τ = 0.25 · 2^{B R C
There are 2B of unit resistors.
The highest resistance is at midpoint. Half resistance will be connected to VDD and half to ground.
For equivalent resistance these resistances are in parallel.}

Calculation of output voltage with equivalent sources:
R is 1 kΩ.
All data inputs can be looked at as voltage sources V_D0...V_D3.
The voltages internally for equivalent sources are V_0L..V_2L, V_outL.
\( V_{0L} = V_{D0} \frac{2 R}{4 R} = V_{D0} \frac{1}{2} \)
\( R_{i0} = 2 R || 2R = \frac{2 R 2 R}{2 R + 2 R} = R \)
\( V_{1L} = (V_{0L} - V_{D1}) \frac{2 R}{4 R} + V_{D1} = V_{0L} \frac{1}{2} + V_{D1} \frac{1}{2} = V_{D0} \frac{1}{4} + V_{D1} \frac{1}{2}\)
\( R_{i1} = 2 R || 2R = \frac{2 R 2 R}{2 R + 2 R} = R \)

\( V_{nL} = (V_{(n-1)L} - V_{Dn}) \frac{2 R}{4 R} + V_{Dn} = V_{(n-1)L} \frac{1}{2} + V_{Dn} \frac{1}{2} = \sum_{i=0}^{n} \frac{V_{Di}}{2^{n-i+1}} \)
Capacitors can be used instead of the resistance R giving a C2C DAC.

Äquivalente Quellen und Zusammenfassen: R2 = 2.5k Die Dateneingänge sind ideale Quellen VD0...VD3. \( V_{0L} = V_{D0} \frac{R0}{R0 + R1} = V_{D0} \frac{1}{2} \) \( R_{i0} = R0 || R1 = \frac{R0 R1}{R0 + R1} = R0 \) \( V_{1L} = (V_{0L} - V_{D1}) \frac{R2}{R2 + R5 + Ri0} + V_{D1} = (V_{0L} - V_{D1}) 0.45 + V_{D1} \) \( V_{1L} = 0.56 \cdot V_{0L} + 0.44 \cdot V_{D1} \) \( R_{i1} = R2 || (Ri0 + R5) = \frac{R2 * (Ri0 + R5) }{R2 + Ri0 + R5} = 1.1 R0 \) \( V_{2L} = (V_{1L} - V_{D2}) \frac{R3}{R3 + R6 + R_{i1}} + V_{D2} \) \( V_{2L} = (V_{1L} - V_{D2}) 0.486 + V_{D2} \) \( V_{2L} = 0.486 V_{1L} + 0.514 V_{D2} \)