Tube Amps / Music Electronics
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|8/28/2005 10:35 AM|
||6L6GC/EL34 plate current rating|
How much current can a 6L6GC or a EL34 can take on its plate?
I can get a power Xformer with 340VAC - 700ma.
So if I use four tubes in push-pull, only two will be working at full peak power, it means that they have to take with almost 350ma-pk each, isn't it? Are these tubes up to the task? Or I need to have six (3 each side) tubes?
BTW, all the transformers produced this days have their current ratings as rms values like Hammond, or it is different from manufacturer to manufacturer?
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|8/28/2005 11:18 AM|
I think there may be a bit of confusion on your part with regard to the electrical characteristics of tubes and transformers. when your power transformer says it's rated for 340v/700mA, it means that at full load of 700mA, it will still be putting out 340vac. this by no means should be taken to mean that whatever you have it powering must necessarily be capable of drawing that much current. at no load, your full-wave rectified DC voltage will be somewhere over 500v, and will drop predictably with increasing load, and ending up at about 480v when loaded at its rated current of 700mA. now as far as the tubes go, 6L6's and EL34's will both run at 480v or even higher with no problems, and how much current they draw will depend entirely on how much negative bias voltage they are given on their control grids and the internal DC resistance of the output transformer. the fact that your transformer can put out 700mA, that does not mean that your power tubes must also be able to handle that much current. the tubes will only draw what you tell them to. the rest of the transformer's capacity will go towards sag reduction. you will end up with a very well regulated supply voltage. you would need about 14 EL34's to fully load down that power transformer at 480V anyway, even at full plate dissipation, 20 at 70%.
|8/28/2005 1:09 PM|
Thanks for your reply!
Althought this is not a confusion. The OT that I'm thinking to use with this supply is a 2000 Ohm A-A 100 Watt toroidal unit. It means that each half of the circuit will have a 500 ohm load. So,the total DC voltage swing will be:
((340V * sqrt(2)) - 10%sag) - Vsat =
= ((489.6 - 10%) - VSat) = 440 - Vsat
If Vsat is 60V (just an example) » 440 - 60 = 380V
So 380V / 500 ohm = 760ma
Maybe I will regulate the supply to lower this peak current.
But first I've got to know if common tubes (6l6GC's or EL34's) will be up to the task.
One other problem, if they can pull 350ma each, is the total voltage required on their grids to achieve such current on their plates. If I set their idle current for 70%, will be 50ma (6l6gc), the max signal current should be increased 700%!!!!
Help needed here!!!
|8/28/2005 12:37 PM|
Tiago, that 2000 ohm figure is AC impedance and not an exact figure anyhow.
AC figures are NOT as easy to calculate with as DC factors!
If your goal is to limit current to avoid saturation in the OT, would it not be easier to 'scope it out and limit the drive? The current would only be high enough to saturate the OT on heavy signal peaks. The average current would be much lower. Only a 'scope could show if there's a problem.
I don't understand your concerns about "signal current". The driver amplifier stages are voltage amplifiers. There virtually is NO signal current involved! Unless you are talking about a class AB2 amplifier. Even then, grid current with the 6L6's is only a few ma. This is trivial compared to the plate current. And if you think a 6L6 can and should handle 350 ma each then I'd love to own the store where you buy your 6L6's!
You're gonna buy a lot of them!
Just what is it you're trying to build or do, anyway? More explanation might make things more clear.
|8/28/2005 4:08 PM|
My goal, is to get the maxium output power of the OT and PT.
The "max signal current" that I'm refering, is the max signal PLATE current.
To calculate AC figures, we just must decide what kind of figure (RMS, Avg, pK, pK-to-pK).
If I have a 100 watt RMS amplifier, it is 144watt peak and 288watt peak-to-peak.
So if we need 144Watt peak, on a 1000 ohm primary, with a B+ of 450V (just an example and no sag) then:
(450V - Vsat) = 450V - 70V = 380V
Now 380V / 1000 Ohm = 380ma-pK
380V * 380ma = 144.4 Watt peak
Is this correct?
The conclusion is that with a load of 1K each side 4K A-A load for a 100Watt RMS amplifier, two tubes each side of the primary, each tube has to take with 190ma-peak! Now If we want to make an amp with those standard 100W OT, most of them are 2K A-A, and each tube will see 500 ohm! So if I try to make an amp with 100Watt RMS, 2000 Ohm OT, 4 tubes, guess what is the current that each tube has to handle!!
|8/28/2005 2:36 PM|
You're making this way too difficult for yourself. Consider the following (from TUT5):
100W with 2k primary:
Load for each half = 2k/4 = 500ohms
Peak power = 2xRMS = 200W
Peak drive voltage required = sqrt(200W*500ohms) = 316Vpeak
Peak drive current = 316V/500ohms = 632mApeak
Supply voltage req'd = 316V + 70Vsat = 386VDC
|8/28/2005 2:57 PM|
a 6L6 can pass 380mA for a very brief period of time at the voltage you're talking about, and it may be necessary to drive the output tubes into grid current (class AB2) and bias them very cold, which can adversely affect tone. there's no reason why your power amp *must* draw 700mA from the PT in order to put out 100W. it is not necessary to have a transformer rated for the peak DC current, as you have filter capacitors to absorb load fluctuations.
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