5
LTC1040
1040fa
The LTC1040 uses sampled data techniques to achieve its
unique characteristics. Some of the experience acquired
using classic linear comparators does not apply to this
circuit, so a brief description of internal operation is
essential to proper application.
The most obvious difference between the LTC1040 and
other comparators is the dual differential input structure.
Functionally, when the sum of inputs is positive, the
comparator output is high and when the sum of the inputs
is negative, the output is low. This unique input structure
is achieved with CMOS switches and a precision capacitor
array. Because of the switching nature of the inputs, the
concept of input current and input impedance needs to be
examined.
The equivalent input circuit is shown in Figure 1. Here, the
input is being driven by a resistive source, R
S
, with a
bypass capacitor, C
S
. The bypass capacitor may or may
not be needed, depending on the size of the source
resistance and the magnitude of the input voltage, V
IN
.
APPLICATIO S I FOR ATIO
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Figure 1. Equivalent Input Circuit
V
IN
R
S
C
S
LTC1040 • AI01
S1
S2
C
IN
≈ 33pF
V
–
LTC1040 DIFFERENTIAL INPUT
+
–
For R
S
< 1Ok
Assuming C
S
is zero, the input capacitor, C
IN
, charges to
V
IN
with a time constant of R
S
C
IN
. When R
S
is too large,
C
IN
does not have a chance to fully charge during the
sampling interval (≈ 80µs) and errors will result. If R
S
exceeds 10kΩ, a bypass capacitor is necessary to mini-
mize errors.
For R
S
> 1OkΩ
For R
S
greater than 10kΩ, C
IN
cannot fully charge and a
bypass capacitor, C
S
, is needed. When switch S1 closes,
charge is shared between C
S
and C
IN
. The change in
voltage on C
S
because of this charge sharing is:
∆V = V
IN
•
C
IN
C
IN
+ C
S
R
IN
=
V
IN
I
IN
=
1
f
S
C
IN
=
1
f
S
• 33pF
This represents an error and can be made arbitrarily small
by increasing C
S
.
With the addition of C
S
, a second error term caused by the
finite input resistance of the LTC1040 must be considered.
Switches S1 and S2 alternately open and close, charging
and discharging C
IN
between V
IN
and ground. The
alternate charge and discharge of C
IN
causes a current to
flow into the positive input and out of the negative input.
The magnitude of this current is:
I
IN
= q • f
S
= V
IN
C
IN
f
S
where f
S
is the sampling frequency. Because the input
current is directly proportional to input voltage, the LTC1040
can be said to have an average input resistance of:
Notice that most of the error is caused by R
IN
. If the
sampling frequency is reduced to 1Hz, the voltage error is
reduced to 66µV.
(see typical curve of Input Resistance vs Sampling Fre-
quency). A voltage divider is set up between R
S
and R
IN
causing error.
The input voltage error caused by these two effects is:
V
ERROR
= V
IN
Example: f
S
= 10Hz, R
S
= 1MΩ,
C
S
= 1µF, V
IN
= 1V
()
C
IN
C
IN
+ C
S
+
R
S
R
S
+ R
IN
V
ERROR
= 1V
= 33µV + 330µV = 363µV.
()
33 • 10
–12
10
6
1 • 10
–6
10
6
+ 3 • 10
9
+
