Need Help With Op Amp Circuit Explanation - MoreDat2024-03-29T06:49:26Zhttp://moredat.ning.com/forum/topics/need-help-with-op-amp-circuit-explanation?commentId=6483656%3AComment%3A938&x=1&feed=yes&xn_auth=noLet’s take each item that you…tag:moredat.ning.com,2012-04-02:6483656:Comment:9382012-04-02T19:53:20.950ZJerryhttp://moredat.ning.com/profile/Jerry
<p>Let’s take each item that you asked about. First the 680K resistor; as shown in your formula for gain, this is your feedback resistor. That means this resistor takes part of the output of the op amp and puts it back into the input of the op amp. But, it goes into the input labeled with a negative sign which is the inverting input. This means that instead of adding the signal from the 680k resistor to the input, it actually is subtracted from the input. That means we have negative…</p>
<p>Let’s take each item that you asked about. First the 680K resistor; as shown in your formula for gain, this is your feedback resistor. That means this resistor takes part of the output of the op amp and puts it back into the input of the op amp. But, it goes into the input labeled with a negative sign which is the inverting input. This means that instead of adding the signal from the 680k resistor to the input, it actually is subtracted from the input. That means we have negative feedback.</p>
<p>Now let’s look at the 100nF Capacitor. It also carries signal from the output of the op amp to the inverting input. Again, we have negative feedback, but with a twist. Now this feedback is dependent on the frequency of the signal. If our input is a DC voltage or a very low frequency AC voltage then the capacitor has high impedance, which is like saying it is a very large resistor. If it is a large resistor then it is only carrying a small amount of signal from the output to the input. Now what if the input signal is increased to a higher frequency, then the capacitor would act like a smaller resistor and carry a larger amount of the output signal back to the inverting input, providing more negative feedback. This would then reduce the gain at higher frequencies causing a faster roll-off of the gain than you would have without the capacitor. This would establish a bandpass for the circuit.</p>
<p>Now there is the question of the 6.8K resistor. Why is it there, what is it needed for? Look at your formula. It’s the input resistance for your op amp. Without an input resistor your input resistance of an ideal op amp is infinite. To help in controling the gain we need some value of resistance here.</p>
<p>Now let’s look at the combination of the 1 micro-Farad capacitor and the 68K resistor. These form a high pass filter. It blocks low frequencies and passes high frequencies. But there is the question of what frequencies are blocked and what frequencies are passed. Use your formula:</p>
<p>1/(2* pi * R * C). You will find that it also is 2.34 Hz. That means this high pass filter passes frequencies of 2.34 Hz and above. The 100 nano Farad capacitor in the feedback network limits how high a frequency it will pass.</p>