E40M Review #2 1 M. Horowitz, J. Plummer, R. Howe
Electrical Device: MOSFETs • Are very interesting devices – Come in two “flavors” – pMOS and nMOS – Symbols and equivalent circuits shown below • Gate terminal takes no current (at least no DC current) – The gate voltage * controls whether the “switch” is ON or OFF R on gate gate pMOS nMOS * actually, the gate – to – source voltage, V GS 2 M. Horowitz, J. Plummer, R. Howe
Building Logic Gates from MOS Transistors • pMOS connect to Vdd; nMOS connect to Gnd – Otherwise need voltage > Vdd, or < Gnd to turn them on • Need to connect output to either Vdd, or Gnd • For the gate shown below – The output is connect to Gnd when A and B are true – The output is connected to Vdd when A or B is false A B 3 M. Horowitz, J. Plummer, R. Howe
Example: CMOS Circuits • Find the node voltages and branch currents in the circuit below. • Assume R on for the transistors is 0Ω. 4 M. Horowitz, J. Plummer, R. Howe
Example: CMOS Circuits • Find the node voltages and branch currents in the circuit below. • Assume R on for the transistors is 0Ω. V B = 0V ∴ i 3 = 0 − 5V i 1 = i 2 = 0 ∴ V A = 5V V C = 5V ∴ i 6 = 0 = − 5mA 1 k Ω i 4 = i 5 = 0 5 M. Horowitz, J. Plummer, R. Howe
Truth Tables & Logic Gates A NOT A B AND !(A) NOT 0 1 0 0 0 1 0 0 1 0 1 0 0 Logic Gate Symbols 1 1 1 (A && B) AND A B OR 0 0 0 0 1 1 1 0 1 1 1 1 (A || B) OR 6 M. Horowitz, J. Plummer, R. Howe
Example: CMOS Logic Gate V dd • Construct a truth table for the CMOS logic gate shown. What does it do? !(B) !(A) A B V out 0 0 B A 0 1 V out 1 0 !(B) B 1 1 !(A) A 7 M. Horowitz, J. Plummer, R. Howe
Example: CMOS Logic Gate V dd • Construct a truth table for the CMOS logic gate shown. What does it do? !(B) !(A) A B V out 0 0 0 B A 0 1 1 V out 1 0 1 !(B) B 1 1 0 XOR Gate !(A) A 8 M. Horowitz, J. Plummer, R. Howe
Binary Numbers Operation Output Remainder Carry 1 1 23/2 7 0 23 11 1 0 1 1 23/2 6 0 23 3 0 0 1 1 23/2 5 0 23 Addition 1 1 1 0 23/2 4 1 7 (14) 23/2 3 0 7 Add by carrying 1s 23/2 2 1 3 23/2 1 1 1 23/2 0 1 0 • Subtract by borrowing 2s. 23 in binary is 00010111 9 M. Horowitz, J. Plummer, R. Howe
Generating Two’s Complement Numbers • Take your number – Invert all the bits of the number • Make all “0” “1” and all “1” “0” – And then add 1 to the result • Lets look at an example: 42 0 0 1 0 1 0 1 0 Flip 1 and 0 1 1 0 1 0 1 0 1 Add 1 1 1 0 1 0 1 1 0 • So -42 in two’s complement is 11010110 10 M. Horowitz, J. Plummer, R. Howe
Example: Base 10: 11 – 23 = -12 6 bits + sign bit: 11 = 0001011 23 = 0010111 2’s complement negative number: -23 = 1101000 (first bit is sign bit) + 1 à 1101001 Sum: 0001011 +1101001 = 1110100 Check: subtract 1 à 1110011;then flip bits: 0001100 11 M. Horowitz, J. Plummer, R. Howe
Time Division Multiplexing • If we use time division multiplexing to drive the LED array – How do you light up the red LEDs? V dd T3 0 T2 Time T1 Time slot #1: Time slot #2: T0 N0 N1 N2 N3 12 M. Horowitz, J. Plummer, R. Howe
Time Division Multiplexing • If we use time division multiplexing to drive the LED array – How do you light up the red LEDs? V dd T3 0 T2 Time T1 Time slot #1: N1 at Gnd. Time slot #2: N0, N1 and N2 at Gnd. T0 N0 N1 N2 N3 13 M. Horowitz, J. Plummer, R. Howe
Duty Cycle Code • For things that take real power – And where some elements can average – Can control output • By changing duty cycle of input • Examples – Motor (inductance) • Switch voltage, current drives motor – Power supply converter (inductance) • Switch voltage, inductance/cap filter – LED (eyes) 14 M. Horowitz, J. Plummer, R. Howe
Frequency Domain Analysis Output Circuit + • If we have a circuit with an input voltage that varies with time, we can figure out what the output of that circuit will be by considering the individual frequency components of the input signal. • Superposition will give us the resulting output. 15 M. Horowitz, J. Plummer, R. Howe
Example: Frequency Decomposition 1 V 0.5 V 0.1 V 60 120 600 F [Hz] 2 See annotated notes 1 0 0 -1 -2 16 M. Horowitz, J. Plummer, R. Howe
Example: Frequency Decomposition 1 V Output voltage: 1 V, 100 Hz and 0.5 V, 200 Hz sinusoids are multiplied by 0.2; the 0.1 V, 600 Hz is unchanged 0.5 V 0.2 V 0.1 V 0.1 V 100 200 600 F [Hz] 0.5 See annotated notes 0.25 0 10 20 30 t [ms] -0.25 -0.5 17 M. Horowitz, J. Plummer, R. Howe
Impedance • Impedance is a concept that is a generalization of resistance: R = V i R is simply a number with the units of Ohms. • What about capacitors and inductors? If V and i are sine waves, ( ) V O sin 2 π Ft Z C = V 1 Z C = V V i = = = j* 2 π FC i CdV /dt ( ) 2 π FCV O cos 2 π Ft ( ) 2 π FLi o cos 2 π Ft Z L = V Z L = V i = Ldi / dt i = j ∗ 2 π FL = i ( ) i o sin 2 π Ft 18 M. Horowitz, J. Plummer, R. Howe
Analyzing RC, RL Circuits Using Impedance C 1 v out v in Z C = Z R = R j ∗ 2 π FC R • If the circuit had two resistors then we would know how to analyze it V out or more generally, V out R 2 Z 2 = = V in R 1 + R 2 V in Z 1 + Z 2 • So we can still use the voltage divider approach with impedances 19 M. Horowitz, J. Plummer, R. Howe
Asymptotic Circuit Analysis i ( t ) i ( t ) = (2 A) cos( 2 πFt) Find the power in the 2A current source when F = 0 Hz and when F à infinity 20 M. Horowitz, J. Plummer, R. Howe
Asymptotic Circuit Analysis: F = 0 Hz F = 0 Hz à current source is 2 A, DC short + open v i short - 21 M. Horowitz, J. Plummer, R. Howe
Asymptotic Circuit Analysis: F = 0 Hz F = 0 Hz à current source is 2 A, DC short + open v i short - V i = (2 A)(5 || 15 Ω) = 7.5 V à P = - (2 A)(7.5 V) = - 15 W 22 M. Horowitz, J. Plummer, R. Howe
Asymptotic Circuit Analysis: F à ∞ Hz F à ∞ Hz open + short v i i ( t ) open - 23 M. Horowitz, J. Plummer, R. Howe
Asymptotic Circuit Analysis: F à ∞ Hz F à ∞ Hz open + short v i i ( t ) open - V i = [ i ( t )] (5 Ω) à P ( t ) = - (2 A) 2 (5 Ω) cos 2 (2 πFt ) = - (20 W) cos 2 (2 πFt ) 24 M. Horowitz, J. Plummer, R. Howe
RC Low Pass Filters R=11K W 1 v out v in V out 1 j ∗ 2 π FC 1 = = = 1 + jF/F c V in 1 1 + j ∗ 2 π FRC R + C=0.1 µ F j ∗ 2 π FC F C = 1/[2 p RC] 1 0.8 0.6 V out /V in 0.4 0.2 0 RC = 1.1 ms 0 500 1000 1500 2000 F c = 1/[2 p RC] =145 Hz F (Hz) 25 M. Horowitz, J. Plummer, R. Howe
Bode Plots • Plot of log (gain = Vout/Vin) vs. log of frequency • Why plot log? – Log converts multiplication to addition – Makes the plots very simple • If gain proportional to F, the slope of the line is 1 • If gain is constant, the slope of the line is 0 • If gain is proportional to 1/F, the slope of the line is -1 • If gain is proportional to 1/F 2 , the slope of the line is -2 26 M. Horowitz, J. Plummer, R. Howe
Plotting dB vs. Frequency • Consider the simple low pass filter we looked at earlier R = 11kΩ Gain = V out 1 1 v out v in = = V in 1 + j* 2 π FRC 1 + jF / F c C = 0.1 µF ⎛ ⎞ 1 Gain Gain dB = 20log 10 ⎜ ⎟ ⎜ ⎟ 1 + jF / F dB ⎝ ⎠ c 0 ( ) − 20log 10 1 + jF / F ( ) = 20log 10 1 c ( ) ≅ 0 − 20log 10 F / F -20 c (assuming F is large and -40 neglecting the phase) Freq RC = 1.1 ms 10 4 10 10 2 10 3 Hz 1 F c = 1/[2 p RC] =145 Hz 27 M. Horowitz, J. Plummer, R. Howe
RL High Pass v out v in j ∗ 2 π F L V out j ∗ 2 π FL R R = = 1 + j ∗ 2 π F L L V in R + j ∗ 2 π FL R Gain 1 20dB/decade F C = dB 2 π L 0 R -20 If R = 1 k Ω and L = 1 mH, then F c ≅ 159 kHz -40 Freq 10 3 10 6 10 4 10 5 10 7 Hz 28 M. Horowitz, J. Plummer, R. Howe
Example: Filter and Bode Plot C 1 V out Z R1 || Z C2 v in v out = V in Z C1 + Z R1 || Z C2 10nF C 2 R 1 Z R1 || Z C2 = Z R1 ∗ Z C2 R 1 = 15nF Z R1 + Z C2 1 + 2 π FR 1 C 2 4.7kΩ Gain R 1 dB V out 1 + 2 π FR 1 C 2 2 π FR 1 C 1 0 = = R 1 V in ( ) 1 1 + 2 π FR 1 C 1 + C 2 + 2 π FC 1 1 + 2 π FR 1 C 2 -20 1 F = 1.35kHz c = ( ) 2 π R 1 C 1 + C 2 -40 ⎛ ⎞ C 1 Gain dB @HighF = 20log ⎟ = − 7.96dB ⎜ ⎟ Freq ⎜ C 1 + C 2 ⎝ ⎠ 10 4 Hz 1 10 10 2 10 3 29 M. Horowitz, J. Plummer, R. Howe
Op-Amps • Are very high gain amplifiers – Have more gain then we need (1M) – Use feedback to get the gain we want • Two ways to figure out output voltage – Write equations for V+ and V- • As a function of input and output voltages • Solve the equations – Vout = A(V+ - V-) – Use ideal Op-Amp equations • Assume A = infinite • Find the output voltage that makes V+ = V- 30 M. Horowitz, J. Plummer, R. Howe
Ideal Op Amps The Two Golden Rules for circuits with ideal op-amps* No voltage difference between 1. op-amp input terminals 2. No current into op-amp inputs * when used in negative feedback amplifiers 31 M. Horowitz, J. Plummer, R. Howe
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