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LTC1040CJ Datasheet(PDF) 5 Page - Linear Technology |
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LTC1040CJ Datasheet(HTML) 5 Page - Linear Technology |
5 / 12 page 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, 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 RS CS LTC1040 • AI01 S1 S2 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 = 1 fS CIN = 1 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 RS + RIN VERROR = 1V = 33 µV + 330µV = 363µV. () 33 • 10–12 106 1 • 10 –6 106 + 3 • 109 + |
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