Cathode bias shift

[Merlinb]

So, if the grid -V relationship to the cathode remains the same (as in fixed bias) and the plate voltage drops, does that not incline the point between cutoff and saturation closer to cutoff?

In a linear valve yes, but not necessarily in a real valve. The distortion of the waveform can cause an increase in the average anode current, so even though you might have kept the bias rock steady, the whole load line can still move vertically upwards. It's called self rectification, BTW.

[loudthud]

At full bore (not max W RMS power), controls set as per stage use plus vol full up (not as per W test) corrected plate voltage drops to 404vdc @ 81mA (less screen current), cathode voltage rises to 110v, grid voltage 89vdc & each tube is now dissipating 32.6W...didn't measure screen current, but let's say it maxes out around 10mA, that's still 28W of plate current.

This is not a valid measurement of plate dissipation. Power dissipated by the load is being neglected. Such a method is only valid for measuring bias dissipation under no signal conditions.

To get the actual plate dissipation, you would have to digitize plate voltage and plate current (cathode current if you can neglect screen dissipation) and multiply the two point by point over one or several complete cycles.

Edit: Note that the plate dissipation will change depending on if a resistive or reactive load is used.

[MWJB]

OK Loudthud, then give us a practical example, with both resistive & reactive loads. Not that I've ever seen anyone gig with a resistive load, or determine whether their amp is correctly biased with one either. Looking for a real world context here, very much looking forward to finding an amp that gets cooler, the more you drive it.

Is it not then true that plate dissipation is reflected by that at the cathode?

[MWJB]

Here's some info gleaned from one of Steve Connor's posts in another thread...

"Let's take an example: One of my homebuilt amps has two KT88s idling at about 45mA each, on a B+ of 475V. 90mA at 475V is a power draw of 43W, of which 21.5W is dissipated in each tube, and none delivered to the load.

When I crank it to full power using a signal generator and dummy load, the current drawn from the B+ goes to somewhere over 300mA. That's 145W drawn from the B+ and I measure something like 80W delivered to the load, therefore the rest must be dissipated by the tubes, which gives us 32.5W per tube. (and an efficiency of about 50%, which is about as good as you can expect from a tube amp with the "latest" beam tetrode technology.)

None of these figures has much to do with the 40W plate dissipation rating of a KT88."

Still appears that dissipation rises with drive. One thing that does strike me though is that the amp's ability to produce 80W from 2xKT88 does seem to have something in common with the 40W rating of a KT88, if not with plate dissipation precisely, in this example.

I left out the load dissipation figures from my example simply because due to the way that the tone stack is set skews the result. In my test the 6G# was dissipating around 15W through the load resistor, which is not unusual for a real world test under these conditions. So 56W disipated by the plates, less 15W via the load = 41W dissipated by the tubes, 20.5W per tube, still rising over idle.

[loudthud]

OK Loudthud, then give us a practical example, with both resistive & reactive loads. Not that I've ever seen anyone gig with a resistive load, or determine whether their amp is correctly biased with one either. Looking for a real world context here, very much looking forward to finding an amp that gets cooler, the more you drive it.

As I mentioned in reply #12 above, the plate dissipation goes down when the amp is driven to heavy clipping. In Steve's example, he was only driving the amp to full power with a sine wave. That is a different situation.

Notice how Steve subtracted the load power from the power drawn from B+ to get the average plate dissipation. That is one of my main problems with the way calculations are being done here. The load power is not being taken into account.

Calcuating the load power: When the output is anything but a pure sine wave, the 0.707 times the peak equals RMS voltage rule doesn't work. A true RMS meter should be used when the output is a square wave. Furthermore, any RMS reading is only valid for a power calculation when the load is purely resistive. With a reactive load and a non-sinewave output, the calulations necessary to compute the load power dissipation become enormous. A computer program would work or some method that would sample current and voltage waveforms at the load and multiply the waveforms on a point by point basis. Then you take the average over one or several complete waves. When the power waveform swings negative, that means power is being reflected by the load back to the tube plates.

A second method to measure plate dissipation under signal conditions would be to acquire the current and voltage waveforms at the power tube, multiply them and take the average.

A large number of speaker attenuators present a resistive load to the amp when large amounts of attenuation are dialed in. The load is usually somewhat higher than the nominal impedance.

Is it not then true that plate dissipation is reflected by that at the cathode?

That's true under no-signal conditions. When a signal is applied, you must subtract the load power from the total power consumed to get the plate dissipation as Steve did. In cathode biased amps you also deduct the power of the cathode resistor. When the output waveform is a non-sine wave or the load is not resistive, calculating the load power becomes complicated.

Link to the thread you quoted: http://music-electronics-forum.com/t8442/

[es345]

In spring this year I made an excel for myself to visualize the plate power dissipation with sinus signal drive
- averaged
- and instantaneous
to help to understand what is happening.

The calculation is far away from an exact simulation. The calculation is in 360 steps for one sinus period.
For each step the instantaneous dissipated power is calculated. If at one step there is an overdrive situation then it is assumed that the plate voltage will be the same as for 100% drive. This for sure is a simplification, but one can see the mechanism what happens.

I have attached the averaged and instantaneous plate dissipation up 100% signal drive and up to 150% signal drive ( basically EL34 PP at 800V 11K load)

You can see that with 150% overdrive the plate dissipation is going down compared to 100% signal

[kg]

there's a much easier way to do this:

run the amp good and hard in a dark room.
if the plates are red, you're biased too hot.

[GregS]

Looking for a real world context here, very much looking forward to finding an amp that gets cooler, the more you drive it.

I found this to be true of a 6V6 cathode biased amp I built last year. I checked the temperature of the power tubes (with one of those temperature sensing devices) immediately after I'd been playing it cranked for a while. To my surprise the temperature of the power tubes was significantly lower than what I'd measured prior to playing it cranked - even lower than at idle IIRC. I was expecting the opposite. I repeated this test more than once and had the same observations.

At the time of these temperature tests I had the power section biased hotter than I do now. I haven't checked this with the cooler bias.

[Dave H]

I have constantly read - when referring to cathode bias of the power amp - that often the biasing "should" (or perhaps I should say "can") be set hotter at idle because when being driven the bias will shift cooler and so to avoid crossover distortion the idle setting can be exempt from the so called 70% rule and be set more like 80% or hotter.

Am I missing something here?

I think what they really mean is - "a (class A/B) cathode biased amp because of its bias shift when driven, can run cooler than a fixed biased amp would under similar drive conditions", not that the cathode biased amp runs cooler under drive than it does at idle.

Dave H.

[loudthud]

You can see that with 150% overdrive the plate dissipation is going down compared to 100% signal

Thanks es345 for your post. I was thinking about that data sheet example where a single pair of EL34s produces 100W with an 800V B+.

I can't verify the math behind your calculations but it does verify a point I made earlier. In graphs 1 and 3 the plate dissipation peaks at around 70% of full amplitude (I'm assuming voltage output). That is the half power point where you reach maximum plate dissipation.

[loudthud]

There are three mechanisms that lead to bias shift....

The third is cut-off clipping, which shifts the average anode current more positive and is equivalent to reducing bias (even in a fixed bias amp).

Merlin, I'm not sure how this works. Could you elaborate?

[imaradiostar]

I found this to be true of a 6V6 cathode biased amp I built last year. I checked the temperature of the power tubes (with one of those temperature sensing devices) immediately after I'd been playing it cranked for a while. To my surprise the temperature of the power tubes was significantly lower than what I'd measured prior to playing it cranked - even lower than at idle IIRC. I was expecting the opposite. I repeated this test more than once and had the same observations.

At the time of these temperature tests I had the power section biased hotter than I do now. I haven't checked this with the cooler bias.

Picture for a moment that you have a tube biased at the recommended 70% idle bias and there is no input to the amp. Also assume the amp is cathode biased. That means it's dissipating 70% of max wattage at all times. Now picture that same amp overdriven heavily. It's now dissipating 100% of max wattage for a half cycle then 0% for the next half cycle. That, given roughly 50% duty cycle, equates to less wattage being dissipated than the amp that is amplifying nothing- the dissipation averages out to only 50%.

This is how (in a perfect world) a pair of El34's can be used to approach 100 watts output in a "class B" amp- each tube can dissipate 50 watts for a half cycle then nothing for the next half cycle, averaging out to 25 watts. In the real world screen grids melt and tubes get overheated and gassy but it's not impossible.

If you have a push-pull cathode biased amp that's being played hard it may get cooler simply from cathode voltage rise biasing the outputs colder. It's also not uncommon to see a "peak" in plate current as signal input goes up. Past a certain point the various methods merlin mentioned cause the outputs to be turned off, allowing plate voltage to rise above normal resting levels. I've seen this with a bunch of old cheap tube stereo stuff I've played with over the years.

jamie

[Chuck H]

Excellent thread. Too bad much of it is going over my head at a glance. Perhaps with some real world measurements I could see it more clearly. But it IS a subject that is commonly ignored WRT guitar amps. I assumed that a cathode biased amp indeed does move the bias point closer to cutoff when the tube is conducting by virtue of the greater negative relationship between the grid and cathode. Merlin explains that when clipping a tube tends to equalize itself. The sheer bredth of the clipped wave form would dictate this. Perhaps with a clean sine wave a tube could be made to exactly amplify 50% of the time at 200% dissapation for a total AB1 circuit power of double the max diss for a pair. But that would be a perfect world and not conducive to clipping I'm sure. I'm still puzzled by how a cathode biased tube whose cathode voltage rises with conduction doesn't move the center bias value closer to cutoff due to the greater -V grid relationship. Isn't this what cathode bias is said to do? self bias? The circuit is always fighting for equalibium in power tube watts, regardless if the bias point must shift closer to cutoff to do it.

[es345]

I can't verify the math behind your calculations but it does verify a point I made earlier

The simplified method I am using is the following ("engineering approach"):

In PP class B each of the tube sees Ra=Raa/4 in the active half. I neglect the resistance of the copper wire in the OT as a good OT should have only a small resistance. When the tube is fully conductive you have a remaining voltage between Cathode and plate - lets call it Uc, normally something like 50 -70 V. In this simplified model the peak voltage at Ra - no overdrive, no SAG, no resistance in the OT - is therefore Ups= Ub- Uc and the corresponding peak plate current is Ips= Up/Ra. What you need to add is the idle current for calculation of the plate dissipation. Next step you modulate Ups and Ips with a sinus, so you get Us(t)=Ups*sinus (t) and Is (t)=Ips*sinus (t)+ I idle

Now you can modell in steps, I have done it in 360 steps for one sinus wave. Resulting negative values of Is(t) is set to 0 which happens during the inactive half of that tube. The instantaneous power dissipation pd(t) is the voltage between plate and cathode multiplied by the corresponding current to the plate Pd(t)=(Ub-Us(t))*Is(t)

If you are now going to overdrive the signal will be clipped. Assume now worst case a rectangular signal. Then the average plate dissipation will be Uc*Ips/2 as the tube is only active in one half. Interesting to see that the average plate current is now at maximum and also the corresponding g2 current.

What do we observe:
in overdrive situation the plate dissipation is not the critical point but the grid dissipation and the cathode current - which fits somewhat to the real world.

[Merlinb]

Merlin, I'm not sure how this works. Could you elaborate?

If the valve is biased cooler than centre then it will hit cutoff before grid current. The anode current can only swing a short way negative before cutoff, but a long way positive. The average anode current is therefore more positive than the quiescent value.
(Someone else also mentioned that the natural nonlinearity of a valve will do this even before clipping, but the effect is not as great)