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Active low pass filter

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Karolina_1

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Hi friends, I wanted to build one circuit from an article but i cant understand it. I think its a active low pass filter but i am not sure. Can someone help me figure out what kind of filter is it. Its a very strange filter indeed. I tried simulating but i couldnt arrive to a logical answer. (Bode plot, AC frequency analysis)

* Why he put 3V reference to the input. Is it going to be 3V. When i built it in real life it wasnt 3V in the positive input of the opamp.

* Connecting the outputs of the the two opamps also seemed illogical.

Its said to be one pole/one zero active filter in the description. Thats all what is known.

Can someone help please Capture.PNG

Capture.PNG
 

hi K,
A quick observation, the OPA outputs are not connected, where they cross is a dashed line ---
E
 

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* Why he put 3V reference to the input. Is it going to be 3V. When i built it in real life it wasnt 3V in the positive input of the opamp.


View attachment 140697

When you operate an opamp from a single supply and want to process an AC signal, it is customary to have a "virtual ground" which is biased halfway between supply and ground.
You could use a simple resistor divider, but to have a a low impedance ground, it is customary to buffer it. That is what OA4 is doing.

Most likely, as esp1 mentioned, you have a wiring error.
 
Capture3.PNG

Capture2.PNG

when i simulate using a ref voltage 3V above ground output of the opamp is always zero.

but when i connect groung to positive input it works fine.

Capture5.PNG

Capture6.PNG

I am confused:
1.Why did he used 1.5k resistor series with 330nF capacitor.
2.Why not directly connect 3V to input(but it attenuates all the signals, output is zero as shown in figure above)
3.Why outputs of both opamps are connected.

When I simulate exact same circuit in the figure it doesnt works.
 

Hi,

Filter:
It is an inverting amplifier circuit with some gain-frequency adjustment.
It´s DC gain is about 14,000 (all inverting)
then at about 0.1Hz gain starts decreasing
then at about 320Hz gain settles to about 0.45

Wiring error:
I don´t think so.
I asume the dashed vertical line just wants to show that it is not connected with the horizontal line. This is not accrding schematic standards.

Why 3V biasing:
Unfortunately we only see a snippet of a complete document. And I assume it is well documented.
The input "PhC" is a signal from a phase comparator. I assume this is digital signal from a CMOS circuit with 0V/6V levels.
Therefore the "VCC = 6V" information.

I assume the complete circuit is from a PLL, a cos(phi) measurement or a lock_in_amplifier circuit. Am I right?

How often did i write "assume"? This could be avoided by posting complete information (document) with post#1.

Klaus
 

Your simulator assumes that your opamp (that has no power supply) has a + and - power supply.
Your simulator is stupid because it assumes that your opamp has no input DC offset voltage and does not calculate that the DC gain of about 14000 might cause the output to try to go to +142V or -142V.
 

    V

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Your simulator assumes that your opamp (that has no power supply) has a + and - power supply.
Your simulator is stupid because it assumes that your opamp has no input DC offset voltage...
The simulator actually assumes nothing. It's the user who selected an ideal OP model. It may be appropriate for some problems, but not for the present one.
 

    V

    Points: 2
    Helpful Answer Positive Rating
Hi,

that the DC gain of about 14000 might cause the output to try to go to +142V or -142V.
Very true.

Therefore this circuit needs to be a part of a feedbacked regulation loop to operate properly.
The high DC gain makes the "filter" to operate similar to an integrator.

Klaus
 

Therefore this circuit needs to be a part of a feedbacked regulation loop to operate properly.
Yes, according to post #1, the circuit is operated in close loop.
 

Lets consider that two opamps are not connected but separated. For the sake of simplicity lets consider only filter and reference voltage circuit.

I simulated the circuit as it was in PSpice Schematics (AC frequency sweep 1Hz to 10kHz) but couldnt get voltage to frequency graph. Its blank.
Capture10.PNG
Capture11.PNG

When i removed R22(resistor in the positive pin of opamp 1 U4) from the circuit output is Zero volts, nothing presents at the output.

Capture12.PNG
Capture13.PNG

Here is the full article. https://arxiv.org/pdf/1003.1290.pdf

I still cant understand why 3V is used in positive input of opamp cant we just ground it directly.

Thanks everybody for their time and help.
 

Hi,

It is already answered before.
* positive input is connected to 3V, because the application needs this. Maybe because of single supply. Maybe because of input signal range.

***
Just for simulation:
* either connect positive input to GND.
* or connect AC_source_bottom to 3V. (This is how the application works)
* or use a series decoupling capacitor.

Klaus
 
Assuming that the opamp is a rail-to-rail type, is powered by +6V and 0V, the (+) input is biased at half the supply voltage and the signal is fed through a coupling capacitor that is missing, then the output can swing equally up and down with the signal averaging +3V.
 

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