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Dual Slope ADC

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How long will it take for the integrator to reach zero? Long enough to exactly give the same magnitude but opposite sign integral we just computed above. Leaving out the signs, VinT/RC = VrefTdown/RC. How convenient -- the RC values cancel. Tdown=VinT/Vref. Since T and Vref are fixed, the time that the down count takes is proportional to the input voltage. If we're clever (and why wouldn't we be?), we can get the down-count time to read out as if the units were volts. If the magnitude of Vref is 5 V, and the input voltage that we're trying to measure is 1.000 V, then we want something that will count to 1000 time units. Suppose we have a clock that puts out one pulse every 10 μs. Then if Tdown is 1000 × 10 μs, we have it. But that makes Tdown=10 ms, so T (the integration time) must be set to 10 ms * 5.000 V/1.000 V = 50 ms. But once we set the initial integration time to 50 ms, the time that it takes to return to 0 will read out in voltage directly. For example, if we integrate 2.345 V for that same 50 ms, then the discharge back to zero will take 2345 ticks of the 10 μs clock, and we have the readout as we wanted it. Here's a sketch of what we just said:

dualslope noiseless

   
 

 

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