w. g. oldham eecs 40 fall 2001 lecture 5 copyright regents of university of california 1 review of...

W. G. OldhamEECS 40 Fall 2001 Lecture 5 Copyright Regents of University of California

1

Review of charging and discharging in RC Circuits

• Lectures 2 and 3: Gate Delay Ideas• Lecture 4: Basic properties of circuit elements

• Today: Finish up circuit elements• How to find delays in circuits which have to charge capacitors (which

every circuit effectively has)


2

BASIC CIRCUIT ELEMENTS

• Voltage Source• Current Source

• Resistor• Capacitor

• Inductor

(like ideal battery)(always supplies some constant given current)

(Ohm’s law)(capacitor law – based on energy storage in electric field of a dielectric)

(inductor law – based on energy storage in magnetic field produced by a current


3

DEFINITION OF IDEAL VOLTAGE SOURCE

Special cases:

upper case V constant voltage … called “DC”

lower case v general voltage, may vary with time

Symbol

V

Note: Reference direction for voltage source is unassociated (by convention)

Examples:

1) V = 3V

2) v = v(t) = 160 cos 377t

Current through voltage source can take on any value (positive or negative) but not infinite


4

IDEAL CURRENT SOURCE

Actual current source examples – hard to find except in electronics (transistors, etc.), as we will see

+ note unassociated

v direction

upper-case I DC (constant) value

lower-case implies current could be time-varying i(t)

“Complement” or “dual” of the voltage source: Current though branch is independent of the voltage across the branch

i


5

CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT SOURCES

Describe a two-terminal circuit element by plotting current vs. voltage

V

i

Assume unassociated signs

Ideal voltage source

V

If V is positive and I is only positive

releasing power

absorbing power (charging)

But this is arbitrary; i might be negative so we extend into 2nd quadrant

But this is still arbitrary, V could be negative; all four quadrants are possible

absorbing power

releasing power


6

CURRENT-VOLTAGE CHARACTERISTICS OF VOLTAGE & CURRENT SOURCES (con’t)

V

iabsorbing power releasing power

Remember the voltage across the current source can be any finite value (not just zero)

If i is positive then we are confined to quadrants 4 and 1:

And do not forget i can be positive or negative. Thus we can be in any quadrant.

i

+

v

_


7

RESISTOR

If we use associated current and voltage (i.e., i is defined as into + terminal), then v = iR (Ohm’s law)

+

v

i

R

Question: What is the I-V characteristic for a 1K resistor?

I (mA)

V1 2 3

1

2

3

Slope = 1/R

Answer: V = 0 I = 0

V = 1V I = 1 mA V = 2V I = 2 mA

etc

Note that all wires and circuit connections have resistance, though we will most often approximate it to be zero.


8

CAPACITORAny two conductors a and b separated by an insulator with a difference in voltage Vab will have an equal and opposite charge on their surfaces whose value is given by Q = CVab, where C is the capacitance of the structure, and the + charge is on the more positive electrode.

We learned about the parallel-plate capacitor in physics. If the area of the plate is A, the separation d, and the dielectric constant of the insulator is , the capacitance equals C = A /d.

A d

Symbol or

But so

where we use the associated reference directions.dt

adQ

i dt

dvCi

i

+a

b

VabConstitutive relationship: Q = C (Va Vb).

(Q is positive on plate a if Va > Vb)


9

ENERGY STORED IN A CAPACITOR

You might think the energy (in Joules) is QV, which has the dimension of joules. But during charging the average voltage was only half the final value of V.

Thus, energy is .2CV2

1 QV

2

1


10

ENERGY STORED IN A CAPACITOR (cont.)

More rigorous derivation: During charging, the power flow is v i into the capacitor, where i is into + terminal. We integrate the power from t = 0 (v = 0) to t = end (v = V). The integrated power is the energy

i

v

+

Vv

0vdq vdt

endt

0

dt

dqend

0vdt ivE

but dq = C dv. (We are using small q instead of Q to remind us that it is time varying . Most texts use Q.)

2CV2

1Vv

0vdv CvE


11

Capacitor Inductor

dt

dVCi

dt

diLv

2CV2

1E

2LI

2

1E

INDUCTORSInductors are the dual of capacitors – they store energy in magnetic fields that are proportional to current.

Switching properties: Just as capacitors demand v be continuous (no jumps in V), inductors demand i be continuous (no jumps in i ). Reason? In both cases the continuity follows from non-infinite, i.e., finite, power flow.

Capacitor Inductorv is continous i is continousI can jump V can jumpDo not short circuit acharged capacitor(produces current)

Do not open an inductorwith current flowing(produces voltage)


12

SWITCHING PROPERTIES OF L, C

Capacitor Inductorv is continous i is continousI can jump V can jumpDo not short circuit acharged capacitor(produces current)

Do not open an inductorwith current flowing(produces voltage)

Rule: The voltage across a capacitor must be continuous and differentiable. For an inductor the current must be continuous and differentiable.

Basis: The energy cannot “jump” because that would require infinite energy flow (power) and of course neither the current or voltage can be infinite. For a capacitor this demands that V be continuous (no jumps in V); for an inductor it demands i be continuous (no jumps in i ).


13

The RC Circuit to Study(All single-capacitor circuits reduce to this one)

• R represents total resistance (wire plus whatever drives the input)

Input node Output node

ground

R

C

• C represents the total capacitance from node to the outside world (from devices, nearby wires, ground etc)


14

RC RESPONSE EXAMPLECase 1 Capacitor uncharged: Apply + voltage step of 2V


ground

R

CVin

Vout+

-

time

Vin

0

0

The capacitor voltage, Vout, is initially zero and rises in response to the step in input.

Its initial value, just after the step, is still zero WHY?

Its final value, long after the step is is 5V WHY?

5

timeV

ou

t

0

0

5

We need to find the solution during the transient.

Time period during which the output changes: transient


15

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 increase in energy stored (1/2CV²), that is, infinite power. (Mathematically, V must be differentiable: I=CdV/dt)


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


16

RC RESPONSE

Case 1 – Capacitor uncharged: Apply voltage step

• 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.

• Vout approaches its final value asymptotically (It never quite gets to V1, but it gets arbitrarily close). Why?

That is, Vout V1 as t . (Asymptotic behavior)

time

Vin

0

0

V1

Vout


ground

R

CVin

Vout+

-

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


17

RC RESPONSE: Case 1 (cont.)

time

Vo

ut

00

V1

time

Vin

00

V1

Exact form of Vout?Equation for Vout: Do you remember general form?

?

Exponential!

Vout Input node Output node

ground

R

CVin

Vout+

-

Vout = V1(1-e-t/)

time

Vo

ut

00

V1


18

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/


19

Further Review of simple exponentials.

Rising Exponential from Zero Falling Exponential to Zero

time

Vout

00

V1 + V2

.37V1 + V2

V2

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

Vout = V1(1-e-t/) + V2 Vout = V1e-t/ + V2

.63V1+ V2

Vout

0

V1 + V2

time0

V2

We can add a constant (positive or negative)


20

Further Review of simple exponentials.Rising Exponential Falling Exponential

Both equations can be written in one simple form:

Thus: if B < 0, rising exponential; if B > 0, falling exponential

Initial value (t=0) : Vout = A + B. Final value (t>>): Vout = A

time

Vout

00

A+B

A

Here B > 0

Vout

0

A

time0

A+B

Here B < 0

Vout = V1(1-e-t/) + V2Vout = V1e-t/ + V2

Vout = A + Be-t/


21

RC RESPONSE: Proof

Vin

RVout

CiCiR

law) ce(capacitan dt

dVCi law) s(Ohm'

R

VVi out

Coutin

R

! ii But CR )VV(RC

1

dt

dV

dt

dVC

R

VV Thus, outin

outoutoutin

or

RC is “time constant” !

Large R or large C => slow rise and fall


22

RC RESPONSE: Proof

RCt/-outin 0)]e(tV-[V

RC

1

)]]eV-0)(t[V(V-[V RC

1 (t))V-(V

RC

1 RCt/-inoutininoutin

RCt/-outin 0)]e(tV-[V

RC

1

)VV(RC

1

dt

dVoutin

out

Vout(t) = Vin + [Vout(t=0)-Vin]e-t/RC

We know

Is a solution? Substitute:

RC/tinout

RC/tinoutin

out e]V)0t(V[RC

1e]V)0t(V[V

dt

d

dt

dV


23

Example


ground

R

CVin

Vout+

-

The simple RC circuit with a step input has a universal exponential solution of the form:

Example: R = 1K, C = 1pF, Vin steps from zero to 10V at t=0:

time

Vin

0

0

10

Vout

1nsec

6.3V

1) Initial value of Vout is 0

2) Final value of Vout is 10V

3) Time constant is 10-9 sec

4) Vout reaches 0.63 X 10 in 10-9 sec

Vout = A + Be-t/RC

Vout = 10 - 10e-t/1nsec

DRAW THE GRAPH

Now just write the equation:


24

Charging and discharging in RC Circuits(The official EE40 Easy Method)


ground

R

CVin

Vout+

-

4) Solve the dc problem for the capacitor voltage after the transient is over. This is the asymptotic value.

5) Sketch the Transient. It is 63% complete after one time constant.

6) Write the equation by inspection.

1) Simplify the circuit so it looks like one resistor, a source, and a capacitor (it will take another two weeks to learn all the tricks to do this.) But then the circuit looks like this:

Method of solving for any node voltage in a single capacitor circuit.

2) The time constant of the transient is = RC.

3) Solve the dc problem for the capacitor voltage before the transient. This is the starting value (initial value) for the transient voltage.

w. g. oldham eecs 40 fall 2001 lecture 5 copyright regents of university of california 1 review of...

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