AMP01 instrumentation amplifier

Although this is not TI's device, the answer is fairly simple. At low frequency the capacitor can be considered an open circuit. So the gain of the circuit is 20*100k/100k = 20. At high frequency, the capacitor is a short circuit, and now the effective value of Rg is 100k||4k = 3.846k and the gain becomes 20*100k/3.846k = 520.

Thank you John.

For the above circuit although ratio of Rs and Rg is same for different combination of Rs and Rg the passband cut-off frequency varies.For ex Rs=100k and Rg=100k and other combination Rs=10k and Rg=10k although ratio is same passband cut-off frequency is different.Can any one tell a relation bet resistor values and cutoff frequency.

Thank you

Regards

Further questions would be better addressed on Analog Device's forum.