Transcript Capacitor Transient Analysis

Lesson 15 – Capacitors
Transient Analysis
Learning Objectives


Calculate capacitor voltage and current as a function
of time.
Explain Capacitor DC characteristics.
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.
Capacitor Current and Voltage

Capacitor v-i relationship
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.
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

Capacitor charging

Capacitor is initially fully discharged
 acts like a short circuit

When switch is closed (position 1), the current instantaneously
jumps to:
E 100V
iC  
 100mA
R 1000
Capacitor charging
vC (t )  E (1  e

As charge is stored in the capacitor, the
voltage across the capacitor starts to rise.

This makes the voltage drop across the
resistor drop, so current in the circuit
drops
 t / RC
)
Capacitor Charging Equations

Voltages and currents in a charging circuit change
exponentially over time
Steady State Condition (Fully Charged)

Circuit is at steady state
 When
voltage and current reach their final values and stop
changing


Capacitor has voltage across it, but no current flows
through the circuit
Capacitor looks like an open circuit
The Time Constant

Rate at which a capacitor charges and discharges depends
on R and C, which is called the TIME CONSTANT:
  RC  1000  20 x106 F   20 msec

Transients can be considered
to last for five time constants
vC (t )  E (1  e  t / )
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.
Capacitor Discharging

Capacitor is initially fully charged
 acts like a open circuit

When switch is moved to discharge, the current
instantaneously jumps to -E/R
E
100V
iC    
 100mA
R
1000
Capacitor Discharging
vC (t )  Ee

As charge flows out of the capacitor, the
voltage across the capacitor drops.

This makes the voltage drop across the
resistor drop, so current in the circuit
drops until the capacitor is fully discharged
 t / RC
Capacitor Discharging Equations

Voltages and currents in a discharging circuit also change
exponentially over time
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.
More complex circuits

Thèvenin's Equivalent of charging circuit:
ETH
RTH
9000


 24V 
  18V
 9000  3000 
 9000 3000  2250
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

You write the charging equation using Thèvenin Voltage.
vC (t )  ETH (1  e
 t / RTH C
)  18(1  e
 t /225 m sec
)
More complex circuits


The discharge portion of the circuit operates the same as we
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
Example Problem 2
The capacitor in the circuit below is initially at steady state
with the switch open and capacitor fully discharged.
After the switch is shut: (CHARGING)
a. determine how long it will take for the capacitor to fully charge
(>99% of final voltage).
b. Write the equation for vc(t). Sketch the transient.
vC (t )  E (1  e
 t /
)
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.
vC (t )  Ee
 t / RC