Hi everyone. I need help on how I can connect the Arduino with wheatstone bridge circuit. I had done the diagram. Hope u all can correct me if Im wrong.
And is there any program that could help me to read an unknown resistance and display it on the Serial Monitor.thank you.
Really in need of your help.
You don't need a Wheatstone bridge to measure resistance with an Arduino just use a voltage divider.
Mark
It does not make much sense to use a Wheatstone Bridge to only measure one varying resistance. 2 or more link resistor such as 2 or 4 strain gauge, That really would make the use of Wheatstone bridge shine.
for simple method, Just use a simple voltage divider such as the pic i attach below.
R1 is a fixed resistor, while R2 is the resistive sensor your trying to measure.
the value of R1 can be any value you want, but for a simpler calculation i suggest you use 1K ohm resistor.
Remember that voltage divider circuit formula is Vout = ( (R2) / (R1+R2) ) * Vin
Vout =AnalogRead( A0);
Vi = 1023;
R1 = 1;
R2 = (Vout/Vin) - 1;
Serial.print ( R2 );
Serial.println("k ohm");
follow this site for complete explaination
If you want a temperature stable measurement stick woith the wheatstone bridge idea
The Wheatstone Bridge
The theory ... just implement the equations in C
For measuring accurately any electrical resistance Wheatstone bridge is widely used. There are two known resistors, one variable resistor and one unknown resistor connected in bridge form as shown below. By adjusting the variable resistor the current through the Galvanometer is made zero. When the current through the galvanometer becomes zero, the ratio of two known resistors is exactly equal to the ratio of adjusted value of variable resistance and the value of unknown resistance. In this way the value of unknown electrical resistance can easily be measured by using a Wheatstone Bridge.
The equations to use..
The bridge voltage is calculated as follows:
VB= Vin*[Rx/(R3+Rx)- R2/(R1+R2)]
If R3=R1, and Rx= R2+delta, then
VB= Vin*[ (R2+delta)/(R1+R2+delta)-R2/(R1+R2)]
Now if we assume delta is much smaller than R1+R2, then
VB= ~ Vin*[delta/(R1+R2)]
So we can see that the bridge voltage is approximately proportional to the error delta, divided by the sum of the resistors on one side.
Given the bridge voltage we can compute the value for an unknown resistor.
(R1+R2)(R3+Rx)VB/Vin= Rx(R1+R2)+ R2(R3+Rx)
Rx*(R1+R2)VB/ Vin + R3 (R1+R2)VB/Vin= RxR1+RxR2 - R2R3- RxR2
RxR1 - Rx(R1+R2)VB/ Vin = R2R3 + R3* (R1+R2)VB/Vin
Rx = (R2R3 + R3 (R1+R2)VB/Vin )/ (R1- (R1+R2)*VB/ Vin)
