Download 2.4 Circuits with Resistors and Capacitors

2.4 Circuits with Resistors
and Capacitors
• response of a series connected resistor and
capacitor 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 is connected to the battery, current
will flow and charge the capacitor. The
charging will continue until the voltage
across the capacitor equals the voltage of
the battery.
switch
i
R
+
V
!
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.
time
0s
1
2
4
C
∞
2.4 : 2/10
capacitor
charge
0 mC
0+0.9
0.9+0.81
1.71+0.73
C
9 mC
VC
0 mC/1 mF = 0 V
0.9 mC/1 mF = 0.9
1.71 mC/1 mF = 1.71
2.43 mC/1 mF = 2.43
C
9V
VR
9V
8.1
7.29
6.57
C
0V
i
0.9 mA
0.81
0.73
0.66
C
0 mA
Response of an RC Circuit to DC (2)
V = VC + VR
Q
V = + iR
C
dV 1 dQ
di
=
+R =0
dt C dt
dt
t
t
1
di
=
−
∫ i RC ∫ dt
0
0
1
voltage (V)
The figure at the right shows the
voltage across the resistor (blue)
and capacitor (red) when a square
voltage pulse is applied the circuit
on the previous slide. The time
dependence of the current is derived
as follows.
RC = 20
0
1
100
200
The time dependence of the two
voltages is given by the following.
Ri = Ri0e−t
VR = Ve −t
RC
RC
VC = V − VR =
2.4 : 3/10
0
100
time (s)
Impedance of an RC Voltage Divider
What is the impedance of an RC
voltage divider when measuring
across the capacitor? Start with
resistance and capacitive reactance,
R
~
V
C
VC
XC
− j 2π f C
=
R + X C R − j 2π f C
solve for the impedance,
Z RC =
− j 2π f C
j 2π f C
1
=
2
R − j 2π f C R + j 2π f C
1 + ( 2π RCf )
and finally, define a characteristic frequency, _______________.
Z RC =
2.4 : 4/10
1
1 + ( f f3dB )
2
The impedance is frequency
dependent with the equation
constant depending upon RC.
Voltage Across the Capacitor
1.2
G ≡ VC V = Z RC
1
RC = 1
f3dB = 0.16
gain
0.8
as f → 0, G → 1
as f → ∞, G → 0
when f = f3dB
0.6
0.4
0.2
0
0
0.25
0.5
0.75
1
1.25
1.5
1.75
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
2
Phase Shift Between V and VC
• put the voltage divider expression into the standard form, a + jb
XC
− j 2π f C
1
2π RCf
=
=
−j
2
R + X C R − j 2π f C 1 + ( 2π RCf )2
1 + ( 2π RCf )
• derive an expression for the phase angle using the arctangent in
the complex plane
⎛ −2π RCf
⎜
2
+
1
2
RCf
π
(
)
Im
⎛
⎞
φ = tan −1 ⎜ ⎟ = tan −1 ⎜
⎜
1
⎝ Re ⎠
⎜⎜
2
RCf
π
+
1
2
(
)
⎝
⎞
⎟
⎟ = tan −1 ( −2π RCf ) =
⎟
⎟⎟
⎠
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
• A is called attenuation for
G < 1 and amplification for
G>1
• for a large range of
frequencies a Bode plot is
easier to interpret than a
linear plot
• one plot represents all
low-pass filters
• A is flat below ___f3dB and
drops by 20 dB per decade
above ___f3dB
2.4 : 7/10
2
⎛ Vout
⎞
R⎞
⎟⎟ =
⎟ = 10log ⎜⎜ 2
⎠
⎝ Vin R ⎠
10
-0.06°
-3dB
0
-0.6°
-5.7°
-45°
-10
-20dB/decade
A(dB)
⎛ Pout
A ( dB ) ≡ 10log ⎜
⎝ Pin
-20
-84°
-30
-40
-89°
-50
-89.9°
-60
-3
-2
-1
0
log(f/f3dB)
1
2
3
RC High-Pass Filter
10
-3dB
0
-10
-20 dB/decade
R
V
VR
A(dB)
~
-20
-30
C
-40
-50
-60
-3
-2
-1
0
1
2
log(f/f3dB)
for a high-pass filter voltage is measured across the resistor
*
⎛ R ⎞⎛ R ⎞
Z = ⎜
⎟⎜
⎟
R
+
X
R
+
X
C ⎠⎝
C ⎠
⎝
as f → 0, G → 0; as f → ∞, G → 1; when f = f3dB, G = 1
2.4 : 8/10
2
3
RC Band-Pass Filter
A band-pass filter can be created
by feeding the output of a lowpass filter into a high-pass filter.
R
R
Vout
Vin
GBP = GLP GHP
ABP ( dB ) =
C
C
bandpass RC filter
-10
-20
gain (dB)
The values of R and C do
not need to be the same for
the two parts of the filter.
This permits the creation of
a "flat" gain between the
low-pass and high-pass 3dB
frequencies (not shown).
0
-30
-40
-50
-60
-2
-1
0
1
log(f/f0)
2.4 : 9/10
2
3
Signal-to-Noise Enhancement
RC = 10 ms
signal + raw data
1
0
200
400
600
time (ms)
800
RC = 20 ms
This is an everpresent trade in
instrument
design.
signal
1
0.5
0
200
2.4 : 10/10
400
600
time (ms)
As the RC time
constant is
increased, the
_____ decreases
and _________
increases.
800
signal
0.5
0.5
0
200
400
600
time (ms)
800
RC = 50 ms
1
signal
signal
1
0.5
0
200
400
600
time (ms)
800