On the Drawing Board: Finding the resistor values in a voltage divider

Am currently designing an instrumentation amplifier circuit that requires two different voltages for level shifting (Vcc/2 and Vcc/10). To obtain these voltages I'm using a simple voltage divider using three resistors. I tried finding the values by via trial and error--plugging the voltage divider equation into a spreadsheet and manually trying different values to get the correct voltages as well as values available for 5% resistors. Well, after some half dozen tries, I threw my hands up in the air. It's just too clumsy to do it that way. So I took pen and paper and started deriving the necessary equations. Here are the results.

Given the above circuit and the value of either A, B, or C, and given f and g, we want to find the values of the other two resistors. f and g are defined as follows:

$\frac{B+C}{A+B+C}=\frac{1}{f} \textup{ \text{\ \ \ \ Eq.1}}$

$\frac{C}{A+B+C}=\frac{1}{g} \textup{ \text{\ \ \ \ Eq.2}}$

You will notice that if we multiply Eq.1 and Eq.2 by source voltage V_S on either side of the equation we get the voltage divider equations. But we don't need V_Sto find the resistor values.

We first solve for A in both equations.

For Eq.1

$f(B+C)=A+B+C$

$A= (f-1)(B+C)\textup{\ \ \ \ Eq.3}$

For Eq.2

$gC=A+B+C$

$A=(g-1)C-B \textup{\ \ \ \ Eq.4}$

Equating Eq.3 and Eq.4 we have

$(f-1)(B+C)=(g-1)C-B)$

We then solve for either B or C. Let's do B first.

$(f-1)B+B=(g-1)C-(f-1)C$

$fB=(g-f)C$

$B=\frac{(g-f)C}{f}\textup{\ \ \ \ Eq.5}$

Thus,

$C=\frac{fB}{(g-f)}\textup{\ \ \ \ Eq.6}$

Using Eq.3 and substituting Eq.6 for C we have

$A=(f-1)(B+\frac{fB}{g-f})$

$A=(f-1)[\frac{(g-f)B+fB}{g-f}]$

$A=\frac{g(f-1)B}{g-f}\textup{\ \ \ \ Eq.7}$

Thus,

$B=\frac{(g-f)A}{g(f-1)}\textup{\ \ \ \ Eq.8}$

Using Eq.3 and substituting Eq.5 for B we have

$A=(f-1)[\frac{(g-f)C}{f}+C]$

$A=(f-1)[\frac{(g-f)C+fC}{f}]$

$A=\frac{g(f-1)C}{f}\textup{\ \ \ \ Eq.9}$

Thus,

$C=\frac{fA}{g(f-1)}\textup{\ \ \ \ Eq.10}$

Eq.3 and Eq.4 gave us A in terms of B and C. Using these sames equations we can also find B in terms of A and C, and C in terms of A and B. The derivation is trivial so only the results are provided below.

$B=\frac{A}{f-1}-C \textup{\ \ \ \ Eq.11}$

$C=\frac{A}{f-1}-B \textup{\ \ \ \ Eq.12}$

$B= (g-1)C-A\textup{\ \ \ \ Eq.13}$

$C= \frac{A+B}{g-1} \textup{\ \ \ \ Eq.14}$

So there you have it--all the equations necessary to find the resistor values given any of them. In my case I still needed the spreadsheet to help me find values that are commercially available. I plugged in a couple of the equations in the spreadsheet, and then by choosing different values for B given f = 2 and g = 10, I eventually got A = 15K, C = 3K, given B = 12K. Those are values readily available for 5% resistors.

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Note: I used Codecogs' LaTeX Equation Editor to produce the images for the equations above. Am a newbie to Latex and have no idea how to pad spaces to move "Eq." towards the right. With a bit of trial and error I found that placing x number of "\ ", that's forward slash and then a blank space, will force x number of blank spaces.

On the Drawing Board

Sunday, January 24, 2010

Finding the resistor values in a voltage divider

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