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Energy Storage of Capacitor • Capacitors do not dissipate power, but store energy when charging and restore it to the circuit when discharging. • How much energy does a capacitor store?
dV E = ∫ Pdt = ∫ VIdt = ∫ VC dt 0 0 0 dt T
T
T
RHS allows us to integrate wrt V instead of T, but we must change the upper limit of integration to the final capacitor voltage VT at t=T VT
E = ∫ CVdv = 0
Lecture 6
1 2 CVT 2
ENG1030 Electrical Physics and Electronics
In general, just use the formula 1
E=
B.Lovell/T.Downs
2
CV 2 1
School of Computer Science and Electrical Engineering
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Parallel Capacitors • Recall that capacitance is given by
Q C= V where Q is the charge and V is the applied voltage. • Thus if capacitors are placed in parallel, the total charge stored will sum and hence the equivalent capacitance will just be the sum of the capacitances of the individual capacitors. QTotal=Q1+Q2+Q3 Q1 V
C1
Q2 C2
Q3 C3
V
N
C1+C2+C3
CEquiv = ∑ Ci i =1
Lecture 6
ENG1030 Electrical Physics and Electronics
B.Lovell/T.Downs
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Series Capacitors • If a current, I, is passed through a set of series capacitors over a period of time, T, each will acquire the same charge Q=IT. • Each capacitor, Ci, will exhibit a voltage, Vi=Q/Ci. V1 =
Q C1
Q C1
V2 =
Q C2
Q C2
V3 =
1 1 1 VTotal = Q + + C1 C 2 C3
Q C3
Q
Q C3
1 CEquiv
1 1 1 = + + C1 C 2 C3
DC
DC
1 CEquiv
Lecture 6
ENG1030 Electrical Physics and Electronics
N
=∑
1 i =1 C i
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Parallel and Series Capacitors • In other words – Capacitors in parallel are combined like the formula for resistors in series – Capacitors in series are combined like the formula for resistors in parallel
• Non-standard capacitance values can be made be combining standard value capacitors in parallel and series.
Lecture 6
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Behavior of Simple RC Circuit • Consider that the switch has been in the position shown for a long time so that the capacitor is completely discharged. • The switch moves to the other position at time t=0. What is the behaviour of the voltage, v, across the capacitor?
R
Vs Lecture 6
ENG1030 Electrical Physics and Electronics
C
v
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Simple RC Circuit • Initially, there will be no charge on the capacitor (hence no voltage across it) and a current of i=Vs/R will flow through the resistor • As the capacitor charges, the voltage across the capacitor will rise and the current will decrease. • Eventually, the capacitor will become completely charged to the voltage Vs and no current will flow at all. R
Vs Lecture 6
ENG1030 Electrical Physics and Electronics
C
B.Lovell/T.Downs
v 6
School of Computer Science and Electrical Engineering
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Differential Equations • Using Kirchoff’s Voltage Law we have
q − Vs + iR + = 0 (1) C where i=i(t) and q=q(t) are the instantaneous current and charge respectively • Differentiating wrt t yields
di 1 di R + i = 0 ⇒ i = − RC dt C dt
( 2)
• Since this equation says that i(t) is simply a scalar, -RC, times di/dt, we guess that i(t) must take the form of an exponential function. (Recall that the derivative of ex wrt x is also ex) t − • Thus to solve (2), i(t) must take the form RC where A is a scalar constant
i = Ae
Lecture 6
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Solution to Equations
i = Ae
t − RC
• We solve for A by substituting back into (1) at time t=0. q − Vs + iR +
C
= 0 (1)
at time t = 0, q = 0, and i = Ae
-
t RC
= A, yielding
− Vs + AR = 0 (1)
• Thus A=Vs/R and the final solution is Lecture 6
ENG1030 Electrical Physics and Electronics
Vs i= e R
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t − RC
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Capacitor Voltage • The voltage across the capacitor, v, is given by Vs minus the voltage across the resistor given by VR=iR R
v = Vs − iR i
Vs
v
C
v t Lecture 6
Vs − RCt = Vs − e R R t − RC = Vs 1 − e t − RC v = Vs 1 − e
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Final Voltages and Currents • The quantity RC is called the time constant of the circuit denoted by the Greek symbol,τ, (Tau) • The time constant has the units of time (seconds).
R
i
Vs
C
v
t − v = Vs 1 − e τ
Vs i= e R Lecture 6
ENG1030 Electrical Physics and Electronics
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t − τ
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Voltage Characteristics • If we plot normalised voltage (fraction of supply voltage) across the capacitor against time expressed in multiples of the time constants, τ =RC, we obtain the following graph.
Voltage
τ 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0
Lecture 6
86.5%
95.0% 98.2%
99.3%
After t=τ time constant we obtain 63.2% of final voltage. After t=5τ, we obtain 99.3% of the final voltage
63.2%
1
2
Time
3
4
5
ENG1030 Electrical Physics and Electronics
Initial slope is Vs/τ. This is a quick way to estimate τ on an oscilloscope. B.Lovell/T.Downs
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Current Characteristics
Current
Similarly, If we plot normalised current (fraction of initial current, i0) through the capacitor against time expressed in multiples of the time constants, τ =RC, we obtain the following graph. 1 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0
Lecture 6
After t=τ time constant we obtain 36.8% of initial current. After t=5τ, we obtain 0.6% of the initial current
36.8% 13.5%
τ
1
2
Time
3
4.98%
4
1.83%
0.6%
5
ENG1030 Electrical Physics and Electronics
Initial slope is i0/τ. This is a quick way to estimate τ on an oscilloscope. B.Lovell/T.Downs
12