Question 1 (2 points each unless noted otherwise) Homework Assignment 09 1. For SPICE, Explain very briefly the difference between the multiplier M and Meg, as in a resistor has value 2M versus a resistor with value 2Meg? Answer: SPICE interprets M as milli and Meg as mega so that 2M is 0.002 Ω" and 2Meg is 2,000,000 Ω. 2. Consider the following drive circuit for an IR remote control. The drive signal is a 0 5 V square wave and VV CCCC = 9 V. The average current through the LED is a) 83 ma b) Need additional information c) 41 ma d) 58.3 ma e) 525 ma Answer: The BJT is off until the LED current reaches a peak of 0.7 12 = 58.3 ma. Then it turns on and prevents the MOSFET from turning on further. With a 70% duty cycles this peak corresponds to an average of 0.7 58.3 41 ma, so (c). 3. The following symbol indicates which type of transistor (circle one)? (a) n-channel MOSFET (b) n-channel JFET (c) p-channel MOSFET (d) n-channel JFET (e) BJT Answer: Option (a) 4. Explain with one word/phrase what every letter in CMOS mean. Answer: Complementary Metal Oxide Semiconductor 5. A MOSFET is biased such that II DD = 1 ma. What should gg mm be so that a 1-mVchange in VV GGGG results in a 1.78 μμa change in II DD? Answer: δδii DD = gg mm δvv GGGG (1.78 10 6 ) = (gg mm )(1 10 3 ) gg mm = 1.78 ma/v 1
6. Below are the characteristics for a MOSFET. What type of FET is this (circle one)? (a) (b) (c) (d) Enhancement PMOS Depletion PMOS Enhancement NMOS Depletion NMOS Answer : The subscript SG indicates a p- channel FET (the more common n-channel MOSFET has GS as in VV GGGG ) and note that increasing VV GGGG increases II DD, so this is an enhancement PMOS, so option (a) is the answer. 7. Which one of the MOSFETs in the circuits below behaves as a non-linear resistor? (1) (2) (3) (a) Only (1) (b) Only (2) (c) Only 3 (d) Both (1) and (3) (e) All (f) None Answer: Option (e) 8. Below are two schematics of current sources implemented with MOSFETs. Which current source has the best compliance voltage? (a) (b) Answer: (a) 2
Question 2 For the circuit shown, ββ =350, and II CC = 2.5 ma. What is the voltage gain AA vv = vv oo vv ii? (2 points) Solution The voltage gain is approximately 1 (this is a follower; RR cc does not affect the gain) Question 3 Consider the following circuit. Assume that VV TTTT = 1.5 V, KK nn = 1.5 ma V 2, and λλ = 0. What is VV DDDD(SSSSSS)? Answer: VV DDDD(ssssss) = VV GGGG VV TTTT = 2 1.5 = 0.5 V Question 4 The figure is a plot of the open-loop gain function for the LT1007 voltage amplifier. An engineer will use the amplifier as a non-inverting amplifier with a mid-frequency voltage gain of 10. (a) What is the GBP of the LT1007? (2 points) (b) Use the plot and estimate the bandwidth of the feedback amplifier. (2 points) (c) Write an expression for the gain AA(ff) for the feedback amplifier. (2 points) (a) The open loop gain is 120 db (1 10 6 ) at ff = 10 Hz, so the GBP is 10 MHz. Alternatively, the BW is about 10 MHz when the open loop gain is 0, so the GBP is 10 MHz. (b) A voltage gain of 10 is equivalent to a gain of 20 log 10 (10) = 20 db. A horizontal line at 20 db intercepts the LT1007 gain curve at 950 khz. Alternatively, from the GBP, with a gain of 10, the bandwidth is 1 MHz. (c) The closed-loop response is AA(ff) = 10 ff 1+jj 1 10 6 3
Question 5 (N. 4.44) The source follower shown has VV TTTT = 0.4 V, KK nn = 0.5 ma V 2, and λλ = 0. The power supply is VV DDDD = 3 V, and CC CC1 = 1 μμf a) Design the circuit (by specifying the resistors) such that RR ii = 300K, II DDDD = 0.25 ma, and VV DDDDDD = 1.5 V. (6 points) b) Perform a quick estimate of the small-signal voltage gain. (1 point) c) Next, perform a small-signal analysis and calculate the small-signal voltage gain. (4 points) d) Perform a small-signal analysis to determine the output resistance RR oo. (4 points) e) Calculate the input resistance RR ii. (1 point) f) Calculate the lower 3-dB frequency. (3 points) g) Perform a Micro-Cap SPICE simulation to check your calculations. For the transistor, use the $GENERIC_N part, and set LAMBDA = 0, W = 200u, L = 20u, KP = 100u, and VTO = 0.4. 1) Perform a Dynamic DC analysis to find II DDDD and VV DDDDDD. 2) Perform an AC analysis to find the gain at 100 Hz and the lower 3-dB frequency. 3) Perform an AC analysis to find RR OO at 100 Hz. 4) Perform an AC analysis to find RR ii at 100 Hz. For the Micro-Cap SPICE analysis, show schematics for (1) (4) and for (2) (4) the simulation outputs. Pay attention to font sizes, line thickness, etc., and use annotation to clarify your work. You must also submit four separate, ready to run, SPICE files to ICON. Summarize your calculated and SPICE values in table. Solution Part (a) VV DDDDDD = 1.5 V implies that the quiescent voltage across RR ss is also 1.5 V. With II DDDD = 0.25 ma this means that RR ss = 1.5 0.25 = 6K. II DD = KK NN (VV TTTT VV GGGG ) 2 0.25 = 0.5(0.4 VV GGGG ) 2 Solving for VV GGGG (using Excel, but one can do this algebraically, a graphing calculator, or Matlab) yields VV GGGG = 1.107 V. Consequently, VV GG = 1.107 + 1.5 = 2.607 V. Next we need to determine values for RR 1 and RR 2 such that RR 1 3 = 2.607, RR 1 + RR 2 and RR 1 RR 2 = 300 K Solving yields RR 1 = 2.29M and RR 2 = 345K. Part (b) This is a source follower so that the voltage gain should be 1. This is because the circuit is a source follower. However, MOSFETs have low gg mm so the voltage gain will be 10 20% less, so that a rough estimate would be AA vv 0.8. 4
Part (c) Shown is a small-signal model for determining the voltage gain. Note that vv ii + vv gggg + vv OO = 0 so that vv gggg = vv ii vv OO. KCL at the source, assuming current flow away from the source gives vv OO RR SS gg mm vv gggg = 0 vv OO RR SS + gg mm (vv ii vv OO ) AA vv = vv OO vv ii = Substituting the numerical values for gg mm and RR SS yields AA vv = (0.7071 ms)(6k) 1 + (0.7071 ms)(6k) = 4.423 1 + 4.423 = 0.809 gg mmrr SS 1 + gg mm RR SS Part (d) Shown is a small-signal model for calculating the output resistance. KCL at the source, using the convention that current flow away from the source is gg mm vv gggg II xx + VV xx RR SS = 0 Note that vv gggg = VV xx so that the KCL equation becomes gg mm VV xx II xx + VV xx RR SS = 0 Solving for RR OO = VV xx II xx yields RR OO = RR SS (1 + gg mm RR SS ). Substituting the numerical values for gg mm and RR SS yields RR OO = 6K (1 + (6K)(0.7071 ms )) = 1.14K Part (e) The gate current is essentially zero so that RR ii = RR 1 RR 2 = 299.8K 300K Part (f) The coupling capacitor CC CC1 sees the input resistance RR ii so that ff 3dB = 1 2ππRR ii CC CC1 = 0.531 Hz 5
Part g (Micro-Cap SPICE simulation) Part 1: Dynamic DC analysis Shown is the output from the Dynamic DC analysis. Note that VV DDDDDD = 1.505 V and II DDDD = 0.251 ma. These values are close to the target design values. Part 2: AC analysis for gain and bandwidth. Left: Micro-Cap SPICE schematic. Right: AC analysis output. The voltage gain is 0.808. The 3-dB frequency is where the gain drops to 0.808 2 = 0.571. The plot shows that this occurs at 0.526 Hz. Part 3: Determining RR OO Left: Micro-Cap schematic for determining RR oo. Right: Simulation output, showing that the output resistance at 100 Hz is 1.152K 6
Part 4: Determining RR ii Left: Micro-Cap schematic for determining RR ii. Right: Simulation output, showing that the input resistance at 100 Hz is 299.74K Calculation Micro-Cap SPICE AA vv 0.809 0.808 ff 3dB 0.531 Hz 0.526 Hz RR oo 1.14K 1.152K RR ii 299.8K 299.74K The voltage gain is 0.808. The 3-dB frequency is where the gain drops to 0.808 2 = 0.571. The plot shows that this occurs at 0.526 Hz. 7
Question 6 Text book problem 3.26 Question 7 Text book problem 3.37 Question 8 Text book problem 4.34 Question 9 Text book problem 4.34 Question 10 Text book problem 4.61 8
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