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amplifier and filter circuit

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jpglotzer

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Hey guys!

I'm building this circuit linked below. I get what some of it does, but I'm trying to fully understand the exact purpose of EVERY component, and why that exact size capacitor or resistor was used.

ece5030 plethysmograph

I get that the 10uF with the 20kohm is a high pass filter. How do I calculate the cut off frequency, and why that frequency was chosen? I think the 100K and 1K resistors determine the gain right? If i'm correct, it's just a ratio, so the gain is 100...

I don't get what the 0.47uF capacitor in parallel with the 100K resistor does..and I don't know why that value was chosen.

And the variable resistor from 1 to ground.. if I just kept that as a 10kohm resistor.. what does that do?

Ok, i think that's pretty much it. I'm sorta struggling with this, I've found similar circuits in textbooks, but they're all slightly different, so it doesn't seem 100% relevant. Any help would be much appreciated! Thanks!
 

chuckey

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the 10Mf +20 kR reduces the throughput when Xc = 20K, i.e. 1/(2 X pi X .0001 X F) = 20,000. This amplifier is to amplify the pulse rate?, so its should (in theory) have a decent response down to 1/60 BpM or 1 Hz
The .47 Mf is to reduce the amplifiers gain (more negative feedback) at higher frequencies, because there is no information there, only noise. Same formula as above except this time the throughput rises (more negative feedback = less gain).
The "gain" control makes up for the probe not getting enough IR, or too much.
Frank
 

Miguel Gaspar

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First you estart with a voltage divider
Vf = 20k/(20k+1/(jw10u)) = 20k / [(jw20k10u+1)/jw10u] = (jw20k10u)/ [jw20k10u+1]
the cut off frequency jw20k10u+1 = 0

w = 1/(20k10u) = 2*3.141592* fo = 5 rad/sec = 2*3.141592*.796 Hertz

For the filter with opamp this configuration is non inverter amplifier, and the gain is G = 1+ 100k/1k = 101

the capacitor in paralell with the resistor gives the frequency response or cut off frequency at f0 = 1/[2*3.141592*100K*.47U] THIS IS FIRST ORDER LOW PASS FILTER
 

jpglotzer

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Thanks guys! But what I'm not quite sure about is that... for the high pass filter for example, the cut off frequency is f=1/(2*pi*R*C). Many different combinations of R and C can give the same cut off frequency. So how do I justify choosing specifically 20K and 10uF?
 

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