Determining Maximum Amplitute of Sinewave that can be applied to an OpAmp

Question

A noninverting amplifier with a closed-loop gain of 1000 is designed using an opamp having an input offset voltage of 5 mV and output saturation levels of ±13 V. What is the maximum amplitude of the sine wave that can be applied at the input without the output clipping? If the amplifier is capacitively coupled in the manner indicated below, what would the maximum possible amplitude be?

The first part of the question I've worked out by rearranging:

\$V_{out} = [1 + \frac{R_2}{R_1}]V_{in} + [1 + \frac{R_2}{R_1}]V_{offset}\$

to

\$V_{in} = \dfrac{V_{out} - [1 + \frac{R_2}{R_1}]V_{offset} }{ 1 + \frac{R_2}{R_1} }\$

\$V_{in} = \dfrac{13 - (1001)(5) }{ 1001 }\$

\$V_{in} = 8 mV\$

So 8 mV is the peak voltage (which according to my professor is the right answer). However I am completely lost to what happens once the capacitors are couple

Answer 1

The key is to think about what happens to the gain at DC when you couple with the capacitors (this is assuming the capacitors are large enough not to attentuate the signal passed through significantly - I'm assuming this probably isn't the point of the question since the values are not shown)

If you apply no signal to the input in each scenario, what is the output voltage?

Answer 2

What you have is a first order, bandpass filter, with gain. The normal way of calculating the saturation/clipping point will only be true substantially within the corner frequencies of the passband.

The addition of C1 with R1 and R2 forms the lowpass. C2 and R3 forms the highpass.

To find out what happens when you reach the corner frequencies, you will have to calculate them first, then find the attenuation point on the slope at a given frequency. Be advised the corners are usually calculated at about a -3dB attenuation already.

Once attenuation begins, a higher amplitude input signal may be applied the farther down the slope you go, before clipping. The lowpass effectively adjusts the gain ratio, while the highpass simply attenuates the input signal at low frequencies.

R3 is often referred to as a ground reference. In fact, without that resistor, and without the capacitor having some charge across it, no input would appear at the non-inverting input since the input impedence is extremely high, and thus nothing would appear on the output.

I once spent an hour checking connections, supply voltages, the oscilliscope, and swapping amps and input sources until I put a resistor in that location.