EE204
Basic Electronics and Electric Power
Course Notes
Resistive – Capacitive Transient Behaviour.
Now let’s look at the transient behaviour of a capacitor when subjected to s step change
in circuit conditions.
First. Let’s again review capacitors in a static DC environment:
Recall from PHY155 that the capacitance of an ideal parallel-plate capacitor, C = ε r ε o
A
,
d
where A = the plate area (m2), d = the distance between the plates (m), εo = permittivity
of free space (8.854 X 10-12 C2/N-m2), and εr is the relative permittivity of the dielectric
between the two capacitor plates (a unit-less number giving the ratio of the permittivity of
the dielectric to that of a vacuum, with a vacuum considered essentially the same as
space).
Also recall that the voltage across an ideal capacitor is proportional to the charge held on
the plates and inversely proportional to the amount of the capacitance: VC =
Q
(Q in
C
coulombs and C in farads will give V in volts). Since the amount of charge is a function
of the current and how long it has been flowing, we can also write: VC =
incremental increase in voltage can be expressed: dVC =
an expression for the current: iC = C
i ⋅t
. An
C
i ⋅ dt
, and re-arranging, we get
C
dVC
(note the “duality” with the expression for
dt
voltage across an inductor?)
In a “steady-state” DC circuit, a capacitor appears like an open circuit, with no current
flowing through it and a voltage across it determined by the other circuit components and
their configuration.
© Denard Lynch 2012
Page 1 of 7
Sep 13, 2012
EE204
Basic Electronics and Electric Power
(
Redraw the schematic VC1 = 72V
(
VC 2 = 72V
)
(
)
(
Course Notes
2Ω
= 16V
2 + 7Ω
)
7Ω
= 56V
2 + 7Ω
)
Alternatively, we could have figured out the current flowing
through R1 and R2 (8A) and determined the voltages across the
capacitors from that.
Capacitors in parallel add simply (like resistors or inductors in series):
CTotal = C1 + C2 + C3 + ...
CTotal = 800 µ F + 60 µ F + 1200 µ F = 2050 µ F
Capacitors in series add like resistors or inductors in parallel:
CTotal =
1
1
1
1
+
+
+ ...
C1 C2 C3
© Denard Lynch 2012
Page 2 of 7
Sep 13, 2012
EE204
Basic Electronics and Electric Power
Course Notes
For the transient case, it is mathematically simpler (and a better parallel to the inductor
case), to consider a simple R – C circuit with a current source instead of a voltage source
in order to derive expressions (a simple source conversion illustrates equivalence):
Using Kirchhoff’s Current Law this time, we can write an expression for the currents in
the top node (rail/wire):
Finally, recalling that R’ = R and I = E/R from the original series equivalent circuit, we
can write this in terms of the final steady-state voltage to derive an expression very
similar on form the that derived for the current through an inductor:
(8)
Now, recalling that iC = C
dv
, and using the expression for voltage across the capacitor
dt
derived above, we can also develop an expression for iC for this same circuit as a function
of time:
−t
−t
⎛ ⎛
⎞⎞
⎛
⎞
τ
d ⎜ E ⎜1 − e τ ⎟ ⎟
d
1
−
e
⎜
⎟
⎜
⎟
dv
⎝
⎠⎠
⎝
⎝
⎠
iC = C
=C
= CE
dt
dt
dt
© Denard Lynch 2012
Page 3 of 7
Sep 13, 2012
EE204
Basic Electronics and Electric Power
Course Notes
solving for iC, we are left with:
where we observe that E/R is the initial, or maximum current that will flow in the circuit
at time, t = 0+ , just after the switch is closed. Note that τ, the time constant, is RC in
this case, where again, R is the Thevenin equivalent resistance seen by the capacitor.
In keeping with previous forms, this can be written in terms of the final value:
(9)
For the decaying transient phase, it can also be shown that the current, iC is:
(11)
Where Vi and Ii are the conditions at t = 0+.. The voltage, vC is given by:
(11)
Again, the time constant, τ ’= R'C , is that of the decaying circuit, which may not be the
same as the charging circuit (always check RTh if in doubt).
© Denard Lynch 2012
Page 4 of 7
Sep 13, 2012
EE204
Basic Electronics and Electric Power
Course Notes
We can now plot these phases over time (“charging and discharging transients”):
Note: In such exponentially rising/decaying circuits, the varying quantity (i or v) is
generally assumed to have reached steady state after 5 time constants have elapsed (i.e.
5τ = 99.3% of ultimate value).
Again, we have ignored the possibility of any pre-existing conditions (like a charge on
the capacitor) in order to simplify the derivation. The implied assumption is that the
voltage was at 0 at time t 0 (i.e. when the circuit conditions for the capacitor changed).
Of course, in many situations the initial voltage is not 0, and this pre-existing condition
must be taken into consideration. The general expression will again always give the
correct expression as long as the starting and ending values are determined correctly:
−t
x( t ) = X f − (X f − X i )e τ ;
© Denard Lynch 2012
Page 5 of 7
Sep 13, 2012
EE204
Basic Electronics and Electric Power
Course Notes
Energy stored in a Capacitor:
In a derivation very similar to that used to determine the energy stored in an inductor
(using the expressions previously derived for the voltage across and current through a
capacitor), we can determine that the energy stored in a capacitor will be:
t
WC = Poweritime = ∫ iC vC dt =
0
CV 2
2
Example:
Let’s look at an example similar to the inductor’s case:
i)
ii)
iii)
Write the expressions for iC(t) and vC(t) for the charging and discharging
phases (after the switch is closed and after it is re-opened respectively).
Assume there is no current or voltage in the circuit elements prior to switch
closure.
Find the voltage across the capacitor 40µs after the switch is initially closed
Sketch the current through and voltage across the capacitor, as a function of
time, for both charging and discharging phases
© Denard Lynch 2012
Page 6 of 7
Sep 13, 2012
EE204
Basic Electronics and Electric Power
Course Notes
Again remember: τ and R may not be the same in the decaying circuit as in the charging
circuit! (And in this case again, they’re not!)
© Denard Lynch 2012
Page 7 of 7
Sep 13, 2012