Download Capacitors: Review

Capacitors: Review
• A capacitor is a device that stores electrical potential
energy by storing separated + and – charges
– 2 conductors separated by insulating medium
– + charge put on one conductor, equal amount of – charge
put on the other conductor
– A battery or power supply typically supplies
the work necessary to separate the charge
• Simplest form of capacitor is the
parallel plate capacitor
– 2 parallel plates, each with same area A,
separated by distance d
– Charge +Q on one plate, –Q on the other
– Looks like a sandwich on a circuit diagram
Capacitors: Review
• The charge Q on and voltage V across a capacitor
are related through the capacitance C
Q
C
V
Units: C / V = Farad (F)
– “Capacity” to hold charge for a given V
– 1 F is very large unit: typical values of C are mF, nF, or pF
– Capacitance depends on the geometry of the plates and
the material (dielectric) between the plates
– “Static” description of capacitors
• A “dynamic” description of capacitor behavior comes
from taking the time derivative of the above:
– Current passed by a capacitor
dV
I C
depends on rate of change of V
dt
Capacitors: Review
• Water–pipe analogy of a capacitor
– Capacitor can be regarded as an enlargement in a water
pipe with a flexible membrane stretched across the
enlargement (see figure below)
– No water actually passes completely through pipe, but a
surge of water flows out of the right–hand pipe
– For capacitor, no DC current flows through, but AC current
does
– A stiff (flexible) membrane
corresponds to small (large)
capacitance
(Introductory Electronics, Simpson, 2nd Ed.)
RC Circuits: Review
(Lab 2–1)
• Consider a circuit with a resistor and an uncharged
capacitor in series with a battery:
V
Vi
V
0.63Vi
Vi
Vi – V
– Voltage across capacitor increases with time according to:
Vi  V
dV
C
I 
V  Vi  Aet / RC  V  Vi 1  e t / RC

dt
R
• A = –Vi since V = 0 at t = 0
• Vi = maximum (battery) voltage (only reached at t = ,
but 99% of Vi reached in t = 5t)


RC Circuits: Review
(Lab 2–1)
• Consider a circuit with a charged capacitor, a
resistor, and a switch:
V
t = RC = time constant
Vi
0.37Vi
– Before switch is closed, V = Vi and Q = Qi = CVi
– After switch is closed, capacitor discharges and voltage
across capacitor decreases exponentially with time:
dV
V
 t / RC
C
I 
V

V
e

i
dt
R
RC Circuits: Differentiators
(Lab 2–2)
• Now consider the series RC circuit as a voltage
divider, with the output corresponding to the voltage
across the resistor:
(The Art of Electronics, Horowitz and
Hill, 2nd Ed.)
d
V
– The voltage across C is Vin – V, so: C Vin  V   I 
dt
R
– If RC is small, then dV / dt  dVin / dt and
dVin V
d
C

V (t )  RC Vin (t )

dt
R
dt
• Thus the output differentiates the input!
– Simple rule of thumb: differentiator works well if Vout  Vin
RC Circuits: Differentiators
(The Art of Electronics, Horowitz and
Hill, 2nd Ed.)
• Output waveform when driven by square pulse input:
• What would happen if t = RC were too big? (See Fig. 1.38 in
the textbook for an indication of what would happen)
RC Circuits: Integrators
(Lab 2–3)
• Now flip the order of the resistor and capacitor, with
the output corresponding to the voltage across the
capacitor:
(The Art of Electronics, Horowitz and
Hill, 2nd Ed.)
dV Vin  V
– The voltage across R is Vin – V, so: I  C

dt
R
– If RC is large, then V  Vin and
dV Vin
1
C

V (t ) 
Vin (t ) dt  constant


dt
R
RC
• Thus the output integrates the input!
– Simple rule of thumb: integrator works well if Vout  Vin
RC Circuits: Integrators
(The Art of Electronics, Horowitz and
Hill, 2nd Ed.)
• Output waveform when driven by square pulse input:
(H&H)
• What would happen if t = RC were too small? (See Fig. 1.33
in the textbook)
Inductors: Review
• Inductors act as current stabilizers
– The larger the inductance in a circuit, the larger the
opposition to the rate of change of the current
– Remember that resistance is a measure of the opposition
to current
• The rate of current change in an inductor depends
on the voltage applied across it
dI
– Putting a voltage across an inductor causes V  L
dt
the current to rise as a ramp
– Note the difference between inductors and capacitors
• For capacitors, supplying a constant current causes the voltage to
rise as a ramp
• An inductor is typically a coil of wire
(hence its appropriate circuit symbol)
Voltage vs. Current in AC Circuits: Review
The RLC Series Circuit: Review
Series circuit consisting of a
resistor, an inductor, and a
capacitor connected to an AC
generator
• The instantaneous current in the circuit is the same
at all points in the circuit
i  I max sin 2ft
• The net instantaneous voltage Dv supplied by the AC
source equals the sum of the instantaneous voltages
across the separate elements
Dv  DvR  DvC  DvL
The RLC Series Circuit: Review
• But voltages measured with an AC voltmeter (Vrms)
across each circuit element do not sum up to the
measured source voltage
– The voltages across each circuit
element all have different phases
(see figure at right)
• We use the algebra of complex
numbers to keep track of the
magnitude and phases of
voltages and currents
V(t) = Re(Ve jwt)
I(t) = Re(Ie jwt)
where w = 2f
V, I are complex representations
j = (–1)1/2 (see Appendix B)
(Phase relations for RLC circuit)
Impedance
• With these conventions for representing voltages
and currents, Ohm’s law takes a simple form:
V = IZ
– V = complex representation of voltage applied across a
circuit = V0e jf
– I = complex representation of circuit current = I0e jf
– Z = total complex impedance (effective resistance) of the
circuit
• For a series circuit: Z1 + Z2 + Z3 + …
• For a parallel circuit: 1 / Z = 1 / Z1 + 1 / Z2 + 1 / Z3 + …
• The impedance of resistors, capacitors, and
inductors are given by:
– ZR = R (resistors)
– ZC = XC = –j / wC = 1 / jwC (capacitors)
– ZL = XL = jwL (inductors)
Complex Representation Example
• The presence of the complex number j simply takes
into account the phase of the current relative to the
voltage
• Example: place an inductor across
the 110 V (rms) 60 Hz power line
– The phase of the voltage is arbitrary,
so let V = V0  V(t) = Re(Ve jwt)
 V(t) = Re(Vcoswt + j Vsinwt) = V0coswt
– For an inductor, ZC = j wL
– So the (complex) current is given by: I = V / Z = V0 / j wL
= –V0 j / wL
– The actual current is then I(t) = Re(Ie jwt)
= Re(Icoswt + j Isinwt) = (V0 / wL)sinwt
 current lags the voltage by 90°
Phasor Diagrams
• Can also use phasor diagrams to keep track of
magnitude and phases of voltages
– x axis represents the “real” part of the circuit impedance
(resistance)
– y axis represents the “imaginary” part of the circuit
impedance (capacitive or inductive reactance)
– Draw vectors to represent the impedances (with their
signs); add the vectors to determine combined series
impedance
– Axes also represent (complex) voltages in a series circuit
since the current is the same everywhere, so voltage is
proportional to the impedance
Phasor Diagrams
• Example: series RC circuit
f = phase angle between
input voltage and voltage
across resistor or between
input voltage and current
f
(Student Manual for The Art of
Electronics, Hayes and Horowitz,
2nd Ed.)
– Total (input) voltage is obtained from a vector sum
– Note that the vectors indicate phase as well as amplitude
– Remember the mnemonic “ELI the ICE man”
• In an inductive circuit (L), the voltage E leads the
current I
• In a capacitive circuit (C), the current I leads the
voltage E
AC Power
• The instantaneous power delivered to any circuit
element is given by P(t) = V(t)I(t)
• Usually, however, it is much more useful to consider
the average power: Pave = Re(VI*) = Re(V*I)
– V and I are complex rms amplitudes
• Example: hook up an inductor to a
1 V (rms) sinusoidal power supply
–
–
–
–
V=1
I = V / ZL = V / jwL = –j V / wL
Pave = Re(VI*) = Re(jV / wL) = 0
Same result holds for a capacitor (this fun activity is free!)
• All power delivered to an AC circuit is dissipated by
2
R  VR2 / R
the resistors in the circuit: Pave  I rms
– In general: Pave  I rms Vin,rms cos f where cosf = power factor
RC Circuits: High–Pass Filters
• Let’s interpret the differentiator RC circuit
as a frequency-dependent voltage divider
(“frequency domain”):
Resistor–only divider:
R1
R2
Vout
(Lab 2–5)
RC differentiator circuit:
R2

Vin
R1  R2
(Student Manual for The Art of
Electronics, Hayes and Horowitz,
2nd Ed.)
C
R
– ZC = –j / wC = –j / 2 f C
 As f increases (decreases), ZC decreases (increases)
– Thus Vout (= voltage across R) increases with increasing f
and Vout / Vin  1
– Circuit passes high-frequency input voltage to output
RC Circuits: Low–Pass Filters
• Now simply switch the order of the resistor (Lab 2–4)
and capacitor in the series circuit (same
order as the integrator circuit earlier):
Resistor–only divider:
R1
Vout
R2

Vin
R1  R2
RC integrator circuit:
R
C
R2
(Student Manual for The Art of
Electronics, Hayes and Horowitz,
2nd Ed.)
– ZC = –j / wC = –j / 2 f C
 As f increases (decreases), ZC decreases (increases)
– Thus Vout (= voltage across C) increases with decreasing f
and Vout / Vin  1
– Circuit passes low-frequency input voltage to output
RC Filter Frequency Response
• The point where the output “turns the corner” is
called the 3dB point
1
f 3dB 
– Output is attenuated by 3dB relative
2RC
to the input
(for both types of filters)
– Special because a signal reduced by 3dB delivers half its
original power
• A graph of Vout (or Vout / Vin) vs. f is called the
frequency response of the RC filter:
(Student Manual for The Art of Electronics, Hayes and Horowitz, 2nd Ed.)
Example Problem #1.25
Use a phasor diagram to
obtain the low-pass filter
response formula (Vout vs. Vin)
on p. 37 of Horowitz and Hill.
R
Solution details given in class.
C
Example Problem:
Additional Exercise #1.3
Design a “rumble filter” for audio. It should
pass frequencies greater than 20 Hz (set
the –3dB point at 10 Hz). Assume zero
source impedance (perfect voltage source)
and 10k (minimum) load impedance (that’s
important so that you can choose R and C
such that the load doesn’t affect the filter
operation significantly).
Solution details given in class.