wjxcom
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Hi all: During designing the Pipeline ADC, the capacitor is the key component to the MDAC. So I beg help from this BBS and want to know How to calualte the capacitor size when design pipeline adc?
For example, when I design a 12bits PipelineADc, The capacitor size for the MDAC is selected to be 50fF based on our desired signal-to-noise ratio (SNR) of 70dB.
For a 12-bit ADC, the ideal SNR is SNR=6.02×12+ 1.76=74 dB. Therefore, it seems reasonable to aim for approximately 70dB SNR in our ADC. 70dB SNR correlates to 10^(70/20)=3162, that is, 3162 parts signal for every 1 part noise. The largest input voltage swing on the capacitors is 1V peak. This equates to a 0.35V RMS swing. Therefore, the total RMS noise must be 0.35/3162= 110μ.
If we assume that half of the noise comes from the amplifier, and half from thermal noise, we are able to determine the necessary size of the capacitors. ‘Half’ of the total RMS noise of 110μV is 110μV/√2=77μ RMS. Given that √KT/C=σ, we find that the total C should be at least 700fF. For 31 levels, we need 31 individual capacitors of 22fF. A capacitor size of 50fF is selected because it is the smallest capacitor that can be used in order to obtain sufficient matching
I do not know this way is right or not. beg your help, thanx you!!
For example, when I design a 12bits PipelineADc, The capacitor size for the MDAC is selected to be 50fF based on our desired signal-to-noise ratio (SNR) of 70dB.
For a 12-bit ADC, the ideal SNR is SNR=6.02×12+ 1.76=74 dB. Therefore, it seems reasonable to aim for approximately 70dB SNR in our ADC. 70dB SNR correlates to 10^(70/20)=3162, that is, 3162 parts signal for every 1 part noise. The largest input voltage swing on the capacitors is 1V peak. This equates to a 0.35V RMS swing. Therefore, the total RMS noise must be 0.35/3162= 110μ.
If we assume that half of the noise comes from the amplifier, and half from thermal noise, we are able to determine the necessary size of the capacitors. ‘Half’ of the total RMS noise of 110μV is 110μV/√2=77μ RMS. Given that √KT/C=σ, we find that the total C should be at least 700fF. For 31 levels, we need 31 individual capacitors of 22fF. A capacitor size of 50fF is selected because it is the smallest capacitor that can be used in order to obtain sufficient matching
I do not know this way is right or not. beg your help, thanx you!!