Memorex CP8 TURBO UNIVERSAL REMOTE CONTROL Manual de usuario Pagina 73
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The capacitor
will
charge or discharge
when
a pulse
is
applied to
it.
The
exact
operation of
the
circuit really depends upon two
key factors, the RC
time constant and the
on time of the
pulse. The on time of the pulse or the pulse duration
is shown
in
Fig. 8. It is less than the period of pulse. Let's examine the
operation
of
the circuit
in Fig. 9 in
detail.
First assume that the capacitor
is initially
uncharged.
The
input
pulse is like that shown in Fig. 8A and is in its zero-volt
state.
Let's further assume
that the
pulse period
is 4 mS. That
makes the pulse period one half this, or 2 mS.
We
can now compute the
time
constant
of the RC circuit.
T =RC =10x
103x.02x 10-6
T= .2x10 -'sec or 200 pS
The capacitor will charge to 63.2% of the applied
voltage,
or
6.32
volts in 200 RS. Five time constants is:
5x200 = 1000µS =1mS
-10V---
- OV -
OV - --
-0.5V
PULSE
DURATION
-12V
--
ON ON
PERIOD
ON
PULSE
DURATION
ON
PERIOD
ON ON
A
+5V
PULSE
DURATION
0 - - - -
C
-5V
PERIOD
Fig.
8 -Among
the types of voltage pulses are positive
logic (A) where "on"
is represented by a positive voltage;
negative logic (B),
where
"on"
is represented by a negative
voltage; and bipolar
(C), where "on" is signified by a
positive
voltage,
and "off"
by a negative voltage.
Vs
+10
0-
PULSE INPUT
C
.02
E
R (
V
lOKS2
a
i
Fig.
9-A differentiator
output can swing
negative
during
discharge
if the input is truly shorted at
zero volts.
As you can
see, the capacitor
will
fully charge
to IO
volts in 1
mS.
Now, if the pulse
switches on, the input
voltage is suddenly
+
10
volts. The capacitor doesn't
respond immediately.
In
tact,
it acts just
like
a
short circuit initially.
The result is that
we
see the full
+10 volts across the output
resistor. However,
this changes and then the capacitor
begins to charge.
The
voltage
across
the capacitor begins to
increase
just
as you
saw
in
Fig. 6. But,
if you look back at Fig. 9, you
will see that in
the differentiator circuit
the output is taken from across the
resistor, not across
the capacitor.
To
understand
what the resistor
voltage
looks like,
we need
to
review some
basic concepts. First,
remember Kirchoff's
voltage
law that says that the
sum of the
voltage
drops
in a
series
circuit
equals the source
voltage. In Fig. 9, the capaci-
tor voltage
(Ve)
plus
the resistor
voltage
(VR)
must equal the
supply voltage
(Vs).
Vs =Vc
+VR
The resistor or output
voltage
then is:
VR =
Vs-Ve
As you can see, as the capacitor
voltage increases, the
resistor
voltage
decreases.
As the capacitor
voltage
reaches
full charge,
the resistor
voltage will
be zero.
Another way
to
look
at this
is to observe the polarities
of
the voltages involved in Fig. 9. The pulse polarity and the
capacitor
voltage
polarity are series
opposing. The capacitor.
because
it stores
a charge,
acts like a DC source as
well. The
source
and capacitor
voltages
add algebraically
to produce
the
composite
voltage that appears across the resistor. Since
the polarities are opposite, the algebraic addition
becomes a
subtraction.
Therefore, the
resistor
voltage
is the difference
between the
source
and
capacitor
voltages.
That's
what
the
formula told us earlier. Anyway,
as the capacitor charges, the
total
voltage
applied to the resistor decreases.
Figure 10 shows
what's
happening.
As you can see, the
output
rises steeply
to
+
10
volts
as the pulse switches on,
then rapidly
decreases
as
the
capacitor charges. The capacitor
is fully charged in 1 mS or one half the pulse
width. The
resulting
output
voltage
is a narrow
positive
-going
pulse.
Now, the input pulse switches from
+
10 to zero
volts. A
+10V
vs
2mS
vc
+IOV
VR
10V
SHORTER
TIME
CONSTANT
Fig.
10- Notice
the
voltage
swing
during
discharge.
That is
due to the change
in the
current
direction upon shorting.
83
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