Please help me find the output function, with \$V_{REF}\$ as reference, and \$b_5-b_0\$ as the digital input for this circuit (1): circ

I'm kinda confused with this Series capacitor that's splitting the array. It it weren't for that capacitor and all parallel ones were scaled normally, the output would be:

\$\displaystyle V_{OUT} = V_{REF}\frac{C\sum\limits_{b=0}^{B-1}b_i 2^i}{C_{TOT}} \$,

I think, but with this series capacitor I'm really not sure.


(1) DAC Design Lecture, EECS dept, UC Berkeley, CA, 2009

  • \$\begingroup\$ Could you provide a citation for the figure, please? I'm assuming that you did not draw it yourself. \$\endgroup\$
    – Joe Hass
    Dec 22 '13 at 11:44

With the 8C/7 capacitor splitting the the array in half, the total value of the left block of capacitors is made equal to the right block LSB-only capacitance by the inclusion of the series capacitor. Does this help?

C + C + 2C + 4C +8C/7 = C (LSB of right hand block)

This means that the left block LSB capacitor is weighted down appropriately.

Here's an interesting document on the theory. I was a bit disappointed more documents were not avaiable.

  • \$\begingroup\$ That's an interesting article, I'd never heard of that topology before. \$\endgroup\$
    – PeterJ
    Dec 22 '13 at 12:16
  • \$\begingroup\$ Thanks, this is very cool :) Spent hours searching for a similar paper unsuccessfully. However, I still can't figure out the output function. I understand that \$V_{out} = V_{ref}\frac{C_{eq}}{C_{tot}}\$, where \$C_{eq}\$ is the sum of all capacitors connected to Vref, with regard to Vref, but I'm not sure how this works out in this case. \$\endgroup\$
    – Vidak
    Dec 22 '13 at 13:49

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