Midterm 1 October 2007
ECSE 330: INTRODUCTION TO ELECTRONICS Midterm Examination #1
October 11th 2007, 10:05 AM – 11:25 AM Instructor: Joshua D. Schwartz
Please note the following before you begin: - This is a closed book examination, no notes permitted. - This exam consists of 4 problems totaling 40 possible points. - Only the faculty standard calculator is permitted. - Write all answers in the answer booklet only. This paper will NOT be collected - Good luck!
Question 1 (10 points) You have a sinusoidal signal source of amplitude 100mV and source resistance Rs = 50 Ω, and you wish to drive a load of 10 Ω with a sinusoidal signal voltage of at least 1V amplitude. The problem is – you do not have any voltage amplifiers! a) Suppose that somebody offers to sell you a voltage amplifier (AVO = 200 V/V, Rin = 50 Ω, Rout = 10 Ω) and tells you that it runs on a split-supply (VDD = +3.3V, VEE = -3.3V). You happen to have these supply voltages handy. I. [2 pts] What overall gain can you realize with this amplifier? You can realize an output of 100mV*(50/100)*200*(10/20) = 5V… a gain of 50 V/V. II. [1 pt] Explain why, even though you have more than enough gain, this amplifier alone is not a good solution for your problem. Given that the output cannot go above 3.3V or below -3.3V due to the supplies, your input signal will result in a clipped output (distortion). III. [2 pts] You convince the seller to throw in a resistor of a certain value. With this resistor, you explain, you can get exactly 1V output across the load. What value of resistor do you need, and where would you put it? You should NOT stick an 80 Ω resistor in the output before your load, since clipping would still occur at the point of gain. Instead, make the input smaller to begin with by adding a 400 Ω resistor between the source’s Rs and the amplifier input (i.e. a voltage divider of 50/500 now occurs and the amplifier gets a 10 mV input instead of 50 mV). Suppose you decide not to buy this voltage amplifier after all, but you discover later that you have in your collection a transresistance amplifier (RMO = 5 V/A, Rin = 100 Ω, Rout = 10 Ω) and a transconductance amplifier (GMS = 5 A/V, Rin = 2 kΩ, Rout = 2 kΩ). b) [3 pts] Show how you can use these two amplifiers to obtain the required signal output, and calculate the overall gain.
100mV * (2000/2050)*5*(2000/2100)*5*(10/20) = 1.16V, a gain of 11.6 V/V c) [1 pt] In an ideal current amplifier, what is the input resistance? What is the output resistance? Zero input resistance, infinite output resistance. 2
3
4
5
Question #3 (9 points) Consider the following circuit. All diodes are identical and have n = 1. Use the Constant Voltage Drop Model for the diodes. Given R1 = 5kΩ, R2 = 10kΩ, RL = 100Ω, and that all the capacitors and inductors are infinitely large with C = ∞ and L = ∞, respectively.
∞
∞ ∞
a) [4 pts] When VDD = 1V, which diodes are ON/OFF. When VDD = 10V, which diodes are ON/OFF. Show all your working and verify that your final answers are consistent. Note, that D3 is decoupled from DC analysis because of C = ∞, and that D2 is included as L = ∞. For VDD = 1V; It should be obvious that with only 1V to share between them, both D1 and D2 cannot be ON together. If you assumed D1 is ON, you would find by KCL that (0.3-Vx)/5k + (1-Vx)/5k = Vx/10k and Vx = 1.3/2.5 = 0.52 which unfortunately means D1 is OFF (current in it would be backwards). With D1 and D2 OFF you get a better answer: current in the remaining branches is 1/15k = 66.6μA and Vx is 0.667, which confirms that both diodes are off. At VDD = 10V; a fair guess is that both diodes are getting the 0.7V they want… trying that, we get by KCL at the VX node that (9.3-Vx)/5k + (10-Vx)/5k = Vx/10k + (Vx0.7)/10k) which simplifies to 19.3 – 2Vx = Vx – 0.35 and then Vx = 6.55 V. All currents are going the right way here, our assumption must be correct. Give your answers in terms of, R1, R2, RL, and the small-signal parameters (for example, rDX for the small-signal resistance of DX). If two resistances (Rx and Ry) appear in parallel, use the notation Rx//Ry in your answers instead of expanding the expression.
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b) [2 pts] Draw the equivalent small-signal circuit assuming that the diodes D1 and D2 are ON. Do not calculate any small-signal values. Note, that D2 is decoupled from AC analysis because of L = ∞, and that D2 although included as C = ∞, does not contribute, as rD3 = ∞ because it had no DC current flowing through it.
Rout R1 + rD1
vin
R3
R0 R2
R1
vout RL
Rin c) [1 pts] Find an expression for the input resistance, Rin. Rin = R0 + [(R1+rD1)//R2//R1//(R3 + RL)] by inspection. d) [1 pts] Find an expression for the output resistance, Rout. Rout = RL // [R3 + (R1//R2//(R1+rD1)//R0)] by inspection. e) [1 pts] The small-signal model for a diode is a resistor. Briefly (in one or two lines) explain why this model is only accurate for “small-signals”. Because it is a linear approximation of the exponential curve.
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Question 4 (9 points) a) [4 pts] For an NPN BJT, plot the IC vs. VCE curves for three increasing values of VBE (VBE1 < VBE2 < VBE3). Make your plot as neat as possible, and you only need to plot for positive values of VCE, but leave some space to show the Early voltage. On this plot, show the following: i. Direction of increasing VBE. ii. Active region. iii. Saturation region (recall what condition on VCB causes you to leave the active mode, and use that to figure out what VCE value this happens at). iv. Location of the Early voltage.
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b) [5 pts] The circuit in the figure below is a BJT “folded cascode” circuit. Assume that β=100 for both NPN and PNP transistors. i. Solve for IB1 and IE1. ii. Solve for IC1 and IE2. iii. Show that both transistors are active. +5
+5
100 +4 20K inf Q1
Q2 5K
inf
20K 2K
1K
i.Simplify the biasing of Q1 by the Thevenin equivalent: Vthev = 5 * 20K/(20K+20K) = 2.5V Rth = 20K//20K = 10K KVL: 2.5 – 10K*Ib1 – 0.7 – 2K*Ie1 = 0 ...(1) Ie1 = (B+1)*Ib1 ...(2) Combining (1) and (2): Ib1 = 8.49 uA From (2): Ie1 = 857.5 uA ii.From Ic1 = B*Ib1: Ic1 = 849 uA KVL: 5 – Ix*100 – 0.7 – 5K*Ib2 – 4 = 0 ...(3) Ix = Ic1 + Ie2 ...(4) Combining (3) and (4): 0.3 = 100*Ic1 + 100*Ie2 + 5K*Ib2 Using Ie2 = (B+1)*Ib2: 0.2151 = 15.1K*Ib2 -> Ib2 = 14.24 uA Thus: Ie2 = 1.439 mA Ic2 = 1.424 mA Ix = 2.288 mA iii. Vc1 = 5 – Ix*100 = 4.77V Vb1 = 2.5 – Ib1*10K = 2.41V Therefore Vc1 > Vb1 and Q1 is active. Vc2 = 1K*Ic2 = 1.424V Vb2 = 5K*Ib2 + 4 = 4.07 Therefore Vb2 > Vc2 and Q2 is active.
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Formula Sheet (you can detach this) Diode Dynamic Resistance: rd = nVt / ID Diode Exponential Model: I D = I S (e
VD nVT
− 1)
BJTs i iB = C
β
i iE = C
α
iB =
iE β +1
α=
β β +1
10
October 11th 2007, 10:05 AM – 11:25 AM Instructor: Joshua D. Schwartz
Please note the following before you begin: - This is a closed book examination, no notes permitted. - This exam consists of 4 problems totaling 40 possible points. - Only the faculty standard calculator is permitted. - Write all answers in the answer booklet only. This paper will NOT be collected - Good luck!
Question 1 (10 points) You have a sinusoidal signal source of amplitude 100mV and source resistance Rs = 50 Ω, and you wish to drive a load of 10 Ω with a sinusoidal signal voltage of at least 1V amplitude. The problem is – you do not have any voltage amplifiers! a) Suppose that somebody offers to sell you a voltage amplifier (AVO = 200 V/V, Rin = 50 Ω, Rout = 10 Ω) and tells you that it runs on a split-supply (VDD = +3.3V, VEE = -3.3V). You happen to have these supply voltages handy. I. [2 pts] What overall gain can you realize with this amplifier? You can realize an output of 100mV*(50/100)*200*(10/20) = 5V… a gain of 50 V/V. II. [1 pt] Explain why, even though you have more than enough gain, this amplifier alone is not a good solution for your problem. Given that the output cannot go above 3.3V or below -3.3V due to the supplies, your input signal will result in a clipped output (distortion). III. [2 pts] You convince the seller to throw in a resistor of a certain value. With this resistor, you explain, you can get exactly 1V output across the load. What value of resistor do you need, and where would you put it? You should NOT stick an 80 Ω resistor in the output before your load, since clipping would still occur at the point of gain. Instead, make the input smaller to begin with by adding a 400 Ω resistor between the source’s Rs and the amplifier input (i.e. a voltage divider of 50/500 now occurs and the amplifier gets a 10 mV input instead of 50 mV). Suppose you decide not to buy this voltage amplifier after all, but you discover later that you have in your collection a transresistance amplifier (RMO = 5 V/A, Rin = 100 Ω, Rout = 10 Ω) and a transconductance amplifier (GMS = 5 A/V, Rin = 2 kΩ, Rout = 2 kΩ). b) [3 pts] Show how you can use these two amplifiers to obtain the required signal output, and calculate the overall gain.
100mV * (2000/2050)*5*(2000/2100)*5*(10/20) = 1.16V, a gain of 11.6 V/V c) [1 pt] In an ideal current amplifier, what is the input resistance? What is the output resistance? Zero input resistance, infinite output resistance. 2
3
4
5
Question #3 (9 points) Consider the following circuit. All diodes are identical and have n = 1. Use the Constant Voltage Drop Model for the diodes. Given R1 = 5kΩ, R2 = 10kΩ, RL = 100Ω, and that all the capacitors and inductors are infinitely large with C = ∞ and L = ∞, respectively.
∞
∞ ∞
a) [4 pts] When VDD = 1V, which diodes are ON/OFF. When VDD = 10V, which diodes are ON/OFF. Show all your working and verify that your final answers are consistent. Note, that D3 is decoupled from DC analysis because of C = ∞, and that D2 is included as L = ∞. For VDD = 1V; It should be obvious that with only 1V to share between them, both D1 and D2 cannot be ON together. If you assumed D1 is ON, you would find by KCL that (0.3-Vx)/5k + (1-Vx)/5k = Vx/10k and Vx = 1.3/2.5 = 0.52 which unfortunately means D1 is OFF (current in it would be backwards). With D1 and D2 OFF you get a better answer: current in the remaining branches is 1/15k = 66.6μA and Vx is 0.667, which confirms that both diodes are off. At VDD = 10V; a fair guess is that both diodes are getting the 0.7V they want… trying that, we get by KCL at the VX node that (9.3-Vx)/5k + (10-Vx)/5k = Vx/10k + (Vx0.7)/10k) which simplifies to 19.3 – 2Vx = Vx – 0.35 and then Vx = 6.55 V. All currents are going the right way here, our assumption must be correct. Give your answers in terms of, R1, R2, RL, and the small-signal parameters (for example, rDX for the small-signal resistance of DX). If two resistances (Rx and Ry) appear in parallel, use the notation Rx//Ry in your answers instead of expanding the expression.
6
b) [2 pts] Draw the equivalent small-signal circuit assuming that the diodes D1 and D2 are ON. Do not calculate any small-signal values. Note, that D2 is decoupled from AC analysis because of L = ∞, and that D2 although included as C = ∞, does not contribute, as rD3 = ∞ because it had no DC current flowing through it.
Rout R1 + rD1
vin
R3
R0 R2
R1
vout RL
Rin c) [1 pts] Find an expression for the input resistance, Rin. Rin = R0 + [(R1+rD1)//R2//R1//(R3 + RL)] by inspection. d) [1 pts] Find an expression for the output resistance, Rout. Rout = RL // [R3 + (R1//R2//(R1+rD1)//R0)] by inspection. e) [1 pts] The small-signal model for a diode is a resistor. Briefly (in one or two lines) explain why this model is only accurate for “small-signals”. Because it is a linear approximation of the exponential curve.
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Question 4 (9 points) a) [4 pts] For an NPN BJT, plot the IC vs. VCE curves for three increasing values of VBE (VBE1 < VBE2 < VBE3). Make your plot as neat as possible, and you only need to plot for positive values of VCE, but leave some space to show the Early voltage. On this plot, show the following: i. Direction of increasing VBE. ii. Active region. iii. Saturation region (recall what condition on VCB causes you to leave the active mode, and use that to figure out what VCE value this happens at). iv. Location of the Early voltage.
8
b) [5 pts] The circuit in the figure below is a BJT “folded cascode” circuit. Assume that β=100 for both NPN and PNP transistors. i. Solve for IB1 and IE1. ii. Solve for IC1 and IE2. iii. Show that both transistors are active. +5
+5
100 +4 20K inf Q1
Q2 5K
inf
20K 2K
1K
i.Simplify the biasing of Q1 by the Thevenin equivalent: Vthev = 5 * 20K/(20K+20K) = 2.5V Rth = 20K//20K = 10K KVL: 2.5 – 10K*Ib1 – 0.7 – 2K*Ie1 = 0 ...(1) Ie1 = (B+1)*Ib1 ...(2) Combining (1) and (2): Ib1 = 8.49 uA From (2): Ie1 = 857.5 uA ii.From Ic1 = B*Ib1: Ic1 = 849 uA KVL: 5 – Ix*100 – 0.7 – 5K*Ib2 – 4 = 0 ...(3) Ix = Ic1 + Ie2 ...(4) Combining (3) and (4): 0.3 = 100*Ic1 + 100*Ie2 + 5K*Ib2 Using Ie2 = (B+1)*Ib2: 0.2151 = 15.1K*Ib2 -> Ib2 = 14.24 uA Thus: Ie2 = 1.439 mA Ic2 = 1.424 mA Ix = 2.288 mA iii. Vc1 = 5 – Ix*100 = 4.77V Vb1 = 2.5 – Ib1*10K = 2.41V Therefore Vc1 > Vb1 and Q1 is active. Vc2 = 1K*Ic2 = 1.424V Vb2 = 5K*Ib2 + 4 = 4.07 Therefore Vb2 > Vc2 and Q2 is active.
9
Formula Sheet (you can detach this) Diode Dynamic Resistance: rd = nVt / ID Diode Exponential Model: I D = I S (e
VD nVT
− 1)
BJTs i iB = C
β
i iE = C
α
iB =
iE β +1
α=
β β +1
10
