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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}\$


\$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

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Here's a hint: With no input signal coming through C2, what is the DC operating point of the amplifier? Why? – Dave Tweed Jan 10 '13 at 22:29
The feedback network is an RC filter. So the closed-loop gain is frequency dependent at low frequencies. The gain is not 1000 down to DC. At DC, the amplifier is a voltage follower because C1 blocks DC, and so you have 100% feedback through R2. Thus the answer is frequency dependent. – Kaz Jan 11 '13 at 1:04
@Kaz, but the input is also AC coupled, so the fact that there is a capacitor on the feedback is probably moot. I am surprised the coupling caps are not labeled infinite to show that they only block DC, but it could be a more realistic problem. If you think of them as infinite capacitance they block DC but all other frequencies see them as a dead short, which I am betting is part of what he is supposed to find out. – Kortuk Jan 11 '13 at 1:18
@Mike, if you consider the non-inverting amplifier gain to be 1000, you should not substitute (1+R2/R1) by 1001. You should substitute by 1000 instead and that would give 8mV. Please correct your statement, I can't do it since that would take two characters only... (stack exchange rules) – gmagno Jan 17 '13 at 0:07
up vote 0 down vote accepted

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?

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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.

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