306 EE F07-02A MIDTERM EXAM #2 CLOSED BOOK + 1 Cheat Sheet Show all work, state any assumptions, and test hypotheses. (1 point) Print Your Name:
Braun
Part 1 (14 Points) Part 2 (8 Points) Part 3 (6 Points) Part 4-5 (6 Points) Total (36 Points)
SOLUTIONS
BRAUN
No unauthorized help given or received. (1 point) Signature: 1. Diode Circuits (14 points)
I D = I S [exp(VD VT ) - 1]
Use CVD for hand analysis
For the circuits shown below, assume diodes obey the ideal diode law and have VZ = 10 V. Each battery supplies 5 V. The output voltage, VOUT, appears between the upper right node of the circuit and ground. The input signal VS (dashed line) for each case is a 5 V triangle wave (10 V peak-to-peak) with a frequency of 1 kHz and no DC offset. For each circuit, plot the output voltage trace on the upper graph and the capacitor voltage trace on the lower graph. For each signal you sketch, clearly indicate maximum and minimum values 0.1 V. Repeated for emphasis: Pick VOUT as the POSITIVE terminal of the capacitor. Pick VOUT as the POSITIVE terminal of the capacitor.
VC 1 F
+
VOUT
+
VOUT
VS
1 F 5V
VC
VS 5V
KVL: VOUT = VC + 5 V 15 10
VOUT [V] 4.3 V
KVL: VOUT = VS + VC 15 10
VOUT [V]
14.3 V
5 0 -5
0V Diode Operation ON OFF
5 0 -5
OFF ONOFF 4.3 V
-10 -15 15 10
VC [V]
-10 Diode Operation 1.5 2 2.5 3 [ms] -15 15 10
-0.7 V VC [V]
0
0.5
1
0
0.5
1
1.5
2
2.5
9.3 V
3 [ms]
5 0 -5
-5.0 V
5 0 -5
5.0 V
-10 -15 0
-10 1 1.5 2 2.5 3 [ms]
.
0.5
-15
0
0.5
1
1.5
2
2.5
3 [ms]
.
List the circuit's function:
Peak Detector
List the circuit's function:
Clamp
2. MOS Biasing F07-02A This problem asks you to analyze the circuit shown to the right and determine the Q-point of the transistor. Assume VTN = - VTP = 1.25 V, KN' = 50 A/V2, KP' = 200 A/V2, = 0 V1/2, = 0 V-1. W/L = 10. (4 points) A) Hypothesize the transistor turns on and operates in the saturation mode of operation. Determine values for VGS, VDS and ID, based on the transistor operating in saturation mode. VDD = 5 V Please enter the values in the table below.
VGS = VG = VDD
R1 120 k =5 V = 3.0 V > VTN R1 + R2 120 k + 80 k
R2 80 k
RD 6 k
K N = K N (W / L) = (50 A/V 2 )(10) = 500 A/V 2 KN (VGS - VTN ) 2 2 500 A/V 2 = (3.0 V -1.25 V) 2 = 0.766 mA 2 VDS = VDD - I D RD ID = = 5.0 V - 0.766 mA (6.0 k) 0.406 = V Test: VDS = 0.406 V < VGS - VTN = 3.0 V 1.25 V = 1.75 V
R1 120 k
X
Doesn't agree with Saturation
VGS [V] VDS [V] ID [mA] (1 point) B) Is hypothesis A correct? YES (3 points) NO
3.0 V 0.406 V 0.766 mA
C) Explain whether hypothesis A is correct or not.
Saturation requires VDS VGS - VTN , so the hypothesis doesn't work. The problem didn't REQUIRE finding the actual Q-point based on the FET operating in LINEAR mode. Doing so yields (3.0 V, 1.08 V, 0.653 mA)
3. JFET Biasing F07-02A (6 points) This problem asks you to analyze the JFET circuit shown to the right and determine its Q-point. Please enter the values in the table below. Assume VP = -2 V, IDSS = 2 mA.
VDD = 1 V
VDS < VDD = 1 V < VGS - VP = 0 V (2 V) = 2 V So the JFET operates in the Linear region 2I V V - VDS I D = DSS (VGS - VP - DS )VDS = I R = DD 2 VP 2 R
2 VDS - [2(VGS - VP ) +
R 1 k
VP2 I DSS R
]VDS +
VP2 I DSS R
VDD = 0
- [2(0 V - (-2 V )) +
(2 V) 2 ] (2 mA)(1 k)
(2 V) 2 1V (2 mA)(1 k) 2V2
-6V
2 VDS - [2(VGS - VP ) +
VP2 I DSS R
]VDS +
VP2 I DSS R
VDD = 0
(6 V) 2 - 4(2 V 2 ) 6V = 0.354 V , 5.65 V VDS = Agrees with linear operation 2 2 Only the lower value provides a physical solution. VDS = 0.354 V < (VGS VP) = 2 V is consistent with the transistor operating in Linear mode. V - VDS 1 V - 0.354 V = = 0.646 mA I D = I R = DD 1 k R
VGS [V] VDS [V] ID [mA]
0V 0.354 V 0.646 mA
Please turn page.
4. BJT Operation F07-02A Using the Gummel-Poon or Transport model, determine the operating points for the following transistors. Assume IS = 10 f A, F = 50, R = 1, and room temperature with VT = 25 mV. 0.1 V 0.70 V
(3 points)
IC [mA] IB [mA] VCE [V]
13.9 mA 0.554 mA 0.1 V
v v I v I C = I S exp BE - exp BC - S exp BC - 1 V V V T T R T 0.7 V 0.6 V 10 fA 0.6 V =10 fA exp 25 mV - exp 25 mV - 1 exp 25 mV - 1 =13.9 mA IB = I S v BE I S v BC - 1 - 1 + exp exp F VT R VT 10 fA 0.7 V 10 fA 0.6 V - 1 + - 1 = exp exp 50 25 mV 1 25 mV = 0.554 mA
(3 points)
5. LED Operation A GaP LED emits light at a wavelength of 565 nm. Please determine the energy of the light in electron volts and its color.
E PH =
1240 [eV nm] 1240 [eV nm] = 565 nm = 2.19 eV, which corresponds to green light.