Directly measuring dissipation with digital oscilloscope

[mrriggs]

I used the 2.5 Ohm load for these measurements.

Idle, Plate on the left, Output transformer on the right

Large signal, Plate on the left, Output transformer on the right

 

[trobbins]

Are you able to capture the last scope plot in post #8 with the zero levels aligned to the bottom axis, and confirm the idle dc cathode current ? As the traces for V and I aren't marked on that plot (afaik) I'd sort of anticipate the minimum power dissipation region relates to minimum instantaneous voltage and max current ?

It is insightful to note that the ellipsoid shape of the loadline (resulting from simple phase shift) shows itself as two different peak dissipation levels during a full signal cycle as seen from post #8 last plot, and hence the signal swing for 2.8W is great enough to cut through the constant dissipation curves as seen on typical datasheet 6V6 plate curves.

 

[peteb]

Because your rules are wrong and he's already proven that.

I'm honestly sorry we haven't been able to explain it to you in a way that makes it "click". We have tried.

I have my own single ended center biased class A amps.

Two blackface fender champs.


The rules that I quote are the rules based on my own amps.

That is how single ended amps are known to behave.

Constant current draw at the power tube under all playing conditions.


What they actually say is ‘full power dissipation at idle’.

It is done. It is in the can.

I can show you a video if you would like.



Your rule is more like a side note. An asterisk. An exception to the rule.


Why would anyone care if a single ended amp drew current with signal? It is interesting because that is not what they normally do. It is kind of like a trivia question for smarty pants.

 

[peteb]

Sorry OP.

If I tell the truth and somebody says that it is not. I have speak up.


I want to ask what the purpose of this thread is.


The title says measuring dissipation with an O-scope.


That has been done.


What is the next step ?

 

[elpico]

I have my own single ended center biased class A amps.

Two blackface fender champs.


The rules that I quote are the rules based on my own amps.

That is how single ended amps are known to behave.

Constant current draw at the power tube under all playing conditions.


What they actually say is ‘full power dissipation at idle’.

It is done. It is in the can.

I can show you a video if you would like.



Your rule is more like a side note. An asterisk. An exception to the rule.


Why would anyone care if a single ended amp drew current with signal? It is interesting because that is not what they normally do. It is kind of like a trivia question for smarty pants.

I suggest you make a separate thread when you want to write stuff like this. You're not "telling the truth", you're repeating the same mistake over and over. Do it in your own thread.

peteb:
The title says measuring dissipation with an O-scope.

That has been done.

Yes it has been done. Plate dissipation was 12.2W at idle but went down to 10W when signal was present. I can't imagine why you'd still deny it?

 

[peteb]

You're not "telling the truth",

You need to stop this.

It’s common beginner amp tech knowledge that class A has constant full plate dissipation whether at idle or full audio power conditions.

 

[elpico]

You need to stop this.

It’s common beginner amp tech knowledge that class A has constant full plate dissipation whether at idle or full audio power conditions.

Literally every person here except for you knows you're wrong. Just take it somewhere else, you're embarrassing yourself.

 

[mrriggs]

Are you able to capture the last scope plot in post #8 with the zero levels aligned to the bottom axis, and confirm the idle dc cathode current ? As the traces for V and I aren't marked on that plot (afaik) I'd sort of anticipate the minimum power dissipation region relates to minimum instantaneous voltage and max current ?

It is insightful to note that the ellipsoid shape of the loadline (resulting from simple phase shift) shows itself as two different peak dissipation levels during a full signal cycle as seen from post #8 last plot, and hence the signal swing for 2.8W is great enough to cut through the constant dissipation curves as seen on typical datasheet 6V6 plate curves.

The zero levels are at the bottom. Well maybe not exactly on those screen shots as I didn't figure out how to set it to exactly -4 divisions until I did the load line stuff. It is hard to tell the traces apart on a monochrome screen. You are correct that the dissipation is lowest when amps are up and volts are down.

 

[YellowBoots]

The zero levels are at the bottom. Well maybe not exactly on those screen shots as I didn't figure out how to set it to exactly -4 divisions until I did the load line stuff. It is hard to tell the traces apart on a monochrome screen. You are correct that the dissipation is lowest when amps are up and volts are down.

One would also expect a valley when voltage is high and current is low. And we do—just not as low as the other end of the swing. To me this indicates a suboptimal design from an efficiency standpoint. Seems to indicate the bias is too warm. What do you think?

 

[mrriggs]

One would also expect a valley when voltage is high and current is low. And we do—just not as low as the other end of the swing. To me this indicates a suboptimal design from an efficiency standpoint. Seems to indicate the bias is too warm. What do you think?

My first thought was that the higher amplitude of the voltage was causing that. Could also be the DC offset. Lets find out!

Here are two out of phase signals from a signal generator. The product of them is a symmetric wave of twice the frequency.

If we make one four times larger than the other then the product wave still looks pretty symmetrical.

However, applying a 13 Volt DC offset to one of the waves gives us the uneven valleys in the product wave.

 

[trobbins]

It would help if we had a basic default loadline to comment on.

Estimating the loadline based on 360V B+ and idle operating point of 300V and 40mA (12W), and a reflected impedance of 6k7 (based on post #10 and assuming 400Vpp and 60mApp ellipse). The plot in post #8 indicates dissipation at min Vak relates to 58mAx40V=2.3W, and at max Vak relates to 12mAx640V=7.7W. The 2.3W calculation doesn't align well with plot dissipation showing circa 1.2W (based on 2W/div), so perhaps some misalignment in 0V axis. On a loadline, that seems a reasonable swing symmetry, especially for loadline being more an ellipse than straightline. The main aspect is that plate dissipation only just reaches 12W for a small portion of the cycle - so a fair amount of thermal headroom.

 

[Bob Womack]

Somehow it seems like dissipation might be better measured with a breathalyzer.

Bob

 

[YellowBoots]

My first thought was that the higher amplitude of the voltage was causing that. Could also be the DC offset. Lets find out!

Here are two out of phase signals from a signal generator. The product of them is a symmetric wave of twice the frequency.

If we make one four times larger than the other then the product wave still looks pretty symmetrical.

However, applying a 13 Volt DC offset to one of the waves gives us the uneven valleys in the product wave.

I hope I'm following you...

My thoughts, spoken with authority, but also could be wrong:
First, in Class A, there is a DC bias to both voltage and current, not just one or the other. That means there is always inefficiency. M (your stand in for watts) will have some raised value just because of the DC offsets. If you DC offset both inputs, the M will return to a normal sine wave, but the M Mean will be raised. Same AC output, but more watts burned.

Second, at no time is voltage negative or is current negative. So the minimum DC offset (bias) for each is the peak AC value. In a real tube, the voltage bias must be more than the peak AC value because voltage can never practically reach 0V due to space charge. Again that means more inefficiency--and also the M curve can never be perfectly sinusoid in a real tube.

Third, this illustrates the relationship between efficiency and bias. To maintain the highest level of efficiency, we should raise the DC bias of voltage in exact proportion to the DC bias of current. If we raise the bias of one by x%, the bias of the other should be raised by x% too. If we don't maintain the proportion, the M curve on your scope gets wonky and the efficiency drops.
Eg. 100Vp and 10mAp swings. This implies a load of 10K. The theoretical most efficient bias would be 100V and 10mA. The practical bias for a beam tetrode under these conditions might be something like 110V and 10mA.

@trobbins makes a good point. We are just fooling around with a scope to prove what we can see on a load line. But it is a fun brain exercise.

 

[YellowBoots]

Estimating the loadline based on 360V B+ and idle operating point of 300V and 40mA (12W), and a reflected impedance of 6k7 (based on post #10 and assuming 400Vpp and 60mApp ellipse). The plot in post #8 indicates dissipation at min Vak relates to 58mAx40V=2.3W, and at max Vak relates to 12mAx640V=7.7W.

Post #10 Vpp seems to be more near 600Vpp.

I am confused by the current RMS readout on the right though. It says RMS, but it seems to be closer to the peak-peak value. Or maybe I'm confused which trace is voltage and which is current.

I wonder if the @mrriggs had the current probe DC coupled and the voltage probe AC coupled. That would throw things off in my mind.

 

[trobbins]

Plot in post #10 could be construed to have 400Vpp with 60mApp, or alternatively 600Vpp with 40mApp. I assumed 400Vpp, as the B+ was 360V, and Vak at idle would be down near 300V, so a swing of 200Vpk (or 400Vpp) was a better fit imho.

I'm liking the plots - experimental results of this type are few and far between. It reminds me of Loudthud's scope plots of anode voltage and cathode current on Music Electronics forum some years back (which I can't link to directly due to the hassles with that forum).

 

[sds1]

I can show you a video if you would like.

lol plz do share it and I hope it's using that janky power monitor thing you have

 

[mrriggs]

Sorry about the confusion with the zero crossing points. In post #8 only the Ch1 and Ch2 traces were zeroed at the bottom of the chart. The Watts trace was just arbitrarily moved down to frame it better in the expanded shot.

In post #10, Ch1 and Ch2 zero points were still set to -4 divisions so zero Volts is one division from the left of the chart.

The probes have been DC coupled to the scope all along.

 

[peteb]

lol plz do share it and I hope it's using that janky power monitor thing you have

No.

Not like that.

It shows that the cathode voltage stays constant over all playing conditions. Idle up to and including all out mayhem.


Constant current draw through the power tube means constant plate dissipation.


The OP had the same result. ‘Constant current over all playing conditions’.



That’s why I know this stuff. I have seen it with my own eyes.

 

[YellowBoots]

That’s why I know this stuff. I have seen it with my own eyes.

Unfortunately your brain isn’t correctly interpreting what your eyes saw. I also saw David Copperfield make the Statue of Liberty disappear when I was a kid. It would require some humility to admit that though. Do you consider yourself a humble person?

 

[peteb]

Another established result:


When playing a class AB amp, playing the amp, adding significant signal, causes the current draw at the wall to increase. That is class AB. Small signal won’t. That’s class A.


When playing a class A amp. The current draw at the wall never changes. No matter how you play your guitar or don’t play it. The current draw at the wall will never change. That is class A.