[jparkhur]

Ok. I know just enough to be dangerous, with that said, what is wrong with my formula.   I'm not going to bug Papa Pass again, so educate me please.

ACA Amp with 10k input impedance. Here is email with help from Jom at diy

Hi Jon,
 
The input impedance will be a little bit more than 10K, but that’s a good figure
for a thumbnail calculation.  If you put 4.7 uF in series with the RCA, you should
get pretty close to 80 Hz, however as the cap gets bigger, the frequency goes
down, not up.
 
Your formula would be 1/RC = 6.3 * F
 
where R is in ohms,  C is in Farads, and F is frequency.
 
I think it will work fine.
 
best,  np
 
 
From: Jon Parkhurst
Sent: Thursday, April 30, 2015 7:14 AM
To: Pass Nelson
 
I need to create a passive inline rca filter for an ACA.
 
6V6 suggested there would be an issue if I did that?
 
There is 68K from input positive to almost ground, and 1K from that node to ground, making about 70K. Start with that as an assumption.

BUT, the 10K and 10uF in series is going to maybe screw up a passive filter in front of it... so at this point I a past my understanding, have to punt and suggest that you should ask Nelson. He usually replies to simple questions in a day or two.



Can I calculate for 10K input and do 5 uF for 80 hz first order
7uF for 110
8.5 for 135 etc.


159155/10000*cap=FReq


Thanks

Jon

[HAL]

Jon,
The equation is F-3dB = 1/(2*PI*R*C), or C = 1/(2*PI*R*F-3dB) for an RC high pass or low pass filter.

If the F-3dB is 80Hz, with R=10K Ohms, C = 0.199uF for a single pole filter 3dB down at 80Hz.

[jparkhur]

So why is NP telling me 4.7.  Because of the internal parts?   

[jcotner]

So why is NP telling me 4.7.  Because of the internal parts?

That would be my guess.
I looked at the ACA schematic and the input is a little different
since he incorporates feedback right at the input of the amp via
R12. C3 could also figure into this.
I'm not an amp specialist but that's my guess.
Your question of why the 4.7 is a legitimate follow on question and
I think it would be reasonable to ask NP for further expansion
(if he is so inclined).

[HAL]

Agree with jcotner, ask NP why the difference.  That formula is basically for an infinite load on the 10K Ohm input resistance. If that is modified be the feedback loop, then it is incorrect.

[Davey]

There's no way that 4.7uF is the correct value.  NP has either misunderstood your question or just made a mistake in his response.
I suspect the 200nF value computed by Hal is essentially correct.

But why not simply test the amplifier to determine the input resistance.  Simply wire a variable resistor in series with the input and adjust it so the output of the amplifier drops by 6db.  Check at a couple of different frequencies to be sure it's linear.  Remove the variable resistor and measure it with an ohmeter.  That is the input resistance of the amplifier.

Dave.

[nickd]

Not sure what model the "ACA" is Jon but not all of NP designs use an input buffer. If he is driving the 1st gain stage transistor directly, the cap value may be different than one used for a standard buffer. He could also be "lifting" the ground with a resistor to lower the noise floor. Not sure if that will effect the cap value.

NP has forgotten more than I will ever learn about amp design.

[jparkhur]

Not sure what model the "ACA" is Jon but not all of NP designs use an input buffer. If he is driving the 1st gain stage transistor directly, the cap value may be different than one used for a standard buffer. He could also be "lifting" the ground with a resistor to lower the noise floor. Not sure if that will effect the cap value.

NP has forgotten more than I will ever learn about amp design.

Nick,

Its just the little 6 watt Amp Camp Amp that he put together and DIY did a mono block kit.   

With that said, I have, with hesitation, emailed him.  I continue to get HAL's figure from his formula and others.  I am guessing that i did not articulate myself well enough and he may have thought i was putting it on the inside or other?

Jon

I will update... Thanks everyone for their guidance.

[richidoo]

ACA = Amp Camp Amp

[rak313]

ACA = Amp Camp Amp

I cant be sure but i thinkthis amp has again of 68/10 and has a virtual gnd after the 10k resister. So 10k is the resistance to use.  Then as was indicated before 1/RC=2•PI•F

[jparkhur]

Hal and others r right thanks guys.  Here is no response to my timid email.

I just couldn't figure out if something was odd on the inside.


Np-----
My apologies, I was off in my math (the formula is correct but the
calculation wasn’t)
 
1/RC = 6.3 * F    where R is in ohms,  C is in Farads, and F is frequency.
 
Let’s run it through:
 
1/(10,000 * C) = 6.3 * 80
 
10,000*C = 1 / (6.3 * 80)
 
C = 1/(6.3 * 10,000 * 80)
 
C = 1/ 5040000
 
C = 2 * 10^-7
 
C = .2 uF

You are correct

[jparkhur]

I need an F6

[HAL]

Cool!

I want a NP SIT amp to try.

[Brad]

I'm with HAL, I want a SIT amp...

[corndog71]

I want to build an F6 too but I need my hand held a bit when it comes to solid state gear.  The finished version is out of my price range.

[Danny Richie]

Hey guys, Here is a web site that has the formula built into a calculator that will let you drop in the cap value and input impedance and it will give you a -3db down.

http://www.carstereo.com/help/Articles.cfm?id=1