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I'm planning to use an LM324 opamp as a signal amplifier for a cell load (strain gauge in a Wheatstone bridge). This particular application is shown in the datasheet (see image). It shows the Wheatstone brige (bottom-left resistor is the strain gauge) and the opamp in a differential amplifier circuit.

bridge amplifier from datasheet

My problem is I can't quite figure out how the formula is calculated.

I get I have to use Thevenin on both sides of the bridge. This gives me:

$$R_{t-}=\frac{R}{2}$$

$$ V_{t-} =\frac{ V_{REF}}{2} $$ for the right side of the bridge and $$R_{t+}\approx{\frac{R}{2}}$$

$$ V_{t+} \approx{\frac{ V_{REF}+\delta V_{REF}}{2}} $$ for the left side. Now the circuit becomes:

wheatstone to thevenin

and the differential amplifier circuit formula gives me:

$$ v_o = \frac{R_f}{R/2}(v_{t+}-v_{t-})=\frac{R_f}{R/2} \frac{\delta V_{REF}}{2} $$

which differs from the datasheet formula for a 1/2 factor.

So, what's wrong with my calculation?

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Your problem resides in the oversimplification of $$V_{t+}$$

If you take directly the tension divider between the R and R(1+delta) resistors you get instead $$V_{t+} = V_{REF}\frac{1+\delta}{2+\delta}$$

Which is approximately the same as what you used, but yields the good formula.

$$v_o = \frac{R_f}{R/2}\frac{\delta V_{REF}}{2(2+\delta)} \approx \frac{R_f}{R}\frac{\delta V_{REF}}{2}$$

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  • \$\begingroup\$ Thanks! Then again the same should apply to Rt+ and the Vo expression get more complicated. \$\endgroup\$ – dodod Oct 27 '17 at 13:52

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