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DAC based digital potentiometer

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http://img851.imageshack.us/img851/2991/captureyd.jpg [Broken]

To start off, I understand that I can compute the current flowing through each of the ladder rungs, as I = Vin/((2^n)*100k), where n is the terminal ladder interested in.

Doing so, I solved part a of the question for when the digital code D= 1111111, (7-bit DAC) was completely all on. Since there are two inverting op-amps with equal resistors in feedback, the ending output is a positive voltage value, which would be the total current flowing into the first op-amp multiplied by the feedback resistor.

Thus I got the total current for the first part as I = 9.921 * 10^-6 Amps, and thus the total output voltage to be Vout = 0.992 volts.

For part b, where D = 0000001, implies that just one of the switches is turned "on" to the op-amp while the others are shut off given the digital bit value of 0. Thus I assumed it is just D6 that is on, and calculated the total current as 5 * 10^-6 Amps, and the total Vout = 0.5 volts.

For part c, I am not entirely sure how to calculate the equivalent decimal resolution of the DAC.

For part d, I got the input impedance as just being equivalent to 2R for a R-2R type ladder circuit, no matter how long (even infinite), so the answer I got was 100k Ohm.

Please tell me if I am on the right track, appreciate the help!
 
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Answers and Replies

  • #2
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I caught an error for part B, i shouldve used the current running through D0 only, so recalculating gives me Vout = 0.00781 volts for D = 0000001.

for part c, I understand there are 2^7 = 128 bits possible. To calculate the decimal resolution would it just be 1volt/128 bits = 0.00781 volt/bit? I am not sure...
 
  • #3
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any insight?
 

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