Determining voltage gain of instrumentation amplifier

[JJBladester]

Homework Statement

The following voltages are applied to the instrumentation amplifier in the figure.
V_in1=5 mV, V_in2=10 mV, and V_cm=225 mV. Determine the final output voltage.

Homework Equations

$$R_{1}=R_{2}=R$$

$$A_{cl}=1+\frac{2R}{R_{G}}$$

$$V_{out}=\left ( 1+\frac{2R}{R_{G}} \right )\left ( V_{in2}-V_{in1} \right )$$

The Attempt at a Solution

This is an odd-numbered problem in the back of my book, so I know the answer should be 1.005 V.

I already know my initial answer (see below) is wrong because it comes out to 1 V and does not take into account V_cm. Since I'm given V_cm but not A_cm, CMRR, or anything else, how do I solve this problem?

Incorrect:
$$V_{out}=\left ( 1+\frac{2\cdot 100k\Omega }{1.0k\Omega} \right )\left ( 10mV-5mV \right )=1mV$$

 

[UltrafastPED]

What does V_cm=225 mV represent?

 

[JJBladester]

I believe it represents a common-mode voltage (a noise signal or some other common signal on both inputs).

 

[NascentOxygen]

$$V_{out}=\left ( 1+\frac{2\cdot 100k\Omega }{1.0k\Omega} \right )\left ( 10mV-5mV \right )= ?$$

Work this out again.

 

[JJBladester]

That's what I get for working 15 hours of overtime this weekend and trying to get homework done.

Vout = 1.005 V as expected... Amazing what sleep will do for you. Thanks for pointing out the math error!