E40M Review #2 1 M. Horowitz, J. Plummer, R. Howe Electrical - PowerPoint PPT Presentation

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

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 5 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 6 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 7 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 8 M. Horowitz, J. Plummer, R. Howe

Example: Base 10: 11 – 23 = -12 9 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 10 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) 11 M. Horowitz, J. Plummer, R. Howe

Frequency Domain Analysis Circuit Output + • 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. 12 M. Horowitz, J. Plummer, R. Howe

Example: Frequency Decomposition 1 V 0.5 V 0.1 V 60 120 600 F [Hz] 2 1 0 -1 -2 13 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 200 600 F [Hz] 100 0.5 0.25 0 10 20 30 t [ms] -0.25 -0.5 14 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 15 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 R 2 or more generally, V out 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 16 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 17 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 - 18 M. Horowitz, J. Plummer, R. Howe

Asymptotic Circuit Analysis: F à ∞ Hz F à ∞ Hz open + short v i i ( t ) open - 19 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) 20 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 21 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 Gain dB 0 -20 -40 Freq RC = 1.1 ms 10 4 10 10 2 10 3 Hz 1 F c = 1/[2 p RC] =145 Hz 22 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 23 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 = = V in R + j ∗ 2 π FL 1 + j ∗ 2 π F L L 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 24 M. Horowitz, J. Plummer, R. Howe

Example: Filter and Bode Plot C 1 v in v out 10nF C 2 R 1 15nF 4.7kΩ Gain dB 0 -20 -40 Freq 10 4 1 10 10 2 10 3 Hz 25 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 = = V in R 1 1 + 2 π FR 1 C 1 + C 2 ( ) 1 + 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 1 10 10 2 10 3 Hz 26 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- 27 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 28 M. Horowitz, J. Plummer, R. Howe

Example - Find The Gain 100k 100k 20k 10k - Vout + Vin 29 M. Horowitz, J. Plummer, R. Howe

Example: Diode in the Feedback Loop Given – the diode current is: - v D /(0.025V) i D = (100 fA) e Find v o as a function of v s 30 M. Horowitz, J. Plummer, R. Howe

Example: Diode in the Feedback Loop - v D /(0.025V) i D = (100 fA) e 31 M. Horowitz, J. Plummer, R. Howe

Adding Capacitors C f • Suppose we add a capacitor in the feedback • We can treat this exactly as we did the earlier circuits by using impedances. • Our earlier analysis showed R f v o = − v s R s 1 Z f = Z s = R s 1 + j ∗ 2 π FC f 1 R f Sinusoidal voltage 1 + j ∗ 2 π FC f ⎛ ⎞ Z f R f R f 1 ∴ v o = − v s = − v s = − ⎜ ⎟ ⎜ ⎟ Z s R s R s 1 + j ∗ 2 π FR f C f ⎝ ⎠ 32 M. Horowitz, J. Plummer, R. Howe

Example: Time Domain Analysis C • What is the transfer function in the time domain? + - v o v in Sinusoidal voltage 33 M. Horowitz, J. Plummer, R. Howe