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Can anyone calculate the cut-off frequency and tell me which formula to be used on the Positive end of OPAMP?

what is the role of R1 here and R7 from the feedback? Please answer in details with resistance calculation, so that I can understand.

I suppose... Fc = 1 / [ ((R5||R1)||R7)+R2 ] * C1 x 2 * Pi

But I am not sure if pull-down resistor (r1) is either pull down here or not... and whether I shud use it for cut off calculation or not?

enter image description here

The AC analysis looks like that of bandpass filter... and can be seen here.. Cutoff finding would be helpful

enter image description here

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    \$\begingroup\$ Stick to the main idea of being polite when asking for help from volunteers. If you're being asked for more info, it means b you're missing something that helps the community to answer the question you're asking. \$\endgroup\$ – MrGerber Feb 2 '18 at 14:17
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    \$\begingroup\$ We take time out of our days to help you. Try not to be rude. If someone wants more information, give it to them. Otherwise you may get a wrong answer \$\endgroup\$ – MCG Feb 2 '18 at 14:19
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    \$\begingroup\$ Also, it's not possible from attached schematic to tell where the resistors are connected, since you're mixing dotted and not dotted wire junctions in the same schematic. \$\endgroup\$ – MrGerber Feb 2 '18 at 14:23
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    \$\begingroup\$ "stick to the main idea of problem-solving rather asking more questions." is what is rude \$\endgroup\$ – MrGerber Feb 2 '18 at 14:45
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    \$\begingroup\$ Then edit your schematic to represent what the circuit is like. Don't expect answerers to read through the comments, piecing together the info \$\endgroup\$ – MrGerber Feb 2 '18 at 17:47
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The circuit was/is unclear, first there was R7 feeding back the ouput of the opamp. Now R7 only feeds a DC voltage of 3.3 V to the input. Why is the 3.3 V source near the opamp's output and crossing the wire while it has nothing to do with it?

Also the opamp is used in such a wrong way that this circuit can never do anything useful, ever. For starters there's a huge DC voltage difference present at the input of the opamp so the output of the opamp will be stuck on one value. The opamp will basically do nothing useful.

To calculate the cutoff point of the input part of the circuit we can reduce the circuit to:

schematic

simulate this circuit – Schematic created using CircuitLab

In my experience the cutoff point is where the impedance of the capacitor is the same as the total resistor in parallel with it, that includes R5 as for small signal it is shorted by V1 so:

R1 // R7 // R5 = 1.31k ohm

Z = 1 / (2*pifC) = 1.31 kohm = 1 / (2*pi*f*1nF) => f = 122.4 kHz

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  • \$\begingroup\$ I did AC analysis and I guess the centre frequncz is 126 kHz \$\endgroup\$ – letmehelpyou Feb 2 '18 at 14:09
  • \$\begingroup\$ My bad, I forgot about R5 will update my answer \$\endgroup\$ – Bimpelrekkie Feb 2 '18 at 14:14
  • \$\begingroup\$ Hi, r1||r5||r7 and did u forget r2? r2 would be in series then, right? \$\endgroup\$ – letmehelpyou Feb 2 '18 at 14:25
  • \$\begingroup\$ Why would R2 be of importance? It is connected to the input of the opamp. Opamp inputs have a very high input impedance so... \$\endgroup\$ – Bimpelrekkie Feb 2 '18 at 14:54
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The input RC filter is low pass indeed. The OP AMP itself may ipose further high frequency attenuation above certain frequency. This implies that the overall frequency response should be flat up to the cut off frequency Fc=1(2pi.Reqv.C), where Reqv =1/(1/R5+1/R1+1/R7) if the input filter is dominant (at lower frequency that the op amp limitation. If the op amp has unity gain at 1 MHz, it will have just gain of 1000 at 1kHz.

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    \$\begingroup\$ The problem with this circuit is that the opamp is used in such a way here that it will be impossible to make it act as an amplifier. There will be a very large DC offset at the input of the opamp so the output will basically be "high" all the time. \$\endgroup\$ – Bimpelrekkie Feb 2 '18 at 15:00

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