By using a DAC you are generating a analog voltage V which is proportion to corresponding digital voltage D

V≈kD where k is some constant and D can be written as

D=a_{1}*2^{-1}+a_{2}*2^{-2}+_________+a_{n}*2^{-n} ................( 1 )

a_{1}, a_{2} and so on are digital voltage which generate analog voltage V

**DAC BLOCK DIAGRAM**

a_{1}, a_{2} and so on are digital voltages(binary inputs) that generate analog voltage V

**DAC BUILDING BLOCK**

1>resistor network

2>analog switches

3>an op-amp

reference voltage Vref and digital data which need to be converted into analog voltage is applied to resistor network. output coming out from this resistor network is goes to analog switches. which in turn goes to op-amp and op-amp generate the output voltage Vo

**4-bit DAC**

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For a set of input voltages [V1,V2,V3,V4] that represents the value of quantity expressed as a 4-bit binary number, the (negative of the) output voltage. all these resistor are connected to voltage source through analog switches a1,a2, a3 and a4

-Vout=(2V_{ref}/R)R_{f}(a_{1*}2^{-1}+a_{2*}2^{-2}+a_{3*}2^{-3}+a_{4*}2^{-4})

where (2V_{ref}/R)R_{f} is constant [a_{4} is most significant bit and a_{1} is least significant bit]

the equation derived above is similar to equation one that we have talked about earlier

**DAC R-2R LADDER NETWORK**

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advantage of using R-2R ladder network over 4-bit DAC is it required only two type of resistor.A DAC utilizes a resister ladder network to divide current with equal current sources, and an operational amplifier to sum these currents and convert them into an output voltage.

**INTEGRATED CIRCUIT DAC(1408)**

8-bit D/A converter is 1408 which has a current output that can be converted to a voltage type using a current to voltage converter op-amp. reference voltage is applied between pin no. 14 and 15.

V0 = (V_{ref}/R_{ref)} *(RF)*{D0/2 + D1/4 + D2/8 + D3/16 + D4/32 + D5/64 + D6/128 + D7/256}

->uses R-2R ladder network

->8 bit convertor

->interfaced through i/o port

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