LTC1040CSW datasheet(5/12 Pages) LINER

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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, RS, with a

bypass capacitor, CS. The bypass capacitor may or may

not be needed, depending on the size of the source

resistance and the magnitude of the input voltage, VIN.

APPLICATIO S I FOR ATIO

Figure 1. Equivalent Input Circuit

VIN

LTC1040 • AI01

CIN

≈ 33pF

V–

LTC1040 DIFFERENTIAL INPUT

–

For RS < 1Ok

Assuming CS is zero, the input capacitor, CIN, charges to

VIN with a time constant of RS CIN. When RS is too large,

CIN does not have a chance to fully charge during the

sampling interval (

≈ 80µs) and errors will result. If RS

exceeds 10k

Ω, a bypass capacitor is necessary to mini-

mize errors.

For RS > 1OkΩ

For RS greater than 10kΩ, CIN cannot fully charge and a

bypass capacitor, CS, is needed. When switch S1 closes,

charge is shared between CS and CIN. The change in

voltage on CS because of this charge sharing is:

∆V = VIN •

CIN

CIN + CS

RIN =

VIN

IIN

fS CIN

fS • 33pF

This represents an error and can be made arbitrarily small

by increasing CS.

With the addition of CS, 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 CIN between VIN and ground. The

alternate charge and discharge of CIN causes a current to

flow into the positive input and out of the negative input.

The magnitude of this current is:

IIN = q • fS = VIN CIN fS

where fS 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 RIN. If the

sampling frequency is reduced to 1Hz, the voltage error is

reduced to 66

µV.

(see typical curve of Input Resistance vs Sampling Fre-

quency). A voltage divider is set up between RS and RIN

causing error.

The input voltage error caused by these two effects is:

VERROR = VIN

Example: fS = 10Hz, RS = 1MΩ,

CS = 1µF, VIN = 1V

()

CIN

CIN + CS

RS + RIN

VERROR = 1V

= 33

µV + 330µV = 363µV.

()

33 • 10–12

106

1 • 10 –6

106 + 3 • 109

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