LTC2483 datasheet(16/20 Pages) LINER

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LTC2450-1

24501fb

APPLICATIONS INFORMATION

Figure 16. Measured INL vs Input Voltage,

CIN = 0.1μF, VCC = 5V, TA = 25°C

For most applications, it is desirable to implement CIN as

a high quality 0.1μF ceramic capacitor and RS ≤ 1k. This

capacitor should be located as close as possible to the

actual VINpackagepin.Furthermoretheareaencompassed

by this circuit path as well as the path length should be

minimized.

In the case of a 2-wire sensor which is not remotely

grounded, it is desirable to split RS and place series

resistors in the ADC input line as well as in the sensor

ground return line which should be tied to the ADC GND

pin using a star connection topology.

Figure 16 shows the measured LTC2450-1 INL vs

Input Voltage as a function of RS value with an input

capacitor CIN = 0.1μF.

In some cases, RS can be increased above these guide-

lines. The input current is zero while the ADC is either in

sleep or I/O modes. Thus, if the time constant of the input

R-C circuit τ = RS • CIN is of the same order magnitude or

longer than the time periods between actual conversions,

then one can consider the input current to be reduced

correspondingly.

These considerations need to be balanced out by the input

signal bandwidth. The 3dB bandwidth ≅ 1/(2π • RS • CIN).

Finally, if the recommended choice for CIN is unacceptable

for the user’s speciﬁc application, an alternate strategy is to

eliminate CIN and minimize CPAR and RS. In practical terms,

this conﬁguration corresponds to a low impedance sensor

directly connected to the ADC through minimum length

traces. Actual applications include current measurements

through low value sense resistors, temperature measure-

ments, low impedance voltage source monitoring and so

on. The resultant INL vs VIN is shown in Figure 17. The

measurements of Figure 17 include a CPAR capacitor cor-

responding to a minimum size layout pad and a minimum

width input trace of about 1 inch length.

Figure 17. Measured INL vs VIN, CIN = 0, VCC = 5V, TA = 25°C

INPUT VOLTAGE (V)

–4

24501 F16

–8

–12

–16

RS = 10k

RS = 1k

RS = 0

INPUT VOLTAGE (V)

–2

–4

–6

–8

24501 F17

123

3.5

0.5

1.5

2.5

4.5

RS = 1k

RS = 10k

RS = 0

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