Thanks for that tip. I am intending to mount the main PSU transformer, T3 concentricly on top of the chassis. If I can source a long stainless bolt, I will use it.
How many watt does this put out into 8 or 32 ohms?
if > 2 watts how do the plates not melt down at > 50w dissapation.
calc 2w into 8 ohm = .5A .5A * 150v = 75w = 38w per tube plus bias wattage dissapation
this is for just 2w RMS !
if > 2 watts how do the plates not melt down at > 50w dissapation.
calc 2w into 8 ohm = .5A .5A * 150v = 75w = 38w per tube plus bias wattage dissapation
this is for just 2w RMS !
Last edited:
2W is just the initial operating point where tube efficiency is lowest (2W/38W=5%). As input is increased the efficiency increased to about 40% (20/50), to output 20V over 8 ohms required 50V drive on each grid. So only 60% power is dissipated in the tubes itself . 20V-p is about 25W rms output, so 62.5W rms is disspated as heat in the tubes, half of this is 31W + 38W =69W watts, which is slightly more than allowed (60W). So when the initial bias is reduced slightly, all is well. There are parallel tubes to increase the tube efficiency, so dissipation is well handled within limit while giving much greater output than one pair. In my amp I use 10x6c19p, the efficiency is approaching 55%. The efficiency is also increased when the load is increased, in this 6c33c pair 50W output over 32 ohms is possible.
Last edited:
Edit: I recalculated as 62.5W is total power, 62.5-25=37.5W dissp. in tube, 37.5/2=18.75+38~=57W (tot. dissp.) not 69W.
I think your estimate of the power dissipation in the tube is a little optimistic. In the attached pdf, I've calculated the average power dissipated in each output tube in the simplified idealisation where exact class B operation is assumed. The dissipation will be somewhat higher in class AB.
Plugging in the numbers for an 8 ohm speaker impedance, a B+ voltage of 150V, and an output power of 25W, it comes out at about 107W average dissipation for each of the upper and the lower output tubes. Of course, 107W considerably exceeds the 60W nominal maximum for the 6C33C. But tube ratings can typically be considerably exceed for brief periods. Also, and crucially, one does not in practice listen to continuous sinewaves at 25W; typical music will have high-power crescendos, but most of the time the power level will be very much lower. So averaged over a reasonable period of time, the power dissipated in the tubes will be a lot less than the power calculated above.
In my OTL I have verified I can just reach 25W into an 8 ohm dummy load. But I am always careful not to keep the power level at 25W for more than a very few seconds while making the measurements.
As mentioned above, the power dissipation in the tubes will be somewhat higher than the figure I calculated above once one allows for the idle current of class AB operation. But still, because the high power levels are demanded only intermittently, one seems to get away with it in practice. I've been using my Tim Mellow OTL on an almost daily basis for several years now, with no problems.
Chris
Last edited:
In that case, as I intend to use my TM otl on 32 ohms load, shall I expect higher power disipated by the tubes ? Shall I take anny precautions ?
At 32 ohms shall be at least 40 Watts (in theory 50...)
Last edited:
For 25W into 32 ohms, the formula I obtained gives about 47W average dissipation in each tube (in ideal class B), which is a lot less than 107W for the 8 ohm case. (The peak current I_0 for 25W into 32 ohms is now only 1.25A.) So indeed 32 ohms is a much easier load than 8 ohms for an OTL. But unless your taste in music demands very high uninterrupted output power levels, it probably doesn't matter too much even at 8 ohms.
Chris
I agree Chris. I'll be driving my mid horns with my amp and they are 115db efficient so I'll be needing 1 wpc or less for loud listening.
For 25W it is right, but I suppose (allready checked on dummy load) at 32 ohms we gett 40 V rms so 40x40/32=50W and 50:40=1,25A
In anny case I don't need so much level continously but some peaks may be at 40-50W ( my ESL63 are low spl ) .
Hi Cnpope,
What is the efficiency of 8 ohms load based on your calculation? If that is 25/214 < 12%, why so low? Beside the internal resistance can be reduced by NFB, if 50% efficiency, the internal resistance should be near 8 ohms also, hence dissipation is same as output, which is also 25W. Just a thought.
I read this somewhere:
"Class B has a maximum theoretical efficiency of π/4 (≈ 78.5%). This is because the amplifying element is switched off altogether half of the time, and so cannot dissipate power."
What is the efficiency of 8 ohms load based on your calculation? If that is 25/214 < 12%, why so low? Beside the internal resistance can be reduced by NFB, if 50% efficiency, the internal resistance should be near 8 ohms also, hence dissipation is same as output, which is also 25W. Just a thought.
I read this somewhere:
"Class B has a maximum theoretical efficiency of π/4 (≈ 78.5%). This is because the amplifying element is switched off altogether half of the time, and so cannot dissipate power."
Yes that is correct but only in theory, with bipolar transistors you can have almost no voltage over the amplifying element and can therefore reach almost theoretical efficiency. For the OTL case the amplifying element will dissipate heat as the tube need relative high voltage between cathode and anode to carry the required current.
An easy way is to calculate efficiency based on the current drawn by the amplifier. For a class B amplifier the average current drawn from the power supply is Ipeak/Pi, for 25W in 8 ohm Ipeak is 2.5A so the average current from the power supply is 2.5/3.14 ~ 0.8A.
With 160V anode voltage the power input to each tube is 160*0.8 = 128 W or for 2 tubes 256W, subtract output power 25W from this and you get 231W so about 115W per tube. As Cnpope explains this is way over the 60W allowed dissipation but it is not a problem when playing music as the average level is much higher than the peaks.
We have run our OTL amplifiers that use 160V on 6C33C for 24 hours at 25W output power and the tubes survive this with no ill effects but the anode is on the brink on glowing dark red.
Going back to efficiency, if you get 25W with 231W input efficiency is about 11%
Sorry to break the flow on this power dissipation topic but I need some advice. Due to a mistake in my layout, I might need to put the meters in the Kathodes of V5 instead of the Anodes of V4. Does anybody see a problem with that?
The idealised maximal efficiency in class B is when the output voltage to the load swings over the entire range from +V to -V, where V is the B+ voltage (and the B- voltage also; 150 V each, in our case). Of course, that is completely unrealistic, especially with vacuum tubes, since there will always be a substantial voltage dropped across the tube even when it is conducting maximally. And in any case, to be in a reasonably linear portion of the the tube's transfer characteristic, the output swing will have to be a lot less than the B+ to B- voltage.
Nevertheless, the more general formula I derived could be applied in the idealised maximally-efficient case, which would mean that the peak current I_0 would be V/R, where V is the supply voltage and R is the load impedance. The average power dissipated in the two output tubes together would then, from my formula, be
P_{diss} = 2V^2/R (1/Pi - 1/4)
The rms power into the load would be P_{load}= V^2/(2R).
The efficiency would then be
P_{load}/(P_{diss} + P_{load}) = Pi/4,
which is the classic result of maximal 78.54% efficiency in class B.
So for the tube OTL amplifier, the very much lower efficiency is because the B+ and B- supply voltages have to be much larger than the voltage swing that can be achieved into the load. In the 8 ohm example we have 20V peak going into the load, but we have B+ and B- being 150V. The OTL won't work with only 20V B+ and B- supplies!
Chris
Last edited:
In fact, for the more general formula I derived, where the output voltage swing is Vout= I_0 R, then it follows that the efficiency in class B will be
eff = (Pi/4) * (Vout/V0),
where V0 is the supply rail voltage and Vout is the peak output voltage.
Chris
Do you not deduct Vout from V0, that's the actual voltage between the plate and cathode, not the full V0. If so, eff = 0.78*(20/130)=~12%. not 0.78*(20/150)~10%.
Feedback won't affect the internal resistance at all. The proof of this is that the output power at lower impedances will be unaffected. If the internal resistance was really in fact reduced, this would not be the case.
Not only will the amp make more power into a higher impedance, the distortion will be lower and the heat produced by the power tubes lower as well. In addition, you might see that it draws less power from the wall.
No.
I just plugged I0=Vout/R into the formula I derived, and calculated the efficiency from that. And in my formula, the necessary allowance for the tube dissipation being current times (anode to cathode voltage) had already been taken into account. So I think what I wrote is correct.
If you modified the efficiency formula to what you are saying, namely (Pi/4) * (Vout/(V0-Vout)), then it would go to infinity in the "ideal" case when Vout=V0, which clearly must be wrong.
Chris
There will be a limited for Vout, ex, just go into clipping after run out of supply voltage where Vout will simply clipped, don't think it goes to infinity. Nevertheless your calculation is off by more than 2%. Let's stop here, thank you.
I'm not sure quite what you are saying. I understood that you were suggesting that my formula (Pi/4) * (Vout/V0) for the efficiency should be modified to (Pi/4) * (Vott/(V0-Vout)). I really don't think that could be right. I illustrated the problem with your proposal by going to the limiting case under which the maximal efficiency of 78.5% is derived, namely when Vout = V0, in which case my formula correctly gives Pi/4, whereas yours would give infinity. But you don't need to go to such an extreme limit to test it, if you don't want. Your formula would say that the efficiency would exceed the standard (Pi/4) for class B whenever the voltage swing Vout was bigger than half the supply voltage V0. But clearly, the maximal efficiency (Pi/4) should in fact only be reached in the extreme case when Vout=V0. The class B efficiency must always be less than the 78.5% figure if the output voltage swing is less than the supply rail voltage.
I am pretty sure that my formula is exactly correct (assuming ideal class B operation always, of course).
Chris
I am pretty sure that my formula is exactly correct (assuming ideal class B operation always, of course).
Chris
Last edited: