math problem ??

[prorms]

1/(2*pi*22k*47nF)
PLEASE SHOW ME THE MATH

 

[Nelson Pass]

Frequency = 1/(2pi R C)

2pi = 6.283

R = 22k = 2.2 x 10 ^ 4

C = 47 nF = 4.7 x 10 ^ -8

2pi RC = 65 x 10 ^ -4

1/(2pi RC) = 154 Hz

 

[Nordic]

While I have you here... how do I convert times into gain if you get what I mean....in db

 

[AndrewT]

db=20*(log (gain))

If gain is less than 1 then db comes out as negative.

gain = 2times >= +6db

gain = 20times >=+26db

gain = 0.5times >=-6db

when comparing power. db=10*(log(power ratio))

 

[prorms]

let make sure i understand you

2*pi*r*c
2*3.14=6.28
6.28*22000=138160
138160*0.000000047=0.00649352


m=1000000
k=100000
hz=100
mF=0.000
uF=0.000000
nF=0.000000000
pF=0.000000000000
HERE MY UNDERSTANDING HOW TO CONVERT
RESISTORS AND CAPACITORS
BUT DONT UNDERSTAND HOW TO CONERT IN TO Hz

 

[Nelson Pass]

Frequency = 1/(2pi R C)

2 X PI X FREQ = 1/RC

R IN OHMS
C IN FARADS

That's all there is

 

[AKN]

Hi,

1M=1000000
1k=1000
1m=0.001
1u=0.000001
1n=0.000000001
1p=0.000000000001

1/0.00649352=154Hz

All at once: 1/(2*pi*22000*0.000000047)=154Hz

 

[AKN]

Hi prorms,

Little more

0.00649352 this is actually the period time at the corner frequency you are calculating. Period time is the time from one zero crossing to another, eg 360 degree or 2pi in radians.

Frequency=1/t

 

[john_ellis]

Hi Prorms

If you turn your equation round, swapping f and R, the impedance of a capacitor is

(R=) Z=1/(2*pi*f*c)

Note the use of Z instead of R because it is reactive even though Z and R are both measured in ohms.

Time can be used instead of frequency, because the time period t=1/f so that we can also write

Z=t/(2*pi*c)

and we can swap C and Z (R) again to give

C=t/(2*pi*Z)

Why bother with all this swapping? Here's the point: the units of capacitance can be stated to be seconds (from the t) /ohm (from the Z) and this means that your answer in the original version f=1/(2*pi*R*c) is already in Hertz because the time-dependency is included in the C.

Cheers
John