From The Bonded Electrical Resistance Strain Gage by Murray & Miller where it says:
For operation with a single gage, and under some conditions with two
gauges, the efficiency can be improved by increasing the bridge ratio.
There is relatively little to be gained, however, but going beyond a
ratio of about 10, which will yield approximately 90% of the ultimate.
Many investigators prefer to use a maximum value of 5 which allows
considerably lower voltage for the power supply with an efficiency
that is above 80%.
What exactly does that mean? If I have a single strain gauge of \$ R_1 = 350 \Omega \$ and a constant voltage supply \$ V_{EX} = 5V \$,
Then for bridge ratio of 5:
\$ I_{\text{ratio of 5}} = {{V} \over {(R_1 + R_2)}} = {5 \over ((350*5) + 350)} = 23.8mA \$
\$ I_{\text{ratio of 1}} = {5 \over (350+350)} = 71.4mA \$ .
Since power dissipation is \$ \propto I^2 \$, the ratio of their efficiency and voltage drop would be
\$ { ({I_{\text{ratio of 5}}})^2 \over ({I_{\text{ratio of 1}}})^2} = {23.8mA^2 \over 71.4mA^2} = 11 \% \$
\$ { ({V_{\text{ratio of 5}}}) \over ({V_{\text{ratio of 1}}})} = {{23.8mA*350 \Omega} \over {71.4mA*350 \Omega}} = 33 \% \$
And for bridge ratio of 10 (\$ R_2 = 10 * R_1 = 3500 \Omega \$), I'll get
\$ { ({I_{\text{ratio of 10}}})^2 \over ({I_{\text{ratio of 1}}})^2} = {1.3mA^2 \over 71.4mA^2} = 0.03 \% \$
\$ { ({V_{\text{ratio of 10}}}) \over ({V_{\text{ratio of 1}}})} = {{1.3mA*350 \Omega} \over {71.4mA*350 \Omega}} = 1.8 \% \$
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If I go for a ratio of 10, the strain gauge \$ R_1 \$ dissipates only 0.03% power than that in a bridge ratio of 1 – hence the circuit has reached an efficiency of 99.07%! That certainly not
relatively little to be gain ... going beyond a ratio of about 10stated in the textbook. I think I'm misunderstanding something here. -
At bridge ratio of 5, the voltage drop of the strain gauge \$ R_1 \$ is 67% lower than its drop in a 1:1 bridge ratio, but that has no implication that it
allows considerably lower voltage for the power supplyother than knowing it will take on a lower voltage drop just because a bigger resistance consumes a larger share of the supply voltage. Again, what does that statement means?
Best Answer
$$Bridge\ Ratio = \frac {R_2} {R_1} = \frac {R_3} {R_4} = a$$
You are confusing electrical efficiency vs efficiency of the strain gage.
From p111:
$$\eta = \frac {a} {1+a}\ \ \ (4.36)$$
So if a = 1 (all resistances are equal), the efficiency will be 50%. a = 10, efficiency = 90.9%, a = 5, 83.3% (above 80%) and a = \$\infty\$, efficiency = 100%.
If \$R_3\$ >> \$R_4\$ (a strain gage), then \$\Delta R_4\$ has smaller impact on current, so essentially constant-current circuit. Essentially, with a constant-current source, the change in voltage \$\Delta V_4\$ = change in strain. A Bridge Ratio of 5 is enough to make the results linear.
The higher the bridge ratio, the more linear the circuit.