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diyMod and Corner Frequency

post #1 of 20
11/22/09 at 5:23am
Thread Starter 
Please help me make sure I'm understanding things:

1. The calculation for corner frequency (~3.4 Hz using 4.7 uF caps and 10K pot for amp) involves only the input impedance of the amp because that's the only resistor involved in the high-pass filter circuit, and the rest of the amp's/headphones' resistance is irrelevant. Is this correct?

2. It's unlikely I'll set the pot at the full 10K. If I listen with it turned to 7K (made-up number) and use the same caps, cutoff frequency will increase to ~4.8 Hz. So I should calculate the necessary capacitance using a resistance value less that the pot's max, right?

3. Finally, what happens if the amp has its own output caps? Would this create a second high-pass filter with the headphones? If so, how would the first cutoff frequency calculation (b/w diyMod caps and amp) relate to the second (b/w amp caps and phones)?

Many thanks
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post #2 of 20
11/22/09 at 5:50am
1) Yes
2) Nope, the input impedance is always around 10K as this is the resistance between the input signal and ground.
3) If the amp has "input caps", those are input coupling caps, those couple the source and the amp; you do not need two pairs of coupling caps. "Output caps" is another story, those are coupling caps between the amp and headphones and will need to be considerably higher, 100uF+ for most headphones, though you should probably consider 200uF+ to be safe.
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post #3 of 20
11/22/09 at 5:54am
Quote:
Originally Posted by dRummie View Post
Please help me make sure I'm understanding things:

1. The calculation for corner frequency (~3.4 Hz using 4.7 uF caps and 10K pot for amp) involves only the input impedance of the amp because that's the only resistor involved in the high-pass filter circuit, and the rest of the amp's/headphones' resistance is irrelevant. Is this correct?
Yes
Quote:

2. It's unlikely I'll set the pot at the full 10K. If I listen with it turned to 7K (made-up number) and use the same caps, cutoff frequency will increase to ~4.8 Hz. So I should calculate the necessary capacitance using a resistance value less that the pot's max, right?
No. The pot splits the resistance to ground. The source sees the 10K regardless of pot position.
Quote:

3. Finally, what happens if the amp has its own output caps? Would this create a second high-pass filter with the headphones?
Yes.
Quote:
If so, how would the first cutoff frequency calculation (b/w diyMod caps and amp) relate to the second (b/w amp caps and phones)?
It doesn't. They are separate circuits. Most headphone amp output caps (those that use them) are ~470uf, enough to get you close to single digits of bass response with 32ohm headphones (~10.6Hz).
Quote:

Many thanks
EDIT: I didn't type as fast as FallenAngel.
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post #4 of 20
11/22/09 at 7:01am
Quote:
Originally Posted by dRummie View Post
Please help me make sure I'm understanding things:

1. The calculation for corner frequency (~3.4 Hz using 4.7 uF caps and 10K pot for amp) involves only the input impedance of the amp because that's the only resistor involved in the high-pass filter circuit, and the rest of the amp's/headphones' resistance is irrelevant. Is this correct?

2. It's unlikely I'll set the pot at the full 10K. If I listen with it turned to 7K (made-up number) and use the same caps, cutoff frequency will increase to ~4.8 Hz. So I should calculate the necessary capacitance using a resistance value less that the pot's max, right?

3. Finally, what happens if the amp has its own output caps? Would this create a second high-pass filter with the headphones? If so, how would the first cutoff frequency calculation (b/w diyMod caps and amp) relate to the second (b/w amp caps and phones)?

Many thanks
Just to be a little different

1. No, the pot isn't the only resistance involved. There usually is a resistor from the opamps input pin to ground. If this value is very high, the effect in your circuit is negligible.

2. If there's a 10k pot and the above resistor value is 10k, you'll get an input resistance of 10k at zero and 5k at max volume.

3. I'm not so clever with filters but I think the corner frequency has to be calculated separately, but the filtering will add to one another and become steeper. You know the 1st order, 2nd order thing.
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post #5 of 20
11/22/09 at 10:56am
FORMULA
F = 1/(2 * pi * C * R)ENTER AMP INPUT IMP IN K =10
F = 1/(2 * pi * 0.000003300 * 25000)ENTER CAPS in uF =4.7
F = 1/(2 * pi * 0.08)CORNER FREQ =3.386275389
F = 1/(0.52)PHASE DISTORTION UP TO IN HZ =33.86275389< SHOULD BE BELOW 20
F = 1.93 Hz
PI =3.14159265
CAPS FORMULA0.0000047
INPUT IMP FORMULA10000
WORKING0.047
WORKING0.295309709
WORKING3.386275389

so with a 10k pot your 4.7uf is inadequate IMO

to avoid phase distortion you would need closer to 8uf


FORMULA
F = 1/(2 * pi * C * R)ENTER AMP INPUT IMP IN K =10
F = 1/(2 * pi * 0.000003300 * 25000)ENTER CAPS in uF =8
F = 1/(2 * pi * 0.08)CORNER FREQ =1.989436791
F = 1/(0.52)PHASE DISTORTION UP TO IN HZ =19.89436791< SHOULD BE BELOW 20
F = 1.93 Hz
PI =3.14159265
CAPS FORMULA0.000008
INPUT IMP FORMULA10000
WORKING0.08
WORKING0.502654824
WORKING1.989436791
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post #6 of 20
11/22/09 at 11:14am
Thread Starter 
Thanks for the responses! A couple things...

Quote:
Originally Posted by diditmyself View Post
Just to be a little different
2. If there's a 10k pot and the above resistor value is 10k, you'll get an input resistance of 10k at zero and 5k at max volume.
So you're disagreeing with the first two posters, or did I misunderstand?

Quote:
Originally Posted by diditmyself View Post
3. I'm not so clever with filters but I think the corner frequency has to be calculated separately, but the filtering will add to one another and become steeper. You know the 1st order, 2nd order thing.
This is what I was asking, sorry I wasn't clear. If I calculate 5Hz for the first filter, and 5Hz for the second, what principles governs the final cutoff frequency my phones will show?
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post #7 of 20
11/22/09 at 11:23am
Quote:
Originally Posted by dRummie View Post
Thanks for the responses! A couple things...
Quote:
Just to be a little different
2. If there's a 10k pot and the above resistor value is 10k, you'll get an input resistance of 10k at zero and 5k at max volume.
So you're disagreeing with the first two posters, or did I misunderstand?
Every amp that's well-designed will either use a very large resistor where he says or better - an input impedance setting resistor. This resistor is so large that its addition as a parallel resistance is negligible - the pot's impedance governs.
Quote:
Quote:
3. I'm not so clever with filters but I think the corner frequency has to be calculated separately, but the filtering will add to one another and become steeper. You know the 1st order, 2nd order thing.
This is what I was asking, sorry I wasn't clear. If I calculate 5Hz for the first filter, and 5Hz for the second, what principles governs the final cutoff frequency my phones will show?
They're separate. You can claim that the effect will be additive, but the impedances in the amp circuit itself will cancel this out. For instance the input to most opamps, for all intents and purposes, is close to infinity - at least compared to the input impedance at the pot. So, by the time it reaches the output at the headphones, you're calculating another RC phenomenon altogether (it's an entirely separate RC current loop, iow).
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post #8 of 20
11/22/09 at 12:05pm
yes, so 4.7uf isnt enough, you need closer to 8uf to avoid any distortions in the audio band. at least with a 10k pot anyway. I use less than 3 uf without any problem because all my amps have 50k pots
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post #9 of 20
11/22/09 at 12:31pm
Quote:
Originally Posted by tomb View Post
Every amp that's well-designed will either use a very large resistor where he says or better - an input impedance setting resistor. This resistor is so large that its addition as a parallel resistance is negligible - the pot's impedance governs.
I've had several amps with lowish resistor values affecting the imput impedance as the volume pot is turned up. Even if you take the most typical values - a 10k pot and a 100k resistor, I regard the effect as significant. You go from 10k to 9.09k from min to max. To me 10% is significant.

Quote:
They're separate. You can claim that the effect will be additive, but the impedances in the amp circuit itself will cancel this out. For instance the input to most opamps, for all intents and purposes, is close to infinity - at least compared to the input impedance at the pot. So, by the time it reaches the output at the headphones, you're calculating another RC phenomenon altogether (it's an entirely separate RC current loop, iow).
I still think it will become a 2nd order filter. Look at the picture below. The red line is before the output cap and the green line after. We're speaking of iMOD caps here, not amplifier input caps which usually are positioned between the pot and the input of the amp.



LL
LL
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post #10 of 20
11/22/09 at 12:36pm
Quote:
Originally Posted by qusp View Post
yes, so 4.7uf isnt enough, you need closer to 8uf to avoid any distortions in the audio band. at least with a 10k pot anyway. I use less than 3 uf without any problem because all my amps have 50k pots
I kind of agree to that. The problems with caps are price and size. What is better a 2.2 uF polypropylene or a 10 uF polyester? To my ears it's the polypropylene, but indeed a 10 uF polypropylene would be even better.
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post #11 of 20
11/22/09 at 12:43pm
Quote:
Originally Posted by dRummie View Post
Thanks for the responses! A couple things...


So you're disagreeing with the first two posters, or did I misunderstand?
Yes, this resistor should be a part of the calculation.


Quote:
This is what I was asking, sorry I wasn't clear. If I calculate 5Hz for the first filter, and 5Hz for the second, what principles governs the final cutoff frequency my phones will show?
I'm not entirely sure but like I said I think you should look at it as a second order filter. Look at the frequency response above. Why do you use output caps? What amp is this? It's very 60's (or very Graham Slee) or is it a tube amp.
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post #12 of 20
11/22/09 at 1:24pm
Quote:
Originally Posted by diditmyself View Post
I've had several amps with lowish resistor values affecting the imput impedance as the volume pot is turned up. Even if you take the most typical values - a 10k pot and a 100k resistor, I regard the effect as significant. You go from 10k to 9.09k from min to max. To me 10% is significant.
First of all, a 10 to 1 ratio is an absolute minimum. Most impedance setting resistors are 1M ... and if you size the input cap to react to that 10%, ... well, that's your problem.

Quote:
I still think it will become a 2nd order filter. Look at the picture below. The red line is before the output cap and the green line after. We're speaking of iMOD caps here, not amplifier input caps which usually are positioned between the pot and the input of the amp.
No one mentioned iMOD in this thread until you did - that's a singular aberration - it's not representative of the hundreds of DIY amp designs out there. If the OP says he's interested in iMOD, then you can have it.

Your links don't appear to work, btw. EDIT: OK, now they do.

Quote:
I'm not entirely sure but like I said I think you should look at it as a second order filter. Look at the frequency response above. Why do you use output caps? What amp is this? It's very 60's (or very Graham Slee) or is it a tube amp.
You're asking what the subject matter is NOW?
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post #13 of 20
11/22/09 at 2:57pm
Tomb, the title of the thread says DIYMOD, thats where he got that from
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post #14 of 20
11/22/09 at 3:11pm
Quote:
Originally Posted by qusp View Post
Tomb, the title of the thread says DIYMOD, thats where he got that from
OK - I guess I'm out of the loop completely on this. If you say "diyMOD" means an iMOD on an iPod, then I cry uncle - but it seems FallenAngel interpreted it the same way I did.
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post #15 of 20
11/22/09 at 3:42pm
iMod is the RedWineAudio modifiication of iPods. Some clever guy came up with the name DiyMod and it has become the official name for unofficial iModding. I think it's an excellent name that says it all, like GainClone.
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