ICL8038CCPD Intersil, ICL8038CCPD Datasheet - Page 5

• Manufacturer Part Number: ICL8038CCPD
• Description: IC OSCILL GEN/VOLT CONTROL 14DIP
• Manufacturer: Intersil

Specifications of ICL8038CCPD

• Rohs Status: RoHS non-compliant

The levels of the current sources can, however, be selected
over a wide range with two external resistors. Therefore, with
the two currents set at values different from I and 2I, an
asymmetrical sawtooth appears at Terminal 3 and pulses
with a duty cycle from less than 1% to greater than 99% are
available at Terminal 9.
The sine wave is created by feeding the triangle wave into a
nonlinear network (sine converter). This network provides a
decreasing shunt impedance as the potential of the triangle
moves toward the two extremes.
Waveform Timing
The symmetry of all waveforms can be adjusted with the
external timing resistors. Two possible ways to accomplish
this are shown in Figure 3. Best results are obtained by
keeping the timing resistors R
controls the rising portion of the triangle and sine wave and
the 1 state of the square wave.
The magnitude of the triangle waveform is set at
V
SUPPLY
FIGURE 2A. SQUARE WAVE DUTY CYCLE - 50%
; therefore the rising portion of the triangle is,
7
8
10
4
R
C
A
ICL8038
11
5
FIGURE 3A.
R
B
FIGURE 3. POSSIBLE CONNECTIONS FOR THE EXTERNAL TIMING RESISTORS
A
5
and R
12
6
82K
9
3
2
B
FIGURE 2. PHASE RELATIONSHIP OF WAVEFORMS
separate (A). R
R
L
V+
V- OR GND
1
/
3
A
ICL8038
t
The falling portion of the triangle and sine wave and the 0
state of the square wave is:
Thus a 50% duty cycle is achieved when R
If the duty cycle is to be varied over a small range about 50%
only, the connection shown in Figure 3B is slightly more
convenient. A 1kΩ potentiometer may not allow the duty cycle
to be adjusted through 50% on all devices. If a 50% duty cycle
is required, a 2kΩ or 5kΩ potentiometer should be used.
With two separate timing resistors, the frequency is given by:
or, if R
t
f
f
1
2
=
=
=
=
--------------- -
t
0.33
---------- - (for Figure 3A)
RC
1
C
------------- -
C
------------ -
1
+
×
1
A
×
FIGURE 2B. SQUARE WAVE DUTY CYCLE - 80%
I
t
V
V
2
= R
R
7
8
A
=
=
=
10
B
4
-----------------------------------------------------------------------------------
2 0.22
------------------------------------------------------
R
------------ 1
0.66
C
------------------------------------------------------------------ -
(
A
= R
C
×
C
1kΩ
1/3
0.22
⎛
⎜
⎝
)
V
----------------------- -
C
SUPPLY
+
ICL8038
×
×
×
1
-------------------------
2R
V
R
11
V
5
1/3V SUPPLY
B
SUPPLY
SUPPLY
A
R
R
FIGURE 3B.
B
–
B
–
R
0.22
B
⎞
⎟
⎠
×
V
----------------------- -
12
R
6
SUPPLY
A
100K
R
9
3
2
=
A
R
----------------- -
0.66
A
=
×
R
A
C
-------------------------------------
0.66 2R A
L
= R
V+
V- OR GND
R
(
A
B
R
.
B
–
C
R
B
)