Homework Help: Digital to Analog converter concept problem

1. Jan 19, 2014

nil1996

1. The problem statement, all variables and given/known data
Hello PF
i am stuck with a concept of weighted resistor D/A converter.

VA={V0/R0+V1/(R0/2)+V2/(R0/4)+V3/(R0/8)}/{1/R0+1/(R0/2)+1/(R0/4)+1/(R0/8)}

I want to find out how the equation is derived but have no clue.

Thanks.

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2. Jan 20, 2014

maajdl

Write the equations of this circuit and solve.
For example:
V0 = R0 i0 + RL iL
...
iL=i0+i1+i2+i3

3. Jan 20, 2014

rude man

You'll have a hard time doing that because your expression omits RL which is clearly necessary.

4. Jan 20, 2014

nil1996

finally i have ended up with this:

VA/RL=V0/(R0+RL) + 2V1/(R0+2RL) +4V2/(R0+4RL) + 8V3/(R0+8RL)

Any way to get rid of RL??

5. Jan 20, 2014

rude man

No. Assume RL=0. What would VA be then?

6. Jan 20, 2014

nil1996

if RL is zero then VA would be grounded and hence it will be zero.

7. Jan 21, 2014

rude man

Right. So VA depends a lot on RL, does it not.

So now let's do it right:
Sum currents at VA to zero: (V3-VA)/(R0/8) + ... = VA/RL.
Fill in the blanks. and solve for VA.

8. Jan 21, 2014

nil1996

the equation --VA={V0/R0+V1/(R0/2)+V2/(R0/4)+V3/(R0/8)}/{1/R0+1/(R0/2)+1/(R0/4)+1/(R0/8)} is right from my text book.

i have got that equation on Post No. 4

9. Jan 21, 2014

nil1996

here is the equation:

(V3-VA)/(R0/8) + (V2-VA)/(R0/4)+(V1-VA)/(R0/2)+(V0-VA)/(R0)=VA/RL

Last edited: Jan 21, 2014
10. Jan 21, 2014

rude man

Well, your textbook is right wrong. See my next post.

11. Jan 21, 2014

rude man

That is wonderful!
Oops, you forgot the divide signs for several terms on the left ....

Last edited: Jan 21, 2014
12. Jan 21, 2014

nil1996

My text book says """ Resistance RL represents load to which divider is connected and RL is considered to be large enough so that it does not load the divider network.

VA={V0/R0+V1/(R0/2)+V2/(R0/4)+V3/(R0/8)}/{1/R0+1/(R0/2)+1/(R0/4)+1/(R0/8)} """

13. Jan 21, 2014

nil1996

so does RL>>R0 affects the equation?

14. Jan 21, 2014

rude man

If RL >> R0 then RL drops out, approximately.
So then, compute VA for RL >> R0 ...

I have the answer if you want to check yours ...

PS R0 drops out too ... which is obvious if you look at your given solution. Multiply numerator and denominator by R0.

Last edited: Jan 21, 2014
15. Jan 21, 2014

nil1996

Can you explain me how RL drops out?
And what does it mean by ""RL is considered to be large enough so that it does not load the divider network.""

16. Jan 21, 2014

rude man

Solve the equation in your post #9 for VA. Then let RL >> R0.

(The divider network is R0, R0/2, R0/4 and R0/8.)

17. Jan 21, 2014

nil1996

solving that i get
(8V3+4V2+2V1+V0)/15=VA

Last edited: Jan 21, 2014
18. Jan 21, 2014

nil1996

thanks a lot!!!
I got it.:thumbs:

(My text book should have stated that it is appox.)

19. Jan 21, 2014

rude man

Good work!
Yes, it should have.
A finite RL reduces the output voltage by a constant: R0/(15R0 + RL) instead of just 1/15.

The idea here is that in a real application there will be some finite value of RL; this computation shows that it should be >> R0 or the output voltage will be a function of RL which makes the d/a converter inaccurate since RL is usually not constant.