| Kaidanovsky |
Hello all,
I've been given a large box of various valves by my grandfather and want to make use of some of them in a headphone amplifier. The snag is that although I know what many of them do in a technical sense (I've looked them up on tdsl and the National Valve museum sites) I don't know my splitters from my pentodes so would be extremely grateful if anyone could give some advice on which of the following valves could be of use. I believe lots of them are government surplus from the war and from RAF receivers and the like.
What I'd like to do is make a very small amplifier, rather like an "earmax", so preferably without output transformers but mains powered (UK 240V). If anyone knows of a schematic (I've checked headwize and liked the Morgan Jones version of the earmax, but thought I'd better see if I could use any of the glass I've got) please throw it my way!
Here goes:
1267
5B255M (x 2)
6H6
6J7GT
6K7
6K8
6SH7
7193
CV51
EABC80
ECC85
EF85
EM81
EZ80 (x 2)
EZ89 (x 2)
MHL4 (X 2)
PCC84
PCF80 (x 2)
PCF802
PCL84 (x 2)
PCL86
PL508
PL509
PL802
PP3/250
PY500A
SC966A
UX281 (x 2)
VP23 (x 2)
VR55 (x 6)
VR57
VR65 (x 4)
VR100
VR101 (x2)
I know a couple of those are tuning indicators, which brings me on to my other question; whether anyone has any suggestions on how to turn one of them into a signal level indicator, as I've seen on another website (URL's slipped my mind) but with a different valve.
Many thanks!;) |
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| whitelabrat |
I'm about halfway done on a 6N1P variation of the Morgan Jones headphone amp.
It's a tiny push/pull amplifier using twin triodes. You would need three twin triode tubes if you want to keep things pretty. You do have a couple of pairs of triode/pentodes:
PCF80 (x 2)
PCL84 (x 2)
You may be able to build these into a a Single Ended output headphone amp, by using the triode as a gain stage and then the Pentode as a SE output stage. I don't have any schematics in mind though. I think I've got a couple of those tubes myself. |
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| Wavebourn |
| 2 x PCL84 are fine for the task. |
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| Kaidanovsky |
What quick replies! Thanks chaps.
whitelabrat: I did think that as I've got quite a few 'ones' of the valves I'd find a workable circuit and if necessary buy the extra few valves. So those PCFs and PCLs are both triodes/pentodes? I assume that means I wire them with or without a common cathode? I'm afraid I am not confident in my ability to design an amp around the valves...would it be a very difficult exercise? Are there resources around or should I stick to either gentle modification of an existing design or something tried and tested? Bear in mind I've only built a preamp and a 300B SE so far :S Could I mod the design of the Morgan Jones to accommodate those twin triodes?
Andy: That looks a great site - I certainly haven't come across it! Thank you very much. |
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| Kaidanovsky |
Wavebourn: Do you know of any diagrams that I could work from to achieve this in as small a space as possible?
Thanks |
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| whitelabrat |
Take a peek at tubedata.org. You'll get a better idea of what you have in your stock. If you're really set on doing the Morgan Jones design, I'd say just go with the ECC88's. I think they're a bit pricey for me so I've gone with a 6N1P design instead and I may use a 6bq7 on the input side just for giggles.
I don't have any good tools for design, aside from my brain which isn't very reliable. ;) You would likely have to do some serious adjustments if using anything other than the ECC88. For example the 6N1P needed a lot more voltage to get it's best performance. That and the heaters draw a lot more power. Bruce Bender's variation on the Morgan Jones design uses a similar topology, but different values on many of the parts. I believe they used pspice to calculate appropriate adjustments? The upside is the much cheaper 6N1P's.
I suppose it may be misleading to suggest that a triode/pentode could be fudged into a Morgan Jones design. Far beyond my skill, but perhaps not impossible. You would need about four PCL84's to do a push-pull I believe. |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
Wavebourn: Do you know of any diagrams that I could work from to achieve this in as small a space as possible?
Thanks |
Let's design one together?
A first, what is impedance of your headphones, and what max power do you need to drive them? |
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| Kaidanovsky |
| I'm not set on anything at the moment, it's just the Jones is the most useful schematic for me I've come across so far. I want to do this on the cheap, with the stock I've got, or at least based on it. I'm afraid my brain, while not exactly reliable either, hasn't the requisite knowledge to infer what does what from the data I've already gleaned from tdsl et al! :D I'll try tubedata though and see if I get further. I'm looking at buying Morgan Jones's book in case that makes things clearer... I wouldn't really want to fudge too much...it's probably a little too much to ask! I expect another basic design would be more appropriate to my needs. I was taken by the small size though... |
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| Kaidanovsky |
| Wavebourn: That's a very cool idea. I wanted to make this because I hope to soon get some Grado headphones and wanted them to feel at home. Probably the SR60s as the SR80s with their harder foam earpieces were rather uncomfortable for me. This gives me an impedance of 32 Ohms and a sensitivity of 98dB/W/m. I wouldn't know what relation that would have to the necessary power...what would you advise? |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
I'm not set on anything at the moment, it's just the Jones is the most useful schematic for me I've come across so far. I want to do this on the cheap, with the stock I've got, or at least based on it. I'm afraid my brain, while not exactly reliable either, hasn't the requisite knowledge to infer what does what from the data I've already gleaned from tdsl et al! :D I'll try tubedata though and see if I get further. I'm looking at buying Morgan Jones's book in case that makes things clearer... I wouldn't really want to fudge too much...it's probably a little too much to ask! I expect another basic design would be more appropriate to my needs. I was taken by the small size though... |
I don't believe you are going to drive Morgan Jones' headphones, I believe you are going to drive yours?
I suggest you to start from research: what power need headphones, and what impedances do they have. Depending on that you may design an optimal topology for set of headphones you are expecting to use, with tubes already available. |
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| Kaidanovsky |
| Of course I intend to drive my own headphones, but since his book has been highly regarded it seemed like the ideal thing to get my sorted on why component x is used at y and how a valve circuit works as a whole... Educational reading? Can't go wrong! ;) |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
Wavebourn: That's a very cool idea. I wanted to make this because I hope to soon get some Grado headphones and wanted them to feel at home. Probably the SR60s as the SR80s with their harder foam earpieces were rather uncomfortable for me. This gives me an impedance of 32 Ohms and a sensitivity of 98dB/W/m. I wouldn't know what relation that would have to the necessary power...what would you advise? |
Be careful with numbers: 98 dB/ 1 mW!
You can blow out your brain applying 1W.
Let's say, you need 20 mW only. So on max power voltage will be square root of 32*0.02 = 0.8V
It means peak voltage = 0.8V * 1.4 = 1.12V
Current = 0.02 / 0.8 = 0.025A,
peak current = 0.025 * 1.4 = 0.035A
What that means?
That means any available tube from your collection when used with output transformer can drive them, but when used without transformer it must work well on idle current of 35 milliamperes (70 milliamperes for a headroom).
Pentode part of PCL84 can do that, that's why I suggested to use it.
You may use a triode part of it for a voltage amplification, while a pentode part may be used as a cathode follower.
Here is it: http://www.r-type.org/pdfs/pcl84-1.pdf
Now, you need to provide power. The simplest way is to use 2 small door bell transformers with 15V secondaries: you will use one transformer to obtain 15V for filament, and another one to obtain from 15V again 120V, but isolated from the outlet. Rectified it will be 170V, enough to drive your tubes. |
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| Kaidanovsky |
Have I got it wrong? I thought sensitivity/efficiency was measured in decibels per watt per metre distance from the speaker. Is that not 98dB/W/m?
Obviously 98dB coming through a headphone earpiece would not be pleasant! 98 was the figure quoted on the Grado website. Are you working backwards from that, assuming I need XdB, which would be Y% of a Watt, P=I^2R etc.?
I apologise for all the questions but I'm trying to learn and understand :angel:
So the PCL84s would be fine with no output transformers, but as an exercise I could find the idle current flow of the others and use 70mA as the necessary figure.
Why is the idle current of the valve the peak current of the headphones, not the same on each (idle current=no sound, peak current=loudest sound)? Or am I getting muddled...
Thank you very much for your continued help - it's much appreciated. |
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| Wavebourn |
Try this one:
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
Have I got it wrong? I thought sensitivity/efficiency was measured in decibels per watt per metre distance from the speaker. Is that not 98dB/W/m?
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No, 98 dB in specs are given for 1 mW output, directly to your ear. :D
Read carefully! It is sometimes dangerous!
I assumed you need up to 120 dB max, a jet airplane motor level of sound.
Now, you have a schemo for your tubes and for your headphones.
Build and try, or discuss with other people on the forum who has many different tastes!
| quote: | Originally posted by Kaidanovsky
Why is the idle current of the valve the peak current of the headphones, not the same on each (idle current=no sound, peak current=loudest sound)? Or am I getting muddled...
Thank you very much for your continued help - it's much appreciated. |
A positive peak will be provided by a tube, while a negative peak will go from the resistor in it's cathode. To get it you have to divide a idle voltage on it minus a negative peak by that resistor's value: more of a negative swing it can't give.
Probably, 70 mA is too much since headphone speakers are less reactive than guitar speakers so unlikely they will draw twice of current.
In the schemo I've designed for you working point and an AC voltage gain are determined by feedbacks (2 loops: one for AC, another for DC)
Tweak R3 to balance between negative current peak and power consumption.
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| whitelabrat |
Wavebourne, very nice.
I don't suppose this could be adapted for other triode/pentodes. :D
6KT8'sand 6GH8's? |
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| Kaidanovsky |
So I've got a 170V DC supply on the right, with the input on the left, output centre right. I would have thought 120dB from a pair of headphones was impossibly loud if not brain-bleedingly unpleasant! :eek:
Power wise, if the current's 70mA then it's outputting 15mW?
Why the LEDs on the cathode of the first stage? Just in lieu of resistors? |
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| Kaidanovsky |
I agree with you there whitelabrat - it's very nice indeed! I'll have a closer look tomorrow, Wavebourn, as it's about bedtime.
Just to clarify my previous question whitelabrat: are these valves triodes or pentodes depending on how they're wired?
Bests,
Josh |
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| Kaidanovsky |
| I made a mistake in my original post - I've actually got 4 PCL84s which makes it just about perfect! However I wonder if it's possible to use the VR valves as I've got so many... |
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| whitelabrat |
Pins 1, 2, and 3 are the triode (the skinny side) and pins 6, 7, 8 and 9 are the pentode (the fat side). The pentode can be operated as a triode as well (I think) depending on the circuit. To illustrate what we're dealing with, imagine taking half of a 12ax7 (twin triode) and a mini EL84 (Pentode) and smushing them into a single tube. The neat thing about the triode/pentodes is that you can drive the pentode side with the triode side and have a single tube mono amplifier as Wavebourn has demonstrated. Two of these tubes would give you stereo.
You'll only need two tubes for Wavebourn's design.
20mw would have you rock'n pretty good. 1w would likely blow your Grados into the next room! |
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| Wavebourn |
| quote: | Originally posted by whitelabrat
Wavebourne, very nice.
I don't suppose this could be adapted for other triode/pentodes. :D
6KT8'sand 6GH8's? |
It may be adapted to any usable triodes and pentodes. 6KT8 may be used (it's triode part) on less current, while it's pentode part is too weak for the output. The same about 6GH8: a triode is better for your application, but a pentode is too weak.
| quote: | Originally posted by Kaidanovsky
So I've got a 170V DC supply on the right, with the input on the left, output centre right. I would have thought 120dB from a pair of headphones was impossibly loud if not brain-bleedingly unpleasant! :eek:
Power wise, if the current's 70mA then it's outputting 15mW?
70 mA / 1.4 = 50 mA for a maximal sinusoidal signal.
50 mA*50 mA * 32 Ohm = 80 mW
Plenty of headroom!
Why the LEDs on the cathode of the first stage? Just in lieu of resistors? |
No, a LED has sharp curve, and very low dynamic resistance. They are used as low noise Zener diodes to stabilize working point so your amp will much less depend on brand and age of tubes than another one without such stabilization.
| quote: | Originally posted by Kaidanovsky
I agree with you there whitelabrat - it's very nice indeed! I'll have a closer look tomorrow, Wavebourn, as it's about bedtime.
Just to clarify my previous question whitelabrat: are these valves triodes or pentodes depending on how they're wired?
Bests,
Josh |
No, depending on how many internal elements they have. A triode has one anode, one cathode, and one grid. Pentode has 2 more grids, so it has 5 electrodes total. Accordingly, tubes with 2 internal electrodes are named diodes, with four -- tetrodes, and so on.
| quote: | Originally posted by Kaidanovsky
I made a mistake in my original post - I've actually got 4 PCL84s which makes it just about perfect! However I wonder if it's possible to use the VR valves as I've got so many... |
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| Wavebourn |
This topology gives several advantages:
1) Safe against tube-rollers :D
2) Deep bass, less dependence on output capacitor used (safe against cap-rollers :D )
3) Quieter pops on switch on/off than without a feedback from a headphone out
4) No so called common-mode distortions than with feedback to cathode that introduces them
5) Longer life of tubes (self-adjusted bias)
6) Nice glow of LEDs when tubes are ready to sound
One disadvantage: reversed polarity. But who hears that?
By the way, for stereo may be used 6N1P (left tube) and 6N6P (right tube); the same number of tubes, but differently arranged.
Something like IRF610 may be used instead of a right tube, so only one 6N1P is needed for a stereo amp. |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
However I wonder if it's possible to use the VR valves as I've got so many... |
VR55 and VR101 may be used as the left triode (voltage amplifier), just ground anodes of both diodes: you don't need them. They need 6.3V filament voltage. MHL4 will need 4V filament voltage, and is probably in demand by owners of hundred years' old radios. VR65 may be tried in the cathode follower (right part). Anyway you will need to tweak that 4.7 Ohm resistor to get about 80 V on cathode of the cathode follower.
But in such case you will have a bar of bottles instead of couple of tiny tubes. How about your small size requirements?
More about tubes: http://www.john-a-harper.com/tubes201/ |
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| salas |
| quote: | Originally posted by Wavebourn
No, depending on how many internal elements they have. A triode has one anode, one cathode, and one grid. Pentode has 2 more grids, so it has 5 electrodes total. Accordingly, tubes with 2 internal electrodes are named diodes, with four -- tetrodes, and so on.
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Well known to tech people, but for ones that wonder why such names, the answer is these are Greek composite words. First part is number*, rest is ''way''. So diode = two way, triode = three way, tetrode = four way, pentode = five way.
*dyo = two
tria = three
tessera = four
pende = five
0dos = way
ano = up (anode=way up)
kato = down (cathode = way down) |
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| Wavebourn |
| quote: | Originally posted by salas
Well known to tech people, but for ones that wonder why such names, the answer is these are Greek composite words. First part is number*, rest is ''way''. So diode = two way, triode = three way, tetrode = four way, pentode = five way.
*dyo = two
tria = three
tessera = four
pende = five
0dos = way
ano = up (anode=way up)
kato = down (cathode = way down) |
I believe "Octal" and "Noval" tube sockets are also made of Greek words? |
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| salas |
| Yes indeed. Octo = 8 in Greek. But noval must have been filtered through Latin. Nine in Italian is nove. |
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| Kaidanovsky |
I did some Greek with my Latin A-Level but wasn't aware of 0dos, ano or kato - I realised the numerical prefix but didn't know the reasoning behind the "ode". It makes a lot more sense now - thank you! I had some hazy idea it was to do with pins so I couldn't understand why my 300Bs are triodes but with four pins! :D I should have guessed, what with diodes and all that that internal element number is much more sensible.
So I'm using each tube to drive itself, effectively, meaning two for stereo? So your diagram actually shows two halves of one valve? If so I'm getting close to understanding. :D
So does this mean that there are for example dual pentodes with 1/2 the valve per pentode, like there are dual triodes?
I wouldn't be using the VRs for my one, I thought if this worked out well I'd knock up another with my brother using them to see what happens... |
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| Tom Bavis |
There are a few dual pentodes in one bottle, but at least 10 pins are needed to bring out all the connections - so some have elements tied together, which reduces your design options.
For anyone else that wants to build this circuit, the 6DX8 (common and cheap) is the same tube as the PCL84 but with a 6.3V heater. With different pin connections, the 6CX8, 6JE8, 6EB8, 6HF8, 6GN8, 6LQ8, 6LY8 should work as well, and maybe the 6AU8, 6BH8 and 6AW8 would do at reduced plate current. |
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| Kaidanovsky |
| So for the power supply I'll need to get a 170V B+ and 15V filament windings. How does one design a power supply? Have I got to use filter capacitors etc? I'm getting the hang of the circuit now, except should the LEDs be any particular values? |
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| Kaidanovsky |
| Thanks Tom! It seems that with a bit of tinkering I could use anything in Wavebourn's circuit! What fun I shall have... :) |
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| wakibaki |
I wouldn't put ANYTHING on my head that was remotely connected to a valve, but there's no accounting for taste.
For anybody contemplating it, you should first read about some of the safety issues as discussed somewhere on this guy's site:-
http://www.tubecad.com/
It's quite interesting generally on the subject of tube design.
w |
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| Kaidanovsky |
There is no accounting for taste, I'm afraid, as I frequently do! (plug my headphones into my preamp, that is). ;)
I've liked tubecad since I came across that article on "missing sonic controls" - very interesting and some fun ideas. |
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| EC8010 |
| quote: | Originally posted by Kaidanovsky
Could I mod the design of the Morgan Jones to accommodate those twin triodes? |
No. That design only works with ECC88/6DJ8 valves. That was what was so clever about the original Earmax design - it used all those properties to optimum effect. |
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| Kaidanovsky |
| Hey thanks EC8010 - it all goes towards the learning quota! I'm looking hard at Wavebourn's circuit at present, possible transformers I might have around etc. I think the Morgan Jones is a must for a later project though...mmm... ;) |
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| Wavebourn |
WARNING!!!
I made an error in mental calculations; R2 and R3 must have 10 times higher values!
I.e. 1.2 KOhm and 100 Ohm.
Sorry for mistake!
| quote: | Originally posted by Wavebourn
Try this one:
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| Kaidanovsky |
| I've ordered the parts - watch this space... :cool: |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
I've ordered the parts - watch this space... :cool: |
Good luck!
By the way, you need 30 VA transformer for B+ and 40VA for 15V (if you did not calculate yet)
I've googled for Morgan Jones, it consumes less because it is loaded on a modulated current source instead of on a plain resistor. But it is more critical to tubes used. Also, a current source load is overmodulated, so plate resistor's value should be decreased a lot. I personally would implement a feed-forward modulation of CCS instead of a feedback one like in Morgan Jones, so the thingy would work much better on the real complex load that differs from a plain resistor. If you don't afraid of adding transistors to tube gear I'll show you how.
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| Kaidanovsky |
| I've got two 50VA ones...is that all right? How do you calculate the necessary rating? I went with what I felt would do the job... :ashamed: |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
I've got two 50VA ones...is that all right? How do you calculate the necessary rating? I went with what I felt would do the job... :ashamed: |
50 VA is more than plenty!
((66.6 mA x 2) +(3 mA x 2) ) x 170V
+
0.3 m A x 2 x 15V
==
~40 VA
Your 50 VA transformers will work colder. |
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| Kaidanovsky |
| So it makes no difference except in terms of the operating heat? If I had a 6VA transformer for example it would do the same job but just run very hot as the valves tried to draw the current? |
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| Wavebourn |
| quote: | Originally posted by Kaidanovsky
So it makes no difference except in terms of the operating heat? If I had a 6VA transformer for example it would do the same job but just run very hot as the valves tried to draw the current? |
Yep, for few last minutes in his life... |
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| Kaidanovsky |
| :D So many things to learn... |
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| Tweeker |
| quote: | | Yep, for few last minutes in his life... |
I kinda doubt it, I suspect it might hit magnetic limits first and be saturated too, not doing the same job. Though it would overheat just fine. |
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| ilimzn |
| quote: | Originally posted by Wavebourn
0.3 m A x 2 x 15V
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That should be 0.3 A * 2 * 15V... that being said, there are variations between various PCL84 regarding heater voltage, but 15V should be OK. |
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| Tweeker |
| VA does not equal watts, at least with a capacitor input supply. It is wise to derate somewhat in this case (by about a third). 50VA is fine though. |
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| Kaidanovsky |
| Do you mean it does equal Watts (as current x p.d. often does!) if there is no load on it? Is it basically a type of rating at an "ideal point" if you like? |
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