2.4 Circuits with and

• response of a series connected and to a dc (steady) voltage • response of a series connected resistor and capacitor to an ac (varying) voltage • decibels and Bode plots • high-pass electronic filter • band-pass electronic filter • using a low-pass filter for signal-to-noise enhancement

2.4 : 1/10 Response of an RC Circuit to DC

Consider the circuit at the right. When the switch switch is connected to the battery, current

will flow and charge the capacitor. The i R charging will continue until the voltage + across the capacitor equals the voltage of V the battery. ! C For V = 9 V, R = 10 kΩ and C = 1 mF, the current can be computed as a function of time by tracking the total capacitor charge.

capacitor

time charge VC VR i 0 s 0 mC 0 mC/1 mF = 0 V 9 V 0.9 mA 1 0+0.9 0.9 mC/1 mF = 0.9 8.1 0.81 2 0.9+0.81 1.71 mC/1 mF = 1.71 7.29 0.73 4 1.71+0.73 2.43 mC/1 mF = 2.43 6.57 0.66 CC C CC ∞ 9 mC 9 V 0 V 0 mA

2.4 : 2/10 Response of an RC Circuit to DC (2)

RC = 20 The figure at the right shows the 1 voltage across the resistor (blue) and capacitor (red) when a square voltage pulse is applied the circuit on the previous slide. The time 0 dependence of the current is derived as follows. voltage (V) VV=+CR V 1 Q ViR=+ 100 0 100 200 C time (s) dV1 dQ di =+=R 0 The time dependence of the two dt C dt dt voltages is given by the following. tt di 1 =− dt −tRC ∫∫iRC RiRie= 0 00 −t RC VVR = e VVCR= −V =

2.4 : 3/10 Impedance of an RC Voltage Divider

What is the impedance of an RC R voltage divider when measuring across the capacitor? Start with C resistance and capacitive reactance, ~ V VC X − j 2π fC C = RX+−C Rj2π fC solve for the impedance,

− jfCjfC22ππ 1 ZRC == R−+ j22 fC R j fC 2 ππ12+ ()RCf

and finally, define a characteristic frequency, ______.

1 The impedanceπ is frequency ZRC = 2 dependent with the equation 1+ ()ff3dB constant depending upon RC.

2.4 : 4/10 Voltage Across the Capacitor

1.2

GVV≡=CRC Z 1

0.8 RC = 1 f3dB = 0.16

0.6 gain

0.4 as f → 0, G → 1 as f →∞, G → 0 0.2

when f = f3dB 0 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 frequency

When voltage is measured across the capacitor, the series combination of a resistor and capacitor is called a low-pass filter. Note that the gain of the circuit does not decrease rapidly. At

frequencies larger than f3dB the functional form approaches _____ .

2.4 : 5/10 Phase Shift Between V and VC

π • put the voltage divider expression into the standard form, a + jb X − j 212π fCπππ RCf C == −j 22 RX+−C Rj2 fC 12++()RCf 12() RCf • derive an expression for the phase angle using the arctangent in φπthe complex plane ⎛⎞−2π RCf ⎜⎟2 −−11⎛⎞Im ⎜⎟12+ ()π RCf − 1 ==tan⎜⎟ tan =−= tan() 2 RCf ⎝⎠Re ⎜⎟1 ⎜⎟2 ⎝⎠12+ ()RCfπ

when f = 0, φ = 0°; when f = f3dB, φ = -45°; as f →∞, φ → -90° • the phase shift is close to 0° at frequencies up to ____f3dB • the phase shift is close to -90° at frequencies above ____f3dB • the phase shift varies significantly from 0.1f3dB to 10f3dB • phase shifts distort the shape of the signal

2.4 : 6/10 Decibels and the Low-Pass Bode Plot

• graphs of amplitude or power gain versus frequency are usually in decibels ⎛⎞2 ⎛⎞PVRout out AdB()≡=10log⎜⎟ 10log⎜⎟ = P ⎜⎟2 ⎝⎠in ⎝⎠VRin

• A is called attenuation for 10 G < 1 and amplification for -0.06° -3dB G > 1 0 -0.6° -5.7° -45° • for a large range of -10 -20dB/decade frequencies a Bode plot is -20

easier to interpret than a -84° linear plot A(dB) -30

• one plot represents all -40 low-pass filters -89° -50 • A is flat below ___f3dB and -89.9° drops by 20 dB per decade -60 -3 -2 -1 0 1 2 3 above ___f 3dB log(f/f3dB)

2.4 : 7/10 RC High-Pass Filter

10 -3dB 0

-10 -20 dB/decade -20 R V V R A(dB) ~ -30 C -40

-50

-60 -3 -2 -1 0 1 2 3

log(f/f3dB)

for a high-pass filter voltage is measured across the resistor

* ⎛⎞⎛⎞RR Z = ⎜⎟⎜⎟ ⎝⎠⎝⎠RX++CC RX

as f → 0, G → 0; as f →∞, G → 1; when f = f3dB, G = 12

2.4 : 8/10 RC Band-Pass Filter

A band-pass filter can be created R R V by feeding the output of a low- out

pass filter into a high-pass filter. Vin

C C GGGBP= LP HP

AdBBP ()=

bandpass RC filter

0 The values of R and C do not need to be the same for -10 the two parts of the filter. -20

This permits the creation of -30

a "flat" gain between the gain (dB) -40 low-pass and high-pass 3dB frequencies (not shown). -50

-60 -2 -1 0 1 2 3

log(f/f0)

2.4 : 9/10 Signal-to-Noise Enhancement

signal + raw data RC = 10 ms

1 1

As the RC time signal 0.5 signal 0.5 constant is increased, the 0 _____ decreases 0 200 400 600 800 and ______200 400 600 800 time (ms) time (ms) increases. RC = 20 ms RC = 50 ms This is an ever- 1 present trade in 1 instrument design. signal signal 0.5 0.5

0 0 200 400 600 800 200 400 600 800 time (ms) time (ms)

2.4 : 10/10