ECE-342 Test 1: Sep 27, 2011 6:00-8:00, Closed Book Name : SOLUTION All solutions must provide units as appropriate. Use the physical constants and data as provided on the formula sheet the last page of the exam). Use of rounded values will result in points deductions. Unless otherwise stated, assume T = 300 K. 1. 10 pts) For the circuit shown below, assume that the input v in t) is a 20 V peak) sinusoid. Use a 0.7 V diode drop model for the Zener diode, which has a breakdown voltage of 3.3 V. Find and sketch the output waveform v out t). If the diode is OFF, we have v out = 50 Ω 250 Ω v in = 1 5 v in 4 V peak sinusoid) This solution is valid whenever.7 < v out < 3.3 or 3.5 < v in < 16.5). For v in < 3.3 V, the diode is turned on, giving v out = 0.7 V. For v in > 16.5 V, the diode is in the breakdown region, giving v out = 3.3 V.
2. 20 pts) The diode shown below is driven by a square-wave current as illustrated, where I 0 is a fixed reference current. The output waveform v o t) is also a square-wave.) a) 10 pts) Use the exponential diode model to derive the average and peak-to-peak output voltages. Show that the peak-to-peak output voltage is independent of I 0 and. b) 5 pts) In general, does the voltage drop across a Silicon diode at room temperature tend to increase or decrease as the temperature of the diode increases? assume a fixed diode current) c) 5 pts) How will the peak-to-peak output voltage change as the temperature of the diode increases? Justify) a) The average output voltage is v oavg = 1 [ ) 5I0 nv T ln + nv T ln 2 I0 The peak-to-peak output is ) 5I0 = nv T ln nv T ln v op p )] I0 The peak-to-peak voltage does not depend on I o or.) ) = nv T [ 1 2 ln 5I 2 o I 2 s )] ) 5 I o = nv T ln ) 5Io = nv T ln Is = nv T ln 5) I o b) As the diode temperature increases, the grows quickly, and the diode voltage drop will decrease. For the circuit above, the average value of v o t) will get smaller.) c) From part 2a, v op p is proportional to the absolute temperature of the diode. Since V T = kt abs /q, v op p = nk q ln5)t abs For the circuit above and an increasing temperature, the peak-to-peak voltage will get larger.)
3. 25 pts) Consider the circuit shown below. a) 10 pts) Assuming that the operational amplifier is ideal, find the transfer function of the cicuit, and draw a Bode diagram showing the magnitude of the frequency response. b) 5 pts) Determine component values to complete the design such that: The high-frequency gain is 10. All resistors are 10 kω. The system pole is at 50 Hz. a) The transfer function of this non-inverting amp configuration is c) 5 pts) Modify the design so that the output voltage is insensitive to the operational amplifier input bias current and yet maintains the high input impedance of the non-inverting amplifier). d) 5 pts) What is the circuit response to the operational amplifier input offset voltage? v o R 2 = 1 + v i R 1 + 1/Cs = 1 + R 1 + R 2 )Cs 1 + R 1 Cs The system has a pole at ω = 1 R, and a zero at ω = 1 1C R 1+R 2)C. The gain at low frequencies ω 0) is 1, and the gain at high frequencies ω ) is R 1 + R 2 )/R 1 = 1 + R 2 /R 1. The magnitude Bode diagram is shown below. 1 b) R 1 = 10 kω, R 2 = 9 kω, C = = 318.3 nf 2π50 Hz)R 1 c) The required resistor is shown on the circuit diagram above in red. The resistor is set to the DC) impedance seen at the inverting terminal when the op-amp output is at the desired 0 V. Since the capacitor is an open circuit at DC, the result is R 3 = R 2 d) The input offset voltage is modeled as an additional DC source voltage V os ) at the non-inverting input terminal of the op-amp. Since the capacitor is an open circuit at DC, the non-inverting op-amp gain at DC is 1, giving v o = V os.
4. 25 pts) a) 10 pts) The circuit shows a differential amplifier except for a component mismatch between two of the resistors. For this circuit, derive the differential gain A d = v o /v id and the common-mode gain A cm = v o /v icm, where v id and v icm are the differential and common-mode components of the input signal. Assume that the operational amplifier is ideal. Give the common-mode rejection ratio in db). b) 5 pts) How would you select an op-amp model if the differential amplifier is to achieve a bandwidth of at least 500 khz? c) 10 pts) Draw a circuit diagram of an instrumentation amplifier with a differential gain of 50, and a common mode gain close to zero. a) Setting v i2 = 0 shows that v i1 is the input of an inverting amplifier with a gain of 50. Setting v i1 = 0 shows that v i2 drives a voltage divider to provide the input to a non-inverting amplifier, giving output 50/51.05)1 + 50)v i2 = 49.951v i2. Superposition gives To find the differential gain, set v i2 = v i1 = v d /2. v o = 49.951v i2 50v i1 v o = 49.951v d /2) 50 v d /2) = 49.97v d A d = 49.97 The common-mode gain is found by setting v i1 = v i2 = v cm. v o = 49.951v cm 50v cm =.049v cm A cm =.049 The resulting common-mode rejection ratio is ) Ad CMRR) db = 10 log 20 A cm ) 49.97 = 10 log 20 = 60.2 db.049 b) The amplifier bandwidth is determined by the gain-bandwidth product or unity-gain frequency) f T of the op-amp. To obtain a 500 khz bandwidth, select an amplifier to satisfy c) f T 1 + R 2 /R 1 = f T 51 500 khz f T 25.5 MHz
5. 20 pts) Silicon has been doped using Phosphorus at a concentration of 10 16 cm 3, and Boron at a concentration of 5 10 17 cm 3. a) 15 pts) Find the free carrier free electron and hole) concentrations. Is this an n-type or p-type semiconductor? Assume 300 K. b) 5 pts) Repeat part 5a if the temperature is increased to 350 K. a) Phosphorus is a donor, and Boron an acceptor, so N D = 10 16 cm 3 and N A = 5 10 17 cm 3. The acceptors dominate the dopant population, so this is p-type material. p N A N D = 4.9 10 17 cm 3 n = n2 i p = 1.5 1010 cm 3 ) 2 4.9 10 17 cm 3 = 459 cm 3 b) As the temperature changes, n i increases to a value of n i 4 10 1 1 cm 3 taken from the plot at the bottom of the formula sheet). Doping levels have not changed, so p N A N D = 4.9 10 17 cm 3 n = n2 i p = 4 1011 cm 3 ) 2 4.9 10 17 cm 3 = 3.3 105 cm 3
ECE-342 Test 1, Fall 2011 Last Page) Open-Loop Op-Amp Characteristics first-order model) Exponential Model: ) i D = e v D/nV T 1 e v D/nV T : Reverse Saturation Current V T : Thermal Voltage 25.8 mv at 300K) n : Ideality Coefficient Closed-Loop Op-Amp Amplifiers first-order op-amp model) A 0 large, low-frequencies: v 0 = 1 + R ) 2 v 1 R 2 v 2 R 1 R 1 }{{}}{{} non-inverting gain inverting gain v D = nv T ln i D /I S ) r D = nv T /I D Diode Breakdown: -3 db bandwidth either input): f BW = f T 1 + R 2 R 1 Finite A 0 Gains reduced by 1 + 1 + R ) 2/R 1 A 0 Periodic Table Segment III IV V B C N Al Si P Ga Ge As In Sn Sb Physical Constants k = 8.62 10 5 ev/k = 1.38 10 23 J/K q = 1.60 10 19 coulomb ɛ 0 = 8.854 10 14 F/cm V T = kt/q = 25.8 mv at 300K Silicon at 300K: E G = 1.12 ev ɛ r = 11.7 n i = p i = 1.5 10 10 cm 3 µ n 1350 cm 2 /V s D n 34.8 cm 2 /s µ p 480 cm 2 /V s D p 12.4 cm 2 /s n i =1.5 x 10 10 at 300K Si Approximation: n i2 doubles every 5 degrees n i 2 T 3 e E G /kt E G =1.12 ev