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Spring 2016
Lecture 11: Switching in RLC Circuits
Question: Why circuits with switches and switching in
general?
Answer:
(1) Circuits with switches, that do not switch,
Are equivalent to circuits that have no SWITCH.
And circuits with no SWITCH to switch,
Cannot compete with those that switch.
(2) Anything a non-switched circuit can do, a switched
circuit can do and do better. ☺
(3) Switching circuits can do thing way beyond the world of
non-switched circuits.
Example 1. R1 = R2 = 10 Ω and C = 0.1 F. Compute
vC (t), t ≥ 0 , supposing that
(1) vin(t) = −20u(−t)+ 20u(t)−10u(t −10) V;
(2) the switch S is closed for a long long time, opens at t = 0
and closes at t = 10 s.
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Step 1. Compute vC (0− ) .
(a) A constant –20 volts has excited the circuit for a
long time before 0. A very long time!!!
(b) The capacitor looks like an open circuit.
(c) Since the switch S is closed during this time frame,
by voltage division, vC (0− ) = −10 V.
Step 2. Draw the equivalent s-domain circuit valid ONLY
for 0 ≤ t ≤10 s. Then compute
VC (s) =VC ,zi(s)+VC ,zs(s)
(a) Equivalent s-domain circuit. Note: over 0 ≤ t ≤10
s, vin(t) = 20u(t) V.
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(b) Zero-state response. Set vC (0− ) = 0 . Then
VC ,zs(s) =
10s
10+ 10s
× 20s= 20
s(s+1)= 20
s− 20
s+1
Conclusion:
vC ,zs(t) = 20u(t)− 20e−tu(t) V for 0 ≤ t <10 s.
(c) Zero-input response. Set vin(t) to zero. Source
becomes a short. Since CvC (0− ) = −1,
VC ,zi(s) =1010
s
10+ 10s
(−1) = − 10s+1
Conclusion: vC ,zi(t) = −10e−tu(t) V.
(d) By superposition, for 0 ≤ t <10 s, the complete
response is:
vC (t) = vC ,zs(t)+ vC ,zi(t) = 20 1− e−t( )u(t)−10e−tu(t) V
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Note that at the next switching time, vC (10− ) ≅ 20 V. Why?
How many years, I mean, time constants have gone by?
Step 2. Draw the equivalent s-domain circuit valid ONLY
for 10 ≤ t s. Then compute
VC (s) =VC ,zi(s)+VC ,zs(s)
(a) Equivalent s-domain circuit. CvC (10− )e−10s = 2e−10s .
(b) Zero-state response. Set vC (10− ) = 0 . Here,
Vin(s) = e−10s 10
s. Then writing a simple node equation,
Spring 2016
(0.1s+ 0.1)VC + 0.1 VC − e−10s 10
s⎛⎝⎜
⎞⎠⎟= 0
in which case
VC ,zs(s) = e−10s 10
s(s+ 2)
and for t ≥10
vC ,zs(t) = 5u(t −10)−5e−2(t−10)u(t −10) V.
(c) Zero-input response. Set voltage source to zero.
CvC (0− ) = 2e−10s. Therefore,
VC ,zi(s) =510
s
5+ 10s
2e−10s( ) = 20e−10s
s+ 2
implying that vC ,zi(t) = 20e−2(t−10)u(t −10) V
(d) By superposition, for 10 ≤ t s,
Spring 2016
vC (t) = vC ,zs(t)+ vC ,zi(t)
= 5 1− e−2(t−10)( )u(t −10)+ 20e−2(t−10)u(t −10) V
Example 2. Compute vout (t) when
(i) R1 = R2 = 2 Ω, L = 1 H,
(ii) vin(t) = 4u(−t)+8u(t)+8u(t − 2) V,
(iii) the switch is initially OPEN and has been open for a
long time;
(iv) the switch CLOSES at t = 0 s, and
(v) OPENS again at t = 4 s.
Step 1. Compute iL(0− ) .
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For t < 0 , the input voltage has been 4 V for a very very
long time; the inductor looks like a short. Hence iL(0− ) = 1
A.
Step 2. Determine the s-domain equivalent circuit valid for
0 ≤ t < 4 s.
(a) Define the input Vin(s) valid for this time interval:
Vin(s) = 8
s+ 8
se−2s
(b) Draw s-domain equivalent.
Step 3. Compute vout (t) , 0 ≤ t < 4 s. By superposition,
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Vout (s) = s
s+ 2Vin(s)− 2s
s+ 2×
iL(0− )s
= 8
s+ 2+ 8e−2s
s+ 2− 2
s+ 2
CONCLUSION: for 0 ≤ t < 4 s,
vout (t) = vout ,zs(t)+ vout ,zi(t)
= 8e−2tu(t)+8e−2(t−2)u(t − 2)− 2e−2tu(t) V
Step 4. Determine s-domain equivalent valid for t ≥ 4 s.
(a) Vin(s) = 16
se−4s
V is now the “new” input seen by
the “new” circuit for t ≥ 4 s, and
(b) since iL(4− ) = −iR1(4− ) ,
iL(4− ) =
vin(4− )− vout (4− )
2
⎛
⎝⎜
⎞
⎠⎟ =
15.852
= 7.93 A
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(c) For the s-domain equivalent circuit we will use the
voltage source model for the inductor. Note, that the voltage
source value needs to be corrected to LiL(0− )e−4s = 7.93e−4s .
Step 5. By superposition
Vout (s) = s+ 2
s+ 4Vin(s)− 2
s+ 4⎛⎝⎜
⎞⎠⎟
LiL(4− )e−4s( )
= 8
s+ 8
s+ 4⎡
⎣⎢⎤
⎦⎥e−4s − 15.85
s+ 4e−4s
CONCLUSION: for 4 ≤ t ,
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vout (t) = 8 1+ e−4(t−4)( )u(t − 4)−15.85e−4(t−4)u(t − 4) V
Appendix—The Boost Converter Example:
DATA: Switch closes at t = 0, opens at t = 1, closes at t = 2,
and the process repeats for ever and ever and ever.
The following circuit represents a CHARGER for a
SUPERCAPACITOR used on a bus in Shanghai for example.
Different parameter values for the real charger and cap, but
the same type of circuit. For this analysis vin (t) = 10u(t) V.
Equivalent Circuit in s-world when switch is closed in which
k represents a non-negative integer consistent with the
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switch opening or closing. The above switch closes on even
integer values and opens on odd integer values.
Remarks:
IL(s) = 1
Ls+ R1
Ke−sk
s+ LiL(k− )e−sk⎡
⎣⎢⎢
⎤
⎦⎥⎥
and the capacitor voltage is unchanged.
Equivalent circuit in s-world when switch is open, odd
integers.
Spring 2016
With L = 2 H, C = 0.5 F, R1 = 1 Ω, and R2 = 3 Ω:
VC (s) = 1
s2 + 2s+1Ke−sk
s+ LiL(k− )e−sk⎡
⎣⎢⎢
⎤
⎦⎥⎥+ 2s+ 4
s2 + 2s+1CvC (k− )e−sk
IL(s) = s
2s2 + 4s+ 2Ke−sk
s+ LiL(k− )e−sk −
vC (k− )e−sk
s
⎡
⎣⎢⎢
⎤
⎦⎥⎥
Solution in MATLAB’s Symbolic Toolbox.
syms t s Vin VC Z1 Z2 IL vc il H
Vin = 10/s; L=2; C = 0.5; R1 = 1; R2 = 3;
Z1 = L*s + R1;
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% Part 1: Analyze Circuit over the interval 0 ≤ t < 1.
IL = Vin/Z1
IL =
10/(s*(2*s + 1))
IL = collect(IL)
IL =
10/(2*s^2 + s)
il = ilaplace(IL)
il =
10 - 10/exp(t/2)
PLOTTING
t = 0:0.01:1;
f = 10*(1 - exp(-0.5*t));
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plot(t,f)
grid
ylabel('iL(t) A')
xlabel('time in sec')
Remark: vC(t) = 0 over this interval.
% Calculate Initial Condition on Inductor at t = 1-:
>> iL1 = 10 - 10/exp(0.5)
iL1 = 3.9347e+00
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% Part 2: Analyze Circuit over the interval 1 ≤ t < 2 NOTE:
we need the initial condition on the inductor and will
assume that the initial capacitor voltage at t = 1- is ZERO.
Also, I will not use the exp(–sT) to compute the responses
over the different intevals but will insert the time shift
later—i.e. I will refer everything to t = 0, and then after
solving, will insert the “t – T”.
% Compute Transfer Function from Vin(s) to VC(s) which
also works for the V-source s-domain model of an initialized
inductor.
H = (1/(C*s))/(R1 + R2 + L*s + 1/(C*s));
H = collect(H)
H =
1/(s^2 + 2*s + 1)
% Compute vc(t)
Spring 2016
VC = H*Vin + L*iL1*H
VC =
1107517733937437/(140737488355328*(s^2 + 2*s + 1)) +
10/(s*(s^2 + 2*s + 1))
Remark: using MATLAB we can simplify the numbers as
follows:
>> 1107517733937437/140737488355328
ans =7.8694
Thus we have
>> VC = 7.8694/(s^2 + 2*s + 1) + 10/(s*(s^2 + 2*s + 1));
>> vc = ilaplace(VC)
vc =
10 - (10653*t*exp(-t))/5000 - 10*exp(-t)
>> 10653/5000
ans = 2.1306
Spring 2016
% Hence, after time shifting to the right by 1 second, the
capacitor voltage is given by the equation:
vC (t) = 10−10e−(t−1) − 2.1306(t −1)e−(t−1)⎡
⎣⎤⎦u(t −1)
REMARK: Because the switch closes at 2, we need vc(2).
Because
of the diode, the capacitor will NOT discharge (our goal is to
charge up the “super cap”
to some desired voltage. Hence vc(3)= vc(2) which is the
needed initial condition on the capacitor when the switch
reopens at t = 3 s.
>> vc2 = 10 -10*exp(-1) -2.1306*1*exp(-1)
vc2 =
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5.5374e+00
>>
>> vc3 = vc2
vc3 = 5.5374e+00
PLOTTING THE CAPACITOR VOLTAGE SO FAR
% Now because the switch closes at t = 2, we need to
compute the inductor current at t = 2 s. This is done by
Spring 2016
considering the impedance seen by the voltage sources in
the s-domain circuit for 1 ≤ t < 2 s. Voltage/Impedance =
Current:
>> Z2 = L*s + R1 +R2 + 1/(C*s);
>> Z2 = collect(Z2)
Z2 =
(2*s^2 + 4*s + 2)/s
>> IL = Vin/Z2 + L*iL1/Z2
IL =
(1107517733937437*s)/(140737488355328*(2*s^2 + 4*s + 2)) +
10/(2*s^2 + 4*s + 2)
>> iL = ilaplace(IL)
iL =
3.9347/exp(t) + 1.0653*t/exp(t))
% I fixed the numbers in the above expression.
Hence, the actual inductor current for 1 ≤ t ≤ 2 s is:
Spring 2016
iL(t) = 3.9347e−(t−1)u(t −1)+1.0653te−(t−1)u(t −1)
>> iL2 = 3.9347/exp(1)+1.0653*1/exp(1)
iL2 =
1.8394e+00
PLOTTING WE HAVE
Spring 2016
Part 3 2 ≤ t < 3: We now analyze the circuit for this new time
period when the switch (a transistor) closes at t = 2,
recharging the inductor with energy.
>> IL = Vin/Z1 + L*iL2/Z1
IL =
2070977142721217/(562949953421312*(2*s + 1)) + 10/(s*(2*s +
1))
>> iL = ilaplace(IL)
iL =
10 – 8.0616/exp(t/2)
% The correct expression shifted in time is:
iL(t) = 10u(t − 2)−8.0616e−0.5(t−2)u(t − 2)
>> iL3 = 10 - 8.0616/exp(0.5)
iL3 = 5.1104e+00
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% Notice that the whole point of the above analysis is to
compute the initial inductor current at t = 3 s which will
then charge the capacitor up to a higher voltage over the
next interval.
PLOTTING
Part 4 3 ≤ t < 4: Now we have a source, a re-charged
inductor, and an initialized capacitor to deal with.
Spring 2016
% Because it is necessary to compute the cap voltage, the s-
domain current source model is used for the capacitor. This
requires the computation of the impedance seen by that
current source. Z3 is that impedance.
>> syms Z3
>> Z3 = ((R1 + R2 + L*s)/(C*s))/(R1 + R2 + L*s + 1/(C*s));
>> Z3 = collect(Z3)
Z3 =
(2*s + 4)/(s^2 + 2*s + 1)
>> VC = H*Vin + L*iL3*H + C*vc3*Z3
VC =
5753790364987759/(562949953421312*(s^2 + 2*s + 1)) +
10/(s*(s^2 + 2*s + 1)) + (389660000182965*(2*s +
4))/(140737488355328*(s^2 + 2*s + 1))
>> vc = ilaplace(VC)
vc = ilaplace(VC)
vc =
5.7582*t/exp(t) – 4.4626/exp(t) + 10
Spring 2016
% Again as before, the actual correct expression must be
shifted in time.
vC (t) = 10u(t − 3)+5.7582(t − 3)e−(t−3)u(t − 3)− 4.4626e−(t−3)u(t − 3)
% Again as before, we need to compute the capacitor
voltage at t = 4 which is the same
as the capacitor voltage at t = 5 because the diode will not let
the capacitor discharge when the switch recloses at t = 4.
>> vc4 = 5.7582/exp(1) - 4.4626/exp(1)+10
vc4 =1.0477e+01
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% The new capacitor voltage is now bigger than 10 V which
is larger than the
value at t = 2 s. Hence the capacitor is charging up as it is
supposed to do.
% As before we need the inductor current at t = 4 so that we
can analyze
the circuit when the switch is closed over the interval 4 ≤ t <
5 s.
>> IL = Vin/Z2 + L*iL3/Z2 + (vc3/s)/Z2
IL =
(6388*s)/(625*(2*s^2 + 4*s + 2)) + 77687/(5000*(2*s^2 + 4*s +
2))
>> iL = ilaplace(IL)
iL =
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(3194*exp(-t))/625 + (26583*t*exp(-t))/10000
% Fix the numbers:
>> 3194/625
ans =5.1104
% Write the expression for iL:
iL =
5.1104*exp(-t) + 2.6583*t*exp(-t)
% The correct expression is given by changing t in the above
to t-3 and
multiplying the expression by u(t-3):
iL(t) = 5.1104e−(t−3)u(t − 3)+ 2.6583(t − 3)e−(t−3)u(t − 3)
>> iL4 =5.1104*exp(-1) + 2.6583*exp(-1)
iL4 =
2.8579e+00
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ETC
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