Fender Champ Power Output Rating

[bobloblaws]

Hi there, first time posting in this section. My question has to do with power ratings for a single 6V6 amp such as a Fender Champ. I worked on a similar amp recently, a Harmony 8418 made by Valco. I calculated the power dissipation of the power tube to be just over 10 watts, which seemed reasonable since 6V6's are rated for 12 watts max dissipation. What confuses me though is that the literature on the Fender Champs says they are 5 watt amps. Am I missing something here? Why would they design the Champ to output only 5 watts if they had up to 12 watts available from the power tube?

B.L.

[mac dillard]

You are confusing tube dissipation with the amp output power..Typically output power will be something less than 60% of the tube dissipation.

[bobloblaws]

Hi there, first time posting in this section. My question has to do with power ratings for a single 6V6 amp such as a Fender Champ. I worked on a similar amp recently, a Harmony 8418 made by Valco. I calculated the power dissipation of the power tube to be just over 10 watts, which seemed reasonable since 6V6's are rated for 12 watts max dissipation. What confuses me though is that the literature on the Fender Champs says they are 5 watt amps. Am I missing something here? Why would they design the Champ to output only 5 watts if they had up to 12 watts available from the power tube?

B.L.

Oh, OK. I thought it was pretty much the same thing. I know how to calculate tube dissapation, is there a way to calculate output power? Or do you have to look at the output waveform on a scope?

A Deluxe Reverb cranks out 11 W per 6V6. Is the difference because one is class A and the other class AB?

[Leo_Gnardo]

Why would they design the Champ to output only 5 watts if they had up to 12 watts available from the power tube?

5 watts - from 400 or 500 Hz on up. You'll be very disappointed at the puny power of the mighty Champ at 100 Hz. Like 1 watt or less, clean sine wave. Of course you can measure some distorted waveform and find a larger power figure but any way you look at it, the power is very limited at lower frequencies.

[Enzo]

Output power is what audio power it can put into a load. A crude method would be to measure the signal voltage across the load (speaker) and use Ohm's Law to calculate power.

[bobloblaws]

So why is the output power of a Deluxe Reverb, for example, so close to that of the tube dissipation while the Champ is roughly only half? Something to do with efficiency and/or amp class?

[Old Tele man]

Po ≈ (Zo')*(gm*Vg)^2

where:
Po = Output power, watts(avg)
gm = tube transconductance, A/V
Vg = grid-signal AC-drive voltage(*)
Zo' = effective plate load impedance, Ω

NOTE: (*) with PP circuits: (1) assume the bias voltage is the limit of the AC-drive voltage (ie: Class-AB1) and (2) Zo = Zoo/4 roughly.

[g1]

Something to do with efficiency and/or amp class?

Yes. Class AB is much more efficient than class A.

Amplifier Efficiency

http://www.electronics-tutorials.ws/...r-classes.html

[Jazz P Bass]

Also, I believe that you are missing the relationship that the output tube has with the output transformer.

[bobloblaws]

Right, that makes sense to me now, thanks all.

[bobloblaws]

Yes. Class AB is much more efficient than class A.

Amplifier Efficiency

http://www.electronics-tutorials.ws/...r-classes.html

Right, that makes sense to me now, thanks all.

[bobloblaws]

Also, I believe that you are missing the relationship that the output tube has with the output transformer.

As in "reflected load" etc.? I read up on that awhile back, I would have to refresh my memory.

[J M Fahey]

Oh, OK. I thought it was pretty much the same thing. I know how to calculate tube dissapation, is there a way to calculate output power? Or do you have to look at the output waveform on a scope?

Preferred method, simply because itīs "real World" is to drive amp to just clipping, checking that on a scope screen , and do the Math.

Now when you are *designing* an amplifier, when there{s nothing to turn on and measure yet, just some drawings and calculations on a piece of paper, the regular method is to calculate it using the tube curves, available from the datasheet. (thatīs what they are there for).

Here{s an example, not exactly of a Champ, but similar principles apply:


Details to be considered:
1) the (incredibly useful) curves , showing:
Current
plate voltage
grid voltage
dissipation
etc. , in all combinations.
personally I like "graphic" design using curves instead of "digital" design using simulators any day of the week, because although the latter can crunch numbers "apparently" very precise (as "0.5384% distortion", "102.257 mA current" ... REALLY????) but graphics show you "all data at once".

Ok, in the example you have a tube with 300V plate , 100mA plate, both at idle, and designer drew a 4000 ohm load curve.
See that plate voltage can swing from 0V (nominally, actually some 60/80V minimum "saturation voltage") to twice +V or 600V and current can swing from 0mA to 200 mA (again , nominal, actual tubes might have difficulty reaching that, specially if worn)

So here you have 300V peak swing, 100mA peak swing, so 300*0.1=30W peak Watts (yes, it{s actually that simple) so 15W RMS possible.

Now I said that actual peak voltage swing will be less, some 300V minus 80V saturation voltage so 300-80=220V peak or 73% what an "ideal" unobtanium tube would achieve.
By the same token, current swing will also be 73% of ideal value, so 73 mA peak.

So letīs recalculate using real world tube values (which are also shown in the datasheet, they are the bent *real* curves) 220V peak * 73mA peak=16W peak , so 8W RMS ... which matches very well with what we can get from a real world "Champ", using an EL34.

By the same token, you will find the 6V6 one can provide some 50W RMS.

As you see, amps *can* be designed, and very accurately, just with curves, pencil, paper and calculator

By the way, whay Leo and others dis, no doubt.
Not many copmputers available way back then

Nor calculators, the calculation tool used by Engineers was the mighty slide rule:


EDIT: now I notice I had only a very quick look at the graph, did all after it in my mind (since I do it *often*) and thought idle current was 100mA.
Itīs actually 80mA so that value should be used instead, lowering end results a little, but too lazy on a Mothersīs day Sunday afternoon to rewrite everything with corrected values, in any case the procedure is correct, Math is correct .... just does not exactly apply to the graph I pulled from the Net.

The *method* is the one I showed in any case.

EDIT2: maybe this will clear OP doubts:
*idle* power dissipation is 300V * 80mA=24W, yet we only get less than 8W RMS
terrible but true

[bobloblaws]

Wow, that is a lot to digest. I come across references to the tube curves quite often but I don't know much about it. I will definitely study this info when time permits.Thanks!

- B.L.

[cjenrick]

push pull amps are like making toast, you press the lever down and the toaster starts to heat up, the toast is done and it pops out and the toaster cools off,
well, a class A amp like the Champ is like having the toaster on all the time, maybe not as hot, but always on, the tube does not like this, it will protest after a while, so we have to limit the plate dissipation of the tube a bit so it does not fail.

this means we can not get as much juice from the Champ as we do in the Deluxe Reverb.
In the Deluxe Reverb, the tubes get to cool off every half cycle.