bass amp tube question

a fine line between stupid and clever

Postby dhuebert » Thu Oct 04, 2007 7:25 am

Next there is a comment that adding more tubes will not help unless the voltage rail is increased. Thats not right - 3 pairs of tubes will deliver 1/2 again as much current into the transformer as 2 pairs - giving you 1/2 again as much power with the SAME voltage.


Here is where my understanding fails. For a given load resistance, in order to deliver more current more voltage must appear. If we parallel tubes we deliver more current to the windings but the voltage on the output winding is set by the input voltage and the winding ratio so we can't deliver more voltage to the load. The only way I can figure is the input current is increased transforming to increased current to the load which causes a greater voltage to be dropped across the load which is reflected back to the input winding causing plate voltages to skyrocket far past B+, but of course this is impossible since the power supply would prevent this by sinking the excess. This is an infinite loop in my head that I would love to see broken...

Don

BTW Thanks Ian for taking the time.
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Postby dhuebert » Thu Oct 04, 2007 8:35 am

I would however put in separate bias controls for each output tube. I will never build another HiFI or Guitar Amp without separate bias controls.


Like I said, Gar was dead set against any bias pots at all. He was repeatedly servicing amps that had pots that overenthusiastic owners had tweaked. I got chewed out for the one pot I was using. It must be remembered, however, that the tubes he bought were actual new Mullards, RCAs etc that he bought in the thousands when tube production was in the 10s of millions. Different days. From my experience with the Sovtek KT88s, individual bias pots might be inevitable. I might try LM317s in the cathodes of my next BFA.

Don
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Postby Gingertube » Thu Oct 04, 2007 6:06 pm

Don,
Think of the output stage in these terms.

The B+ supply goes to the cetre tap of the output tranny. On each of the push pull sides there is a tube which acts like a variable resistance, variable with the audio signal that is. The amount of current drawn depends on that resistance. Add another tube and that same current is drawn through that tube too (since it has the same resistance). Add a 3rd tube and you have 3 times the current ect. - all without changing the B+ voltage.

Bias controls are a must for accommodationg different tubes and allowing the use on unmatched pairs quads etc.. I follow the arguement that if you provide a control someone will fiddle with it and stuff it up - may have some merit in a commercial mass production design BUT we are DIY'ers so lets do it right.

Cheers,
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Postby gerryc » Fri Oct 05, 2007 10:17 am

Gingertube wrote:Hi guys,
There are a couple of comments / opinions in the thread above which are just plain wrong and I can throw a bit of light elsewhere.
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.
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Next there is a comment that adding more tubes will not help unless the voltage rail is increased. Thats not right - 3 pairs of tubes will deliver 1/2 again as much current into the transformer as 2 pairs - giving you 1/2 again as much power with the SAME voltage.
Cheers,
Ian


I think you didn't totally understand my point or I failed to be clear enough. What I stated was:

The 1900 ohm of the 1650T or W is a bit limiting to give you more power without raising the B+ voltage. If you're at the limit at how much power a quad of tubes will give you at a given P-P impedance and B+, adding another pair won't buy you anything. With the same B+ voltage you could parallel 10 pairs of tubes and you won't get any more power. Either B+ must be raised or OT impedance must be decreased. Then you can add more tubes.

The amount of output power is determined by the amount of voltage swing into the output transformer primary (let's assume a resistive load on the secondary for simplicity), or P = E^2 / R, where E is the RMS value into the load and R is the primary impedance. The peak voltage swing is limited to the B+ voltage (under load of course) minus the saturation voltage of the tube. If the tube can not sink enough current to impress the maximum theoretical peak voltage swing into the output transformer (meaning that it can not sink enough current to reach saturation), then adding another in parallel will indeed increase the output power. But once you reach the point where the maximum voltage swing is reached, you're at the end of the road with adding more tubes alone to increase output power. You need more voltage swing into the same impedance or you need to reduce the impedance at the same B+ voltage (in which case you need to add more tubes in parallel to sink enough current into the lower impedance).

It's like connecting 8 D-cells in series to get 12 volts and connecting that across a 120 ohm resistor. You'll get 100 ma of current and 1.2 watts of power dissipated. If you instead use a 12 volt car battery, capable of hundreds of amps and connect it to the same 120 ohm resistor, you'll still only get 100 ma of current and 1.2 watts of power. You need more voltage or need to reduce the resistance to get more power.
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Postby dhuebert » Fri Oct 05, 2007 5:47 pm

It's like connecting 8 D-cells in series to get 12 volts and connecting that across a 120 ohm resistor. You'll get 100 ma of current and 1.2 watts of power dissipated. If you instead use a 12 volt car battery, capable of hundreds of amps and connect it to the same 120 ohm resistor, you'll still only get 100 ma of current and 1.2 watts of power. You need more voltage or need to reduce the resistance to get more power.


My objection exactly. I have yet to hear a good explanation why adding more tubes automatically adds more power. The only thing I can think of is if you lower primary impedance the internal resistance of the voltage source (Rs) causes the voltage generated by the source to sag. (Remember your current and voltage equivalent circuits from school) Adding voltage sources in parallel reduces the internal resistance by the parallel law, allowing the same voltage to be dropped across a lower impedance increasing power.

For the above example if you have a cell that makes 12 volts but has an internal resistance of 100 ohms you will only get 6 volts out of it when connected to a 100 ohm load. But if you connect two of these cells in parallel the internal resistance is halved to 50 ohms and we now get 8 volts across the load and so on for more paralleled voltage sources. The more sources we have the more power we can generate.


Don
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Postby gerryc » Fri Oct 05, 2007 6:03 pm


For the above example if you have a cell that makes 12 volts but has an internal resistance of 100 ohms you will only get 6 volts out of it when connected to a 100 ohm load. But if you connect two of these cells in parallel the internal resistance is halved to 50 ohms and we now get 8 volts across the load and so on for more paralleled voltage sources. The more sources we have the more power we can generate.

Don


Yes, you're on the right track. If you keep going, if you connect four in parallel, now you have 25 ohms internal resistance. You can now drive more current into the resistor and thus the power will indeed increase. However, in the limit as N gets larger (actually as it approaches infinity), you asymptotically approach 0 ohms, and the most amount of power you get is 12*12/100 = 1.44 watts. Time to lower the resistance or increase the voltage for more power.
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Postby 4-CHANNEL » Thu Feb 28, 2008 8:39 pm

dhuebert wrote:My objection exactly. I have yet to hear a good explanation why adding more tubes automatically adds more power. The only thing I can think of is if you lower primary impedance the internal resistance of the voltage source (Rs) causes the voltage generated by the source to sag. (Remember your current and voltage equivalent circuits from school) Adding voltage sources in parallel reduces the internal resistance by the parallel law, allowing the same voltage to be dropped across a lower impedance increasing power.

For the above example if you have a cell that makes 12 volts but has an internal resistance of 100 ohms you will only get 6 volts out of it when connected to a 100 ohm load. But if you connect two of these cells in parallel the internal resistance is halved to 50 ohms and we now get 8 volts across the load and so on for more paralleled voltage sources. The more sources we have the more power we can generate.
Don


That's wrong!! If you parallel 2 12V batteries together, the potential voltage is still 12v, nothing to do with internal resistance. Ohm's Law doesn't deal with internal resistance, only deal with loaded resistance.

By parallel tubes in output transformer, it doubles the flow current, of course, more power will get. P = I X V , with voltage unchanged, current flows double, what power would you get??? Depends on the power supply can provide enough current, this is the only factor to affects the output power.

The internal resistance inside a battery will vary when the battery being use, with the percentage of power been used, the internal resistance will increase. Yellow_Light_Colorz_PDT_16 Yellow_Light_Colorz_PDT_16
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Postby dhuebert » Fri Feb 29, 2008 8:20 am

Draw the circuit and do the math. A 12 volt battery with a 100 ohm internal resistance and a 100 ohm load. Do the math for a single cell, then two in parallel then three and so on and see what you get for load voltage and current, ie power.

The point here is: the only way to get more power from a given voltage source and a given load is to reduce the internal resistance of that voltage source, which for tube amps means paralleling tubes.

Don
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yabbut......

Postby EWBrown » Fri Feb 29, 2008 11:44 am

Maybe it's mixing pears with apples, but I'm running a little SE CAD sim, using 6BQ5s in pseudo-triode mode, and a 5K OPT (125ESE). B+ = 365V, I = 35mA, RK = 400 ohms per tube (pretty much standard values).

Paralleling tubes will reduce plate resistance (Rp) the by half and will double the transconductance (gm). The overall gain (mu) remains the same, as it is a product of Rp and gm. This holds true for signal or power tubes. Since Rp is halved, feeding two paralleled tubes into an OPT with half the impedance will (nearly) double the output power. The slight disparity is due to the effects of the OPT's DC resistance.

The 125ESE has a tapped secondary which allows for 2.5, 5 and 10K impedance being reflected back into the primary. I chose this trannie 'cuz it can handle 80 mA, and 70 mA worth of plate current shouldn't smoke it Yellow_Light_Colorz_PDT_06 .

single 6BQ5 into 10K: 2.05 W @ 2.5% 2nd harmonic, 0.1 % 3rd harmonic

2 PSE 6BQ5 into 10K: 2.44W @ 1.65% 2nd harmonic, 0.1 % 3rd harmonic

single 6BQ5 into 5K: 2.91 W @ 5.5% 2nd harmonic, 0.7 % 3rd harmonic

2 PSE 6BQ5, into 5K: 4.05W @ 2.6% 2nd harmonic, 0 % 3rd harmonic

2 PSE 6BQ5 into 2.5K: 5.69 W @ 5.5% 2nd harmonic, 0.7% 3rd harmonic


I didn't bother with a single 6BQ5 into 2.5K 'cuz it would sound like a yowling alleycat with strep throat Yellow_Light_Colorz_PDT_07

What this illustrates that twp parallel tubes into a given OPT impedance will offer some slight power gain, but in order to double the output, (or nearly doubled) the power with 2 paralleled tubes , the OPT primary impedance has to also be halved.

2 paralleled tubes into a given impedance gives only a slight power increase, but also offers a significant THD decrease.

I don't yet have push-pull CAD, so I had to rely on SE CAD

FWIW, according to SE CAD, a triode connected 6BQ5 and half of a 6BL7 have nearly identical operating characteristics.

/ed B in NH
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Postby dhuebert » Fri Feb 29, 2008 1:58 pm

What I see is that two tubes into a given impedance doubles the power for the same THD as a single tube into twice the impedance. I think this bears up what I have been saying very nicely.

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