On 31 Mar 2005 19:46:33 -0800, "
wrote:
KISS 125 by Andre Jute
snip truly unbelievable wads of irrelevant waffle
Although we commonly say that "a Class A power tube draws no current
on its grid", a power tube requires current from the driver to
overcome various capacitance which loiter with malicious intent in your
amp. A useful shortcut to half a good answer is slew rate current
(which tells us how fast a capacitance is charged up and discharged)
and as usual experience provides the other half of the answer. The
first half of the answer can be calculated with slew rate formulae and
one empirical assumption generously suggested to me by Gordon Rankin
(who isn't responsible for what I do with the information) on 15
October 1996 while I was working on an 845 amp:
***
The Slew Rate
SR = 2*Pi*Bandwith*Vmax in V/uSecond
Where
Bandwidth = 20kHz or whatever your design will be good for
Vmax is the maximum voltage the stage will deliver
This requires some clarification and correction:
For this calculation to be correct, 'Vmax' is Vpk, i.e. 0.5*Vpk-pk or
1.414*Vrms. The result is of course in Volts/sec, not Volts/usec.
The input capacitance
Cin = (A + 1) Cgp + Cgc
Where
A = the gain of stage for which the input capacitance is being
determined
Cgp is the Capacitance of the Grid to Plate
Cgc is the Capacitance of the Grid to Cathode
The Slew Rate Current
SRC = Cin*SR
Erno Borbely, Jung and Ron Gunzler (according to Gordon Rankin) suggest
that the stage current should be 5 times the slew rate current to
overcome the input capacitance of the next stage.
The Stage Current
Scur = SRC * K
where
K = 5
***
Bear with me while I put in some numbers:
Desired Stage Current is therefore 5*Cin*SR.
For a 300B,
Cin =9 + (1 + 3.85)*15 = 81.85pf
For a fullrange amp with 80V signal voltage,
Slew Rate = 2*3.14*20000*80 = 10,048,000
therefore
Desired Stage Current = 5*81.85*10,048,000
and after moving the decimal we get
4.1mA
If anyone wants to know how to put the decimal in the right place
(i.e. to be sure that it's 4 and not 40mA), do the calculations in
Farads, Hertz and Volts. For those without scientific calculators, you
won't run out of digits if you work in MHz and nanoFarads, where
40kHz=0.04 and 82pF=0.082. The answer will be in mA.
snip of more reams of waffle where we find that a fudge factor is
introduced for the sole purpose of getting to the 8mA or so which
experienced designers know is about right for a 300B
This is the 'ultrafidelista' design process? Just make up numbers
until you get to the value that every reasonable designer uses?
You can use the capacitances loitering in your computer constructively
if you think laterally. For instance, the bandwidth of an amp should be
balanced. Any extension below a rather high level, say 50Hz, should be
matched by an extension at the top end. If the bottom end is being
deliberately sloped off early to protecthorn drivers (which rapidly
become unloaded below Fs), the HF should not be designed out to
infinity, whatever the iron may be capable of;
Why? What has this to do with bass extension?
The question arises, why do you want an amp that has HF extension so
far beyond your hearing? What is this "balance" good for? Does
something lie beyond the achievement of "balance"? The answer is
that there is a subliminal effect, which ultrafidelista sometimes refer
to as "speed" or the "fast amp" syndrome, and which novices
think is only about undersizing power supply caps. Unfortunately
building a "fast" amp is far, far more complicated than that and
definitely requires attention at both ends of frequency scale.
Utter hooey, as expected. The simple fact of the matter is that an amp
with a -3dB point of 20kHz will sound dull in the treble, whereas one
with a -3dB point of 40kHz will be only 1dB down at 20kHz, probably
inaudible in comparison with one which is ruler flat.
--
Stewart Pinkerton | Music is Art - Audio is Engineering
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