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'[EE]:Formula for unbalanced bridge Rx-> V'
2001\02\26@073600 by mike

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Any maths wizards out there....?
Given a conventional resistance bridge :
Vin
+----+
R1  R2
+----|--- Vout (differential)
|    +---
R3  Rx
+----+
GND

What is the formula to determine Rx (Or, better,  the ratio Rx/R3)
given the ratio Vout/Vin. R1=R2 if that makes anything easier.

My algebra isn't quite up to juggling it round - I got as far as the
following before my brain started to hurt :
Rx/(Rx+R2) = V-(R3/(R1+R3)
where V is the ratio Vout/Vin

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2001\02\26@085052 by o-8859-1?Q?K=FCbek_Tony?=

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Hi,
Mike Harrison wrote:
{Quote hidden}

Well I'm definately not an math wizard but this is semi-related to
'my' field.
As you are asking about an wheatstone bridge circuit I can
elaborate a bit, it might be of help.
Given the following circuit(note resistor naming):

Vin
+----+
R1  R4
+----|--- Vout (differential)
|    +---
R2  R3
+----+
GND

You have the following:
Where V is bridge unbalance for an variation of any R called dR.

V = Vout/Vin = ( R1+dR1 / R1+dR1+R2+dR2 ) - ( R4+dR4 / R3+dR3+R4+dR4 )

If one assumes this is an strain gauge ( you did not mention )
then normally dR is much smaller than R ( i.e. dR << R ) and
that R1=R2, R1=R4 ( or even that R1=R2=R3=R4 ), thus we can simplify
further:

V = Vout/Vin = 1/4 ( dR1/R1 - dR2/R2 + dR3/R3 - dR4/R4 )

Normally the error caused by this simplification is well below
pratical technical limits.

Now if R1=R2 and R3=R4 the bridge is balanced and output is zero.

You did not specify 'your' conditions so I don't know if these
ramblings are valid. Also straight from the head so there might be an
error :).

Please post again if you would like to elaborate.

/Tony





Tony Kübek, Flintab AB            
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2001\02\26@090438 by wzab

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On Mon, Feb 26, 2001 at 12:40:15PM +0000, Mike Harrison wrote:
{Quote hidden}

(available at http://www.ma.utexas.edu/users/wfs/maxima.html )
The "C3" line contains the answer...

wzab@myhost:~$ maxima
GCL (GNU Common Lisp)  Version(2.3) pi± wrz 22 14:49:21 CEST 2000
Licensed under GNU Library General Public License
Contains Enhancements by W. Schelter
Maxima 5.4 pi± wrz 22 14:48:33 CEST 2000 (with enhancements by W. Schelter).
Licensed under the GNU Public License (see file COPYING)
(C1)  V1 : Vin*R3/(R1+R3);

                                   R3 Vin
(D1)                                -------
                                   R3 + R1
(C2) V2 : Vin*Rx/(R2+Rx);

                                   Rx Vin
(D2)                                -------
                                   Rx + R2
(C3) solve(Vout=V1-V2,Rx);

                          (R2 R3 + R1 R2) Vout - R2 R3 Vin
(D3)               [Rx = - --------------------------------]
                              (R3 + R1) Vout + R1 Vin
(C4) --                                 HTH
                             Wojciech M. Zabolotny
       http://www.ise.pw.edu.pl/~wzab  <--> .....wzabKILLspamspam@spam@ise.pw.edu.pl

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