Repeat Ex. 30.16 if the comparators switch at 305 mV (a 55 mV offset) and 675 mV

Question

Repeat Ex. 30.16 if the comparators switch at 305 mV (a 55 mV offset) and 675 mV (a -75 mV offset).

Accepted Answer

[latex] \begin{aligned} &v_{in} \quad\quad\quad \quad\quad\quad\quad a_{1.5x}b_{1.5x} \quad v_{out} \quad\quad\quad\quad\quad\quad\quad\quad ext{Digital out}\ & 600 \ \mathrm{mV}(N-1=7) \quad\quad 01 \quad \quad700 \ \mathrm{mV} \quad\quad\quad\quad\quad\quad b_7=\overline{a_{1.57}} b_{1.57} \oplus a_{1.56} \oplus c_6=0 \ \ & c_7=\overline{a_{1.57}} b_{1.57} a_{1.56}+c_6\left(\overline{a_{1.57}} b_{1.57}+a_{1.56} ight)=1 \\ & 700 \ \mathrm{mV}(N-2=6) \quad\quad 11 \quad\quad -100 \ \mathrm{mV} \quad\quad\quad\quad\quad\quad b_6=\overline{a_{1.56}} b_{1.56} \oplus a_{1.55} \oplus c_5=0 \\ & c_6=\overline{a_{1.56}} b_{1.56} a_{1.55}+c_5\left(\overline{a_{1.56}} b_{1.56}+a_{1.55} ight)=0 \\ & -100 \ \mathrm{mV}(N-3=5) \quad\quad 00 \quad\quad 300 \ \mathrm{mV} \quad\quad\quad\quad\quad\quad b_5=\overline{a_{1.55}} b_{1.55} \oplus a_{1.54} \oplus c_4=0 \\ & c_5=\overline{a_{1.55}} b_{1.55} a_{1.54}+c_4\left(\overline{a_{1.55}} b_{1.55}+a_{1.54} ight)=0 \\ & 300 \ \mathrm{mV}(N-4=4) \quad\quad 00 \quad\quad 1.1 \ \mathrm{V} \quad\quad\quad\quad\quad\quad b_4=\overline{a_{1.54}} b_{1.54} \oplus a_{1.53} \oplus c_3=1 \\ & c_4=\overline{a_{1.54}} b_{1.54} a_{1.53}+c_3\left(\overline{a_{1.54}} b_{1.54}+a_{1.53} ight)=0 \\ & 1100 \ \mathrm{mV}(N-4=3) \quad\quad 11 \quad\quad 700\ \mathrm{mV} \quad\quad\quad\quad\quad\quad b_3=\overline{a_{1.53}} b_{1.53} \oplus a_{1.52} \oplus c_2=1 \\ & c_3=\overline{a_{1.53}} b_{1.53} a_{1.52}+c_2\left(\overline{a_{1.53}} b_{1.53}+a_{1.52} ight)=0 \\ & 700 \ \mathrm{mV}(N-6=2) \quad\quad 11 \quad\quad -100 \ \mathrm{mV} \quad\quad\quad\quad\quad\quad b_2=\overline{a_{1.52}} b_{1.52} \oplus a_{1.51} \oplus c_1=0 \\ & c_2=\overline{a_{1.52}} b_{1.52} a_{1.51}+c_1\left(\overline{a_{1.52}} b_{1.52}+a_{1.51} ight)=0 \\ & -100 \ \mathrm{mV}(N-7=1) \quad\quad 00 \quad\quad 300\ \mathrm{mV} \quad\quad\quad\quad\quad\quad b_1=\overline{a_{1.51}} b_{1.51} \oplus c_0=0 \\ & c_1=\overline{a_{1.51}} b_{1.51} a_{1.50}+c_0\left(\overline{a_{1.51}} b_{1.51}+a_{1.50} ight)=0 \\ & 300 \ \mathrm{mV}(N-8=0) \quad\quad 00 \quad\quad 1.1 \ \mathrm{V} \quad\quad\quad\quad\quad\quad b_0=\overline{a_{1.50}} b_{1.50} =0\\ & c_0={a_{1.50}}=0 \\ & \end{aligned} [/latex]

and [latex]b_{8}=a_{1.57}\oplus c_7=1[/latex]. Again, we can reorder the bits so the MSB is on the left, the LSB is on the right, yielding 1 0001 1000 and subtract 0 0111 1111 yielding

1 0001 1000 (280)
-0 0111 1111 (127)
0 1001 1001 (153)

This is the exact result given in Ex. 30.10 (1001 1001, or decimal 153). In this case the 0.5 LSB round-off worked in our favor.

While the comparator performance can be extremely poor, the circuit of Fig. 30.42 must subtract and amplify to an accuracy set by the least significant bit of the converter. When we calculated values in Ex. 30.16 and here in this example we assumed subtractions of exactly 0, [latex]V_{\scriptscriptstyle C M}[/latex], and 2[latex]V_{\scriptscriptstyle C M}[/latex]followed by a multiplication of exactly two.

Finally, notice that the negative output of -100 mV (single-ended) and the 1.1 V output that is greater than VDD (= 1 V here) are easily accommodated when using fully-differential op-amps.

Question 30.17: Repeat Ex. 30.16 if the comparators switch at 305 mV (a 55 m......

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