Active Bandpass filter using Op-amp

[@reshinmk2076]

09:12 can we make bandpass filter by using single op amp(from this circuit) ?

[@reshinmk2076]

Can you explain this circuit

[@ElectronicsforYoututorials]

Yes, we can design bandpass filter using single op amp. Choose value of Q between 10 to 20, also choose fc first. Then follow the procedure as mentioned in video.

[@eleazarmfonobong4941]

You didnt explain the action of the feedback capacitor that made the system pass low frequencies to become a bandpass filter. Because originally, thr system is a highpass filter.

[@ElectronicsforYoututorials]

I already did in my previous videos. The filter is a combination of LPF and HPF with different cut-off frequencies. The action is dependent on formula/equation Xc = 1/ 2(pi)fC. As frequency is inversely proportional the reactance, it will decrease with increase in frequency. The decrease in feedback resistance simply means decrease in gain and thus it blocks high freq. Plz watch videos on LPF and HPF for more details.

[@eleazarmfonobong4941]

@@ElectronicsforYoututorials I understand LPF and HPF, my problem is specifically on the use of capacitor at the feedback loop. I also know that high frequency lowers the reactance of the capacitor. In the feedback loop, the same high frequency pass through the capacitor in the feedback which also lowers it's reactance and suppose to also pass high frequency. I think the reactance of the capacitor is in response to the frequency of the applied signal? If so, then this makes both capacitors to respond alike and pass only high frequencies. How is the feedback capacitor allowing low frequency when in most cases, C1 and Cf are taken to be equal?

[@rk_study9]

❤❤ excellent

[@ElectronicsforYoututorials]

Thanks

[@arifahmed2492]

Sir i have got a problem that there is 2nd order low pass filter attached with 1st order high pass filter,low pass filter is on the left side and high pass filter is on the right side,there is told to determine upper cut off and lower cut off frequency and gain,how to calculate if u kindly help me with reply plz

[@ElectronicsforYoututorials]

Such problems are very easy to calculate. First use the formula, for lower cutoff freq. Consider any value of c between 0.01 micro farad to 1 micro farad and calculate R1. Similarly, for higher cutoff freq consider a value of capacitor and calculate R2. Use formula of gain. Consider Rf anywhere in between 10k to 100k and clculate Ri.

[@arifahmed2492]

@@ElectronicsforYoututorials Thank you so much...❤️

[@michealnicholosananth]

Send pdf format in the description sir

[@mr.jyotiranjankalta8098]

Thank you sir

[@ElectronicsforYoututorials]

Welcome

[@AshleyDada-tk2wk]

❤

[@louisdesipio3343]

how about some values

[@ElectronicsforYoututorials]

Depends on frequency range. First chose any value of capacitor between 0.001 uF to 0.1 uF. Then using formula of cut off frequency calculate the resistance.

[@pavanpeter8895]

direct formulae😴

[@gantidevisaidevid7372]

Bhai Tu pele samaj le ache se

[@ElectronicsforYoututorials]

Ok beta 16 sal se padha raha hu aur 3-4 sal baith leta hu college me