Why is the frequency cut-off 0.707?

Nirvana

Joined Jan 18, 2005
58
Well If you are talking about an AC component under AC conditions then the formula for the Capacitive Reactance is:
XC = 1/2π f C,
Im not sure which example you are talking about, but I'll go on anyway.
If we are going to make the Resistance (R) which is in Parallel with Capacitor C, then the formula becomes;
XC = 1/2π f C = R, then to determine the frequency f the formula is transposed to give; f = 1/2πRC,
Think of it like this, as XC = R the voltages across each component will be equal, plotting this out on a Phasor diagram will show that the angle of VR and VC are 45 degrees , therefore the voltage VR is equal to Vin cos(45) =Vin/(root2) which is 0.707. Basically you need to draw phasor diagrams, using complex numbers will help.

Nirvana.

 

aliashar86

Joined Nov 23, 2006
71
plz explain

i have never seen the formula for calculating cut-off, instead just know abt it tht it is the breakdown voltage for transistor , MOSFET . can u plz explain again tht why it is cut off at 0.7 why not more or less than tht
 

Papabravo

Joined Feb 24, 2006
14,701
Cutoff frequency and breakdown voltage are two entirely different things. Can we backup for a moment and decide which one we are talking about.

BTW are you aware of the following:
Rich (BB code):
1 / SQRT(2) = SQRT(2) / 2 = 0.707 = sin(45) = cos(45)
 

gort

Joined Dec 4, 2006
3
me tink it was 0.7071 wen i were in school a long time ago, look it up in a sine table[ 4 fig ] to reveal the hidden angle or inverse sine on a calculator, if i remember rightly we were fiddling with rms value
 
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