9/15/2004 EE 42 fall 2004 lecture 7

Lecture #7 Graphical analysis, Capacitors•Finish graphical analysis

•Capacitors

•Circuits with Capacitors

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

Reading: Malvino chapter 2 (semiconductors)

9/15/2004 EE 42 fall 2004 lecture 7

Power of Load-Line MethodWe have a circuit containing a two-terminal non-linear element “NLE”, and some linear components.

The entire linear part of the circuit can be replaced by an equivalent, for example the Thevenin equivalent. This equivalent circuit consists of a voltage source in series with a resistor. (Just like the example we just worked!).

So if we replace the entire linear part of the circuit by its Thevenin equivalent our new circuit consists of (1) a non-linear element, and (2) a simple resistor and voltage source in series.

If we are happy with the accuracy of graphical solutions, then we just graph the I vs V of the NLE and the I vs V of the resistor plus voltage source on the same axes. The intersection of the two graphs is the solution. (Just like the problem on page 6)

9/15/2004 EE 42 fall 2004 lecture 7

The Load-Line Method

We have a circuit containing a two-terminal non-linear element “NLE”, and some linear components.

1V+-

250K

S

Non-linear element

9A1M

D

+- 2V

200K

S

NLE

D

Then define I and V at the NLE terminals (typically associated signs)

First replace the entire linear part of the circuit by its Thevenin equivalent (which is a resistor in series with a voltage source). We will learn how to do this in Lecture 11.

ID

VDS

+ -

9/15/2004 EE 42 fall 2004 lecture 7

Example of Load-Line method (con’t)

And have this connected to a linear (Thévenin) circuit

+- 2V

200K The solution !

Given the graphical properties of two terminal non-linear circuit (i.e. the graph of a two terminal device)

Whose I-V can also be graphed on the same axes (“load line”)

The intersection gives circuit solution

+- 2V

200K

S

NLE

D ID

VDS

+ -

VDS

ID A)

10

(V)1 2

IDD

S

+ -

VDS

9/15/2004 EE 42 fall 2004 lecture 7

Load-Line method

But if we use equations instead of graphs, it could be accurate

The method is graphical, and therefore approximate

It can also be use to find solutions to circuits with three terminal nonlinear devices (like transistors), which we will do in a later lecture

+- 2V

200K The solution !

VDS

ID A)

10

(V)1 2

IDD

S

9/15/2004 EE 42 fall 2004 lecture 7

Power Calculation Review

For any circuit or element the dc power is I X V and, if associated signs are used, represents heating (or charging) for positive power or extraction of energy for negative signs.

For example in the last example the NLE has a power of +1V X 5A or 5W . It is absorbing power. The rest of the circuit has a power of - 1V X 5A or - 5W. It is delivering the 5W to the NLE.

Power is calculated the same way for linear and non-linear elements and circuits.

So what it the power absorbed by the 200K resistor?

Answer: I X V is + 5mA X (5mA X 200K) = 5mW. Then the voltage source must be supplying a total of 10mW. Can you show this?

9/15/2004 EE 42 fall 2004 lecture 7

Storing charge

• If you try to store charge on a piece of metal, like a Van de Graaff generator, the work required to put more and more charge on the piece of metal gets very large, and thus the voltage gets high fast.

++

++

+

+

+

9/15/2004 EE 42 fall 2004 lecture 7

Capacitor

• If you want to store more charge, at less voltage, then you can put two plates of metal close together, and store positive charge on one, and negative charge on the other.

- - - - - - - - - - - - -+ + + + + + + + + + + +

9/15/2004 EE 42 fall 2004 lecture 7

Capacitance

• The voltage is proportional to the charge which is stored on the plates

• +Q on one• -Q on the other• Current is the flow of charge per

unit time• Since charge is proportional to

voltage, we define a proportionality constant C called the Capacitance

Time

ChargeI

dt

dQI

CVQ

dt

dVCI

9/15/2004 EE 42 fall 2004 lecture 7

Capacitance of parallel plates

• The bigger the area of two plates which are close together, the more charge they can hold for a given voltage (proportionally)

• The further apart, the bigger the voltage is for the same charge. (also proportional)

VQ Separation

Area

The variable ε is a property of the material, called the dielectric constant

Separation

AreaC

9/15/2004 EE 42 fall 2004 lecture 7

Capacitors

• A battery is a complex device, chemical reactions generate current (charges) as needed to maintain the voltage

• A capacitor is a much simpler device, it just stores charge by keeping positive charges close to the negative charges on separate conductors

• As charge is added, the repulsion and therefore the voltage increases

• As charge is removed, the voltage drops

9/15/2004 EE 42 fall 2004 lecture 7

CAPACITORS

C

i

dt

dV So

capacitance is defined by

dt

dVCi

Ci(t)

| (V

+

9/15/2004 EE 42 fall 2004 lecture 7

CAPACITORS IN SERIES

, C

i

dt

dV ,

C

i

dt

dV So

2

2

1

1

dtdV

Cdt

dVCi 2

21

1

Equivalent capacitance defined by

dt

)Vd(VC

dt

dVCi and VVV 21

eqeq

eq21eq

Clearly, SERIES IN CAPACITORS CC

CC

C1

C1

1C

21

21

21

eq

C1

V1

i(t)C2

| ( | (V2+ +

Ceqi(t)

| (Veq

+

Equivalent to

eq21

eq

C

i)

C

1

C

1i(

dt

dV so

9/15/2004 EE 42 fall 2004 lecture 7

CAPACITORS IN PARALLEL

C1i(t) C2

| ( | (

+

V

dtdV

CdtdV

C)t(i 21

Equivalent capacitance defined by

dt

dVCi eq

Ceqi(t)

| ( +

V(t)

Clearly, PARALLELIN CAPACITORS CCC 21eq

9/15/2004 EE 42 fall 2004 lecture 7

Charging a Capacitor with a constant current

C

i

dt

dV(t)

Ci

| (V(t) +

C

i

dt

dV(t) t

0

t

0 dtdt

C

ti

C

iV(t)

t

0

dt

9/15/2004 EE 42 fall 2004 lecture 7

Discharging a Capacitor through a resistor

RC

V(t)

C

i(t)

dt

dV(t)

Ci

V(t) +

R

This is an elementary differential equation, whose solution is the exponential:

// 1

dt

d tt ee /0)( teVtV Since:

i

9/15/2004 EE 42 fall 2004 lecture 7

Voltage vs time for an RC discharge

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4

Voltage

Time