E40M RC Filters M. Horowitz, J. Plummer, R. Howe 1 Reading - PowerPoint PPT Presentation

E40M RC Filters M. Horowitz, J. Plummer, R. Howe 1

Reading • Reader: – The rest of Chapter 7 • 7.1-7.2 is about log-log plots • 7.4 is about filters • A & L – 13.4-13.5 M. Horowitz, J. Plummer, R. Howe 2

EKG (Lab 4) • Concepts – Amplifiers – Impedance – Noise – Safety – Filters • Components – Capacitors In this project we will build an – Inductors electrocardiagram (ECG or EKG). This is a noninvasive device that measures the – Instrumentation and electrical activity of the heart using Operational Amplifiers electrodes placed on the skin. M. Horowitz, J. Plummer, R. Howe 3

RC Circuit Analysis Approaches 1. For finding voltages and currents as functions of time , we solve linear differential equations or run EveryCircuit. For finding the response of circuits to sinusoidal signals, * we 2. use impedances and “frequency domain” analysis * superposition can be used to find the response to any periodic signals M. Horowitz, J. Plummer, R. Howe 4

Key Ideas on RC Circuit Frequency Analysis - Review • All voltages and currents are sinusoidal • So we really just need to figure out – What is the amplitude of the resulting sinewave – And sometimes we need the phase shift too (but not always) • These values don’t change with time – This problem is very similar to solving for DC voltages/currents M. Horowitz, J. Plummer, R. Howe 5

Key Ideas on Impedance - Review • Impedance is a concept that generalizes resistance: – For sine wave input mag ( V ) Add j to represent Z = 90 o phase shift mag ( i ) • Z for a resistor is just R – It does not depend on frequency, it is simply a number. • What about a capacitor? ( ) V O sin 2 π Ft Z C = V V = = i CdV /dt ( ) 2 π FCV O cos 2 π Ft Z C = V 1 i = j ∗ 2 π FC M. Horowitz, J. Plummer, R. Howe 6

Analyzing RC Circuits Using Impedance - Review • The circuit used to couple sound into your Arduino is a simple RC circuit. • This circuit provides a DC voltage of V dd /2 at the output. • For AC (sound) signals, the capacitor will block low frequencies but pass high frequencies. (High pass filter). • For AC signals, the two resistors are in parallel, so the equivalent circuit is shown on the next page. M. Horowitz, J. Plummer, R. Howe 7

Analyzing RC Circuits Using Impedance – Review (High Pass Filter) C=0.1 µ F V out R j ∗ 2 π FRC = = v out v in 1 V in 1 + j ∗ 2 π FRC R + j ∗ 2 π FC R=110k W RC = 11ms; 2 p RC about 70ms 1 0.8 0.6 V out /V in 0.4 0.2 0 F (Hz) 0 50 100 150 200 M. Horowitz, J. Plummer, R. Howe 8

RC FILTERS M. Horowitz, J. Plummer, R. Howe 9

RC Circuits Can Make Other Filters • Filters are circuits that change the relative strength of different frequencies • Named for the frequency range that passes through the filter – Low pass filter: • Passes low frequencies, attenuates high frequency – High pass filter • Passes high frequencies, attenuates low frequencies – Band pass filter • Attenuates high and low frequencies, lets middle frequencies pass M. Horowitz, J. Plummer, R. Howe 10

RC Low Pass Filters R=11 k W • Let’s think about this before we do any v out v in math C=0.1 µ F • Very low frequencies à RC = 11 x 10 3 x 0.1 x 10 -6 s = 1.1 ms • Very high frequencies à 2 π RC = 6.9 ms 1/(2 π RC ) = 145 Hz M. Horowitz, J. Plummer, R. Howe 11

RC Low Pass Filters R=11K W 1 v out v in V out 1 j ∗ 2 π FC 1 = = = 1 + jF/F c 1 V in 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) M. Horowitz, J. Plummer, R. Howe 12

RC Filters – Something a Little More Complicated C • Let’s think about this before we v out v in do any math 10C R • Very low frequencies à • Very high frequencies à V out /V in capacitive divider F M. Horowitz, J. Plummer, R. Howe 13

RC Filters – Something More Complicated 1 C Z 1 = v in v out j ∗ 2 π FC 1 R Z 2 = = 1 1 + j ∗ 2 π F10RC R 10C R + j ∗ 2 π F10C R Z 1 V out 1 + j ∗ 2 π F10RC = R 1 V in + Z 2 1 + j ∗ 2 π F10RC j ∗ 2 π FC j ∗ 2 π FRC j ∗ 2 π FRC = = ( ) 1 + j ∗ 2 π F11 RC j ∗ 2 π FRC + 1 + j ∗ 2 π F10RC M. Horowitz, J. Plummer, R. Howe 14

RC Filters – Something More Complicated V out à Simplify using Fc = 1 / [2 π R11C] = 13 Hz = V in 0.1 V out /V in 0.08 0.06 0.04 C = 0.1µF, R =11 k W 0.02 F (Hz) 0 0 100 200 300 400 500 M. Horowitz, J. Plummer, R. Howe 15

What If We Combine Low Pass and High Pass Filters? • What do you think it will C 3 =0.1 µ F R 1 =11K do? v out v in • We’ll use a filter that operates like this in the C 2 =0.1 µ F R 4 =110K ECG lab project. V out /V in F M. Horowitz, J. Plummer, R. Howe 16

Analysis Options: Nodal Analysis i 1 i 3 Z 3 • Let’s first solve it using Z 1 -Z 4 and nodal v 1 v out analysis v in Z 1 i 2 i 4 Z 4 Z 2 i 3 = i 4 ∴ V out − V = V out Z 4 1 ∴ V out = V 1 Z 3 Z 4 Z 3 + Z 4 + V 1 − V out i 1 = i 2 + i 3 ∴ V 1 − V in = V 1 Z 1 Z 2 Z 3 • We have 2 equations in 2 unknowns (V 1 and V out ). So we could solve this for V out /V in in terms of the impedances. M. Horowitz, J. Plummer, R. Howe 17

Analysis Options: Using R, C and Voltage Dividers For convenience, let s = j*2πF C 3 =0.1 µ F R 1 =11K v out v 1 sR 4 C 3 V out R 4 v in = = 1 V 1 + sR 4 C 3 R 4 + 1 sC 3 C 2 =0.1 µ F R 4 =110K We can replace R 4 , C 3 and C 2 with Z eqv 1 + sR 4 C 3 1 1 Z eqv = = = 1 sC 3 ⎛ ⎞ sC 2 ∗ C 3 + sC 2 + sC 2 + 1 + sR 4 C 3 ⎜ ⎟ 1 ⎜ ⎟ 1 + sR 4 C 3 C 2 R 4 + ⎝ ⎠ sC 3 1 + sR 4 C 3 ⎛ ⎞ sC 2 ∗ C 3 + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ C 2 ∴ V 1 + sR 4 C 3 ⎝ ⎠ 1 = = 1 + sR 4 C 3 V in ⎛ ⎞ 1 + sR 4 C 3 + sR 1 C 2 ∗ C 3 R 1 + + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ sC 2 ∗ C 3 C 2 ⎝ ⎠ + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ C 2 ⎝ ⎠ M. Horowitz, J. Plummer, R. Howe 18

Output Response C 3 =0.1 µ F V out R 4 sR 4 C 3 R 1 =11K v out v 1 = = v in V 1 1 + sR 4 C 3 1 R 4 + sC 3 1 + sR 4 C 3 C 2 =0.1 µ F R 4 =110K ⎛ ⎞ sC 2 ∗ C 3 + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ C 2 1 + sR 4 C 3 V ⎝ ⎠ 1 = = 1 + sR 4 C 3 V in ⎛ ⎞ 1 + sR 4 C 3 + sR 1 C 2 ∗ C 3 R 1 + + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ ⎛ ⎞ sC 2 ∗ C 3 C 2 ⎝ ⎠ + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ C 2 ⎝ ⎠ V out V 1 V out sR 4 C 3 sR 4 C 3 = = 1 + s(R 4 C 3 + R 1 C 2 + R 1 C 3 ) + s 2 R 1 C 2 R 4 C 3 V in ⎛ ⎞ 1 + sR 4 C 3 + sR 1 C 2 ∗ C 3 V 1 V in + 1 + sR 4 C 3 ⎜ ⎟ ⎜ ⎟ C 2 ⎝ ⎠ Or, V out j ∗ 2 π FR 4 C 3 = 2 R 1 C 2 R 4 C 3 V in ( ) 1 + j ∗ 2 π F(R 4 C 3 + R 1 C 2 + R 1 C 3 ) + j ∗ 2 π F M. Horowitz, J. Plummer, R. Howe 19

Output Response C 3 =0.1 µ F R 1 =11K v out v 1 V out j ∗ 2 π FR 4 C 3 v in = 2 R 1 C 2 R 4 C 3 V in ( ) 1 + j ∗ 2 π F(R 4 C 3 + R 1 C 2 + R 1 C 3 ) + j ∗ 2 π F C 2 =0.1 µ F R 4 =110K 1.2 Gain of Filters 1 0.8 0.6 V out /V in 0.4 0.2 0 0 200 400 600 800 1000 F (Hz) Low Pass High Pass Band Pass M. Horowitz, J. Plummer, R. Howe 20

So What Are The Answers To These Questions? How do we design circuits that What determines how fast CMOS respond to certain frequencies? circuits can work? Why did you put a 200 µF capacitor between Vdd and Gnd on your Arduino? M. Horowitz, J. Plummer, R. Howe 21

Learning Objectives • Become more comfortable using impedance – To solve RC circuits • Understand how to characterize RC circuits – Which are low pass, high pass and bandpass filters • Be able to sketch the frequency dependence of an RC circuit by reasoning about how capacitors behave at low and high frequencies M. Horowitz, J. Plummer, R. Howe 22