Appendix A
In this appendix, the required resistor mismatch to meet DNL better than 0.5 LSB for the conventional resistor string DAC is derived. According to the conditions defined by (5), after incorporating resistance and its error value \(\Updelta {R_i}, \) we can obtain the following:
$$ \left[ \frac{\left( {{2^{N - 1}}} \right)R + \sum\nolimits_{i = 1}^{{2^{N - 1}}} {\Updelta {R_{N,i}}} }{{{2^N}R + \sum\limits_{i = 1}^{{2^N}} {\Updelta {R_{N,i}}} }} - \frac{\left( {{2^{N - 1}} - 1} \right)R + \sum\nolimits_{i = 1}^{{2^{N - 1}} - 1} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{{2^N}} {\Updelta {R_{N,i}}} } - \frac{1}{2^N}\right] \cdot {V_{REF}} \le \frac{V_{REF}}{2^{N + 1}} $$
(29)
where N is the bit numbers of conventional resistor string DAC. \(\Updelta{R_{N,i}}\) is resistor mismatch error of unit resistor in the resistor string.
By re-formulating (29), we can obtain:
$$ \left( \frac{\frac{1}{2} + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^{N - 1}}} {\frac{\Updelta {R_{N,i}}}{R}} }{1 + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^N}} {\frac{\Updelta {R_{N,i}}}{R}} } - \frac{\frac{1}{2} - \frac{1}{2^N} + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^{N - 1}} - 1} {\frac{\Updelta {R_{N,i}}}{R}} }{1 + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^N}} {\frac{\Updelta {R_{N,i}}}{R}} } - \frac{1}{2^N}\right) \cdot {V_{REF}} \le \frac{V_{REF}}{2^{N + 1}} $$
(30)
We can then apply Taylor’s series approximation again:
$$ \left[ \left(\frac{1}{2} + \frac{1}{2^N}\sum\limits_{i = 1}^{{2^{N - 1}}} \frac{\Updelta {R_{N,i}}}{R}\right)\left(1 - \frac{1}{2^N}\sum\limits_{i = 1}^{{2^N}} \frac{\Updelta {R_{N,i}}}{R} \right) - \left(\frac{1}{2} - \frac{1}{2^N} \right. +\left. \frac{1}{2^N}\sum\limits_{i = 1}^{{2^{N - 1}} - 1} \frac{{\Updelta {R_{N,i}}}}{R}\right)\left(1 - \frac{1}{2^N}\sum\limits_{i = 1}^{2^N} \frac{\Updelta {R_{N,i}}}{R} \right) - \frac{1}{2^N}\right] \cdot {V_{REF}} \le \frac{V_{REF}}{2^{N + 1}} $$
(31)
After combining and re-formulating the above formulas, we can obtain:
$$ \left( \frac{{2^N} - 1}{2^{2N}}\sum\limits_{i = {2^{N - 1}}}^{{2^{N - 1}}} {\frac{\Updelta {R_{N,i}}}{R}} - \frac{1}{2^{2N}}\sum\limits_{i = 1}^{{2^{N - 1}} - 1} {\frac{\Updelta {R_{N,i}}}{R} } -\right. \left.\frac{1}{{{2^{2N}}}}\sum\limits_{i = {2^{N - 1}} + 1}^{{2^N}} {\frac{\Updelta {R_{N,i}}}{R}} \right) \cdot {V_{REF}} \le \frac{V_{REF}}{2^{N + 1}} $$
(32)
And re-write (32) as a probability inequality
$$ P\left\{ \left(\frac{{2^N} - 1}{2^{2N}}\sum\limits_{i = {2^{N - 1}}}^{2^{N - 1}} {\Updelta {R_{N,i}}} - \frac{1}{2^{2N}}\sum\limits_{i = 1}^{{2^{N - 1}} - 1} {\Updelta {R_{N,i}} } -\right. \left.\frac{1}{2^{2N}}\sum\limits_{i = {2^{N - 1}} + 1}^{2^N} {\Updelta {R_{N,i}}} \right) \le \frac{R}{2^{N + 1}}\right\} \ge Y $$
(33)
Define X as follows:
$$ X \equiv \frac{{2^N} - 1}{2^{2N}}\sum\limits_{i = {2^{N - 1}}}^{{2^{N - 1}}} {\Updelta {R_{N,i}}} - \frac{1}{2^{2N}}\sum\limits_{i = 1}^{{2^{N - 1}} - 1} {\Updelta {R_{N,i}} } - \frac{1}{2^{2N}}\sum\limits_{i = {2^{N - 1}} + 1}^{{2^N}} {\Updelta {R_{N,i}}} $$
(34)
Then, the variance of X can be expressed as:
$$ \sigma _X^2 \equiv \frac{{\sigma _{\Updelta R}^2}}{2^{4N}}\left[{\left({2^N} - 1\right)^2}\sum\limits_{i = {2^{N - 1}}}^{2^{N - 1}} 1 + \sum\limits_{i = 1}^{{2^{N - 1}} - 1} {1 + } \sum\limits_{i = {2^{N - 1}} + 1}^{2^N} 1 \right] $$
(35)
By re-formulating (35), we can obtain:
$$ {\sigma _X} \equiv \frac{\sigma _{\Updelta R}}{2^{2N}}\sqrt{{2^{2N}} - {2^N}} $$
(36)
Using (34) and the standard normal distribution function, \(\Upphi, \) (33) can be rewritten as:
$$ P\left\{ {X \le \frac{R}{2^{N + 1}}} \right\} = \Upphi \left({\frac{R \mathord{\left/{ {{2^{N + 1}}}} \right. }}{\sigma _X}}\right) \ge Y $$
(37)
Replacing (36) in (37) and applying the inverse standard normal distribution function, (37) can be expressed as:
$$ {\Upphi ^{ - 1}}\left( Y \right)\frac{\sigma _{\Updelta R}}{R} \le \frac{2^{2N}}{{2^{N + 1}}\sqrt {{2^{2N}} - {2^N}} } $$
(38)
Appendix B
The required resistor mismatch to meet INL better than 0.5 LSB for the conventional resistor string DAC based on ideal reference line is derived in this appendix. According to the conditions defined by (19), after incorporating resistance and its error value \(\Updelta {R_i}, \) we can obtain the following:
$$ \left[ \frac{\left( {2^{N - 1}} \right)R + \sum\nolimits_{i = 1}^{2^{N - 1}} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{{2^N}} {\Updelta {R_{N,i}}} } - \frac{1}{2}\right] \cdot {V_{REF}} \le \frac{1}{2^{N + 1}} \cdot {V_{REF}} $$
(39)
By re-formulating (39), we can obtain:
$$ \left( \frac{\frac{1}{2} + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^{N - 1}}} {\frac{\Updelta {R_{N,i}}}{R}} }{1 + \frac{1}{2^N}\sum\nolimits_{i = 1}^{2^N} {\frac{\Updelta {R_{N,i}}}{R}} } - \frac{1}{2}\right) \cdot {V_{REF}} \le \frac{1}{2^{N + 1}} \cdot {V_{REF}} $$
(40)
We can then apply Taylor’s series approximation again:
$$ \left[ \left(\frac{1}{2} + \frac{1}{2^N}\sum\limits_{i = 1}^{2^{N - 1}} \frac{\Updelta {R_{N,i}}}{R}\right) \cdot \left(1 - \frac{1}{2^N}\sum\limits_{i = 1}^{2^N} \frac{\Updelta {R_{N,i}}}{R} \right) - \frac{1}{2}\right] \cdot {V_{REF}} \le \frac{1}{2^{N + 1}} \cdot {V_{REF}} $$
(41)
After combining and re-formulating the above formulas, we can obtain:
$$ \left( \frac{1}{2^{N + 1}}\sum\limits_{i = 1}^{2^{N - 1}} {\frac{\Updelta {R_{N,i}}}{R}} - \frac{1}{2^{N + 1}}\sum\limits_{i = {2^{N - 1}} + 1}^{2^N} {\frac{\Updelta {R_{N,i}}}{R}} \right) \cdot {V_{REF}} \le \frac{V_{REF}}{2^{N + 1}} $$
(42)
And re-write (42) as a probability inequality
$$ P\left\{ \left(\frac{1}{2^{N + 1}}\sum\limits_{i = 1}^{2^{N - 1}} {\Updelta {R_{N,i}}} - \frac{1}{2^{N + 1}}\sum\limits_{i = {2^{N - 1}} + 1}^{2^N} {\Updelta {R_{N,i}}} \right) \le \frac{R}{2^{N + 1}}\right\} \ge Y $$
(43)
Define X as follows:
$$ X \equiv \frac{1}{2^{N + 1}}\sum\limits_{i = 1}^{2^{N - 1}} {\Updelta {R_{N,i}}} - \frac{1}{2^{N + 1}}\sum\limits_{i = {2^{N - 1}} + 1}^{2^N} {\Updelta {R_{N,i}}} $$
(44)
Then, the variance of X can be expressed as:
$$ \sigma _X^2 = \frac{\sigma _{\Updelta R}^2}{2^{2(N + 1)}}\left[\sum\limits_{i = 1}^{2^{N - 1}} 1 + \sum\limits_{i = {2^{N - 1}} + 1}^{2^N} 1 \right] $$
(45)
By re-formulating (45), we can obtain:
$$ \sigma _X = \frac{\sigma _{\Updelta R}}{2^{N + 1}}\sqrt{2^N} $$
(46)
Using (44) and the standard normal distribution function, \(\Upphi, \) (43) can be rewritten as:
$$ P\left\{ {X \le \frac{R}{2^{N + 1}}} \right\} = \Upphi \left({\frac{R \mathord{\left/ { {{2^{N + 1}}}} \right.}}{\sigma _X}}\right) \ge Y $$
(47)
Replacing (46) in (47) and applying the inverse standard normal distribution function, (47) can be expressed as:
$$ {\Upphi ^{ - 1}}\left( Y \right)\frac{\sigma _{\Updelta R}}{R} \le \frac{1}{\sqrt {2^N} } $$
(48)
Appendix C
The required resistor mismatch to meet INL better than 0.5 LSB for the conventional resistor string DAC based on “end-point” reference line is derived in this appendix. The V
LSB,Ref can be written as:
$$ {V_{LSB, Ref}} = \frac{{V_O}\left( {11 \cdots 11} \right)}{{2^N} - 1} = \frac{\left( {{2^N} - 1} \right)R + \sum\nolimits_{i = 1}^{{2^N} - 1} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{2^N} {\Updelta {R_{N,i}}} } \cdot \frac{1}{{2^N} - 1} \cdot {V_{REF}} $$
(49)
Therefore,
$$ {V_{O,Ref}}(10\cdots 00) = {V_{LSB,Ref}} \cdot ({2^{N - 1}}) = \frac{\left( {{2^N} - 1} \right)R + \sum\nolimits_{i = 1}^{{2^N} - 1} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{2^N} {\Updelta {R_{N,i}}} } \cdot \frac{2^{N - 1}}{{2^N} - 1} \cdot {V_{REF}} $$
(50)
Thus, we can rewrite the equation as follows to achieve INL better than 0.5 LSB.
$$ \left[ \frac{\left( {2^{N - 1}} \right)R + \sum\nolimits_{i = 1}^{2^{N - 1}} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{2^N} {\Updelta {R_{N,i}}} } - \frac{\left( {{2^N} - 1} \right)R + \sum\nolimits_{i = 1}^{{2^N} - 1} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{2^N} {\Updelta {R_{N,i}}} } \cdot \frac{2^{N - 1}}{{2^N} - 1}\right] \cdot {V_{REF}} \le \frac{1}{2} \cdot \frac{\left( {{2^N} - 1} \right)R + \sum\nolimits_{i = 1}^{{2^N} - 1} {\Updelta {R_{N,i}}} }{{2^N}R + \sum\nolimits_{i = 1}^{2^N} {\Updelta {R_{N,i}}} } \cdot \frac{1}{{2^N} - 1} \cdot {V_{REF}} $$
(51)
By re-formulating (51), we can obtain:
$$ \left[ \frac{{\frac{1}{2}} + \frac{1}{2^N}\sum\nolimits_{i = 1}^{2^{N - 1}} {\frac{\Updelta {R_{N,i}}}{R}} }{1 + \frac{1}{2^N}\sum\nolimits_{i = 1}^{2^N} {\frac{\Updelta {R_{N,i}}}{R}} } - \frac{\left( {1 - \frac{1}{2^N}} \right) + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^N} - 1} {\frac{\Updelta {R_{N,i}}}{R}} }{1 + \frac{1}{2^N}\sum\nolimits_{i = 1}^{2^N} {\frac{\Updelta {R_{N,i}}}{R}} } \cdot \frac{2^{N - 1}}{{2^N} - 1}\right] \cdot {V_{REF}} \le \frac{1}{2} \cdot \frac{\left( {1 - \frac{1}{2^N}} \right) + \frac{1}{2^N}\sum\nolimits_{i = 1}^{{2^N} - 1} {\frac{\Updelta {R_{N,i}}}{R}} }{1 + \frac{1}{2^N}\sum\nolimits_{i = 1}^{2^N} {\frac{\Updelta {R_{N,i}}}{R}} } \cdot \frac{1}{{2^N} - 1} \cdot {V_{REF}} $$
(52)
We can then apply Taylor’s series approximation again:
$$ \left[ \left( {\frac{1}{2}} + \frac{1}{2^N}\sum\limits_{i = 1}^{2^{N - 1}} \frac{\Updelta {R_{N,i}}}{R}\right) \cdot \left(1 - \frac{1}{2^N}\sum\limits_{i = 1}^{2^N} \frac{\Updelta {R_{N,i}}}{R} \right) - \left[\left( {1 - \frac{1}{2^N}} \right) + \frac{1}{2^N}\sum\limits_{i = 1}^{{2^N} - 1} \frac{\Updelta {R_{N,i}}}{R}\right] \cdot \left(1 - \frac{1}{2^N}\sum\limits_{i = 1}^{2^N} \frac{\Updelta {R_{N,i}}}{R} \right) \cdot \frac{2^{N - 1}}{{2^N} - 1}\right] \cdot {V_{REF}} \le \frac{1}{2} \cdot \left[\left( {1 - \frac{1}{2^N}} \right) + \frac{1}{2^N}\sum\limits_{i = 1}^{{2^N} - 1} \frac{\Updelta {R_{N,i}}}{R} \right] \cdot \left[1 - \frac{1}{2^N}\sum\limits_{i = 1}^{2^N} \frac{\Updelta {R_{N,i}}}{R} \right] \cdot \frac{1}{{2^N} - 1} \cdot {V_{REF}} $$
(53)
After combining and re-formulating the above formulas, we can obtain:
$$ \left[ \frac{{2^{N - 1}} - {2^{ - \left(N + 1\right)}} - 1}{{2^N} - 1}\frac{1}{2^N}\sum\limits_{i = 1}^{2^{N - 1}} {\frac{\Updelta {R_{N,i}}}{R}} - \left(\frac{1}{2^N}\sum\limits_{i = {2^{N - 1}} + 1}^{{2^N} - 1} {\frac{\Updelta {R_{N,i}}}{R}} \right) \cdot \frac{{2^{N - 1}} - {2^{ - \left(N + 1\right)}}}{{2^N} - 1} + \left(\frac{1}{2^N}\sum\limits_{i = {2^N}}^{2^N} {\frac{\Updelta {R_{N,i}}}{R}} \right) \cdot \frac{{2^{ - 1}} - {2^{ - (N + 1)}}}{{2^N} - 1}\right] \cdot {V_{REF}} \le \frac{V_{REF}}{2^{N + 1}} $$
(54)
And re-write (54) as a probability inequality
$$ P\left\{ \frac{{2^{N - 1}} - {2^{ - \left(N + 1\right)}} - 1}{{2^N} - 1}\frac{1}{2^N}\sum\limits_{i = 1}^{2^{N - 1}} {\Updelta {R_{N,i}}} - \left(\frac{1}{2^N}\sum\limits_{i = {2^{N - 1}} + 1}^{{2^N} - 1} {\Updelta {R_{N,i}}} \right) \cdot \frac{{2^{N - 1}} - {2^{ - \left(N + 1\right)}}}{{2^N} - 1} + \left(\frac{1}{2^N}\sum\limits_{i = {2^N}}^{2^N} {\Updelta {R_{N,i}}} \right) \cdot \frac{{2^{ - 1}} - {2^{ - \left(N + 1\right)}}}{{2^N} - 1} \le \frac{R}{2^{N + 1}}\right\} \ge Y $$
(55)
Define X as follows:
$$ X \equiv \frac{{2^{N - 1}} - {2^{ - \left(N + 1\right)}} - 1}{{2^N} - 1}\frac{1}{2^N}\sum\limits_{i = 1}^{2^{N - 1}} {\Updelta {R_{N,i}}} - \left(\frac{1}{2^N}\sum\limits_{i = {2^{N - 1}} + 1}^{{2^N} - 1} {\Updelta {R_{N,i}}} \right) \cdot \frac{{2^{N - 1}} - {2^{ - (N + 1)}}}{{2^N} - 1} + \left(\frac{1}{2^N}\sum\limits_{i = {2^N}}^{2^N} {\Updelta {R_{N,i}}} \right) \cdot \frac{{2^{ - 1}} - {2^{ - \left(N + 1\right)}}}{{2^N} - 1} $$
(56)
Then, the variance of X can be expressed as:
$$ \sigma _X^2 = \frac{\sigma _{\Updelta R}^2}{{\left[\left({2^N} - 1\right){2^N}\right]}^2}\left[ { \left({2^{N - 1}} - {2^{ - \left(N + 1\right)}} - 1\right)^2}\sum\limits_{i = 1}^{2^{N - 1}} 1 + {\left({2^{N - 1}} - {2^{ - \left(N + 1\right)}}\right)^2}\sum\limits_{i = {2^{N - 1}} + 1}^{{2^N} - 1} 1 + {\left({2^{ - 1}} - {2^{ - \left(N + 1\right)}}\right)^2}\sum\limits_{i = {2^N}}^{2^N} 1 \right] $$
(57)
By re-formulating (57), we can obtain:
$$ \sigma _X = \frac{\sigma _{\Updelta R}}{\left[\left({2^N} - 1\right){2^N}\right]}\left[{\left({2^{N - 1}} - {2^{ - \left(N + 1\right)}} - 1\right)^2}\left({2^{N - 1}}\right) +\right. \left. {\left({2^{N - 1}} - {2^{ - \left(N + 1\right)}}\right)^2}\left({2^{N - 1}} - 1\right) + {\left({2^{ - 1}} - {2^{ - \left(N + 1\right)}}\right)^2}\right]^{\frac{1}{2}} $$
(58)
Using (56) and the standard normal distribution function, \(\Upphi, \) (55) can be rewritten as:
$$ P\left\{ {X \le \frac{R}{2^{N + 1}}} \right\} = \Upphi \left(
{\frac{R \mathord{\left/{ {{2^{N + 1}}}} \right. }}{\sigma _X}}
\right) \ge Y $$
(59)
Replacing (58) in (59) and applying the inverse standard normal distribution function, (59) can be expressed as (60):
$$ {\Upphi ^{ - 1}}\left( Y \right)\frac{\sigma _{\Updelta R}}{R} \le \frac{{2^N} - 1}{2} \cdot \frac{1}{\sqrt {{{({2^{N - 1}} - {2^{ - (N + 1)}} - 1)}^2}({2^{N - 1}}) + {{({2^{N - 1}} - {2^{ - (N + 1)}})}^2}({2^{N - 1}} - 1) + {{({2^{ - 1}} - {2^{ - (N + 1)}})}^2}}} $$
(60)
