Hoffman Amplifiers Tube Amplifier Forum
Amp Stuff => Tube Amp Building - Tweaks - Repairs => Topic started by: panhead on July 22, 2011, 01:14:34 pm
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I’ve often seen amplifiers rated in terms of “input power”, but never paid much attention to it. For instance “…250W input power with a Class A 845 triode…”
Can someone clue me in on this terminology?
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The RMS input current (see note, below) must therefore equal the RMS output current, and we can now calculate the input power using the first equation above. The DC applied is 20V, so the input power is 20 * 1.768A, or 35.36W. The 10.36W difference is dissipated as heat in the output devices (5.18W each), and is removed using a heatsink. We do not need to take the entire supply voltage (40V) into account, since only half of it is used at any point in time.
from here
http://sound.westhost.com/efficiency.htm (http://sound.westhost.com/efficiency.htm)
Kagliostro
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If you have an amp with 4x 6L6's and 6x 12A_7's, you have 4*0.9A + 6*0.15A= 3.6A + 0.9A = 4.5A of heater current. At 6.3vac, that's 4.5A * 6.3vac = 28.35w of power used just to light up the tubes. It's also power that doesn't come out of the speaker.
So is the power to light the pilot light, power dissipated across resistors and preamp tubes, etc.
That stuff is listed on older Fender tube charts as "power consumption" and indicates what amount of current the a.c. outlet needs to be able to supply (or circuit, after considering all the other loads on the circuit).
Your situation might refer specifically only to power input to the output section, or it might not. If it does, input power corresponds to the product of the peaks of plate current and voltage. If you then consider that the audio output power delivered to the speaker is not being dissipated by the output tubes, then the heat dissipation of the tubes while in use is equal to the difference of the input power and audio output power.
For this reason, class A amps idle at or near 100% dissipation, and when they produce an output power, the input power has not increased beyond the condition at idle; therefore, the output tube runs coolest while it is making maximum output power.
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> terms of "input power"
Usually means you don't know the output power.
In radio transmitters, DC meters easily read what is going in, but MHz measurements of the output are difficult. Also the output power depends a lot on the load-match which is tunable. Also on stray losses which are very difficult to measure. So it is not unusual for tube makers to specify a maximum safe input power to get you started. Since losses rise with frequency, the max input power may be given as a curve that drops with freq.
> 250W input power with a Class A 845 triode
845 is rated 100 Watts dissipation. There's another 32.5 watts of power in the filament. So anybody putting 250W "in" an 845 is confused or going to be replacing a lot of $100 845s.
FWIW, for 70-100 Watts in you can get 15-30 Watts OUT of 845 in SE class A.
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So "input power" more or less means the power consumed by the tube as in Va x Ia + heater current?
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The short answer is it's the total power being sucked off the wall source. You can take the name plate rating and divide it by source voltage to calculate amps. It includes all loads, HT, filaments, pilot light, etc. Assume you have a 1:2 PT rated at 120VA. If your Source voltage is 120V and you're running it at 100%, then your primary side is 120V @ 1A and your secondary side is 240V at .5A . Of course you need to factor in filaments and everything else, this is just a simple example. Name plate ratings are required on HVAC and motorized equipment so Joe Electrician can calculate the breaker & wire sizes. It's spilled over into pretty much anything that can be considered an "appliance." It's useful for calculating the fuse size in an amplifier but is somewhat deceptive for Joe Geetar Player who sees 200 watts on the name plate but is surprised that it doesn't sound like a 200 watt amp.
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The more time I have read or heard about input power is, as PRR told, related to the input power at the final tubes of a transmitter
However i think you can assume it also as the whole consumption of the amp vs the output power to the speaker
Kagliostro