Memorex CP8 TURBO UNIVERSAL REMOTE CONTROL Manual de usuario descargar pdf (Pagina 96)

106

E -Z MATH

(Continued from page 84)

To be really useful, the integrator time constant

must be

very

long compared to the input pulse on and off times.

With

that relationship, the

capacitor will never fully charge or

discharge before

the

pulse

switches states. The result is

shown in Fig. 14. Assume the

same + 10

volt,

2 mS

wide

pulse as before.

But assume a time constant of 20 mS. During

the 2 mS input pulse, the capacitor

will charge up to:

V =

Vs(1 -e-t/RC)

V =10(1-e-2'20)

V= 10(I -e )

V =10(1 -.905) = 10(.095)

V =.95 V.

When

the input pulse

shuts off, the capacitor

will

discharge

toward zero and

will

be about

+ .86 volts before the

input

pulse goes

positive.

Examine the resulting

waveform in Fig. 14. The

output

amplitude is

very

low because

the pulse switches

so quickly

that the capacitor never

really has enough time

to charge to a

higher

value. But look at the shape

of the output

waveform. It

is

a triangular

shaped

wave with nearly linear (straight

line)

rises and falls of

voltage.

While

the

capacitor charge

and

discharge curves

are certainly

not linear,

we

are

only looking

at

a

very short portion

of those curves. Over such a

short

duration, the curve

is nearly linear

so the output is a

respecta-

ble triangular

wave

although

its output amplitude

is severely

limited. As you can

see, the integrator

converted our input

square

wave

into a triangular

wave.

+10V

IN

ov-

OUT

+.95V

4

oV

2mS

2mS

CHARGE CURVE

s-

Fig. 14 -An RC integrator circuit

has an

output

whose value

is directly proportional to

the

area

under the curve formed

by the input

signal's

waveform. As

you

move left on the top

curve (the

input) the area under the curve

increases

steadily

for

2

mS. Thus the

output

increases steadily.

The key

to the

integrator's

usefulness

is in making its

time

constant

very

long compared to the input pulse duration.

For

best

results, the time constant should be ten or more

times the

pulse duration.

Or:

T> 10td

Practice Problems

Its time to review the key points

in

this

section with some

exercises.

5. The duration of the on

/off

times

of a square

wave

is

150nS.

What should the time constant of a differentiator

be?

If R is 470

ohms,

what

value

of capacitor should be

used?

6. The duration

of the

on/off pulses

is 3µS.

What

is the

minimum time constant of an RC integrator?

Given C = .001

µF,

what

is R?

Answers

1. The time constant (T

= RC) is

about

150 nS. So t is two

time constants

of 300 nS. RC is 150 nS

so the

exponent

is just

t/RC = 300/150 = 2.

The equation is:

v =

5(I -e-2) =

4.3 volts.

2. It takes five time constants to

fully

charge

or discharge

or about

5

x

150 = 750 nS.

3.

If the capacitor is fully charged to 5

volts,

the

equation is

v =

5e-

2°°nso

= 5e-133

= 5(.26)

=1.3

volts.

4.

T=

RC =(lOx

103)(01

x

10-6)

=1

x

10-4 second =100

µS

t = -1

x

10-41n(1 -10/15)

t = -1

x

10-41n(1 -.66)

t = -1

x

10- 41n(.34)

t = -1

x

10-4( -1.08) =1.08

x

10-4 =1081.S

5. For a differentiator, the time constant (T)

should be less

than

one tenth the pulse duration, or

150/10 =15 nS

maximum.

Since T = RC, then C =

T /R. If R is 470, then

C =15

x

10-9'470

= .032

x

10-9

= 32

x

10-12= 32 pF.

6.

In

an

integrator,

the time constant (T)

should be ten or

more times the pulse

duration or, in this example,

10x3 =30µS

or greater. If C = .00111F, then:

R = T/C

= 30

x

10-6

/.001

x

10-6.

= 30K ohms.

if

4

"How

deep did you make that hole."

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