That will not change the current of the tubes, self bias protect and prolong life of the tubes and give a smooth sound, fixed bias is bad for the tubes and give a more "punchy" sound, but they both make the same current in the tubes ...
Normally, a "standby" switch is a DPDT that cut the high voltage at the ground center tap of the power transformer and supply all the tubes heaters at half voltage ... Then the tubes stay warm but when the power come back, the current is much lower and raise gradually until the heaters get at their normal operating temperature ...
When the heater is colder, the cathode furnishes less electrons then, the current is lower ... That is easy to see with the "lifeline indicator test" of tubes testers and also when you lower the heater supply voltage, the needle drop in the "bad" area of the scale. I just test a 6AQ5 on mine and it test 104% with the heater at 6,3V but only 21% with the heater at 3,3V ...
Very few peoples seem to know that on audio DIY Internet forums desipite all goods old tubes electronic books said so ... When you "turn on" a tubes amplifier, it take many seconds before the tubes warm up and reach their idle current.
Cheers,
Alain.
Like said this datasheet : http://www.shinjo.info/frank/sheets/127/6/6CA7.pdf
The maximum grid leak resistor for a 6CA7 is 700K in class "A" and "AB" and 500K in class "B", but generally, it is better not to use more than a 100K grid leak resistor with fixed bias ...
The reactance of a bypass capacitor should be at least 20 times less than the cathode bias resistor resistance at the lower audible frequency ... The formula for reactance is :
Xc = 1 / ( 2 x pi x F x C )
For example, with a 470 ohms resistor, the reactance should be 470 / 20 = 23,5 ohms at 20 Hz ...
Then : 23,5 = 1 / ( 2 x pi x 20 x C )
And : 1 / 23,5 = ( 125,664 x C ) = 0,0425532
So : 0,0425532 / 125,664 = 0,000338626 Farad
Or : 0,000338626 x 1000000 = 338,627 uF
Then a standard 470uF is all right ...
To design an amplifier, don't take parts values from any schematics you found on Internet, you have to know exactly what you are doing and calculate everything or you running into serious problems !
Cheers,
Alain.
There is a difference between "standby" and "mute" . guitar amplifier makers confuse them as you see. I always prefer to have actual standby, cause i make single ended amps and try to have longer life.
I use some schematics as a guide, almost never took a design as it is posted, for example.
The point of stand by switch is just protecting power tubes. Experience has told me that turning on a tube amp and stand by switch at the same time damages output tubes.
I thought cathode bias made tubes draw more quiescent current, so they last less than in fixed bias. I know the sound is warmer in cathode bias, and there's no need of a lotta watts, with high SPL speakers the amp will be loud enough for local gigging, and more "juice" from output tubes can be obtained, will be easier to crank it!
I know, but as far as guitarists are concerned the standby is used as a muting function (as well as giving them the warm but false feeling that the tubes lasts longer), and if such a switch is not included they look for the panic button, and if that's not there all hell breaks lose.
People often say things like this, and it is almost true yet also rather misleading. The real issue is to ensure that the reactance is significantly less than the parallel combination of cathode resistor and 1/gm, because it is this which should appear as R in the formula. In most cases 1/gm is the smaller so it dominates. A better statement would be "reactance at least 3-5 times smaller than 1/gm at the lowest required frequency".Alain Poitras said:
Like said costis_n, a real standby prolongate the life of the tubes and also act as a mute ... And this is very easy to do, just add a DPDT swich and wire it properly between the center tap of the power transformer and the ground and also to switch from the 6,3V heater supply on the power transformer to the 3,15V at the center tap ... And if there is no center tap on the 6,3V winding, the switch can just put a power resistor in series with all the heaters ...
Suppose your amplifier have two 6CA7 and three 12AX7, that is about 3,9A current, just like a 1,6 ohms resistor ... If you add a standard 1,5 ohm resistor in serie with those heaters, there will be about half the voltage for them. 6.3V / ( 1,5 + 1,6 ) = 2A, then about 2 x 2 x 1,5 = 6W resistor dissipation, a 20W model should be safe. I write "about" because when the heaters are colder, their resistance is lower and the voltage across them will be probably lower than 3 volts and the drop across the resistor higher ... The resistor value can be adjust consequently.
Cheers,
Alain.
It is preferable because there will be a small loss of power in the cathode resistor and it will act as a negative feedback and more input voltage at the grid will be needed to have the same output power ... But the distorsion will be lower ...
I should make a spice simulation about that to see the differences ... But one thing is sure, you have to use a high enough value for the coupling capacitor or no capacitor at all, otherwise, you will get a loss of power at low frequency.
Cheers,
Alain.
No, because when you turn on the power switch, the current is very low because the tubes heater are still colds, what can destroy a tube is mostly over dissipation and the dissipation is very low when the current is low ...
The only thing is when using a tube rectifier, the voltage at the plates can be momentary higher than the maximum the tubes can stand, over that voltage, inter-electrodes arcs can occur ... But with silicon diodes rectifiers and lower plate voltage amp, there is no problems ...
Of course the tube or diodes rectifiers have to be strong enought to stand the momentary in rush current in the filter capacitor but the power transformer winding resistance (internal impedance) help limiting this current, some current limiting resistors can be added in series between the transformer and the rectifiers if necessary ...
Alain.
It is preferable because there will be a small loss of power in the cathode resistor and it will act as a negative feedback and more input voltage at the grid will be needed to have the same output power ... But the distorsion will be lower ...
I should make a spice simulation about that to see the differences ... But one thing is sure, you have to use a high enough value for the coupling capacitor or no capacitor at all, otherwise, you will get a loss of power at low frequency.
Cheers,
Alain.
Ok, I'll try later with and without cap. I thought it would have more output with bypass cap. And about coupling cap, probably 100n, but don't want blocking distortion...
That is only an approximation easy to understand for most DIYers ... The results are about the sames ... They can use a higher value capacitor but not a smaller one ... Some Internet sites authors and most electronic books said 10 times less but I prefer 20 times less ...
The gm is not the same at all operation point and in the databook given in micro-mhos, so it is better to said 1 / gm x 1000000 ... In the case of the 6CA7, the gm is 11000 micro-mhos, then 1 / 11000 x 1000000 = 91 ohms ... I suppose in a push-pull amplifier, just one tube at the time is concern ... You said between 3 and 5 time less this value, 18,2 ohms and 30,3 ohms.
Let use the formula : 1 / ( Xc x 2 x pi x F ) = C
1 / ( 18,2 x 2 x pi x 20 ) = 0,000437 Farad or 437 uF ...
1 / ( 30,3 x 2 x pi x 20 ) = 0,000263 Farad or 263 uF ...
In my example, I calculate 338,627 uF and propose the standard value of 470 uF ...
That's is really about the same ... We are both right ...
You know, gm = /\ ib / /\ ec and the cathode resistance is directly related to the current ( ib ), the bias, then the maximum input signal ( ec ) ... That's why the simpler method is also working well ...
Cheers,
Alain.
100n for this coupling cap is really low, because the power tubes "grid leak" resistor is generally 470K ...
The same "simplified" formula as for the bypass capacitor can be use ... About 20 times less then the resistor ... 470K / 20 = 23,5K
The formula is : 1 / ( Xc x 2 x pi x F ) = C
Then 1 / ( 23500 x 2 x pi x 20 ) = 0,000000338 Farad x 1000000 = 0,338uF ...
So the best is a standard 0,33uF ( 330nF ) or more ...
Using the formula Xc = ( 2 x pi x F x C ), a 100n capacitor at 20 Hz have a reactance of :
1 / ( 2 x pi x 20 x 0,0000001 ) = 79577,5 ohms ...
With the 470K, it form approximately a voltage divider of :
470000 / ( 470000 + 79577,5 ) = 0,8552 ... That's about a 1,4 dB voltage loss ...
If you have a power of 30 watts at 1Khz, you will have only 21,7 watts at 20 Hz ...
Now, if you use a 0,33uF or 330nF :
1 / ( 2 x pi x 20 x 0,00000033) = 24114,4 ohms
470000 / ( 470000 + 24114,4 ) = 0,9512 ... That's about a 0,45 dB voltage loss ...
If you have a power of 30 watts at 1Khz, you will have only 27 watts at 20 Hz ...
It is not perfect but much better ... Some people don't hesitate using a first quality 1uF 400V paper capacitor for a coupling like that but they must be more rich then I am ...
Alain.
Only in American data. Over here we measure gm in mA/V. Putting the 10^6 explcitly in the formula should only be necessary for people who blindly plug numbers into formulas. For everyone else they should use consistent units, ohms and farads and hertz.Alain Poitras said:
Your suggestion for running standby with HT off and heaters cool is a good one, provided people do both. Doing either one on its own could reduce valve life.
100n for this coupling cap is really low, because the power tubes "grid leak" resistor is generally 470K ...
If you have a power of 30 watts at 1Khz, you will have only 27 watts at 20 Hz ...
It is not perfect but much better ... Some people don't hesitate using a first quality 1uF 400V paper capacitor for a coupling like that but they must be more rich then I am ...
Alain.
Remember I'm going to build a guitar tube amp. I have blocking distortion in some cases with 100n coupling caps, had to reduce them. BTW, I always use orange drops, MKT's or Phillips. Don't know if I canget some of those paper caps, but they must worth the money!
Forget about all that ... This method is not accurate ...100n for this coupling cap is really low, because the power tubes "grid leak" resistor is generally 470K ...
The same "simplified" formula as for the bypass capacitor can be use ... About 20 times less then the resistor ... 470K / 20 = 23,5K
The formula is : 1 / ( Xc x 2 x pi x F ) = C
Then 1 / ( 23500 x 2 x pi x 20 ) = 0,000000338 Farad x 1000000 = 0,338uF ...
So the best is a standard 0,33uF ( 330nF ) or more ...
Using the formula Xc = ( 2 x pi x F x C ), a 100n capacitor at 20 Hz have a reactance of :
1 / ( 2 x pi x 20 x 0,0000001 ) = 79577,5 ohms ...
With the 470K, it form approximately a voltage divider of :
470000 / ( 470000 + 79577,5 ) = 0,8552 ... That's about a 1,4 dB voltage loss ...
If you have a power of 30 watts at 1Khz, you will have only 21,7 watts at 20 Hz ...
Now, if you use a 0,33uF or 330nF :
1 / ( 2 x pi x 20 x 0,00000033) = 24114,4 ohms
470000 / ( 470000 + 24114,4 ) = 0,9512 ... That's about a 0,45 dB voltage loss ...
If you have a power of 30 watts at 1Khz, you will have only 27 watts at 20 Hz ...
It is not perfect but much better ... Some people don't hesitate using a first quality 1uF 400V paper capacitor for a coupling like that but they must be more rich then I am ...
Alain.
In fact, a 100nF coupling capacitor before a 470K grid resistor is not so bad !
The basic formula to calculate the voltage across the resistor is :
Vr = 1 / ( R ^ 2 + ( 1 / ( 2 x pi x F x C )) ^ 2 ) ^ 0,5 x R x Vrc
If Vr = 1V and F = 20Hz :
A 100nF give me 0,985967418V or - 0,12275dB
A 330nF give me 0,998686378V or - 0,01142dB
A 1uF give me 0,999856695V or - 0,001245 dB
If the power output is 30W at 1Khz, at 20 Khz it will be :
With a 100nF : 29,164W
With a 330nF : 29,921W
With a 1uF : 29,991W
But ... My Spice simulator give me those results :
With a 100nF : 0,96441495V - 0,3147dB 27,9W
With a 330nF : 0,97629798V - 0,20835dB 28,6W
With a 1uF : 0,97745605V - 0,19805dB 28,66W
The results are differents but there is not much difference using a 100n, a 330n or a 1 uF ... There is one thing for sure, it is much more easy to use the simulator to calculate circuits than a pocket calculator ...
Sorry for my wrong calculations ...
Alain.
Of course ... You use Siemens ... S = A / V
This is much simpler but since most peoples use american tubes datasheets from company like RCA, Sylvania or GE ... Some use micro-Siemens too, like Svetlana ... Philips and JJ specify mA/V for S ... I agree the gm in american datasheets are a bit confusing !
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