Continue to Site

Welcome to EDAboard.com

Welcome to our site! EDAboard.com is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

inverting Low pass filter design

Status
Not open for further replies.

ahmedalaa

Newbie level 3
Joined
Apr 16, 2018
Messages
3
Helped
0
Reputation
0
Reaction score
0
Trophy points
1
Location
Cairo, Egypt
Activity points
26
hi all
i want to design a low pass filter but i have some questions i simulated it in pspice but there is a Capacitor named here Cin what is its function i try to to remove it and the circuit gives the same result.
another technical question why we don't use inductors in filters and if you have any another comment on this circuit please tell me
NB: the two circuits are attached as well as the graph
Capture.JPG
Capture.PNGs.PNG
 
Last edited:

Hi,

From the circuit it looks like an audio application.

i try to to remove it and the circuit gives the same result.
For sure you should see the difference with and without the capacitor in your frequency plot.
With capacitor it is a bandpass, without it is a lowpass.

The circuit attenutes the signal... do you want this?

If you don´t need the "inverting" function of the circuit, then you can do witout OPAMP.

***
Why no inductors:
They are not as precise as capacitors, they don´t have the linearity and thus inductors will create more distortions.
And they may pick up AC noise - maybe from the transformer nearby.

Klaus
 
I think you should connect the LM324's V+ and V- pins to +10V and -10V in simulation if you connect the non-inverting input of the OPAMP to GND. I am not sure the simulator can calculate the small signal model accurately if you just put a source between the V+ and V- pins
 

Hi,

I think you should connect the LM324's V+ and V- pins to +10V and -10V in simulation if you connect the non-inverting input of the OPAMP to GND.
Indeed I didn´t recognize this. I wonder how the simulation could work without a GND referenced power supply.


Klaus
 

There's a certain chance that the initial transient solution sets by chance an operation point in LM324 linear range. But the filter can't work under real conditions.

- - - Updated - - -

The Cin question wasn't answered yet, I think.

It's an coupling capacitor necessary for single supply operation. It creates a high pass corner, but Cin is usually selected for a corner frequency below the frequency band of interest. As a drawback, the filter can be only used for AC signals, e.g. audio.

You can answer the question yourself by trying different values for Cin. You'll see that fhp = 1/(2pi R1 Cin)
 

Thank you for all your responses, i tried to work on the circuit that is without the Cin and gives me that on the oscilloscope but there is some distortion here don't know why, sorry for the quality of the image.Capture.JPG
For the function of Cin i asked someone he told me that it blocks high frequencies to enter i didn't get it is that true and why?
 

if you are talking about the cutoff frequency i think that Cin doesn't enter the equation after simplifications the equation is f = 1/(2pi R2 C1) C1 will affect it not Cin. am i true?
 

The second circuit will not work properly without the power supply referenced to ground.
Without a ground there's no path for the op amp output current to flow back to the power supply
 

For the function of Cin i asked someone he told me that it blocks high frequencies to enter i didn't get it is that true and why?
Not true. It blocks low frequencies and DC. Useful to eliminate DC offset in cascaded stages, set high-pass filtering, and in single supply circuits to set normal operating point for the circuit.
At lower frequencies the capacitor has got bigger impedance, reactance (X_Cin(f)=1/(j*2*pi*f*Cin)) so the gain decreases, because without C1 the Gain is:

Gain(f)=-R2/(R1+X_Cin(f))

X_Cin(f) is bigger at lower frequencies (at lower f values) and it increases the denominator of the Gain equation.
 
Last edited:

Status
Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to EDABoard.com

Sponsor

Back
Top