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26th March 2008, 05:38 AM  #1 
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Join Date: Apr 2002
Location: Belleville, IL.

Current to overcome 6.5 uuF input capacitance?
How much current would be needed to overcome a input capacitance of 6.5 uuF in order not to roll off the highs below say 30KHz?
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26th March 2008, 06:26 AM  #2 
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To properly answer that question you would need to provide two more specs....
What is the source impedance..ie, what is the output impedance of where the signal is comming from ....also if it has output caps..provide those values... Also you need to know the input impedance that is associated with this cap... Is there a input tube grid at this input also??? If so the miller cap needs to be accounted for as well... All these R's and C's need to properly put togather into a equivelent circuit model..then you can properly calculate the frequency response.... The current you are refering to is proportional to the source output impedance..... Chris 
26th March 2008, 06:54 AM  #3 
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The source impedance does play a role in that it provides a voltage divider, but it has nothing to do with his question does it?
The current does however depend on the amount of voltage you need to slew at 30kHz. So if it at the output stage where you can have more than 100volts with low mu tubes, or at the input where you likely have less than 1volt at teh input you will have different currents (obviously). Isn't the formula something like iC=(2*Pi*f*Amplitude)? But remember the capacitance is not just the datasheet capacitance, but the stage's gain multiplied with the capacitance. The output caps have nothing to do with this either. 
26th March 2008, 06:59 AM  #4 
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Hmmm...it's been ages since I bothered to calculate such, so not sure I got this right, probably should have kept my mouth shut.
Think it is iC=(2*Pi*f*Amplitude*C) Btw, uuF is today called pF. 
26th March 2008, 09:18 AM  #5 
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The slew rate limiting equation (5 time constants) when fully derived is equal to the 3dB POLE x 3.3 , I did the proof many years ago...
Lets say you want 20kHz to be free and clear from the slew rate limiting distortion...or it will be the boundary.... By putting the 3dB high frequency pole at 3.3 x 20kHZ , at 66kHz...puts 20kHz in the clear.... The only reason I wanted to know the output caps was for determining the low frequency rolloff... The ouput impedance plays a major role in the high frequency rolloff... It is directly proportional to output current drive.... For the same given output amplitude...the lower Z has to source more current... For proper calculation you really need to include the interconnects, since these have inductance and capacitance per foot that can be modeled as a lumped parameter Transmission Line... Chris 
26th March 2008, 10:22 AM  #6 
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Yes u are right. Its been a while since I last contemplated the issue. (It was interesting to me when designing with MOSFETs and after switching to tubes I havent worried about it).
The formula I mentioned for figuring the needed current for the given slew rate, is just a variation to the formula for capacitor impedance. Impedances and capacitances go into making low/high cutoffs so it just makes into a different view  same problem. Ohms law works either which way it is approached... So basically he needs to know the max source impedance at 3.3*30kHz = 99kHz? Also find te actual capacitance, is there gain etc.. 
26th March 2008, 12:36 PM  #7 
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Well the source is a Cambridge audio 540P MM phono pramp. That doesn't really help does it? Here are the only specs they publish:
http://www.cambridgeaudio.com/specif...0preamplifier I suppose I could write them a email and ask for a little more info. The tube is a 6AS7.
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26th March 2008, 12:51 PM  #8 
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So u are coupling a 6AS7 to a 39dB gained phonostage huh?
At least u wont be bothering the neighbours Should sound great thru headphones perhaps. 
26th March 2008, 12:53 PM  #9 
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Seriously G. U need not worry about the 6.5pF when u also have cables interconnects and such that totally dominate there.
What is it u are trying to achieve? 
26th March 2008, 12:59 PM  #10  
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Quote:
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