# Digital to Analog Converter math question

I'm having a hard time understanding how to calculate this. I'm getting an answer of 3.33K which is not listed in the answers. Am I doing something wrong here or is the answer not given? • Can you show us your work? What is your approach to solving the problem? – Joe Hass Jan 29 '14 at 3:35
• The correct answer is not on the list (I get the same answer as others) BUT your answer cannot possibly be correct. Inverting node is driven to ground as + node is at ground. Current through 37.5k = 5/37.5 0.13 mA and you can calculate total current into node (Clue: Note relative resistor values - 1 + 1/2 + 1/4 + 1/8 = ?)Max I in is less than 2x I_37.5k. BUT at -8.25V your 3.33k Rx will carry R = V/I = -8.25/3.33k = 2.4 mA. Too high. – Russell McMahon Jan 29 '14 at 6:22
• Test are written to uncover many kinds of shortcomings. Looks like this one checks on your sense for orders of magnitude. All the answers but the first are off by a factor of 10-ish. You don't need anything but simple arithmetic for this. Rx=37.5K gives unity gain on D3 or -5V, -2.5 on D2, -1.25 on D1, -0.625 on D0. Add em up you get -9.375V. Not the right answer, but the others are WAY off. 300K gives -5 for D0, -10 for D1, -20 for D2, -40 for D3, and -75V just isn't going to work. Then again, it might be a mistake by the test writer. Where is it from? – C. Towne Springer Jan 29 '14 at 7:57