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Draw and explain R-2R ladder network DAC for 3 bit-input using a suitable example.
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Overview -

• Resistor ladder networks provide a simple, inexpensive way to perform digital to analog conversion (DAC)

• The most popular networks are the binary weighted ladder and the R-2R ladder. Both devices will convert digital voltage information to analog, but the R-2R ladder has become the most popular due to the network’s inherent accuracy superiority and ease of manufacture

• Figure shown below is a diagram of the basic R-2R ladder network with N bits. The “ladder” portrayal comes from the ladder-like topology of the network

• Note that the network consists of only two resistor values; R and 2R (twice the value of R) no matter how many bits make up the ladder. The particular value of R is not critical to the function of the R-2R ladder

• R-2R ladder networks provide a simple means to convert digital information to an analog output

• Although simple in design and function, applying an R-2R resistor network to a real application requires attention to how the device is specified. Output errors due to resistor tolerances are often overlooked in the design of the digital to analog conversion (DAC) circuit and in the selection of the R-2R ladder itself

Working -

• It uses Kirchhoff’s current law, which states that the sum of currents entering a node must be equal to the sum of the currents leaving a node

• In the ladder, at each node, the current is split in half. By switching the currents into each node, the total current flowing is binary weighted

• Using the principle of superposition, when you add more current into a resistance the total voltage appearing is the sum of the voltages caused by all the individual currents i.e. as each bit is activated, so the voltage increases at the output. This results in the conversion of the input digital stream in analog value

• For example, consider the applied binary word to be 101, with reference voltage $V_R$=10V and resistance R=10kΩ

• The output current $I_o$ will be given as

$I_o=\frac{(-10V)}{Ω} [\frac{1}{(2^0×10^3 )}+\frac{0}{(2^1×10^3 )}+\frac{1}{(2^2×10^3 )}]=-0.125$ A

• The output voltage will be given as

$V_o=-R_f I_o=-(5×10 )×(-0.125 A)$=6.25V

• Resolution R=LSB=$\frac{V_{ref}}{2^n} =\frac{10}{2^3}$ =1.25V
• Therefore, for the binary input of $101_2$ or $5_10$, we have output voltage to be 5.625V

625V i.e. $V_{out}= V_{ref}×\frac{D}{2^N} =10× 5/2^3$ =6.25 V

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