written 5.8 years ago by |

The circuit- diagram for R-2R ladder type DAC is shown. Here only two values of resistors are required. This method is suitable for the integrated circuit realization. The value of R can be any where between 2.5k$\Omega$ to 10k$\Omega$

The circuit consists of a resistive ladder network whose series arm is having resistor R and the shunt arm consists of resistor 2R. The ladder n/w is followed by an amplifier.

The shunt arms are connected with the SPDT switch which connects the shunt arm either to logic 1 or logic 0. Here the i/p voltage will get weighted rather than weighted resistor.

Operations of R-2R ladder DAC with 3- inputs

i) When the i/p binary ward id $d_{1}d_{2} d_{3}$=100

$v_{o}=\frac{-R_{F}}{R_{i}}Vin =\frac{-2R}{R}. \frac{-VR}{4}=\frac{VR_{1}}{2}=\frac{V_{FS}}{2}$

For a binary i/p at 100 the analog o/p is $\frac{VR}{2}$

ii) Whent he i/p binary word is $d_{1} d_{2} d_{3}=010$

$V_{o}=\frac{-R_{F}}{R_{i}}V_{B}=\frac{-2R}{R}(\frac{-1}{8}V_{R})=\frac{V_{R}}{4}=$

For a binary i/p of 101 the analog o/p =$\frac{V_{R}}{4}$

iii) When the i/p binary ward is $d_{1}d_{2}d_{3}$=001

$v_{o}=\frac{\frac{10}{11}}{r+\frac{10}{11}}.V_{D}=\frac{10}{21}(\frac{-42}{128}VR)=\frac{-5}{32}$

$v_{B}=\frac{\frac{2}{3}R}{R+\frac{2}{3}R}.(\frac{-5}{32}VR)=(\frac{-5}{32})(\frac{2}{5})VR=\frac{-1}{16}VR$

$V_{o}=\frac{-R_{F}}{R_{i}}V_{B}=\frac{-2R}{R}(\frac{-1}{16}VR)=\frac{VR}{8}$

For a binary i/p of 001 the analog o/p is $\frac{V_{R}}{8}$