Hi,
Mike Harrison wrote:
{Quote hidden}>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
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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