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In Reply to: RE: output caps posted by Constrained 1 on December 18, 2008 at 08:41:15
Why did they use a 100uF electrolytic in the first place?
Follow Ups:
This will give you some idea on how the low the frequency limit will be:
f= 1/2pi*R*C or 0.159/R*C
R = load impedance/resistance
C = coupling capacitor
The output circuit feeding the coupling cap and the load is basically a high pass filter.
Please don't provide information this inaccurate.
nt.
Better check your math next time before making yourself look bad!
1/(2*Pi)=0.15915494309189533576888376337251
Sad to see audiophile not knowing math as shown above.
What you should have written:
f = 1/(2*pi*R*C)
You do even need to show * for multiplication or the brackets. When was the last time you use * to show that it means multiply by 8 = 2*4, or 8 = (2*4)no need for bracket unless 8 = (2+2)* 2. Lokk at you math book. Was I right about 0.159?
More more!
"Live free or die"
Why the fascination with simple algebra? The real world has moved on to automated and useful tools. Why since I got my Shure reactance ruler over thirty years ago, I haven't had to think at all. Now that's technology!
Regards Rick
The 0.159 value is correct if you include all the terms in the denominator. However, your expression [f= 1/2pi*R*C or 0.159/R*C] could be interpreted differently. The '/' symbol has the same logical priority as the '*' symbol, so it is unclear whether you meant the division to be only 1/2 or 1/(2*pi*R*C).
"since you fall asleep in the math class" is doubly-embarrassing for you, as, apparently, you managed to sleep through your computer programming and English grammar classes.
nt
The only time I use the formular the way you show is when I use the excel spread sheet to do some simple math calculation.
Regarding to English Grammar, Enlish is my second language and it is still a learning progress for me everyday. So what else are you going to critizied me since you are so smart. When you can speak and write Thai as good as I can then you can talk.
Maybe budm slept through that part of math class.
Look here for some sample: http://megavoltz.net/capacitive_reactance.aspx
If you still do not get it where the 0.159 comes from.
.
Since no one gets it, I will explain it slowly.Take 2*P1, =, then hit the 1/x key, then hit M+
This stores 0.15915494309189533576888376337251 in the memory.
Hit clear.
Hit MR
Hit /
Enter a cap value in µF.
Hit /
Enter the frequency in Khz.
Hit =
The answer is in Kilohms.
Start out with a reasonable cap value that you can actually buy, you can always get a 1% resistor value you need.
Example:
.159/.1µF/.1Khz=15.9K
By using Kkz and Kilohms and µF, you can use a simple four-funtion calculator and don't need exponential notation (or having to enter a whole bunch of zeros).
Edits: 12/21/08
Thank djk, one thing that get me is the so called 'new math' system they are using in the US to solve algebra equation. I am still having the hard time with English words when it comes to math on how to explain how I get the answer.
I'm not doing this right. Help. The input of the amp is 100k. What's the smallest cap I can get away with? The amps bandwidth is 15Hz to 80KHz so I don't need to go lower than 15hz.
since at fc, Xc = R = 100K load
f = 15Hz
since Xc = 0.159/f*C
Then C = 0.159/fc*Xc
result C = 0.159/15*100K = .000000106 farad = .106uF (micro Farad)
So 1uF is more than enough?
0.47uF should do the job too. This is a commonly used value in many output stages. Hey, now you have even more choices!
Just make sure that the voltage rating is the same (or greater) as the electrolytic that you pull.
Thanks Quiet Earth. You must be a Genesis fan.
Yes,more than enough. I would use film cap, not lytics.
because the do not know the impedance of the amplifier input to which it will be connected, so they are covering all the options.
With any sane or real Zin (> 10K?) it is way over-kill and could be reduced to 2uF or 5uF with no audible loss of bass.
My amp has input impedance at 100k ohms and rated bandwidth is 15Hz to 80KHz before mods(if anything I made the upper bandwidth higher with smaller grid stoppers and no change to the lower). I wonder if I could get away with 1uF output caps on the cd player? I have those on hand.
Find the frequency at which 1uF has an impedance of 100K. Easy to find the formula!
The signal will then be divided by 2, or -3dB.
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