Savage Rohr 15 class A?

[John Clifford]

I know this question has been asked here before, but either it wasn’t definitively answered, or I didn’t understand the answer. I’m trying to understand how my Savage Rohr 15 amp operates, and specifically whether the output section operates in “class A” mode, as the maker claims. It is a push-pull configuration with a long-tail pair phase inverter feeding two EL84 cathode biased power tubes. I gather this would typically suggest class AB operation, but as I understand it the deciding factor is whether the output tubes are biased “hot” enough to pass current continuously at all signal levels (class “A”), or “cold” enough to go into cutoff when the signal goes negative. Since I don’t want to risk connecting my oscilloscope to the plates, I have no way of directly testing this. However, with no signal input I measure 10 volts across the 100ohm common cathode resistor, which I think means 50mA of current through each tube. Plate voltage measured about 312V and screen voltage is about 316V. The grids are close to 0V, so -10V relative to cathodes. As far as I can tell, this puts the operating point at or above the 12W maximum power curve, which strongly suggests class A operation to me. But I’m a novice. What do you all think?


[PRR]

Watch that cathode voltage as you go from silent to LOUD!

If it rises, it is going toward B. If not, not.

Why do you care? It might distort; is that bad?

[pdf64]

Monitoring anode voltage won’t reveal anything in that regard.
Class A is defined in terms of anode current, rather than voltage.

[Latole]

 Current go with voltage on same 100 ohms no ?

[John Clifford]

Watch that cathode voltage as you go from silent to LOUD!

If it rises, it is going toward B. If not, not.

Why do you care? It might distort; is that bad?

Thanks PRR. I don't care about class A or AB in terms of my enjoyment of the amp. I'm just trying to understand how it works and why it sounds the way it does. I'll try the test you suggest. Would you mind explaining why rising cathode voltage would indicate "going toward B?" I'm assuming you mean heading toward cutoff, but I would think current (and cathode voltage) would be dropping in that case. Again, I'm new at this and just trying to understand. I appreciate your help.

[John Clifford]

OK, I tried it. Cathode voltage rose from 10V at idle to about 14V at full throttle. I don't know how to interpret that in terms of A or AB operation.

[pdf64]

Current go with voltage on same 100 ohms no ?

It’s shared by 2 cathodes in push pull, and probably bypassed. So in regard of this determination, no.
Change it to 200ohms per cathode, bypassed, then insert a further 1ohm current sensing resistor each between those and 0V.
Monitor the waveform across a current sensing resistor at max unclipped output. If it bottoms out at 0V for more than just the tip of the lower peak, then it’s AB.

Anode dissipation at idle is unrelated to class A operation.

[John Clifford]

Thanks pdf64. Making those modifications is a little more than I want to take on right now (and I don't have the parts), but I appreciate the advice.

[pdf64]

OK, I tried it. Cathode voltage rose from 10V at idle to about 14V at full throttle. I don't know how to interpret that in terms of A or AB operation.

If the signal level pushes the output stage much beyond clipping, then class of operation becomes somewhat invalid.

[PRR]

Like any decent audio amp which is self-biased by current in a resistor: it can't get far past class A without going strange.

Ideally the current would not rise. Real tubes have curvature and a 1.2X rise may be "clean", a 1.3X rise may be "clean enough". At 1.4X rise you are probably clipping, flat-topping. A fully extended class B amp the current may rise 3X or more, but a 3X rise in cathode resistor voltage would starve it at full power, so we use fixed bias to go there.

In perfect A, the idle dissipation is twice the maximum sine power output. Because of curvature, a little more with real tubes. So two tubes at 12W heat each, pushed hard, may be 15 Watts audio output and still be sine-looking.

[John Clifford]

If the signal level pushes the output stage much beyond clipping, then class of operation becomes somewhat invalid.

That's what I'm thinking. At full crank, the output looks like a square wave, so both tubes would be in cutoff regardless. I guess the label you put on it (A or AB) doesn't really matter. I suppose the question is how hard the tubes are working at maximum clean (no clipping) volume.

[John Clifford]

Thanks PRR, that makes sense and seems consistent with what I'm seeing.

[pdf64]

…. I suppose the question is how hard the tubes are working at maximum clean (no clipping) volume.

Dunno, bear in mind at fair chunk of power is being delivered to the load, rather than being dissipated by the valves.
In class A, anode dissipation is theoretically at its lowest at max power output, highest at idle.

[NotSure]

John,


Per Gerald Weber, Kendrick Amplifiers, Class A is defined by the following:


Ø  The tube always has some current flow and never goes into cutoff.
Ø  The input signal driving the tube is intentionally kept low enough to prevent the tube from going into either cutoff or saturation.
Ø  The tube is biased such that the no signal idle current is set right in the middle of cutoff and saturation.
 All preamp tubes are operated in Class A. This is because all preamp tubes are single-ended. The reason a single-ended amplifier must be Class A is because if the tube ever goes into cutoff, then there will be a piece of the waveform missing (the bottom half of the wave will be cut off when the tube goes into cutoff) and in audio, we don't want to clip off part of the waveform.


In a Class AB amplifier, the tubes go into cutoff but for less than 180 degrees of the input cycle. All Class AB guitar amplifiers are operated in what is called a push-pull configuration. In a push-pull configuration, there are at least two output tubes driven by a phase inverter.


Hope this helps,


NotSure

[John Clifford]

Thanks NotSure. Are you a fan of the movie “Idiocracy?”

[pdf64]

…
Per Gerald Weber, Kendrick Amplifiers, Class A is defined by the following:


Ø  The tube always has some current flow and never goes into cutoff.
Ø  The input signal driving the tube is intentionally kept low enough to prevent the tube from going into either cutoff or saturation.
Ø  The tube is biased such that the no signal idle current is set right in the middle of cutoff and saturation.
 All preamp tubes are operated in Class A. …

The latter 2 points are ‘nice to haves for best linearity’, a SE class A stage need not be centre biased or have its input signal amplitude limited.
eg preamps with SE stages get overdriven, may be designed to accommodate overdrive, yet still, by any reasonable analysis, operate in class A.
Their stages don’t change their class of operation, eg to class C, by virtue of an increase in signal level.
The class A, AB and B definitions can’t really be assessed / applied beyond the linear range.

[PRR]

Yes, these "classes" define "distortionless" operation.

The concept is fuzzy until you study SE class B and class C (which audiopeople never do). At class B half the wave is cut off. At C 80% may be missing.

And yes, in speech/music work the Input Level is subject to operator control and discretion, can't be designed-in.

[John Clifford]

I’m revisiting this thread because I finally got around to making a high-voltage AC probe that allows me to directly monitor the signal at the plates of each of the EL84 power tubes in my Rohr 15. Photos and schematic of my probe are attached, in case anyone is interested.

So here’s what I did:  Signal generator sine wave at 1khz and 100mV into amp input. With speaker output across dummy load on the scope, I adjusted amp volume until the signal started to clip, then backed it off just enough to restore a clean sine wave, so maximum clean volume. Then I connected my probe to the plate (pin 7) of one of the EL84s and put that on the scope - same full sine wave as the speaker output. Same with the other EL84.

Based on this, I am satisfied that insofar as these classifications are meaningful, the Savage Rohr 15 does indeed operate in class A. If anyone finds my method faulty, please let me know. I’m just learning this stuff.

By the way, the probe works great for listening to an actual guitar signal at any point in the circuit. You just plug a guitar cord into the probe and another amp.

[pdf64]

…I connected my probe to the plate (pin 7) of one of the EL84s and put that on the scope - same full sine wave as the speaker output. Same with the other EL84.
…

I’m afraid those tests are no use in identifying class of operation. A, AB or B would all give the same result.
It’s the anode current (under those same conditions) which is the determining metric.

The method I outlined in post #7 would enable that to be done, without the need for an expensive current probe.

If you could explain what you thought an AB waveform would look like, we can try to set you right 

Whatever, well done for pursuing your investigations 

[John Clifford]

I’m afraid those tests are no use in identifying class of operation. A, AB or B would all give the same result.
It’s the anode current (under those same conditions) which is the determining metric.

Ahh, OK. Thanks for setting me straight. I'm sure you're right and I'm not arguing with you, but I don't quite understand why I would still see a full waveform at the plate if the tube is going into cutoff during part of the cycle. Would you mind explaining that? Thanks!

[John Clifford]

Nevermind, I think I've answered my own question. The two plates are connected across the same primary transformer winding, so the voltage swing would be there regardless of which tube(s) are doing the pushing and pulling, right?

[PRR]

> The two plates are connected

 

[John Clifford]

I appreciate you guys being patient with me while I stumble through this stuff. It’s a steep climb for someone with no math or science background!

[pdf64]

No problem, I’m sure we’re all pleased to help, or at least try to 
The key point here is that due to transformer action, the anode of the valve that’s in cut off will be a mirror image voltage of the anode of the valve that’s being pushed towards Vg1-k=0.
ie as one anode / primary leg gets pulled down below the HT voltage, the other will swing up above the HT voltage.