9/20/2004 EE 42 fall 2004 lecture 9 1

Lecture #9 Example problems with capacitors

•Next we will start exploring semiconductor materials (chapter 2).

Reading: Malvino chapter 2 (semiconductors)

9/20/2004 EE 42 fall 2004 lecture 9 2

Simplification for time behavior of RC Circuits 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).

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

time

volt

age

input

time

volt

age

output

Long after the input change occurs things “settle down” …. Nothing is changing …. So again we have a dc circuit problem.

9/20/2004 EE 42 fall 2004 lecture 9 3

General RC Solution• Every current or voltage in an RC circuit has the following form while the sources are unchanging:

• 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.

)RC/()0tt(efx)0t(xfx)t(x

9/20/2004 EE 42 fall 2004 lecture 9 4

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.

9/20/2004 EE 42 fall 2004 lecture 9 5

Finding the Time ConstantTo 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.

9/20/2004 EE 42 fall 2004 lecture 9 6

Example

Find the current I(t).

t = 0

I

R2 = 5 kVs = 5 V + C = 1 mF

R1 = 10 k

9/20/2004 EE 42 fall 2004 lecture 9 7

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.

9/20/2004 EE 42 fall 2004 lecture 9 8

Example: Initial Condition

• 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.

t = 0

I

R2 = 5 kVs = 5 V + C = 1 mF

R1 = 10 k

9/20/2004 EE 42 fall 2004 lecture 9 9

Example: Initial Condition

• 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 k = 0.5 mA

t = 0

I

R2 = 5 kVs = 5 V + C = 1 mF

R1 = 10 k

+

0 V

_

+ 5V -

9/20/2004 EE 42 fall 2004 lecture 9 10

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 open-circuit 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.

9/20/2004 EE 42 fall 2004 lecture 9 11

Example: Final Value

• 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

t = 0

I

R2 = 5 kVs = 5 V + C = 1 mF

R1 = 10 k

9/20/2004 EE 42 fall 2004 lecture 9 12

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.

We will discuss how to combine capacitors in series and in parallel in the next lecture.

9/20/2004 EE 42 fall 2004 lecture 9 13

Example: Time Constant

• 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 k resistor is not “in” the circuit after the switch.• R = 10 k, C = 1 mF so = 10 s

t = 0

I

R2 = 5 kVs = 5 V + C = 1 mF

R1 = 10 k

9/20/2004 EE 42 fall 2004 lecture 9 14

Example: Final Answer

• 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

t = 0

I

R2 = 5 kVs = 5 V + C = 1 mF

R1 = 10 k

9/20/2004 EE 42 fall 2004 lecture 9 15

Hard Example

• Find V1(t) and V2(t).

• Find the energy absorbed by the resistors for t > 0.

5 V 2 F 2 F

1 k 1 kt = 0

1 k

+V1_

+V2_

9/20/2004 EE 42 fall 2004 lecture 9 16

Initial Conditions

• 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.

5 V 2 F 2 F

1 k 1 kt = 0

1 k

+V1_

+V2_

9/20/2004 EE 42 fall 2004 lecture 9 17

Time Constant

• How long does it take to converge after switch?• What are R and C?

5 V 2 F 2 F

1 k 1 kt = 0

1 k

+V1_

+V2_

9/20/2004 EE 42 fall 2004 lecture 9 18

Final Value

• 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…

5 V 2 F 2 F

1 k 1 kt = 0

1 k

+V1_

+V2_

9/20/2004 EE 42 fall 2004 lecture 9 19

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.

9/20/2004 EE 42 fall 2004 lecture 9 20

Application: Transmission on a wire (not too long a wire)

• 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)

Relevance to digital circuits:

We communicate with pulses

We send beautiful pulses outtime

volt

ag

e

But we receive lousy-looking pulses and must restore them

time

volt

ag

e

9/20/2004 EE 42 fall 2004 lecture 9 21

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.

A pulse originating at node I will arrive delayed and distorted at node O because it takes time to charge C through R

Input node Output node

ground

R

CVin

+

-

I O

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.)

Thus the most basic model circuit for studying transients

consists of a resistor driving a capacitor.

9/20/2004 EE 42 fall 2004 lecture 9 22

RC RESPONSE

• Vin “jumps” at t=0, but Vout cannot “jump” like Vin. Why not?

Case 1 – Rising voltage. Capacitor uncharged: Apply + voltage step

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

Input node Output node

ground

R

CVin

Vout+

-

V does not “jump” at t=0 , i.e. V(t=0+) = V(t=0-)

time

Vin

0

0

V1

Vout

Therefore the solution before the transient tells us the capacitor voltage at the beginning of the transient.

9/20/2004 EE 42 fall 2004 lecture 9 23

RC RESPONSECase 1 Continued

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)

time

Vin

0

0

V1

Vout

Input node Output node

ground

R

CVin

Vout+

-

Again the DC solution (after the transient) tells us (the asymptotic limit of) the capacitor voltage during the transient.

9/20/2004 EE 42 fall 2004 lecture 9 24

Review of simple exponentials.

Rising Exponential from Zero Falling Exponential to Zero

at t = 0, Vout = 0 , and

at t , Vout V1 also

at t = , Vout = 0.63 V1

8 at t = 0, Vout = V1 , and

at t , Vout 0, also

at t = , Vout = 0.37 V1

8

time

Vout

00

V1

.63V1

Vout

V1

time00

.37V1

Vout = V1(1-e-t/) Vout = V1e-t/