 1 125 Points}: VCC =15V LL. Vx
5V .2; V1
1. I C2
1”” 12 1mA
£3
»VEE = 15 V Source iin is a small signal ac current source. Find the bias values of VCE and Vx. Assume that VBE = 0.7
V and [3 = 100. Do not ignore base currents. \lE :1 ovubgzaau—wﬁo 2%} (9
\lg/ \lcc '13», ‘z \50 "QMMBQ: m) Problem 2 (25 Points): VCC =15V RC i1” (.3
—VEE = 15 V Assume B = 100. Calculate a numerical value for the gain VO/im. 56m, [1d, lit—6.... :[email protected] \j—c "' 613‘? mm) a YE, 1‘ (K‘R‘ACB)
:; "\fb‘éom “‘9‘ @ Page40f9 Problem 3: (25 Points)
We would like to turn on and off an LED using the circuit below: VCC = 5V
:3
RC
U1A
RB '
W t: 02 fé 3;.
7400 Assume:
0 When the logic gate output is high, the output voltage can be anywhere in the range of 3.5 V to
5 V.
0 When the output of the logic gate is high, the most current it can source is 5 mA.
0 For the BJT, 350 2 B Z 50.
o The LED forward voltage drop is 1.5 V.
0 VCC = 5 0 Assume the BJT saturation voltage is 0.2 V.
o This bullet was put here by accident. We want to design the circuit for the following operation.
0 When the LED is on, 30 mA of current ﬂows through the LED.
0 When the logic gate output is high, the LED should be off. 0 When the output of the logic gate is high, we want the logic gate to supply as little current as
possible and still meet the speciﬁcations. (You can’t just say R3 = 10 because that will supply
more than enough current.) Choose the largest values for R3 and RC that meet the above speciﬁcations and work for any logic high
output and any value of B. xx)“va TM SwV’W/l’x “u DEF, M Rowe $0 ..... , .2 ‘10
9—93 quvdcb‘w} 5‘) O : (’“W‘A C9 «a ’32? m
H 1C \.\ («MAB q 1C?
Ckougz, ’12, 2: ° a“: I SO : O M ﬁw “(\4. L&( cu d’ Page 7 of 9 Problem 4: (25 Points) A new diode has been discovered with the properties shown below: + Id Vd ‘ i __ IOU/(1)15 f0?” Vd Z 0 Where Id — {O for Vd < 0 a) In the circuit below, bias the diode so that the diode voltage Vd is 5 V. Use p = 0.001 A/V 1‘5 . (7
points)
VgC = 15 V
L2
R6 R5 V0 b) Using the value for of R5 determined in part a, ﬁnd a numerical value for the small signal voltage
gain VO/Vm. (18 Points) QC, C2610 \T "MSG
.__e‘59
5x)
+
RS U325d Page 8 of 9 Page 9 of 9 ...
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 Spring '08
 Herniter
 Logic gate, logic gate output

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