Urgent help needed please!!!

[benopeth]

I'm currently doing a re-assessment paper on 3-bit DACs, that use binary weighted resistors.

I'm quite lost on exactly how the resistor values are manipulated to binary values (1, 2, 4, etc) and on just how to work out the analogue outputs.

I have to compare theoretical results against ACTUAL results.

For a 3-bit DAC with an input voltage of 5V+, the results are...

MSB LSB (ACTUAL) (THEORETICAL)
1/P1 I/P2 I/P3
0 0 0 0V 0V
0 0 1 1V 1.25V
0 1 0 2V 2.5V
0 1 1 3V 3.75V
1 0 0 4V 5V
1 0 1 5V 6.25V
1 1 0 6V 7.5V
1 1 1 7V 8.75V

I/P 1 has a 15kOhm resistor, I/P 2 has 2 15kOhm resistors in series and I/P3 has 2 120kOhm resistors in parallel. All these lead to a 741Op Amp, with a 12kOhm Rf above it.

I have been using, to find out the theoretical output voltages...

Vo = V1/1 + V2/2 + V3/4 (obviously according to the logic states as listed above).

If anyone can shed some light on what I am doing wrong then I'd be so grateful, this is so important!

 

[Roff]

You are ignoring the gain of the op amp.

Gain=-Rf/Rs

Where Rf is the 12k resistor. You have 3 Rs resistors: 15k, 30k, and 60k.
This looks like homework. Given this new information, what do you think the output voltages will be?

 

[philba]

so is R off a negating of R on?

 

[ljcox]

The formula for the output of the resistor ladder is:-

Vo = V1/2 + V2/4 + V3/8

You then multiply this by the gain of the amplifier.

 

[Roff]

so is R off a negating of R on?

Sorta kinda. See this thread. Pay attention!