Digital to Analog converter concept problem

[nil1996]

Homework Statement

Hello PF
i am stuck with a concept of weighted resistor D/A converter.
For a 4-bit resistive ladder,

V_A={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.

 

[maajdl]

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

 

[rude man]

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

 

[nil1996]

finally i have ended up with this:


V_A/R_L=V₀/(R₀+R_L) + 2V₁/(R₀+2R_L) +4V₂/(R₀+4R_L) + 8V₃/(R₀+8R_L)

Any way to get rid of R_L??

 

[rude man]

Any way to get rid of R_L??

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

 

[nil1996]

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

thanks for the reply

if R_L is zero then V_A would be grounded and hence it will be zero.

 

[rude man]

thanks for the reply

if R_L is zero then V_A would be grounded and hence it will be zero.

Right. So VA depends a lot on RL, does it not.
So your given answer can't be right.

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.

 

[nil1996]

Right. So VA depends a lot on RL, does it not.
So your given answer can't be right.

the equation --V_A={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 textbook.

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.

i have got that equation on Post No. 4

 

[nil1996]

here is the equation:

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

 

[rude man]

the equation --V_A={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 textbook.



i have got that equation on Post No. 4

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

 

[rude man]

here is the equation:

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

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

 

[nil1996]

That is wonderful!

My textbook 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.


For a 4-bit resistive ladder,

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

 

[nil1996]

so does RL>>R0 affects the equation?

 

[rude man]

so does RL>>R0 affects the equation?

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.

 

[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.""

 

[rude man]

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.""

Solve the equation in your post #9 for VA. Then let RL >> R0.
That will answer both your questions.

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

 

[nil1996]

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

 

[nil1996]

thanks a lot!
I got it.:thumbs:
(My textbook should have stated that it is appox.)

 

[rude man]

thanks a lot!
I got it.:thumbs:

(My textbook should have stated that it is appox.)

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

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