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# ENGG1000 Engineering Design and Innovation Capacitors

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Capacitors

1.0    The Capacitor

Capacitors are a fundamental circuit element, which physically consist of two
conducting plates very close to each other (separated by a dielectric material), as
suggested by their circuit symbol, shown in figure 1. The plates are not connected (i.e.
there is an open circuit between them), however if positive charges were to
accumulate on one of the plates, then due to the proximity of the other plate, negative
charges would be attracted to it, producing a current. The charge accumulation effect
occurs when a voltage is applied to the capacitor, which creates an electric field
between the plates, encouraging charge accumulation at the plates. If charges are
stored at the plates and then the voltage is removed, the charges will flow away from
the plates again. This behaviour leads us to think of the capacitor as a charge storage
device.

Figure 1.

The relationship between charge stored q and the voltage across a capacitor V is given
by q = CV , where C is the capacitance of the device, in Farads (F). Since the flow of
dq
charge is current I =     , this leads us to the voltage-current relationship for the
dt
capacitor:
dV (t )
I (t ) = C
dt

Where I(t) and V(t) are the capacitor current and voltage as a function of time t. What
does this mean ? If the voltage across the capacitor is constant, then no current flows.
If the voltage changes linearly, then a constant current flows.

2.0    First-order Capacitive Circuits

An example of capacitive behaviour can be seen by analyzing the circuit of figure 2.

R1      A     B
+
+   I
Vin               V              R2
-
-

Figure 2.

ENGG1000 Engineering Design and Innovation                                                 1
Suppose initially the switch (       ) is in position B, so no current is flowing in the
circuit and the capacitor voltage V = 0. Now the switch is moved to position A, and

dV (t )
Vin = R1 I (t ) + V (t ) = R1C           + V (t ) .
dt

     −
t

This   is   a differential       equation with             solution   V (t ) = Vin 1 − e R1C  .   Since
          
          
t
dV (t )                   V −
I (t ) = C         , we find I (t ) = in e R1C . Let’s look at the capacitor voltage and
dt                        R1
current at the time the switch closes (t = 0) and a long time afterwards (t → ∞):

t = 0 : V(0) = 0                       t → ∞ : V(∞) = Vin
V
t = 0 : I(0) = in                      t → ∞ : I(∞) = 0
R1

We find that initially, the voltage is zero, then it charges quickly and approaches Vin
as t approaches infinity. Note that the capacitor voltage change is continuous, i.e. it
does not change instantaneously. Initially, the capacitor current becomes quite large,
in order to charge up the capacitor, but then drops as the capacitor voltage approaches
Vin. At t = ∞, when V = Vin, the current is zero.

Suppose we let a long time T pass, so that V = Vin, and flip the switch back to position
V
B again. At this instant, the only current flowing is in , through R2. Now
R2
dV (t )
V (t ) = − R2 I (t ) = − R2C           . This is another differential equation, with solution
dt
t −T                                                      t −T
−
dV (t )                     Vin − R2C
V (t ) = Vin e   R2C
. Since I (t ) = C         , we find I (t ) = − e         . Let’s look at the
dt                        R2
capacitor voltage and current at the time the switch closes (t = T) and a long time
afterwards (t → ∞):

t = T : V(T) = Vin                     t → ∞ : V(∞) = 0
V
t = T : I(T) = − in                    t → ∞ : I(∞) = 0
R2

Note that now the current flows in the opposite direction; the capacitor is acting like a
voltage source supplying R2, in order to keep the capacitor voltage V constant when
the switch opens again. The voltage simply drops continuously from Vin back to zero
during the discharge. The voltage and current waveforms are shown in figure 3.

ENGG1000 Engineering Design and Innovation                                                                  2
capacitor charging up            capacitor discharging

V(t)
Vin

T                                 t
Vin
I(t)
R1

T                                 t
Vin
−
R2

Figure 3.

To summarise:

   The voltage across a capacitor cannot change instantaneously
   The following formula can be used to determine the capacitor voltage for
circuits like figure 2 undergoing a change in voltage from Vinitial to Vfinal at
time t = 0:
t
−
V (t ) = V final − (V final − Vinitial )e       RC

Note that R is the charging/discharging resistance, which needs to be
determined based on the circuit surrounding the capacitor. In our example,
during charging R = R1, during discharging R = R2.

Please note: These notes are intended to give the key basic knowledge required to get started in
ENGG1000. They are certainly not comprehensive, and very likely you will need to consult other
sources even just to complete your project in this course.

ENGG1000 Engineering Design and Innovation                                                         3

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