The R/2R ladder network shown on the right can be most easily (and therefore mos

Question

The R/2R ladder network shown on the right can be most easily (and therefore most accurately) analyzed using the superposition method, since the circuit is linear. That is, you can calculate the output voltage due to each bit independently (with all the other voltage sources turned off), and sum the voltages from each bit to obtain the final result. Here each voltage source (V0~V3) represents the voltage input from each digital bit and shall have values of either VDD (the supply voltage) or zero. In calculating the output voltage of each bit you should start with the most significant bit (MSB), solve the circuit for output (Vanalog) and then do the next most significant bit. You may then write an equation for the less significant bits using mathematical induction. The final result will have the form

Vanalog = a0*V0 + a1*V1 + a2*V2 + a3*V3

If I were calculating V analog for source V3, how exactly do I know what ratio to multiply V3 to get its independent voltage?

The R/2R ladder network shown on the right can be most easily (and therefore most accurately) analyzed using the superposition method, since the circuit is linear. That is, you can calculate the output voltage due to each bit independently (with all the other voltage sources turned off), and sum the voltages from each bit to obtain the final result. Here each voltage source (V0~V3) represents the voltage input from each digital bit and shall have values of either VDD (the supply voltage) or zero. In calculating the output voltage of each bit you should start with the most significant bit (MSB), solve the circuit for output (Vanalog) and then do the next most significant bit. You may then write an equation for the less significant bits using mathematical induction. The final result will have the form Vanalog = a0*V0 + a1*V1 + a2*V2 + a3*V3 If I were calculating V analog for source V3, how exactly do I know what ratio to multiply V3 to get its independent voltage?

Explanation / Answer

I just solved this circuit for my lab, and you can turn the circuit upside down and start using superposition and combining resistors into equivalent resistors where you can. When you can not combine any more resistors use KCL at the nodes and solve for the node voltages. Each voltage will be halved. Starting with (1/2)V3+...+(1/16)V0. The equivalent circuit drawings and equations for the unknown should take a total of for pages, or you can use Thevenin's Theorem.

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