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Old 31st July 2013, 10:47 PM   #1
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Default Fixing an amp head - bad tubes?

I'm fixing an Orange AD140TC amp head for a friend. This is supposedly a 140 watt amp, and his complaint is that it's got low output. My initial assessment is that compared to my 40 watt Hot Rod Deluxe, it falls a bit short even with the volume turned way up.

All the voltages seem normal, and all the tubes have good emission on my tube tester. The only two things that stuck out are that it came with 4 EL34, two of one brand and two of another brand. Not a good sign, but of those, 1 of them is showing twice as much cathode current as the the other 3. I moved the tubes around and confirmed it stayed with that tube. Then it dawned on me, the 3 good tubes are only showing a cathode current of 7mA, and the suspect one is about 15mA. This seems awfully low. The grid bias is -43 volts, and the plate is 545 volts. All the grid, cathode, and screen resistors check out fine. Unfortunately, the schematic doesn't show what I should be reading on the cathode resistor test points.

I already recommended replacing the 4 with 4 new ones of the same type, so I was going to order them. So does it sound like these 4 tubes are shot?
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Old 31st July 2013, 10:53 PM   #2
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Probably not. The bias should be between -32 and -38volts. See what happens. One of each make should be opposing the quad. Set the current for around 40mA per valve and you will have a sweet sound.
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Old 31st July 2013, 10:54 PM   #3
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A matched quad would help when set up.
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Old 31st July 2013, 10:58 PM   #4
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Thanks, that's where the schematic isn't all that helpful. It just has you measure the grid bias voltage and set it for -43 volts. There are test points on each cathode resistor (10 ohm), but it doesn't say what they should be.
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Old 1st August 2013, 12:37 AM   #5
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I'm not sure why your tester showed "good emission" but the EL34s show such different currents? Looks like the tester is too simple to give a reliable result.
Does your amp has separate bias trimmer for each tube? If yes, then try to get all tubes to same/similar current, with your B+ then I think 25mA/tube should be a good starting point.
If no, then at least take out the pair that has the "suspected" EL34, keep only the good pair on the amp, and measure power to get your idea.
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Old 1st August 2013, 02:58 AM   #6
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The right way to set cathode current in my opinion, is to design backwards from the tubes max power spec, which is 25 watts for each EL34. With a plate voltage of 545VDC, ohms law says P=EI, or P/E = I. So 25W / 545V would give you the highest cathode current that would be considered reliable over time. It equals 46mA. You could argue that this would be the proper biasing for class A operation (it is). BUT, since it's almost certainly a class AB pushpull circuit, I'd back that off to more like 35mA per tube and look at the output waveform on a scope to be sure there's no spurious oscillation or excessive distortion present (all the way up to and including clipping). Quiescent current (P diss) isn't the same as average current (P diss). If the amp is going to be used for anything resembling heavy metal (sustained high output), the extra safety margin is needed for reliability. The average power dissipation should never exceed 25 watts per tube.

I'd look for a way to bias each tube independently so you can get it right. Cheap-*** designs will use one resistor for biasing multiple tubes at the expense of reliability and sound quality. Different brands of tubes isn't necessarily an issue, but often is to some extent. That's another reason to have each tube independently biased to the right number. Speaker efficiencies vary significantly too. That may be a contributor to the apparent low power. If the speaker is 3dB less efficient, it's like the power amp driving it has half the power. And then there's the impedance matching issue between the output tranny and the speaker. The voltage swing on the 8 ohm output tap will be about twice as big as at the 4 ohm tap. Voltage is a squared term in the power formula, so every volt lost is a significant power loss. (P = E squared / R). Hope this helps.
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