band pass calculation

[usernet]

hi i have develop a circuit from glolab worldwideweb.glolab.com/glmda/GLMDA.pdf

*so surprised that i cant add link if my post < 3:-(

from the manual , it says that C4,C3,R3,R4 serve as bandpass (10Hz) as well amplifier gain, with gain 100

The problem is when i calculate the bandwith frequency using 1/(2nRC) i get different values...

and, does c4 and R4 perform low pass filter n C3, R3 perform high pass filter?So, by combining these 2 together it become bandpass filter... Pls correct me if im wrong

one more thing that make me thinking, why do we need such filter if we are using DC source....honestly, im not realy good at electronic so can some1 pls help me?:grin:

 

[ejeet]

The amplification of IC1A is:

A(f)=Z(C4//R4) / (Z(C4//R4)+Z(C3+R3)

or in words, the impedance of C4 parallell with R4 DIVIDED BY the impedance of C4 parallell with R4 plus the impedance of C3 in series with R3.

C3 is high pass and C4 is low pass.

Set up this equation and plot the response for an interval of frequencies in a program like matlab.

 

[usernet]

The amplification of IC1A is:
A(f)=Z(C4//R4) / (Z(C4//R4)+Z(C3+R3)
Set up this equation and plot the response for an interval of frequencies in a program like matlab.

Thk ejeet for the reply...

do ur equation is same as below

A(f) = (X4//R4) / [ (X4//R4) + (C3//R3) ]

and X represent impedence of the capacitor which equal to 1/(2nfc).....and since u write A(f) does it means the eqn will be a function of frequency where at the right hand side of the eqn will have expression of f

in other word, the eqn gives me gain IC1A as a function of frequency....

since i dont have matlab right now...could i simply plot gain vs frequency using excel.....

again...sorry for being noob

 

[ejeet]

Thk ejeet for the reply...

do ur equation is same as below

A(f) = (X4//R4) / [ (X4//R4) + (C3//R3) ]

and X represent impedence of the capacitor which equal to 1/(2nfc).....and since u write A(f) does it means the eqn will be a function of frequency where at the right hand side of the eqn will have expression of f

in other word, the eqn gives me gain IC1A as a function of frequency....

since i dont have matlab right now...could i simply plot gain vs frequency using excel.....

again...sorry for being noob

Your equation is a bit off, C3 and R3 are not parallell, they are series coupled. Also it should be the impedance of C3, not capacitance. Using Z = impedance gives:
A(f) = (Z4//R4) / [ (Z4//R4) + (Z3+R3) ]

Your equation for impedance is right, if "n" means pi.

 

[FvM]

could i simply plot gain vs frequency using excel

Yes, if you understand some elementary AC network calculations, particularly how parallel and series circuit of reactive and resistive impedances behave.

 

[ejeet]

Sorry, usernet. I just realized I made a mistake before. The equation should be:

A(f) = [(Z4//R4)+(Z3+R3)] / (Z3+R3) or similarilly A(f) = 1 + [(Z4//R4) / (Z3+R3)]

Sorry again for the confusion.

 

[usernet]

ejeet...ur right...actually i just take ur formula and rewrite it based on my understanding just to comfim it....that parallel symbol was typo error...thk a lot

Yes, if you understand some elementary AC network calculations, particularly how parallel and series circuit of reactive and resistive impedances behave.

does it means i just plot A(f) vs freq....i think it is possible..thks FvM

 

[ejeet]

Yes, and do not forget to update the formula!

A(f) = 1 + [(Z4//R4) / (Z3+R3)]

 

[FvM]

And consider the "geometrical" addition of real and imaginary impedance magnitudes

Z4//R4 = Z4*R4/SQRT(Z4²+R4²)

Z3+R3 = SQRT(Z3²+R3²)

For the same reason A(f) = 1 + [ ] can't be exact, [] is actually a complex value.

 

[usernet]

thk a lot...ejeet for the correction

n thk for the reminder FvM

 

[usernet]

hm...i think i might know why there are filters....from research i've done plus some thorough reading on the pir datasheet i found out that the detector produces AC output....

**broken link removed**