Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: JayB on December 10, 2010, 12:47:45 pm
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Been trying to get this in my head and can't figure it out. It would help understand the frequency response of a parallel capacitor and resistor.
I'm trying to fine tune.
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Basic RC, with F = 1 / (2 * Pi * R * C)
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In the past I've used a .001 w/ a 100k which should equal 1592 Hz
And on my recent build I've used a 220k w/ same cap which equals 723 Hz
This is a high pass filter correct? I did this along w/ a .005 immediately preceding these to slightly drop a tiny bit of signal and also for filtering out lower frequency more so bass wouldn't be quite as muddy at higher volume levels (while using it in place of the tone stack while in bypass mode). But the numbers seem opposite to support my thinking on this?
Could I get a quick explanation of what the two values mean when saying this is there frequency response? I want to make sure if my understanding is okay or not.
Thanks.
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Basic RC, with F = 1 / (2 * Pi * R * C)
I believe that is the typical RC. The typical Cap and resistor to ground or resistor and cap to ground. That's not a problem for me.
I'm trying to understand the frequency of a resistor and then paralleling that resistor with a cap. The resistor blocks all frequencies but the cap would pass the high end over that resistor. I'm trying to determine where that knee begins?
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Jay,
If I'm reading your post correctly if used in series it's a high pass filter at the frequency calculated allowing all frequencies higher to pass and if it's tied to ground then it becomes a low pass filter as the cap passes all highs at the caculated frequency & above to ground.
The knee should be at the calculated frequency for either shouldn't it?
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I'm trying to understand the frequency of a resistor and then paralleling that resistor with a cap. The resistor blocks all frequencies but the cap would pass the high end over that resistor. I'm trying to determine where that knee begins?
OK, then it's basic reactance Xc = 1 / (2 * Pi * F * C) for the cap and with a pure resistor R a composite Z = (R * Xc) / SQR (R^2 + Xc^2) for the RC // network. Quite easy to model in a spreadsheet.
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Jay,
If I'm reading your post correctly if used in series it's a high pass filter at the frequency calculated allowing all frequencies higher to pass and if it's tied to ground then it becomes a low pass filter as the cap passes all highs at the caculated frequency & above to ground.
The knee should be at the calculated frequency for either shouldn't it?
With out the resistor to ground. Am I confusing? :smiley: I'm using a grid blocking resistor but by passing it with a pf cap.
Trying to get in my head the formula so that I don't have to test to much.
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I'm trying to understand the frequency of a resistor and then paralleling that resistor with a cap. The resistor blocks all frequencies but the cap would pass the high end over that resistor. I'm trying to determine where that knee begins?
OK, then it's basic reactance Xc = 1 / (2 * Pi * F * C) for the cap and with a pure resistor R a composite Z = (R * Xc) / SQR (R^2 + Xc^2) for the RC // network. Quite easy to model in a spreadsheet.
I had a spread sheet for all this but I lost it. I remember the typical RC, parallel resistors and caps, even cathode bypass caps.
So what is F? LOL I hated all that math in college even though I made a 4.0. I screwed off in the easy classes.
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F= frequency?
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F= frequency?
Yes, we're dealing with impedances : an ideal cap blocks DC and shows decreasing impedance with rising frequency. An inductor does the reverse.
A real life cap shows an equivalent series resistance (ESR) and begins to also behave like an inductor at high frequencies, it response looks like a flattened U.
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I'm using a grid blocking resistor but by passing it with a pf cap.
Do you mean something like the typical 1.5k grid stopper on an output tube or 68k in a preamp tube, but bypassing it with a cap?
If so, you might as well have no resistor or cap. The point of a grid stopper is to roll off extreme highs (which are now passed by the parallel cap), and to limit chragig of the coupling cap due to grid current (which would now happen through the parallel cap, albeit in a different way).
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I'm trying to understand the frequency of a resistor and then paralleling that resistor with a cap. The resistor blocks all frequencies but the cap would pass the high end over that resistor. I'm trying to determine where that knee begins?
OK, then it's basic reactance Xc = 1 / (2 * Pi * F * C) for the cap and with a pure resistor R a composite Z = (R * Xc) / SQR (R^2 + Xc^2) for the RC // network. Quite easy to model in a spreadsheet.
Sorry for being slow - does "SQR" mean the square root? And you'd have to substitute the "standard" formula for "Xc" in your "composite Z" formula?
I've wanted to understand how to calculate what the 10pf cap in parallel with the 3.3 meg resistor produces in the AB763 circuit: http://www.el34world.com/charts/Schematics/files/fender/DELUXE_REVERB_AB763.pdf
IOW we've got 10pf || 3.3 meg, but there's also a 470k || 220K ground leg forming a voltage divider. I assume it that ground leg also affects the frequency response of the high pass filter formed with the 10pf cap. How the heck would you calculate that?
My reason for asking is to try to understand what an equivalent cap||resistor combination would need to be if your "mixing" resistor is 100K or 500K instead of 3.3 meg.
Cheers,
Chip
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does "SQR" mean the square root? And you'd have to substitute the "standard" formula for "Xc" in your "composite Z" formula?
Yes and yes.
IOW we've got 10pf || 3.3 meg, but there's also a 470k || 220K ground leg forming a voltage divider. I assume it that ground leg also affects the frequency response of the high pass filter formed with the 10pf cap. How the heck would you calculate that?
Work node by node. DC behavior is pretty simple: (470K//220K) = 150K, 3M3 T 150K => Vout = 4.348% of Vin.
My reason for asking is to try to understand what an equivalent cap||resistor combination would need to be if your "mixing" resistor is 100K or 500K instead of 3.3 meg.
You may scale RC values while keeping the same RC time constant, ie. 10K//1µ = 100K//100n = 1M//10n. But a 100K series R won't have the same effect as a 3M3.
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Thank you FYL.
Chip
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I'm using a grid blocking resistor but by passing it with a pf cap.
Do you mean something like the typical 1.5k grid stopper on an output tube or 68k in a preamp tube, but bypassing it with a cap?
If so, you might as well have no resistor or cap. The point of a grid stopper is to roll off extreme highs (which are now passed by the parallel cap), and to limit chragig of the coupling cap due to grid current (which would now happen through the parallel cap, albeit in a different way).
Just like the Slo has a large grid stopper, 470k, before the second stage. I'm wanting to understand the frequency relationship of bypassing the resistor. I know the miller effect of the tube plus the grid current will change it somewhat. Just trying to get in the ball park with out a lot of guess work. Then again, I always end up switching caps in and out then listen with my ears. I'm just trying to understand it.
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> frequency response of a parallel capacitor and resistor.
A 2-terminal network without context does not have a frequency response.
There's another resistor here somewhere.
> a grid blocking resistor but by passing it with a pf cap.
Then there's a source impedance, and your grid impedance. The grid is over 100Meg (|| any grid resistor) and some capacitance.
Draw the WHOLE CIRCUIT.
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PRR - could you look at the AB763 reverb mixing circuit please? 3.3 meg in parallel with 10 pf as the upper leg of a voltage divider. 150K as the lower leg of the voltage divider. Do we simply calculate the frequency "cutoff" point of this high-pass filter as 10pf with 150K, or does the 3.3 meg resistor in parallel with the cap make a difference.
I recognize that there's also a low-pass filter formed by the 3.3 meg resistor and the Miller capacitance of the following 12AX7 gain stage. It occurs to me that may be the reason for the 10pf cap in the first place...
TIA
Chip
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Can I ask a quick question..?
In any of the formulas presented in this post, is "C" the capacitance in Farads or micro Farads..?
Thank You
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In any of the formulas presented in this post, is "C" the capacitance in Farads or micro Farads..?
The formula for Xc (capacitive reactance) uses Farads, so you need to add all the zeros to move the decimal point to the left. So 1uF is 0.000001 F.
The similar formula for inductive reactance uses Henries, which is not so problematic because we work closer to whole units of those (except when operating at RF).
All of the other formulas that FYL listed are derived from the formula for reactance or for resonance, and again use whole Farads or Henries.
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In any of the formulas presented in this post, is "C" the capacitance in Farads or micro Farads..?
The formula for Xc (capacitive reactance) uses Farads, so you need to add all the zeros to move the decimal point to the left. So 1uF is 0.000001 F.
The similar formula for inductive reactance uses Henries, which is not so problematic because we work closer to whole units of those (except when operating at RF).
All of the other formulas that FYL listed are derived from the formula for reactance or for resonance, and again use whole Farads or Henries.
Yeah, I should have known better than to ask. Your answer is the only one that makes sense.
Thank You
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In any of the formulas presented in this post, is "C" the capacitance in Farads or micro Farads..?
When no units are specified use default SI units: Hz, ohm, farad, henry, ampere, etc.
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I have been working on that frequency thing all month changing cathode bypass cap values in my bandmaster Reverb
I have found what i think is the best Fender bass to EQ tone mix. 1.5K / 2.2uf
22-25uf is just too much some times. I am also getting a nice mild harmonic string feedback using the 2.2uf cap
IT'S ALIVE!!!!!!!!!
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In any of the formulas presented in this post, is "C" the capacitance in Farads or micro Farads..?
When no units are specified use default SI units: Hz, ohm, farad, henry, ampere, etc.
That would be another good "rule of thumb" to keep in mind....
Thanks
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Doesn't anybody have a crayon?
Draw the WHOLE CIRCUIT.
(http://i55.tinypic.com/34ren89.jpg)
> 150K as the lower leg
Yup, 470K||220K = 150K.
> the Miller capacitance of the following 12AX7 gain stage. It occurs to me...
Yup. Use "100pFd" for rough estimates.
Untangle Fender's ziggy drawing. Extract the essentials. I include the ~~39K of the stage before, but compared to 3Meg that's nothing. You can look at the stage driving the 470K, but that works out to 154K at the node, which is beneath notice.
At DC and low frequency, we have 3Meg to 150K or 20:1 ratio. (Actually 23:1, NBD.)
At infinite frequency capacitance trumps all. 10pFd and 100pFd is 10:1. (11:1 on a sharp abacus; it may really be more like 140pFd of Miller plus stray capacitance in the long wire or 15:1.)
20:1 at low frequency and 10:1 at high frequency is significant but not a lot.
Where is it?
A full *exact* analysis of this is tedious and not necessary. Take each leg separately.
3.3Meg and 10pFd is 5KHz.
150K and 100pFd is 10KHz.
So "nothing" happens over the fundamental guitar band (80Hz-2KHz). Without the 10pFd response would droop above 10KHz. (Possibly as low as 7KHz.) Fender's habit in this time was to design for flat response far past 5KHz.
To compensate just this section you would use about 6 or 7 pFd. Odd value. Over-compensating with 10pFd gives a rise which cancels some of the droop in the many other stages of this amp.
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Doesn't anybody have a crayon?
LMAO :grin:
Thanks PRR, both for the humor and for the education!
I am going to lift one leg of that 10pf cap just to listen to the result.
Chip