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What is the role of transistor Q1 ?? My circuit is not working because my I/p at pin 2 varies from 0.7-2.7V and o/p from 6.0-9.7mV....what's happening?? how does vs2 affect the output???
op amp circuit
- Thread starter Seema01
- Start date
? can you re-post the revised circuit showing it's INPUT OUTPUT and ALL the power supplies (your previous schm. at post #1 did not show the op Amp supply) . . .
? is there an INPUT or it suppose to be a function generator of some sort . . . which i doubt as the non-inverting input of the op amp is at the signal ground (and there is a DC path from output of the op amp to it's inverting input pin + a DC-path from there to signal ground)
the elements of the feedback chain
https://en.wikipedia.org/wiki/RC_time_constant + https://www.electronics-tutorials.ws/rc/rc_1.html
22k 1nF ~ "5+"RC ~ 2πRC = 138µs (has an integrating effect below ! 3.6kHz !)
500p in series with (2x 100n) = 499pF 100Ω ~ 313ns (1.6MHz)
33kΩ
22kΩ + 2x LED (2x 1.6V = 3.2V ~300µA on) "switches on" above 3.2V+22k·300µ ~ 9.8V but, starts conducting signifficantly likely from 10...100µA 0.7V+3.2V = 3.9V
...
i missed the C19 4.7nF -- it integrates the currents at inverting input . . . sets the "primary AC responce" gain to about 1:10 (along with the 500pF) basically it attenuates any output noise . . . but it also may cause oscillations at some conditions
if the input signal entry point and shape and amplitude are unknown it's hard to predict anything else
? is there an INPUT or it suppose to be a function generator of some sort . . . which i doubt as the non-inverting input of the op amp is at the signal ground (and there is a DC path from output of the op amp to it's inverting input pin + a DC-path from there to signal ground)
the elements of the feedback chain
https://en.wikipedia.org/wiki/RC_time_constant + https://www.electronics-tutorials.ws/rc/rc_1.html
22k 1nF ~ "5+"RC ~ 2πRC = 138µs (has an integrating effect below ! 3.6kHz !)
500p in series with (2x 100n) = 499pF 100Ω ~ 313ns (1.6MHz)
33kΩ
22kΩ + 2x LED (2x 1.6V = 3.2V ~300µA on) "switches on" above 3.2V+22k·300µ ~ 9.8V but, starts conducting signifficantly likely from 10...100µA 0.7V+3.2V = 3.9V
...
i missed the C19 4.7nF -- it integrates the currents at inverting input . . . sets the "primary AC responce" gain to about 1:10 (along with the 500pF) basically it attenuates any output noise . . . but it also may cause oscillations at some conditions
if the input signal entry point and shape and amplitude are unknown it's hard to predict anything else
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for the ~AC.sine chain :
±Vs — [ r ] — Vx — [ R ] — --||-C---|> GND
applies :
\[r=aR\ ,\ A={\left({a+1}\right)}^2-1\\ \frac{Vx}{Vs}=\frac1{\sqrt{1+\frac A{1+\frac1{{\left({2\pi RCf}\right)}^2}}}}\]
the derivative by by \(f\) (frequency) gives you the frequency of the fastest change of the above function say it's \(f\left({f}\right)\) then
\[f'\left({f}\right)=\frac{-A}{{\left({1+\frac A{1+\frac1{{\left({2\pi RCf}\right)}^2}}}\right)}^{\frac32}·{\left({1+\frac1{{\left({2\pi RCf}\right)}^2}}\right)}^2·{\left({2\pi RC}\right)}^2·f^3}\]
±Vs — [ r ] — Vx — [ R ] — --||-C---|> GND
applies :
\[r=aR\ ,\ A={\left({a+1}\right)}^2-1\\ \frac{Vx}{Vs}=\frac1{\sqrt{1+\frac A{1+\frac1{{\left({2\pi RCf}\right)}^2}}}}\]
the derivative by by \(f\) (frequency) gives you the frequency of the fastest change of the above function say it's \(f\left({f}\right)\) then
\[f'\left({f}\right)=\frac{-A}{{\left({1+\frac A{1+\frac1{{\left({2\pi RCf}\right)}^2}}}\right)}^{\frac32}·{\left({1+\frac1{{\left({2\pi RCf}\right)}^2}}\right)}^2·{\left({2\pi RC}\right)}^2·f^3}\]
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How did you conclude that c19 is integrating current??? So its a current amplifier. How do you differentiate between a current and a voltage amplifier. The pin 7 of op amp is connected to 12V. Input varies from 0.05-0.5V at pin 2.
. . . sums the Coulomb charge through it . . . "integrates" . . . basically . . . Δq=I(t)·dt=C·dU
versus http://tinyurl.com/y9n2sgtl
versus http://tinyurl.com/y9n2sgtl
