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#1
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Is plate dissipation really plate current x plate voltage?
I'm about to bias one of my guitar amps. By now I used the "crossover notch"
method (...and it worked for me), but as this is said to be inaccurate I'm willing to improve on that. I've read http://www.tonelizzard.com http://www.aikenamps.com http://aga.rru.com on this issue, but still have a question. The reason on biasing output tubes is afaik to operate the tubes in the appropriate range and not drive them beyond their limits. The important parameter is the max. plate dissipation of the tube which must not be exceeded. I've read that in a class AB amp the maximum value of the plate dissipation is "normaly" reached at an output of about 1/2 of the nominal wattage. So it is considered to be a good rule of thumb to adjust the idle current so that about 70% of the max. allowed dissipation is reached. But: Is it really correct to multiply plate current and plate voltage to determin that bias point? In my case the amp has: - dual 6V6GT tubes - 305V Plate voltage - 26mA Plate current per tube - a kathode resistor of 270 Ohm The date sheet for the 6V6GT says: max plate diss. 12W; idle current 70mA at a plate voltage of 285V in a class AB amp (Bias at -19V). Now 70mA x 285V are about 20W!?! What I read in my amp is 26mA / Tube at -15V bias (26mA x 305V = 18,3W) That confuses me... Claus -- Claus Misfeldt www.frozenfrog.de |
#2
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The data You saw are correct: they are only referred to A COUPLE of tubes as
it was common practice for PP pairs. As per the dissipation: - first, the tube only sees V difference among cathode and anode: therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the tube works actually with (305-19)=286V - second, when fitted with cathode bias the system is already self-regulating. It seems Your amp has a common cathode resistor for both tubes. In this case, the current cannot be 26 mA per tube: being the cathode bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube D/S. Plate dissipation is consequently (70/2)*286=10W, which is correctly conservative for guitar use. As per Your basic question, dissipated power can be normally calculated as V*I at the "base" point (zero signal) because the two halves of the signal tend to compensate each other thanks to plate thermal inertia (the tube heats up during the positive half-cycle and cools down during the negative one). This is true for A1 and AB1 class as normally used for audio, while for high-power tubes operating in B class or after the onset of grid current (AB2 class) things are a bit more complicated and the mean value of the integral of I*dV during the whole operating cycle should be calculated. Ciao Fabio "Claus Misfeldt" ha scritto nel messaggio ... I'm about to bias one of my guitar amps. By now I used the "crossover notch" method (...and it worked for me), but as this is said to be inaccurate I'm willing to improve on that. I've read http://www.tonelizzard.com http://www.aikenamps.com http://aga.rru.com on this issue, but still have a question. The reason on biasing output tubes is afaik to operate the tubes in the appropriate range and not drive them beyond their limits. The important parameter is the max. plate dissipation of the tube which must not be exceeded. I've read that in a class AB amp the maximum value of the plate dissipation is "normaly" reached at an output of about 1/2 of the nominal wattage. So it is considered to be a good rule of thumb to adjust the idle current so that about 70% of the max. allowed dissipation is reached. But: Is it really correct to multiply plate current and plate voltage to determin that bias point? In my case the amp has: - dual 6V6GT tubes - 305V Plate voltage - 26mA Plate current per tube - a kathode resistor of 270 Ohm The date sheet for the 6V6GT says: max plate diss. 12W; idle current 70mA at a plate voltage of 285V in a class AB amp (Bias at -19V). Now 70mA x 285V are about 20W!?! What I read in my amp is 26mA / Tube at -15V bias (26mA x 305V = 18,3W) That confuses me... Claus -- Claus Misfeldt www.frozenfrog.de |
#3
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On Dienstag, 31. August 2004 11:22 Fabio Berutti wrote:
The data You saw are correct: they are only referred to A COUPLE of tubes as it was common practice for PP pairs. As per the dissipation: - first, the tube only sees V difference among cathode and anode: therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the tube works actually with (305-19)=286V ....agreed - second, when fitted with cathode bias the system is already self-regulating. It seems Your amp has a common cathode resistor for both tubes. In this case, the current cannot be 26 mA per tube: being the cathode bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube D/S. -19V Bias is from the data sheet. I measured by two ways: 1.) 267 Ohm (measured) / 14,9V (mesured drop)= 56mA; 56/2=28mA 2.) Resistance of prim. leads of OT: 109 Ohm; Voltage drop on that, when idle 2,78V = I= 26mA Would you agree in saying, that this current seems to be a bit too low? Regards Claus -- Claus Misfeldt www.frozenfrog.de |
#4
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Claus Misfeldt wrote: On Dienstag, 31. August 2004 11:22 Fabio Berutti wrote: The data You saw are correct: they are only referred to A COUPLE of tubes as it was common practice for PP pairs. As per the dissipation: - first, the tube only sees V difference among cathode and anode: therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the tube works actually with (305-19)=286V ...agreed - second, when fitted with cathode bias the system is already self-regulating. It seems Your amp has a common cathode resistor for both tubes. In this case, the current cannot be 26 mA per tube: being the cathode bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube D/S. -19V Bias is from the data sheet. I measured by two ways: 1.) 267 Ohm (measured) / 14,9V (mesured drop)= 56mA; 56/2=28mA 2.) Resistance of prim. leads of OT: 109 Ohm; Voltage drop on that, when idle 2,78V = I= 26mA Would you agree in saying, that this current seems to be a bit too low? 28mA would be about right for Ik at Ea = 285v. Patrick Turner. Regards Claus -- Claus Misfeldt www.frozenfrog.de |
#5
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On Dienstag, 31. August 2004 15:50 Patrick Turner wrote:
Claus Misfeldt wrote: On Dienstag, 31. August 2004 11:22 Fabio Berutti wrote: The data You saw are correct: they are only referred to A COUPLE of tubes as it was common practice for PP pairs. As per the dissipation: - first, the tube only sees V difference among cathode and anode: therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the tube works actually with (305-19)=286V ...agreed - second, when fitted with cathode bias the system is already self-regulating. It seems Your amp has a common cathode resistor for both tubes. In this case, the current cannot be 26 mA per tube: being the cathode bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube D/S. -19V Bias is from the data sheet. I measured by two ways: 1.) 267 Ohm (measured) / 14,9V (mesured drop)= 56mA; 56/2=28mA 2.) Resistance of prim. leads of OT: 109 Ohm; Voltage drop on that, when idle 2,78V = I= 26mA Would you agree in saying, that this current seems to be a bit too low? 28mA would be about right for Ik at Ea = 285v. ....but by now I only get about 6W out of a pair of 6V6 while it should be about 14W. Seems there is still something wrong. Claus -- Claus Misfeldt www.frozenfrog.de |
#6
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Claus Misfeldt wrote: On Dienstag, 31. August 2004 15:50 Patrick Turner wrote: Claus Misfeldt wrote: On Dienstag, 31. August 2004 11:22 Fabio Berutti wrote: The data You saw are correct: they are only referred to A COUPLE of tubes as it was common practice for PP pairs. As per the dissipation: - first, the tube only sees V difference among cathode and anode: therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the tube works actually with (305-19)=286V ...agreed - second, when fitted with cathode bias the system is already self-regulating. It seems Your amp has a common cathode resistor for both tubes. In this case, the current cannot be 26 mA per tube: being the cathode bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube D/S. -19V Bias is from the data sheet. I measured by two ways: 1.) 267 Ohm (measured) / 14,9V (mesured drop)= 56mA; 56/2=28mA 2.) Resistance of prim. leads of OT: 109 Ohm; Voltage drop on that, when idle 2,78V = I= 26mA Would you agree in saying, that this current seems to be a bit too low? 28mA would be about right for Ik at Ea = 285v. ...but by now I only get about 6W out of a pair of 6V6 while it should be about 14W. Seems there is still something wrong. You need the powers of observation from experience. In time that comes along. Patrick Turner. Claus -- Claus Misfeldt www.frozenfrog.de |
#7
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Yes indeed, I'd say that the tubes You've got in there are a bit "cooked".
I don't like very much shared cathode R, because if one tube goes low and draws less current the overall bias decreases, the other tube is overloaded and wears out sooner. If You want to tweak a bit the amp, provide separate Rs with separate bypass capacitors as recommended by the manufacturers for many high transconductance tubes. Being a guitar amp a 47uF/63V should be OK as bypass cap. If You want new 6V6-GTs, I built a HiFi amp using Russian 6P6S (the black ones made by Reflektor). They can be found for about 2$ on the Net and I found them as good as any other, but if You need a matched couple You need to get at least some 5 or 6 of them and sort'em out... tolerances are not according to German standards, unfortunately. Anyway, it's quite an easy tube to get. I saw JJ-Tesla recently launched its "new" 6V6, must be good. Ciao Fabio "Claus Misfeldt" ha scritto nel messaggio ... On Dienstag, 31. August 2004 11:22 Fabio Berutti wrote: The data You saw are correct: they are only referred to A COUPLE of tubes as it was common practice for PP pairs. As per the dissipation: - first, the tube only sees V difference among cathode and anode: therefore, if the feed ing voltage is 305V and the cathode bias is 19V, the tube works actually with (305-19)=286V ...agreed - second, when fitted with cathode bias the system is already self-regulating. It seems Your amp has a common cathode resistor for both tubes. In this case, the current cannot be 26 mA per tube: being the cathode bias given by V=RxI it follows 19=270xI - I=70.37 mA per pair as per tube D/S. -19V Bias is from the data sheet. I measured by two ways: 1.) 267 Ohm (measured) / 14,9V (mesured drop)= 56mA; 56/2=28mA 2.) Resistance of prim. leads of OT: 109 Ohm; Voltage drop on that, when idle 2,78V = I= 26mA Would you agree in saying, that this current seems to be a bit too low? Regards Claus -- Claus Misfeldt www.frozenfrog.de |
#8
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"Fabio Berutti" said:
If You want new 6V6-GTs, I built a HiFi amp using Russian 6P6S (the black ones made by Reflektor). They can be found for about 2$ on the Net and I found them as good as any other, but if You need a matched couple You need to get at least some 5 or 6 of them and sort'em out... tolerances are not according to German standards, unfortunately. I have a box full of them, and they're good for almost *everything*. They sound good, too. I'm happy to have a tube tester, but you can also match your tubes in the amp itself. That's only a statical match, though, and you'll need to insert separate cathode resistors as Fabio explained. -- Sander deWaal "SOA of a KT88? Sufficient." |
#9
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Claus Misfeldt wrote: I'm about to bias one of my guitar amps. By now I used the "crossover notch" method (...and it worked for me), but as this is said to be inaccurate I'm willing to improve on that. I've read http://www.tonelizzard.com http://www.aikenamps.com http://aga.rru.com on this issue, but still have a question. The reason on biasing output tubes is afaik to operate the tubes in the appropriate range and not drive them beyond their limits. The important parameter is the max. plate dissipation of the tube which must not be exceeded. I've read that in a class AB amp the maximum value of the plate dissipation is "normaly" reached at an output of about 1/2 of the nominal wattage. So it is considered to be a good rule of thumb to adjust the idle current so that about 70% of the max. allowed dissipation is reached. But: Is it really correct to multiply plate current and plate voltage to determin that bias point? In my case the amp has: - dual 6V6GT tubes - 305V Plate voltage - 26mA Plate current per tube - a kathode resistor of 270 Ohm The date sheet for the 6V6GT says: max plate diss. 12W; idle current 70mA at a plate voltage of 285V in a class AB amp (Bias at -19V). Now 70mA x 285V are about 20W!?! What I read in my amp is 26mA / Tube at -15V bias (26mA x 305V = 18,3W) That confuses me... Claus The idle dissipation in a tube is the anode to cathode voltage x anode current, PLUS the screen to cathode voltage x screen current, which is usually close to cathode current x anode to cathode voltage. So for 300v a-k, the max total Ia and Ig2 idle current you can use is 12 / 300 = 40 mA For guitar amp use, 25 mA is all I would use. The only time the max allowable idle current would be where the amp is intended for real class A use, with a high value RL, and with 24 watts of idle dissipation for two tubes, all you get at 40% maximum efficiency is 9.6 watts of pure class A. That's never enough for musos, mostly only happy when there is 30 watts from two such tubes when sending a grossly overloaded signal with 50% THD to a forgiving speaker. I have a Fisher amp here with 6BQ5 outputs rated at 12 watt each, and with 350v for Ea and Eg2, I have to trim the idle current down to 34 mA max, and this is for a hi-fi amp which does not have to withstand the abuse and lower value load to get the higher power. If Ea was 250v, then Ik could be only 48 mA max for class A, and lower for AB. Patrick Turner. |
#10
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Sorry, Claus, I forgot to specify that Mr. Turner is right when he says that
it would be better to use a lower dissipation ( - a higher, more negative) bias in a guitar amp. In a push-pull pair of valves operating in AB1 class You actually get MORE power out if You reduce the idle current (increase the bias), because efficiency increases. The counterpart is an increase in cross-over distortion, because the "overlap" between "push" and "pull" phases becomes smaller and the output transformer has its troubles to tie the 2 halves together without leaving a "notch" in between. This is why in HiFi use the PP pairs work closer to class A, while Marshall amps dedicated to heavy metal headbangers use nearly class B biasing. A further advantage of a higher bias is that the tube runs cooler and last longer. Anyway, I'd plug in 2 new tubes, 'cause 26 mA @ -14V bias and 290V are a bit low By the way, are the tubes connected in UL or in pentode? And if they're pentode, how high is the G2 voltage? If it is lower than specified it will reduce dramatically the tube efficiency! Maybe there's only a higher R in the G2 supply line... Ciao again Fabio "Claus Misfeldt" ha scritto nel messaggio ... I'm about to bias one of my guitar amps. By now I used the "crossover notch" method (...and it worked for me), but as this is said to be inaccurate I'm willing to improve on that. I've read http://www.tonelizzard.com http://www.aikenamps.com http://aga.rru.com on this issue, but still have a question. The reason on biasing output tubes is afaik to operate the tubes in the appropriate range and not drive them beyond their limits. The important parameter is the max. plate dissipation of the tube which must not be exceeded. I've read that in a class AB amp the maximum value of the plate dissipation is "normaly" reached at an output of about 1/2 of the nominal wattage. So it is considered to be a good rule of thumb to adjust the idle current so that about 70% of the max. allowed dissipation is reached. But: Is it really correct to multiply plate current and plate voltage to determin that bias point? In my case the amp has: - dual 6V6GT tubes - 305V Plate voltage - 26mA Plate current per tube - a kathode resistor of 270 Ohm The date sheet for the 6V6GT says: max plate diss. 12W; idle current 70mA at a plate voltage of 285V in a class AB amp (Bias at -19V). Now 70mA x 285V are about 20W!?! What I read in my amp is 26mA / Tube at -15V bias (26mA x 305V = 18,3W) That confuses me... Claus -- Claus Misfeldt www.frozenfrog.de |
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