AN013 [INTERSIL]
Everything You Always Wanted to Know About the ICL8038; 你一直想知道的关于ICL8038
| 型号： | AN013 |
| 厂家： | 
Intersil |
| 描述： | Everything You Always Wanted to Know About the ICL8038 |
| 文件： | 总4页 (文件大小：89K) |
| 中文： | 中文翻译 | 下载： | 下载PDF数据表文档文件 |
Everything You Always Wanted to Know
About the ICL8038
Application Note
November 1996
AN013.1
Author: Bill O’Neil
Answer
Introduction
First of all, the voltage difference need only be a few hundred
millivolts so there is no danger of damaging the 8038. One
way to get this higher potential is to lower the supply voltage
on the 8038 and external resistors. The simplest way to do
The 8038 is a function generator capable of producing sine,
square, triangular, sawtooth and pulse waveforms (some at
the same time). Since its introduction, marketing and appli-
cation engineers have been manning the phones explaining
the care and feeding of the 8038 to customers worldwide.
This experience has enabled us to form articulate responses
to the most frequently asked questions. So, with data sheet
and breadboard in hand, read on and be enlightened.
this is to include a diode in series with pin 6 and resistors R
A
and R . See Figure 1. This technique should increase the
B
sweep range to 1000:1.
+15V
Question 1
1N457
I want to sweep the frequency externally but can only get a
range of 100:1 (or 50:1, or 10:1). Your data sheet says
1000:1. How much sweep range can I expect?
0.1µF
DUTY
CYCLE
15K
1K
Answer
4.7K
R
B
4.7K
Let’s look at what determines the output frequency. Start by
examining the circuit schematic at pin 8 in the upper left hand
corner. From pin 8 to pin 5 we have the emitter-base of NPN
R
A
5
4
6
Q and the emitter-base of PNP Q . Since these two diode
1
2
9
3
2
drops cancel each other (approximately), the potential at pins
8, 5, and 4 are the same. This means that the voltage from V+
to pin 8 is the same as the voltage across external resistors
100K
FREQUENCY
LOG POT
8
7
R
and R . This is a textbook example of a voltage across
A
B
10
11 12
two resistors which produce two currents to charge and dis-
charge a capacitor between two fixed voltages. This is also a
linear system. If the voltage across the resistors is dropped
from 10V to 1V, the frequency will drop by 10:1. Changing
from 1V to 0.1V will also change the frequency by 10:1.
Therefore, by causing the voltage across the external resis-
tors to change from say 10V to 10mV, the frequency can be
made to vary at least 1000:1. There are, however, several fac-
tors which make this large sweep range less than ideal.
10M
DISTORTION
100K
0.0047µF
-15V
FIGURE 1. VARIABLE AUDIO OSCILLATOR, 20Hz TO 20kHz
Question 4
O.K., now I can get a large frequency range, but I notice that
the duty cycle and hence my distortion changes at the low-
est frequencies.
Question 2
You say I can vary the voltage on pin 8 (FM sweep input) to
get this large range, yet when I short pin 8 to V+ (pin 6), the
ratio is only around 100:1.
Answer
This is caused partly by a slight difference in the V s of Q
BE
2
Answer
and Q . In trying to manufacture two identical transistors, it is
3
This is often true. With pin 8 shorted to V+, a check on the
potentials across the external R and R will show 100mV
not uncommon to get V
differences of several millivolts or
BE
A
B
more. In the standard 8038 connection with pins 7 and 8 con-
nected together, there are several volts across R and R and
or more. This is due to the V
mismatch between Q and
BE
Q (also Q and Q ) because of the geometries and current
1
A
B
2
1
3
this small mismatch is negligible. However, in a swept mode
with the voltage at pin 8 near V+ and only tens of millivolts
levels involved. Therefore, to get smaller voltages across
these resistors, pin 8 must be raised above V+.
across R and R , the V
BE
mismatch causes a larger mis-
A
B
match in charging currents, hence the duty cycle changes. For
lowest distortion then, it is advisable to keep the minimum
voltage across R and R around 100mV. This would of
Question 3
How can I raise pin 8 above V+ without a separate power
supply?
A
B
course, limit the frequency sweep range to around 100:1.
1-888-INTERSIL or 321-724-7143 | Copyright © Intersil Corporation 1999
1
Application Note 013
Finally, trimming the various pins for lowest distortion
deserves some attention. With pins 7 and 8 connected
together and the pot at pin 7 and 8 externally set at its maxi-
Question 5
I have a similar duty cycle problem when I use high values of
R
and R . What causes this?
A
B
mum, adjust the ratio of R and R for 50% duty cycle. Then
A
B
Answer
adjust a pot on pin 12 or both pins 1 and 12 depending on
minimum distortion desired. After these trims have been
made, set the voltage on pin 8 for the lowest frequency of
interest. The principle error here is due to the excess current
There is another error term which becomes important at very
low charge and discharge currents. This error current is the
emitter current of Q . The application note on the 8038 gives
7
of Q causing a shift in the duty cycle. This can be partially
a complete circuit description, but it is sufficient to know that
7
compensated for by bleeding a small current away from pin
5. The simplest way to do this is to connect a high value of
resistance (10MΩ to 20MΩ) from pin 5 to V- to bring the duty
cycle back to 50%. This should result in a reasonable com-
promise between low distortion and large sweep range.
the current charging the capacitor is the current in R which
A
flows down through diode Q and into the external C. The
9
discharge current is the current in R which flows down
B
through diode Q . Adding to the Q current is the current of
8
8
Q which is only a few microamperes. Normally, this Q cur-
7
7
rent is negligible, but with a small current in R , this current
B
Question 7
This waveform generator is a piece of junk. The triangle wave
is non-linear and has large glitches when it changes slope.
will cause a faster discharge than would be expected. This
problem will also appear in sweep circuits when the voltage
across the external resistors is small.
Answer
Question 6
You’re probably having trouble keeping the constant voltage
How can I get the lowest distortion over the largest
frequency sweep range.
across R and R really constant. The pulse output on pin 9
A
B
puts a moderate load on both supplies as it switches current on
and off. Changes in the supply reflect as variations in charging
current, hence non-linearity. Decoupling both power supply pins
to ground right at the device pins is a good idea. Also, pins 7
and 8 are susceptible to picking up switching transients (this is
especially true on printed circuit boards where pins 8 and 9 run
side by side). Therefore, a capacitor (0.1µF or more) from V+ to
pin 8 is often advisable. In the case when the pulse output is not
required, leave pin 9 open to be sure of minimizing transients.
Answer
First of all, use the largest supply voltage available (±15V or
+30V is convenient). This will minimize V
mismatch prob-
BE
lems and allow a wide variation of voltage on pin 8. The
potential on pin 8 may be swept from V (and slightly
CC
is the total voltage
higher) to 2/3 V
+2V) where V
CC
CC
across the 8038. Specifically for ±15V supplies (+30V), the
voltage across the external resistors can be varied from 0V
to nearly 8V before clipping of the triangle waveform occurs.
Question 8
What is the best supply voltage to use for lowest frequency
drift with temperature?
Second, keep the maximum currents relatively large (1mA or
2mA) to minimize the error due to Q . Higher currents could
be used, but the small geometry transistors used in the 8038
7
could give problems due to V
and bulk resistance, etc.
Answer
CE(SAT)
Third, and this is important, use two separate resistors for R
The 8038AM, 8038AC, 8038BM and 8038BC are all temper-
A
and R rather than one resistor with pins 4 and 5 connected
ature drift tested at V
= +20V (or ±10V). A curve in the
B
CC
together. This is because transistors Q and Q form a differ-
lower right hand corner of Page 4 of the data sheet indicates
frequency versus temperature at other supply voltages. It is
important to connect pins 7 and 8 together.
2
3
ential amplifier whose gain is determined by the impedance
between pins 4 and 5 as well as the quiescent current. There
are a number of implications in the differential amplifier con-
nection (pins 4 and 5 shorted). The most obvious is that the
Question 9
Why does connecting pin 7 to pin 8 give the best temperature
performance?
gain determines the way the currents split between Q and
2
Q . Therefore, any small offset or differential voltage will
3
cause a marked imbalance in the charge and discharge cur-
rents and hence the duty cycle. A more subtle result of this
connection is the effective capacitance at pin 10. With pins 4
and 5 connected together, the “Miller Effect” as well as the
Answer
There is a small temperature drift of the comparator thresh-
olds in the 8038. To compensate for this, the voltage divider
at pin 7 uses thin film resistors plus diffused resistors. The
different temperature coefficients of these resistors causes
compound transistor connection of Q and Q can produce
3
5
several hundred picofarads at pin 10, seriously limiting the
highest frequency of oscillation. The effective capacitance
would have to be considered important in determining what
value of external C would result in a particular frequency of
oscillation. The single resistor connection is fine for very sim-
ple circuits, but where performance is critical, the two sepa-
o
the voltage at pins 7 and 8 to vary 0.5mV/ C to maintain
overall low frequency drift at V
= 20V. At higher supply
CC
voltages, e.g., ±15V (+30V), the threshold drifts are smaller
compared with the total supply voltage. In this case, an
externally applied constant voltage at pin 8 will give reason-
ably low frequency drift with temperature.
rate resistors for R and R are recommended.
A
B
2
Application Note 013
Question 10
Question 13
Your data sheet is very confusing about the phase relationship
of the various waveforms.
How can I buffer the sine wave output without loading it down?
Answer
Answer
The simplest circuit is a simple op amp follower as shown in Fig-
ure 3A. Another circuit shown in Figure 3B allows amplitude and
offset controls without disturbing the 8038. Either circuit can be
DC or AC coupled. For AC coupling the op amp non-inverting
input must be returned to ground with a 100kΩ resistor.
Sorry about that! The thing to remember is that the triangle
and sine wave must be in phase since one is derived from
the other. A check on the way the circuit works shows that
the pulse waveform on pin 9 will be high as the capacitor
charges (positive slope on the triangle wave) and will be low
during discharge (negative slope on the triangle wave).
Question 14
The latest data sheet corrects the photograph Figure 7 on Page
5 of the data sheet. The 20% duty cycle square wave was
inverted, i.e., should be 80% duty cycle. Also, on that page
under “Waveform Timing” the related sentences should read
Your 8038 data sheet implies that all waveforms can operate up
to 1MHz. Is this true?
Answer
Unfortunately, only the square wave output is useful at that
frequency. As can be seen from the curves on page 4 of the
data sheet, distortion on the sine wave and linearity of the tri-
angle wave fall off rapidly above 200kHz.
“R controls the rising portion of the triangle and sine-wave and
A
the 1 state of the square wave.” Also, “the falling portion of the
triangle and sine wave and the 0 state of the square wave is:”
Question 11
Under Parameter Test Conditions on Page 3 of your 8038
data sheet, the suggested value for Min and Max duty cycle
adjust don’t seem to work.
Question 15
Is it normal for this device to run hot to the touch?
Answer
Answer
Yes. The 8038 is essentially resistive. The power dissipation
is then E /R and at ±15V, the device does run hot. Extensive
life testing under this operating condition and maximum
ambient temperature has verified the reliability of this prod-
uct.
2
The positive charging current is determined by R alone
since the current from R is switched off. (See 8038 Applica-
tion Note AN012 for complete circuit description.) The nega-
tive discharge current is the difference between the R
current and twice the R current. Therefore, changing R
will affect only the discharge time, while changing R will
A
B
A
B
B
Question 16
A
affect both charge and discharge times. For short negative
going pulses (greater than 50% duty cycle) we can lower the
How stable are the output amplitudes versus temperature?
Answer
value of R (e.g., R = 50kΩ and R = 1.6kΩ). For short
B
A
B
The amplitude of the triangle waveform decreases slightly
with temperature. The typical amplitude coefficient is
positive going pulses (duty cycles less than 50%) the limiting
values are reached when the current in R is twice that in
A
o
o
-0.01%/ C, giving a drop of about 1% at 125 C. The sine
R
(e.g., R = 50kΩ). This has been corrected on the latest
B
B
output is less sensitive and decreases only about 0.6% at
data sheet.
o
125 C. For the square wave output the V
goes from
CE(SAT)
o
o
0.12V at 25 C to 0.17V at 125 C. Leakage current in the “1”
state is less than a few nanoamperes even at 125 C and is
Question 12
I need to switch the waveforms off and on. What’s a good
way to strobe the 8038?
o
usually negligible.
Answer
+15V
With a dual supply voltage (e.g., ±15V) the external capaci-
tor (pin 10) can be shorted to ground so that the sine wave
and triangle wave always begin at a zero crossing point.
Random switching has a 50/50 chance of starting on a posi-
tive or negative slope. A simple AND gate using pin 9 will
allow the strobe to act only on one slope or the other, see
Figure 2. Using only a single supply, the capacitor (pin 10)
can be switched either to V+ or ground to force the compara-
tor to set in either the charge or discharge mode. The disad-
vantage of this technique is that the beginning cycle of the
next burst will be 30% longer than the normal cycle.
15K
R
R
B
A
4
5
6
9
2
8
7
8038
11
1N914
1N914
10
2N4392
STROBE
C
OFF
100K
+15V (>0V)
-15V
-15V (< -10V)
ON
FIGURE 2. STROBE-TONE BURST GENERATOR
3
Application Note 013
V+
V+
R
R
B
A
R
R
B
A
AMPLITUDE
100K
4
5
6
4
5
6
2
7
7
8038
8038
+
-
+
-
8
2
8
20K
11
10
11
10
4.7K
C
C
V-
V-
FIGURE 3A.
FIGURE 3B.
FIGURE 3. SINEWAVE OUTPUT BUFFER AMPLIFIERS
Schematic Diagram
V+
All Intersil semiconductor products are manufactured, assembled and tested under ISO9000 quality systems certification.
Intersil semiconductor products are sold by description only. Intersil Corporation reserves the right to make changes in circuit design and/or specifications at any time with-
out notice. Accordingly, the reader is cautioned to verify that data sheets are current before placing orders. Information furnished by Intersil is believed to be accurate and
reliable. However, no responsibility is assumed by Intersil or its subsidiaries for its use; nor for any infringements of patents or other rights of third parties which may result
from its use. No license is granted by implication or otherwise under any patent or patent rights of Intersil or its subsidiaries.
For information regarding Intersil Corporation and its products, see web site http://www.intersil.com
4
相关型号：
©2020 ICPDF网 联系我们和版权申明