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+--- Thread: Basic Wheatstone Full Bridge Circuit (/thread-8248.html)
Basic Wheatstone Full Bridge Circuit - Eddie W. Shore - 04-26-2017 01:02 PM
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;Link: http://edspi31415.blogspot.com/2017/04/hp-prime-and-ti-84-plus-basic.html
Source:
Strain Gages and Instruments. Tech Note TN-514 Vishay Precision Group. February 1, 2013. Link:
RE: Basic Wheatstone Full Bridge Circuit - KeithB - 04-26-2017 02:58 PM
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.
RE: Basic Wheatstone Full Bridge Circuit - Vtile - 04-27-2017 09:22 PM
There is a small mistake as current is obviously measured in amperes if not measured over known resistance.
RE: Basic Wheatstone Full Bridge Circuit - KeithB - 04-28-2017 02:47 PM
He meant volts In volts.
RE: Basic Wheatstone Full Bridge Circuit - Vtile - 04-29-2017 11:15 PM
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.
RE: Basic Wheatstone Full Bridge Circuit - Vtile - 05-04-2017 12:45 PM
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.
\[Ue0=-\mathrm{Rb3}\cdot U \frac{\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rx}}{((\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}) \frac{\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rx}}{\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{Rx}}+\mathrm{Rx}) (\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{Rx})}+\mathrm{Rx}\cdot U \frac{\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}}{((\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}) \frac{\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rx}}{\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{Rx}}+\mathrm{Rx}) (\mathrm{Rb1}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb2}\cdot \frac{\mathrm{Rm}}{\mathrm{Rb1}+\mathrm{Rb2}+\mathrm{Rm}}+\mathrm{Rb3}+\mathrm{Rx})}\]
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