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MAX11040 Datasheet(PDF) 29 Page - Maxim Integrated Products |
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MAX11040 Datasheet(HTML) 29 Page - Maxim Integrated Products |
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29 / 31 page ![]() 24-Bit, 4-Channel, Simultaneous-Sampling, Cascadable, Sigma-Delta ADC ______________________________________________________________________________________ 29 Source Impedance and Input Sampling Network The source impedance that drives the analog inputs affects the sampling period. Low-Impedance Sources Minimize the source impedance to ensure the input capacitor fully charges during the sampling phase. The required source resistance is defined by the equation below: where K = 1.5 and RINT = 2600 Ω. For example, the required source resistance to achieve 0.1% accuracy is: High-Impedance Sources If the source impedance is greater than RSOURCE_MAX, as defined in the Low-Impedance Sources section, place a 0.1µF bypass capacitor between AIN_+ and AIN_- to provide transient charge. The average switched-capacitor load with a proper bypass capacitor and XIN clock frequency = 24.576MHz is equivalent to a 130k Ω resistor connect- ed between AIN_+ and AIN_-. This resistance is inde- pendent of the value of the 0.1µF bypass capacitor. If another XIN clock frequency is chosen, this resistance is directly proportional to the XIN clock period. Although the addition of a bypass capacitor helps charge the MAX11040 input capacitor, some gain error due to resistive drop across the source resistance still remains. Calculate this gain error using the following equation: Analog Filtering The analog filtering requirements in front of the MAX11040 are considerably reduced compared to a conventional converter with no on-chip filtering. The internal digital filter has significant rejection of signals higher than the Nyquist frequency of the output data rate that would alias back into the sampled signal. The internal digital filter does not provide rejection close to the harmonics of the 3.072MHz modulator fre- quency. For example, assuming an output data rate of 16ksps if the XIN clock is set to 24.576MHz, then the band between 3.0686MHz and 3.0750MHz is not explicitly filtered. Since this unfiltered band is very small compared to its actual frequency, very little broadband noise enters through this mechanism. If focused narrowband noise in this band is present, a simple analog filter can create significant attenuation at this frequency because the ratio of passband to stop- band frequency is large. In addition, because the device’s common-mode rejec- tion extends out to several 100kHz, the common-mode noise susceptibility in this frequency range is substan- tially reduced. Providing additional filtering in some applications ensures that differential noise signals outside the fre- quency band of interest do not saturate the analog modulator. The modulator saturates if the input voltage exceeds its full scale (±2.2V). The digital filter does not prevent a large signal in the filter stopband from saturating the modulator. If signals outside the band of interest cause violation of this full scale while accurate conversion of passband signals is desired, then additional analog fil- tering is required to prevent saturation. Compensating for the Rolloff of the Digital Filter in Typical FFT Analysis To calculate FIR_GAIN(fAIN_): 1) Decide the number of evenly spaced frequencies between DC and the Nyquist frequency of the output data rate at which correction factors are desired, which is usually the same as the FFT result. 2) Create an array with a length that is 2x the number of the desired frequencies. (Again, the result is likely to correlate with the time domain array that is loaded into an FFT algorithm.) 3) Fill this array with the filter coefficients provided in the Digital Filter section. Fill the rest of the array with zeros. 4) Take an FFT of this array. The result represents the response of the MAX11040 built-in FIR filter. Δ Ω Gain R RR R Rk SOURCE SOURCE LOAD SOURCE SOURCE = + = + 130 R ns xpF x ns xpF x ns xpF x SOURCE MAX _ . ln .% . ln . . < ⎛ ⎝⎜ ⎞ ⎠⎟ = () == − − − 120 15 4 1 01 2600 120 1 5 4 1000 2600 120 15 4 6 91 2600 294 Ω Ω ΩΩ R t Kx C x In Error R ns xpF x In Error SOURCE MAX SAMP SAMP INT _ . < ⎛ ⎝⎜ ⎞ ⎠⎟ = ⎛ ⎝⎜ ⎞ ⎠⎟ − − 1 120 15 4 1 2600 Ω |
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