Basic Wheatstone Full Bridge Circuit
04-26-2017, 01:02 PM
Post: #1
 Eddie W. Shore Senior Member Posts: 1,009 Joined: Dec 2013
Basic Wheatstone Full Bridge Circuit
The program WHEATSTONE (HP Prime) deals with the full Wheatstone circuit.

Formulas

EA = E * (1 – R4/(R4 + R3))
EB = E * (1 – R1/(R1 + R2))
eO = EA – EB

Variables used: A = R1, B = R2, C = R3, D = R4, L = EA, R = EB, O = eO

Resistance is measured in ohms (Ω), current in volts (V).

HP Prime Program WHEATSTONE

Code:
EXPORT WHEATSTONE() BEGIN // 2017-04-25 EWS LOCAL K,A,B,C,D,L,R,O; INPUT({A,B,C,D,E}, "Basic Wheatstone", {"R1: ","R2: ","R3: ","R4: ", "E: "}); L:=E*(1-D/(D+C)); R:=E*(1-A/(A+B)); O:=L-R; PRINT(); PRINT("EA: "+L); PRINT("EB: "+R); PRINT("e0: "+O); END;

Source:

Strain Gages and Instruments. Tech Note TN-514 Vishay Precision Group. February 1, 2013. Link:
04-26-2017, 02:58 PM
Post: #2
 KeithB Member Posts: 205 Joined: Jan 2017
RE: Basic Wheatstone Full Bridge Circuit
Isn't the point of the bridge to see the variation in E0 with a change in one of the resistances?

Your code does not make that obvious.
04-27-2017, 09:22 PM (This post was last modified: 04-28-2017 09:38 AM by Vtile.)
Post: #3
 Vtile Senior Member Posts: 384 Joined: Oct 2015
RE: Basic Wheatstone Full Bridge Circuit
There is a small mistake as current is obviously measured in amperes if not measured over known resistance.
04-28-2017, 02:47 PM
Post: #4
 KeithB Member Posts: 205 Joined: Jan 2017
RE: Basic Wheatstone Full Bridge Circuit
He meant volts In volts.
04-29-2017, 11:15 PM
Post: #5
 Vtile Senior Member Posts: 384 Joined: Oct 2015
RE: Basic Wheatstone Full Bridge Circuit
Yep. If one measures the voltage ie. with multimeter there will be fifth resistance in bridge which is not even mentioned in most of the cases. For balanced circuit it doesn't matter, but my head says it do have "significant " error factor if the measurement is done as a unbalanced state and then unknown value is calculated. I need to try have time tomorrow to derive the equation just for own interest to see with a few extreme cases how big the error is. I suspect it is pretty big in worst case scenario. Then comes the temperatures from unbalanced current flow, total resistance drop in unbalanced situation (throug 5th resistor, which is in place of e0) as it goes more and more parallel to the bridge. Let see if I have time.
05-04-2017, 12:45 PM (This post was last modified: 05-04-2017 01:01 PM by Vtile.)
Post: #6
 Vtile Senior Member Posts: 384 Joined: Oct 2015
RE: Basic Wheatstone Full Bridge Circuit
Here is Wheatstone bridge with 5 Resistors, where the measuring devices resistance is taken account when calculating e0. Hopefully I (or Xcas) didn't make any silly mistakes.

$Ue0=-\mathrm{Rb3}\cdot U \frac{\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rx}}{((\mathrm​{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}) \frac{\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rx}}{\mathrm{R​b1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{R​x}}+\mathrm{Rx}) (\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{R​x})}+\mathrm{Rx}\cdot U \frac{\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}}{((\mathr​m{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}) \frac{\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rx}}{\mathrm{R​b1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{R​x}}+\mathrm{Rx}) (\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{R​x})}$

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