Datasheet LTC1040 (Analog Devices) - 5
| Manufacturer | Analog Devices |
| Description | Dual Micropower Comparator |
| Pages / Page | 12 / 5 — APPLICATIO S I FOR ATIO. For RS > 1Ok. Figure 1. Equivalent Input … |
| File Format / Size | PDF / 227 Kb |
| Document Language | English |
APPLICATIO S I FOR ATIO. For RS > 1Ok. Figure 1. Equivalent Input Circuit. For R. S < 1Ok
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LTC1040
U U W U APPLICATIO S I FOR ATIO
The LTC1040 uses sampled data techniques to achieve its
For RS > 1Ok
Ω unique characteristics. Some of the experience acquired For R using classic linear comparators does not apply to this S greater than 10kΩ, CIN cannot fully charge and a bypass capacitor, C circuit, so a brief description of internal operation is S, is needed. When switch S1 closes, charge is shared between C essential to proper application. S and CIN. The change in voltage on CS because of this charge sharing is: The most obvious difference between the LTC1040 and C other comparators is the dual differential input structure. ∆V = V IN IN • C Functionally, when the sum of inputs is positive, the IN + CS comparator output is high and when the sum of the inputs This represents an error and can be made arbitrarily small is negative, the output is low. This unique input structure by increasing CS. is achieved with CMOS switches and a precision capacitor With the addition of CS, a second error term caused by the array. Because of the switching nature of the inputs, the finite input resistance of the LTC1040 must be considered. concept of input current and input impedance needs to be Switches S1 and S2 alternately open and close, charging examined. and discharging CIN between VIN and ground. The The equivalent input circuit is shown in Figure 1. Here, the alternate charge and discharge of CIN causes a current to input is being driven by a resistive source, R flow into the positive input and out of the negative input. S, with a bypass capacitor, C The magnitude of this current is: S. The bypass capacitor may or may not be needed, depending on the size of the source IIN = q • fS = VIN CIN fS resistance and the magnitude of the input voltage, VIN. where fS is the sampling frequency. Because the input current is directly proportional to input voltage, the LTC1040 CIN can be said to have an average input resistance of: ≈ 33pF R S1 S + V 1 1 R IN IN = = = V I f f IN CS IN S CIN S • 33pF S2 – V– (see typical curve of Input Resistance vs Sampling Fre- LTC1040 DIFFERENTIAL INPUT quency). A voltage divider is set up between RS and RIN causing error. LTC1040 • AI01
Figure 1. Equivalent Input Circuit
The input voltage error caused by these two effects is: C R V ( IN + S )
For R
ERROR = VIN
S < 1Ok
CIN + CS RS + RIN Assuming CS is zero, the input capacitor, CIN, charges to Example: f V S = 10Hz, RS = 1MΩ, IN with a time constant of RS CIN. When RS is too large, C C S = 1µF, VIN = 1V IN does not have a chance to fully charge during the sampling interval (≈ 80µs) and errors will result. If RS 33 • 10–12 106 V ( + ) exceeds 10kΩ, a bypass capacitor is necessary to mini- ERROR = 1V 1 • 10 –6 106 + 3 • 109 mize errors. = 33µV + 330µV = 363µV. Notice that most of the error is caused by RIN. If the sampling frequency is reduced to 1Hz, the voltage error is reduced to 66µV. 1040fa 5
U U W U APPLICATIO S I FOR ATIO
The LTC1040 uses sampled data techniques to achieve its
For RS > 1Ok
Ω unique characteristics. Some of the experience acquired For R using classic linear comparators does not apply to this S greater than 10kΩ, CIN cannot fully charge and a bypass capacitor, C circuit, so a brief description of internal operation is S, is needed. When switch S1 closes, charge is shared between C essential to proper application. S and CIN. The change in voltage on CS because of this charge sharing is: The most obvious difference between the LTC1040 and C other comparators is the dual differential input structure. ∆V = V IN IN • C Functionally, when the sum of inputs is positive, the IN + CS comparator output is high and when the sum of the inputs This represents an error and can be made arbitrarily small is negative, the output is low. This unique input structure by increasing CS. is achieved with CMOS switches and a precision capacitor With the addition of CS, a second error term caused by the array. Because of the switching nature of the inputs, the finite input resistance of the LTC1040 must be considered. concept of input current and input impedance needs to be Switches S1 and S2 alternately open and close, charging examined. and discharging CIN between VIN and ground. The The equivalent input circuit is shown in Figure 1. Here, the alternate charge and discharge of CIN causes a current to input is being driven by a resistive source, R flow into the positive input and out of the negative input. S, with a bypass capacitor, C The magnitude of this current is: S. The bypass capacitor may or may not be needed, depending on the size of the source IIN = q • fS = VIN CIN fS resistance and the magnitude of the input voltage, VIN. where fS is the sampling frequency. Because the input current is directly proportional to input voltage, the LTC1040 CIN can be said to have an average input resistance of: ≈ 33pF R S1 S + V 1 1 R IN IN = = = V I f f IN CS IN S CIN S • 33pF S2 – V– (see typical curve of Input Resistance vs Sampling Fre- LTC1040 DIFFERENTIAL INPUT quency). A voltage divider is set up between RS and RIN causing error. LTC1040 • AI01
Figure 1. Equivalent Input Circuit
The input voltage error caused by these two effects is: C R V ( IN + S )
For R
ERROR = VIN
S < 1Ok
CIN + CS RS + RIN Assuming CS is zero, the input capacitor, CIN, charges to Example: f V S = 10Hz, RS = 1MΩ, IN with a time constant of RS CIN. When RS is too large, C C S = 1µF, VIN = 1V IN does not have a chance to fully charge during the sampling interval (≈ 80µs) and errors will result. If RS 33 • 10–12 106 V ( + ) exceeds 10kΩ, a bypass capacitor is necessary to mini- ERROR = 1V 1 • 10 –6 106 + 3 • 109 mize errors. = 33µV + 330µV = 363µV. Notice that most of the error is caused by RIN. If the sampling frequency is reduced to 1Hz, the voltage error is reduced to 66µV. 1040fa 5
