Download Capacitors Transient Analysis

Lesson 12:
Capacitors Transient Analysis
1
Learning Objectives
• Calculate capacitor voltage and current as a function of time.
• Explain DC characteristics.
• Calculate inductor energy stored.
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Transients in Capacitive Networks:
The Charging Phase
• The placement of charge on the plates of a
capacitor does not occur instantaneously.
• Instead, it occurs over a period of time determined
by the components of the network. This period of
time is called the Transient Phase.
3
Capacitor Voltage and Current
Relationship
• Capacitor v-i relationship:
4
Capacitor Current and Voltage
• The charge on a capacitor is given by:
Q  CVC
• Current (iC) is the rate of flow of charge:
dq d
dVC
iC 
  CVC   C
dt dt
dt
(A)
• Current through a capacitor is equal to C times the
rate of change of voltage across it.
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Circuit Analysis
(for Physics Majors)
• Using KVL:
vR  vC  E
• Substituting in using ohm’s law and the capacitor current
relationship:
Ric  vC  E
 dvc 
RC
  vC  E
 dt 
• Using Calculus:
vC  E 1  e
 t / RC

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Capacitor Charging
• Capacitor is initially fully discharged:
− acts like a short circuit.
• When the switch is closed to position 1, the current
instantaneously jumps to E :
R
E 100V
iC  
 100mA
R 1000
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Capacitor Charging
• As charge is stored in the capacitor, the
voltage across the capacitor starts to rise.
vC (t )  E (1  e
 t / RC
• This reduces the voltage across the
resistor, therefore current in the circuit
drops.
8
)
Capacitor Charging Equations
• Voltages and currents in a charging circuit change
exponentially over time.
Voltages Increases
Exponentially Over Time
Current Decreases
Exponentially Over Time
9
Steady State Condition
(Fully Charged)
• When a circuit is at steady state:
− The voltage and current reach their final values and stop
changing.
• A capacitor, at steady state, will have voltage across
it, but no current will flows through the capacitor.
• Capacitor ‘looks’ (electrically) like an open circuit.
10
The Time Constant
• The rate at which a capacitor charges and discharges depends
on resistance (R) and capacitance (C).
• This discharge rate is called the TIME CONSTANT (τ):
 
V   Q   V   Q 
   *    *    R *C
 I  V   Q  V 
 t 
  RC  (1000  20 x106 F   20 msec
• As the voltage increases, there will
be transients, which can be considered to
last for five time constants.
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sec
Example Problem 1
The capacitor in the circuit below is initially uncharged.
After the switch is shut:
a. determine how long it will take for the capacitor to reach a steady-state
condition (>99% of final voltage).
b. Write the equation for vc(t).
c. Sketch the transient.
a)   RC
  RC  1k  100F   100 msec
b)
vC (t )  E (1  e  t / )
vC (t )  24(1  e
 t /100 m sec
c)
24V
)
5
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Capacitor Discharging
• When a capacitor is initially fully charged:
− It acts like an open circuit.
• When the switch is moved to the discharge position, the
E
current instantaneously drops to  R .
E
100V
iC    
 100mA
R
1000
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Capacitor Discharging
• As charge dissipates from the capacitor,
the voltage reduces across the capacitor.
vC (t )  Ee
• This makes the voltage across the
resistor go down, so current in the circuit
reduces until the point that the capacitor
is fully discharged (V=0 and I=0).
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 t / RC
Capacitor Discharging Equations
• Voltages and currents in a discharging circuit also change
exponentially over time.
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More Complex Circuits
• If the circuit does not look like the simple charge-discharge
circuit, then you will need to use Thèvenin's Equivalent to
make it into the simple circuit.
• The circuit below does not have the same charging equation
as the previous circuits, since the voltage drop across the
capacitor is controlled by the voltage divider circuit.
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More Complex Circuits
• Thèvenin's Equivalent of charging circuit:
ETH
9000


 24V 
  18V
 9000  3000 
RTH  9000 3000  2250
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More Complex Circuits
• Now you can calculate the charging time constant using the
Thèvenin Equivalent resistance:
  RTH C
 2250 100 x10
6
F
  225 msec
• Now you can write the charging equation using the Thèvenin
Voltage.
vC (t )  ETH (1  e
 t / RTH C
18
)  18(1  e
 t /225 m sec
)
More Complex Circuits
• The discharge portion of the circuit operates the same as
previously analyzed.
• The steady-state (fully charged) voltage across the capacitor
can be determined by the VDR (this is the Thèvenin voltage
found earlier).
vC (t )  ETH e  t / R2C  18e  t /900 m sec V
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Example Problem 2
The circuit below will not have the same charging equation due to the voltage
divider in the system.
After the switch is shut: (CHARGING)
a. Transform the circuit into the Thèvenin equivalent as seen by the
capacitor.
b. Determine the charging constant.
c. Write the charging equation for vc(t).
1
a) RTH
1 
 1


  1.82k 
 20k  2k  
 20k  
VTH  44V * 
  40V
22
k



b)   RTH C  1.82k   60 F 
1.82kΩ
  109.2 msec
40V
c) vC (t )  ETH (1  e  t / )  40V (1  e  t /109 m sec )
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Example Problem 2b
The capacitor is now fully charged and at steady-state condition. The switch is
opened to start the discharge cycle.
After the switch is open: (DISCHARGING)
a. determine how long it will take for the capacitor to fully discharge .
b. Identify the direction of current flow.
c. Write the equation for vc(t). Sketch the transient.
a)  d  R2C  20k   60 F 
 d  1.2 sec
b) see circuit to the right
2
c) vC (t )  ETH (e t / )  40V (e t /1.2sec )
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20kΩ
QUESTIONS?
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