Memorex CP8 TURBO UNIVERSAL REMOTE CONTROL Manual de usuario descargar pdf (Pagina 72)

100

v=12(1.12)

Now,

subtract

.12 from

get

.88.

Finally,

multiply

the

source

voltage

the above value

get

the capacitor voltage:

v =12(.88)=10.56

volts

Now,

that wasn't

too

hard, was

it?

Refer back

Fig.

5. Assume we

allowed

the capacitor

fully

charge

volts.

Now, we

throw the switch

the B

position.

The capacitor

will

discharge

through

the resistor.

The voltage

across

the capacitor will

decline

until it is

even-

tually

zero. It

takes one

time

constant

(T =RC)

for the

capacitor

to discharge

to a

voltage

of 36.8%

of the

initial

applied voltage.

Using

the same values

earlier, we

can say

that in

.47 seconds,

the

capacitor voltage

drops

.368

= 4.4 volts.

takes about

5 time

constants

or 2.35

seconds

for

the capacitor

to fully

discharge.

The

discharge

curve,

therefore,

looks like

that in Fig.

7. Its

equation

is:

v =

V,e

-vRc

Here, V,

the initial voltage

the capacitor.

Assume we

let

the capacitor

charge

to the

full 12

volts.

Then we

flipped

the

switch

to the B position. What

the

voltage

after 1.5 sec-

onds'? In

the formula,

t =1.5 second.

V =12e-1'5/

First,

compute

the

exponent,

or:

1.5/.47 =

3.19

Make it

negative,

then press

INV

and

e". You should

get

.0411.

Now

the

equation looks

like this:

12(.0411)

Making

the final

multiplication

gives you

the

voltage

on the

capacitor

after 15

seconds

about .49

volts.

Exercise

Problems

Try your

hand

at this

now;

Assume a

capacitor

of 220pF,

resistor

of 680 ohms,

and

a supply voltage

volts.

What

is the voltage

on the capacitor

after

2 time

con-

stants'?

2. How

long

does it

take for the

capacitor

to fully

charge

(or

discharge)?

3. If the

capacitor

is fully

charged, what

is its voltage

after

200

nS?

Rearranging

A lot

of times

you want

to know what

the time

(t)

is at a

stated

charge

discharge voltage

(v). That

easy to find,

but

you

have

to rearrange

the

above

equations

to solve

for

t in

terms

of v.

won't

bore

you with

the messy

translation

here.

but

if you're

so inclined,

feel free

take a swipe

the

algebra

yourself.

Otherwise,

just use

the equations

below.

Charge: t =

RCIn(1 v /Vs)

Discharge:

t = RCIn(v /V;)

So,

let's

work

on a

couple of problems.

Look

back at Fig.

How much

time

does it take

for the

capacitor

to charge

let's say

volts?

t = RCIn( Iv/V,)

TABLE

1 -FARAD

UNIT

CONVERSIONS

GuNVERT

µF

INTO

µF

THEN

106

1012

10-6

106

10-12

-6

2 3

TIME

CONSTANTS

(T)

Fig.

7 -A

capacitor

discharges in

a fashion

similar

to the

way it charges.

quickly at first.

but slowly as time

continues. Its

pace continues to

slow as time goes

on.

RC is our

time constant

which

previously computed

.47.

The instantaneous voltage

6 volts.

The supply voltage

is 12 volts.

So:

t = .471n(I -6/12)

t = .471n(.5)

To find

the natural log

of .5 on your calculator,

just key

.5, then

press the LN

key. You should get 0.693.

Multiplying

this

by .47 gives:

t =.47(.693)

t = .326

seconds

So after .326

seconds of

charging, the voltage will

be 6

volts.

Finding the discharge

time is just

as easy. Assume

a full

charge

of 12

volts.

How long

does it take

the capacitor

voltage

drop to 3

volts?

t =RCIn(v/V;)

t = .47In(3/12)

t =.471n(.25)

t =.47(-1.39)

t = .65 seconds

Exercise Problem

By now you

have probably

mastered that,

but just to be

sure,

check yourself

on this problem.

4. How long

does it take

a .01µF capacitor

to charge

through a 10K ohm

resistor

to IO

volts

if the supply voltage

volts?

RC Pulse

Shaping

One common

application of RC networks is pulse shaping.

A pulse

is a

voltage

(or current)

that

switches

rapidly between

two levels as shown in

Fig. 8. At A, the pulse switches

between zero

and

+10 volts.

At B, the pulse switches

be-

tween 0.5

and 12

volts.

In Fig. 8C, the two levels

are

and 5

volts.

The

pulses at A and B are DC, the one in

AC. When

the pulse on and off times are

equal,

call it a

square wave.

It is frequently

necessary or desirable to change the shape

the pulse. One example is to create narrower pulses at the

points where

the pulse switches on or off. Such a circuit for

doing that is shown in Fig. 9. It is called a differentiator

circuit. It is nothing more than a simple series RC network.

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