Lecture #9 Example problems with
capacitors
•Next we will start exploring semiconductor materials
(chapter 2).
Reading:
Malvino chapter 2 (semiconductors)
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EE 42 fall 2004 lecture 9
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Simplification for time behavior of RC Circuits
We call the time period
during which the output
changes the transient
We can predict a lot about the
transient behavior from the pre- and
post-transient dc solutions
input
time
voltage
Long after the input change
occurs things “settle down” ….
Nothing is changing …. So
again we have a dc circuit
problem.
voltage
Before any input change occurs we have a dc circuit
problem (that is we can use dc circuit analysis to relate the
output to the input).
output
time
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EE 42 fall 2004 lecture 9
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General RC Solution
• Every current or voltage in an RC circuit has the following
form while the sources are unchanging:

(
t

t
)
/(
RC
)



0
x( t )  x f   x( t0 )  x f e


• x represents any current or voltage
• t0 is the time when the source voltage switches
• xf is the final (asymptotic) value of the current or voltage
All we need to do is find these values and plug in to solve for any
current or voltage in an RC circuit.
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EE 42 fall 2004 lecture 9
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Solving the RC Circuit
We need the following three ingredients to fill in our equation for
any current or voltage:
• x(t0+) This is the current or voltage of interest just after the
voltage source switches. It is the starting point of our transition,
the initial value.
• xf This is the value that the current or voltage approaches as t
goes to infinity. It is called the final value.
• RC This is the time constant. It determines how fast the current
or voltage transitions between initial and final value.
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Finding the Time Constant
To find the response of a circuit with voltage sources, resistances and
capacitances, we just need to find the starting point, the long term steady
state, and the RC constant.
It seems easy to find the time constant: it equals RC.
But what if there is more than one resistor or capacitor?
R is the Thevenin equivalent resistance with respect to the capacitor
terminals.
Remove the capacitor and find RTH. It might help to turn off the voltage
source. Use the circuit after switching.
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EE 42 fall 2004 lecture 9
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Example
R1 = 10 kW
t=0
I
+
Vs = 5 V 
R2 = 5 kW
C = 1 mF
Find the current I(t).
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Finding the Initial Condition
To find x(t0+), the current or voltage just after the switch, we use the
following essential fact:
Capacitor voltage is continuous; it cannot jump when a switch occurs.
So we can find the capacitor voltage VC(t0+) by finding VC(t0-), the voltage
before switching.
We assume the capacitor was in steady-state before switching. The
capacitor acts like an open circuit in this case, and it’s not too hard to find
the voltage over this open circuit.
We can then find x(t0+) using VC(t0+) using KVL or the capacitor I-V
relationship. These laws hold for every instant in time.
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EE 42 fall 2004 lecture 9
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Example: Initial Condition
R1 = 10 kW
t=0
I
+
Vs = 5 V 
•
•
•
•
•
R2 = 5 kW
C = 1 mF
We first find capacitor voltage right after the switch, (at t=0+) and use it to
find the current I at t=0+.
To do this, we look at the circuit before switching, because the capacitor
voltage will remain the same after switching.
Assuming the circuit has been “unswitched” for a long time, the capacitor
acts like an open circuit connected to a resistor.
The capacitor voltage before switching (at t=0-) is 0 V.
The capacitor voltage after switching (at t=0+) is 0 V.
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EE 42 fall 2004 lecture 9
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Example: Initial Condition
R1 = 10 kW
+ 5V +
Vs = 5 V 
t=0
I
R2 = 5 kW
+
0V
_
C = 1 mF
• We know the capacitor voltage is 0 V right after the
switch.
• By KVL, the resistor gets all 5 V that the source
puts out.
• So by Ohm’s law, I(0+) is 5 V / 10 kW = 0.5 mA
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EE 42 fall 2004 lecture 9
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Finding the Final Value
To find xf , the asymptotic final value, we assume that the circuit
will be in steady-state as t goes to infinity.
So we assume that the capacitor is acting like an open circuit. We
then find the value of current or voltage we are looking for using
this open-circuit assumption.
Here, we use the circuit after switching along with the opencircuit assumption.
When we found the initial value, we applied the open-circuit
assumption to the circuit before switching, and found the capacitor
voltage which would be preserved through the switch.
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EE 42 fall 2004 lecture 9
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Example: Final Value
R1 = 10 kW
t=0
I
+
Vs = 5 V 
R2 = 5 kW
C = 1 mF
• After the circuit has been switched for a long time,
the capacitor will act like an open circuit.
• Then no current can flow—eventually, I goes to
zero.
• If = 0 A
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Finding the Time Constant
It seems easy to find the time constant: it equals RC.
But what if there is more than one resistor or
capacitor?
R is the Thevenin equivalent resistance with respect
to the capacitor terminals.
Remove the capacitor and find RTH. It might help to
turn off the voltage source. Use the circuit after
switching.
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EE 42 fall 2004 lecture 9
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Example: Time Constant
R1 = 10 kW
t=0
I
+
Vs = 5 V 
R2 = 5 kW
C = 1 mF
• How long does it take for the current to converge after
switching?
• Look at the circuit after switching to find the time constant.
• The 5 kW resistor is not “in” the circuit after the switch.
• R = 10 kW, C = 1 mF
so t = 10 s
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Example: Final Answer
R1 = 10 kW
t=0
I
+
Vs = 5 V 
R2 = 5 kW
C = 1 mF
• Plugging in, we get:
I(t) = If + (I(0+) – If) e-t/RC
I(t) = 0 + (0.5 mA – 0) e-t/10 = 0.5 e-t/10 mA
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EE 42 fall 2004 lecture 9
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Hard Example
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW
• Find V1(t) and V2(t).
• Find the energy absorbed by the resistors for t > 0.
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EE 42 fall 2004 lecture 9
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Initial Conditions
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW
• Consider the circuit before the switch—if it had
been that way for a long time. Also assume second
capacitor is discharged at t=0.
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Time Constant
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW
• How long does it take to converge after switch?
• What are R and C?
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Final Value
1 kW
5V
t=0
+
V1
_
1 kW
+
V2
_
2 F
2 F
1 kW
• After a while, both capacitors are open circuits.
• They can have nonzero voltage, but voltages must be equal.
• Equate V = Q/C for capacitors to gain insight into final
value…
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EE 42 fall 2004 lecture 9
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Application: DRAM
• Dynamic Random Access Memory (DRAM) cells are really just
capacitors. Each capacitor is one bit.
• The circuit we just is similar to the writing and reading of a DRAM cell
(capacitor 1).
• A logic gate that needed to use the contents of the DRAM cell would be
represented by an RC circuit, and the DRAM capacitor would need to
charge the logic gate’s natural capacitance.
• DRAM capacitors are connected to electronic switches, not physical
switches. So there is always a little current flowing.
• This drains the charge from the DRAM capacitors. They need to be
“refreshed” periodically. Hence the term dynamic.
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EE 42 fall 2004 lecture 9
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Application:
Transmission on a wire (not too long a wire)
Relevance to digital circuits:
We communicate with pulses
voltage
• Before we analyze real electronic circuits - lets study RC circuits
• Rationale: Every node in a circuit has capacitance to ground,
like it or not, and it’s the charging of these capacitances that
limits real circuit performance (speed)
We send beautiful pulses out
But we receive lousy-looking pulses
and must restore them
voltage
time
time
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What environment do pulses face?
• Every real wire in a circuit has resistance.
• Every junction (node) has capacitance to ground and to other nodes.
• The active circuit elements (transistors) add additional resistance in
series with the wires, and additional capacitance in parallel with the
node capacitance.
Thus the most basic model circuit for studying transients
consists of a resistor driving a capacitor.
Input node
+
I
Vin
-
R
Output node
O
C
A pulse originating at node I will
arrive delayed and distorted at
node O because it takes time to
charge C through R
ground
If we focus on the circuit which distorts the pulses produced by Vin,
its most simple form consists simply of an R and a C. (Vin represents
the time-varying source which produces the input pulse.)
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RC RESPONSE
Case 1 – Rising voltage. Capacitor uncharged: Apply + voltage step
V1
Vin
Input node
Output node
+
Vout
Vin
-
0
0
R
time
Vout
C
ground
• Vin “jumps” at t=0, but Vout cannot “jump” like Vin. Why not?
 Because an instantaneous change in a capacitor voltage would require
instantaneous change in the charge on the capacitor, and an
instantaneous change in voltage would require an infinite current
V does not “jump” at t=0 , i.e. V(t=0+) = V(t=0-)
Therefore the solution before the transient tells us the capacitor voltage
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EE 42 fall 2004
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at the beginning
oflecture
the transient.
RC RESPONSE
Case 1 Continued
V1
Vin
Input node
+
Vout
Vin
-
0
0
R
time
Output node
Vout
C
ground
After the transient is over (nothing changing anymore) it means d(V)/dt
= 0 ; that is all currents must be zero. From Ohm’s law, the voltage
across R must be zero, i.e. Vin = Vout.
 That is, Vout  V1 as t  .
(Asymptotic behavior)
Again the DC solution (after the transient) tells us (the asymptotic limit
of) the capacitor voltage during the transient.
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EE 42 fall 2004 lecture 9
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Review of simple exponentials.
Rising Exponential from Zero
Falling Exponential to Zero
Vout = V1(1-e-t/t)
Vout = V1e-t/t
Vout = 0 , and
at t = 0,
Vout = V1 , and
at t  ,
8
Vout  V1 also
at t  ,
at t = t,
Vout  0, also
Vout = 0.63 V1
at t = t,
Vout = 0.37 V1
8
at t = 0,
Vout
Vout
V1
V1
.63V1
.37V1
0
0
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EE 42 fall 2004 lecture 9
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time
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