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ALD500RAU-10PE Datasheet(PDF) 11 Page - Advanced Linear Devices |
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ALD500RAU-10PE Datasheet(HTML) 11 Page - Advanced Linear Devices |
11 / 12 page ALD500RAU/ALD500RA/ALD500R Advanced Linear Devices 11 EQUATIONS AND DERIVATIONS Dual Slope Analog Processor equations and derivations are as follows: 1 60Hz 2.0 20x10-6 = ~ = ~ = ~ DESIGN EXAMPLES We now apply these equations in the following design examples. Design Example 1: 1. Pick resolution = 16 bit. 2. Pick tINT = 4x = 4 x 16.6667 msec. 3. Pick clock period = 1.08507 µs and number of counts 4. Pick VINMAX value, e.g., VINMAX = 2.0 V I BMAX = 20µA RINT = = 100 k Ω 5. Applying equation (3) to calculate CINT: 6. Pick CREF and CAZ ≥ CINT: CREF CAZ 0.33 µF 7. Pick tDINT = 2 x tINT = 133.3333 msec 8. Calculate VREF VINTMAX . CINT . RINT tDINT MAX 4 x 0.33 x 10-6 x 100 x 103 133.3333 x 10-3 1.00V = 0.0666667 sec. V V 1 RINT . CINT VIN(t)dt = ∫ 0 tINT tDINT VIN = VREF . tINT 1 tINT . VIN = VREF . tDINT (2a) (2) (1) CINT = VINT tINT . IB (3) RINT = VINMAX IBMAX (4) For VIN(t) = VIN (constant): From equation (2a), OR Rearranging equations (3) and (4): At VINT = VINT MAX, equation (6) becomes: Combining (6a) and (7): In equation (5b), substituting equation (8) for tINT: For tDINT MAX = 2 x tINT, equation (9) becomes: VIN MAX . tINT tDINT MAX VIN . tINT tDINT (5a) (5b) tINT = CINT . VINT IB (6) IBMAX = VINMAX RINT (7) VREF = VREF = ... tINT = CINT . VINTMAX . RINT VINMAX (8) VIN MAX . tDINT MAX VREF = CINT . VINTMAX . RINT VIN MAX = CINT . VINTMAX . RINT tDINT MAX (9) VREF = CINT . VINTMAX . RINT 2tINT (10) RINT . CINT RINT . CINT VREF . tDINT RINT . CINT ... tINT = CINT . VINTMAX IBMAX and At VINMAX, the current IB is also at a maximum level, for a given RINT value: VIN IB = (6a) CINT = (0.0666667)(20x10-6)/4 where VINT = 4.0V 0.33 µF Design Example 2: 1. Select resolution of 17 bit. Total number of counts during tINT is131,072. 2. We can pick tINT of 16.6667 msec. x 5 = 83.3333 msec. or alternately, pick t INT equal 16.6667 msec. x 6 = 100.00 msec. (for 60 Hz rejection) which is t INT = 20.00 msec. x 5 Therefore, using t INT = 100 msec. would achieve both 50 Hz and 60 Hz cycle noise rejection. For this example, the following calculations would assume t INT of 100 msec. Now select period equal to 0.5425 µsec. (clock frequency of 1.8432 MHz) = 66.6667ms = ~ = 100.00 msec. (for 50 Hz rejection) 0.0666667 1.08507x10-6 over tINT = = 61440 = = |
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