i am building a 2a3pp amp and i am getting too high a voltage from the supplied transformer and the company i purchased the kit from is no longer answering my questions. i need 365v DC at L11 and L12 but i am getting 465. i was told to add a 2k hi wattage resistor in series with the choke on L11 and 12 but i must be doing something wrong as the voltage doesnt change. the amp is
2A3 PP Push-Pull Tube Amplifier 15W Complete Kit (Stereo)_Power Amplifier Kit_Tube Amplifier Kit_Analog Metric - DIY Audio Kit circuit dia` is included.
i am relatively new to electronics and learning as i go and really pleased with what i am learning so far. i would greatly appreciate any help with the above issue as its holding up my build. many thanks
2A3 PP Push-Pull Tube Amplifier 15W Complete Kit (Stereo)_Power Amplifier Kit_Tube Amplifier Kit_Analog Metric - DIY Audio Kit circuit dia` is included.
i am relatively new to electronics and learning as i go and really pleased with what i am learning so far. i would greatly appreciate any help with the above issue as its holding up my build. many thanks
It will make a big difference if your voltage measurements are taken with the tubes installed and running or without them. If you are simply measuring the output of the power supply unloaded then you can expect a quite higher voltage. So if your supply is operating into a no load condition, adding resistors won't show any voltage drop because virtually no current is flowing. So which one is it? Also, a schematic would be helpful. The link doesn't provide one.
Here's one, on the left side: 2A3 PP Push-Pull Tube Amplifier 15W Complete Kit (Stereo) - Analog Metric - DIY Audio Kit
If the voltage doesn't change when adding a series 2k resistor than there must be no current flowing through it.
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I don't get why they provide 2 HV supplies, a tube rectified one and another with a tube.
Are they optional?
FWIW all tubes shown run from one and only (unspecified) "VCC" .
I would love to know what's the AC voltage on the transformer secondary.
And yes, voltages must be read under load.
But getting 465V on 450V rated caps , and you will, always, until tubes warm up, is a recipe for disaster.
Looks like any Chinese with a high school degree (hopefully) and $1000 capital can start a Company and sell to gullible Occidentals, just by catering to their superstitious beliefs.
2A3 and 6SN7 amps in 2014?
Really? 🙄 .
Are they optional?
FWIW all tubes shown run from one and only (unspecified) "VCC" .
I would love to know what's the AC voltage on the transformer secondary.
And yes, voltages must be read under load.
But getting 465V on 450V rated caps , and you will, always, until tubes warm up, is a recipe for disaster.
Looks like any Chinese with a high school degree (hopefully) and $1000 capital can start a Company and sell to gullible Occidentals, just by catering to their superstitious beliefs.
2A3 and 6SN7 amps in 2014?
Really? 🙄 .
In addition to having tubes installed, you need the output transformers connected or you will have no current flow through the output tubes. The output tubes draw most of the power, and if not operating you will have very high voltage like you are measuring.
links work for me.
links work for me.
hi all, and thankyou very much for your input. as mentioned i am pretty new to electronics so have followed the manufacturers instructions which was to insert the rectifier tube and check voltages at L11/12. having now learnt a fundamental? that a resistor will not change the voltage till a current is flowing. the board asks for two 280v supplies but the transformer secondaries are 100/320 and 380. not overly thrilled to be called a "gullible occidental and superstitious" on my first thread as it comes across as smug and you can argue till blue in the face about valve hifi being out of date. Valve Amps: Valve Amps
if you are here to help then many thanks but if here to dazzle with your superior intellect by belittling others then kindly keep it to yourself. thanks
if you are here to help then many thanks but if here to dazzle with your superior intellect by belittling others then kindly keep it to yourself. thanks
Actually … the irony was aimed at the Chinese manufacturer. They took the effort to add fancy "slowly zooming in, zooming out" functionality to their website, yet the engineering schematics and drawings are pretty much crâp. NOT your fault. Theirs.
Yes, you have learned that E = I R … or Volts = Current times Resistance. If there is no current [ I = 0 ], then zero times any resistor value remains "zero" on the E side. It is not clear what the point is of having both a conventional vacuum-tube based rectifier channel AND another semiconductor-based one in parallel. Some of us don't get it.
It may well be this: to preserve the external symmetry of the tube array. There's no easy way to modify the circuit-board layout to squeeze in another big rectifier tube … without challenging the symmetry and harmony of their first layout. That'd be my guess.
I have to agree with FAHEY - that running your electrolytics for any length of time at "over voltage" is a bit dangerous (to them). You might want to purchase a couple of 25 watt, 8,000 or 10,000 Ω resistors, to act as power-supply loads. Safer than running over voltage. Just saying.
Lastly, again: try noodling with Ohm's Law and the Power Law somewhat, to see what you can do. Once you kind of envelop these two equations, you'll feel a lot more confident that you understand what you're seeing.
Ohm's: E = I R where [E = volts] [I = current] and [R = resistance].
Amps: Are usually measured in A = amps, mA = milliamps or uA = microamps.
Resistance: measured in mΩ (milli ohms) Ω ohms, kΩ kilo-ohms, MΩ ohms.
Volts = volts. or mV millivolts. kV kilovolts. A capitalized V.
Resistance "modern way": Since the Ω Ω / omega greek character is a pain in the be-hind to type in, now-a-days you'll see this convention: R33 = 0.33Ω or 330 mΩ. Sometimes you'll see that as 0R33 where the R acts as a decimal point. 3R3 is 3.3 ohms. 33R is 33 ohms. 0k₃₃ is 0.33 kΩ which is the same as 330 Ω. It also could be 330R, which now should be pretty easy to see. 4M7 is 4.7 MΩ = 4700 kΩ = 4,700,000 Ω. And so on. Even the capacitor people are doing the same thing. The only reason I've figured so far is that "no one likes decimal points" because the Americans and many English types use '.' whereas much of the rest of the world uses ','.
So 2u₂ is 2.2 μF (where μ is another hard symbol, but because it looks like a 'u', well, 'u' works). 22u is 22 μF. What about a really big capacitor like 0.47°F? I've rarely seen 0F₄₇, but it would work. the problem comes down to values like 4m₇ which is either 4.7 milli-ohms, or 4.7 milli-farads, or 4.7 milli-henries or … But capacitors have a LONG history of being stated either in μF or pF. (and oddly, not mF or nF as much). So, that 0.47 farad cap would be 0F₄₇ or 470000u. A 16 nF cap might be 0u₀₁₆ or 16000p or without too much ambiguity ¹⁶n. Let's NOT talk about inductors. They are so rare in signal paths, that they're usually spelled out completely. 2.7 μH or μH . 33 mH. 1.5 H. (henries)
E = I R with algebraic rearrangement
I = E / R and
R = E / I
POWER:
P = I E (pie! so simple to remember) and
I = P / E and
E = P / I by rearrangement.
But if you're noodling with the equations, then you know you can substitute…
Remember above E = I R and P = I E?
Well substitute the IR for the E in the PIE equation:
P = I E = I ( I R ) = (I²) R = I²R
power is current squared times resistance. Or, of you know what power you want to dissipate (like my recommendation), AND you know the current you want to sink, then this becomes awfully handy for figuring out the resistance.
You can also substitute in the E=IR and P=IE equations to find it the other way
P = E²/R
Another useful equation.
Anyway, have fun, and don't be put off by the experts.
We tend to sit in front of our bazaar stands, gnaw old bones, and are noisome.
And say things that sound … 'pedantic' or less savory.
Yours,
GoatGuy
Yes, you have learned that E = I R … or Volts = Current times Resistance. If there is no current [ I = 0 ], then zero times any resistor value remains "zero" on the E side. It is not clear what the point is of having both a conventional vacuum-tube based rectifier channel AND another semiconductor-based one in parallel. Some of us don't get it.
It may well be this: to preserve the external symmetry of the tube array. There's no easy way to modify the circuit-board layout to squeeze in another big rectifier tube … without challenging the symmetry and harmony of their first layout. That'd be my guess.
I have to agree with FAHEY - that running your electrolytics for any length of time at "over voltage" is a bit dangerous (to them). You might want to purchase a couple of 25 watt, 8,000 or 10,000 Ω resistors, to act as power-supply loads. Safer than running over voltage. Just saying.
Lastly, again: try noodling with Ohm's Law and the Power Law somewhat, to see what you can do. Once you kind of envelop these two equations, you'll feel a lot more confident that you understand what you're seeing.
Ohm's: E = I R where [E = volts] [I = current] and [R = resistance].
Amps: Are usually measured in A = amps, mA = milliamps or uA = microamps.
Resistance: measured in mΩ (milli ohms) Ω ohms, kΩ kilo-ohms, MΩ ohms.
Volts = volts. or mV millivolts. kV kilovolts. A capitalized V.
Resistance "modern way": Since the Ω Ω / omega greek character is a pain in the be-hind to type in, now-a-days you'll see this convention: R33 = 0.33Ω or 330 mΩ. Sometimes you'll see that as 0R33 where the R acts as a decimal point. 3R3 is 3.3 ohms. 33R is 33 ohms. 0k₃₃ is 0.33 kΩ which is the same as 330 Ω. It also could be 330R, which now should be pretty easy to see. 4M7 is 4.7 MΩ = 4700 kΩ = 4,700,000 Ω. And so on. Even the capacitor people are doing the same thing. The only reason I've figured so far is that "no one likes decimal points" because the Americans and many English types use '.' whereas much of the rest of the world uses ','.
So 2u₂ is 2.2 μF (where μ is another hard symbol, but because it looks like a 'u', well, 'u' works). 22u is 22 μF. What about a really big capacitor like 0.47°F? I've rarely seen 0F₄₇, but it would work. the problem comes down to values like 4m₇ which is either 4.7 milli-ohms, or 4.7 milli-farads, or 4.7 milli-henries or … But capacitors have a LONG history of being stated either in μF or pF. (and oddly, not mF or nF as much). So, that 0.47 farad cap would be 0F₄₇ or 470000u. A 16 nF cap might be 0u₀₁₆ or 16000p or without too much ambiguity ¹⁶n. Let's NOT talk about inductors. They are so rare in signal paths, that they're usually spelled out completely. 2.7 μH or μH . 33 mH. 1.5 H. (henries)
E = I R with algebraic rearrangement
I = E / R and
R = E / I
POWER:
P = I E (pie! so simple to remember) and
I = P / E and
E = P / I by rearrangement.
But if you're noodling with the equations, then you know you can substitute…
Remember above E = I R and P = I E?
Well substitute the IR for the E in the PIE equation:
P = I E = I ( I R ) = (I²) R = I²R
power is current squared times resistance. Or, of you know what power you want to dissipate (like my recommendation), AND you know the current you want to sink, then this becomes awfully handy for figuring out the resistance.
You can also substitute in the E=IR and P=IE equations to find it the other way
P = E²/R
Another useful equation.
Anyway, have fun, and don't be put off by the experts.
We tend to sit in front of our bazaar stands, gnaw old bones, and are noisome.
And say things that sound … 'pedantic' or less savory.
Yours,
GoatGuy
Don't worry, I'm here to learn first, and if possible help somebody along the way.
And yes, I'm quite pissed off at seeing not only Electronics but 1000 other vital Industries being served in a silver platter to China, same as happened in 60s to 90s to Japan.
And in exchange of nothing !!!!!
Oh well.
And of course, it was not personal at all. 🙂
And yes, I'm quite pissed off at seeing not only Electronics but 1000 other vital Industries being served in a silver platter to China, same as happened in 60s to 90s to Japan.
And in exchange of nothing !!!!!
Oh well.
And of course, it was not personal at all. 🙂
hi again. and thanks for the info goatguy. your monica reminds me of a fabulous bill hicks routine "randy pan the goatboy" going to take me many reads to start getting to grips with all that info but superb you took the time. sorry for jumping to conclusions fahey, i also love mountain bikes and get a lot narky chat on forums so a little wary and took it wrong.
all good and putting some strong coffee on to get the grey matter working!
all good and putting some strong coffee on to get the grey matter working!
Dude there no way that's right.....if you have it all hooked up/setup right.... sure they sent you/and its right Xfmer?/ turns/secondary?/rec etc? being that high doesn't sound right at all.....
I didn't read the full source web site, only looked at schematics and photos, so they may have told you in their instructions, if not, ...
Never power up an amp with tubes installed and output transformers wired to the tubes WITHOUT a load on the transformer, be it speakers or dummy loads.
Dummy loads are simply resistors used as loads during testing.
Pick up a couple of 10 ohm 5 Watt resistors to use in place of speakers during you initial testing of the amp.
Using resistors keeps from damaging speakers if something goes wrong (e.g. Bad high frequency oscillation which might damage tweeters, etc.)
The resistor value is not critical, and can be 22 0hms as you are only trying to insure there is a load to protect the output transformer.
Without a load on the output of the amp, a transient on the output causes extremely high voltage to be generated, resulting in a breakdown of insulation within the transformer, and ultimately a short which destroys the transformer.
Never power up an amp with tubes installed and output transformers wired to the tubes WITHOUT a load on the transformer, be it speakers or dummy loads.
Dummy loads are simply resistors used as loads during testing.
Pick up a couple of 10 ohm 5 Watt resistors to use in place of speakers during you initial testing of the amp.
Using resistors keeps from damaging speakers if something goes wrong (e.g. Bad high frequency oscillation which might damage tweeters, etc.)
The resistor value is not critical, and can be 22 0hms as you are only trying to insure there is a load to protect the output transformer.
Without a load on the output of the amp, a transient on the output causes extremely high voltage to be generated, resulting in a breakdown of insulation within the transformer, and ultimately a short which destroys the transformer.
on the 320 secondary i am getting 345. i have only got as far as plugging in the rectifier tube as in the instructions it says to do so first off and then measure voltage at the choke connections. on a side note as per above why on earth there is a valve rect` and then a solid state one next to it is weird.
it would have been good to have the correct voltage as its otherwise stopping the project and the seller doesnt want to know now.
it would have been good to have the correct voltage as its otherwise stopping the project and the seller doesnt want to know now.
I suspect the reason for the two rectifier/filter scheme is that the SS section is used for the preamp and driver tubes, while the tube rectified output is used for the 2A3 supply.
The SS supply would provide higher B+ and lower ripple.
The SS supply would provide higher B+ and lower ripple.
MerlinB asks a good question. Here's His thinking: if the 6.3 (or whatever) filament outputs are likewise (465 - 365) ÷ 365 = … 28% higher than expected, then you have a lines / transformer mismatch. Or: transformers are multiplier devices, thus if the output of HI volts is 1.277× too much, then we almost expect the filament to be 1.277× too much, too. There are 5 outcomes.
[1] Filament is also 28% higher.
[2] Filament is slightly higher, but more or less right
[3] Filament is right on the dot
[4] Filament is slightly lower, but almost right
[5] Filament is way lower than expected.
In the case of [1], you're looking at a transformer which is slightly mismatched between the Chinese power line voltage, and your country's (and your neighborhood's) line tension. What to do? Get a different transformer. Components are specified to ± 10% voltage tolerance. 28% is too much. It isn't reasonable for all the filaments to drag down their heater winding as much as the tube plates are also going to drag down their supply. (You could get a cheap "variac" transformer on E-Bay, which allows bringing down the input supply "variably" to whatever value is right! They are wonderful electronics-bench devices, and you'll never lose it in a dusty corner from lack of use.)
Number [2] is more problematic. IF the filament voltage is only slightly higher than expected, then (as a designer), I would expect that the Hi-V output will substantially drop when all the tubes are plugged in, if that is what it is designed to do. Remember that push-pull amplifiers actually can and often are designed to draw a substantial amount of current when they're "doing nothing" or idle. That mode is called "AB", where the tubes are basically half-biased, and never entirely cut off on either positive or negative input signals swings. Some audiophile fops consider it to be the ultimate (of seventy nine thousand possible) amplifier configuration.
The middle case (filament is "on the spot") gets the highest vote for either of two positions: the previous paragraph's conjecture or the transformer is whack, and its Hi-V secondary is miswound. If it is miswound, then there are only 3 things you can do to "tame the shrew". [1]: replace it, [2] drop the output voltage a bit or [3] drink imprudently. While I can't recommend the third approach for its effectiveness, it sure beats the heck out of chucking the cat, then the amplifier out the window.
If [4] improbably, the filament is slightly low … then you really do have a miswound transformer. If the difference, the shortcoming, is pretty small (not more than 5% low) then all the solutions in the previous paragraph apply. Tubes won't care if they get 5% too low filament/heater voltage. They are designed for such consumer travails.
If entirely improbably, the filament voltage is way low, then you really need to get a different transformer. Chalk it up to crâppy Chinese suppliers. It happens. Remember, in China, a deeply seated and long-held ideal is the parable of a clever Han that makes a money-printing machine (which of course no one else has). The key is "printing".
Your Chinese made circuit board almost perfectly fits that vision. It is the "magic" that allows the supplier to box up a bunch of other stuff using employees who don't know the difference between transformers other than they have the right number (and colors) of wires hanging out. Call it "ignorance" or "naïve", it would account for #4 and #5 cases, and it is not an uncommon complaint.
However, being a betting goat, I'd place my bet on "the filament windings are also 25+ % high", which means you just have a mains-spec to transformer-input mismatch. Get a Variac. When you finally figure out (with the variac) what the right primary (lines-side) voltage should be, then using the voltage-drop and the inductor's version of ohm's law, you might just consider putting a series-inductor on the primary side of the transformer.
It will act like a low (almost no) loss series resistor, which will do its job at reducing your lines voltage, whether there is a load on the main transformer or not. Inductors are serious voodoo in the A/C world. But they ain't cheap.
Anyway, I wrote too much.
The coffee clearly has kicked in.
Good luck.
Buy a VARIAC, like this one:
MASTECH 2KW Transformer Variac 2000VA 0 250V 220V Input | eBay
2 kilowatts. 200 to 250 volts input. 0 to 250 volts output, variable. Less than $100 including shipping. And I would take a bet … it won't gather dust on YOUR bench.
GoatGuy
[1] Filament is also 28% higher.
[2] Filament is slightly higher, but more or less right
[3] Filament is right on the dot
[4] Filament is slightly lower, but almost right
[5] Filament is way lower than expected.
In the case of [1], you're looking at a transformer which is slightly mismatched between the Chinese power line voltage, and your country's (and your neighborhood's) line tension. What to do? Get a different transformer. Components are specified to ± 10% voltage tolerance. 28% is too much. It isn't reasonable for all the filaments to drag down their heater winding as much as the tube plates are also going to drag down their supply. (You could get a cheap "variac" transformer on E-Bay, which allows bringing down the input supply "variably" to whatever value is right! They are wonderful electronics-bench devices, and you'll never lose it in a dusty corner from lack of use.)
Number [2] is more problematic. IF the filament voltage is only slightly higher than expected, then (as a designer), I would expect that the Hi-V output will substantially drop when all the tubes are plugged in, if that is what it is designed to do. Remember that push-pull amplifiers actually can and often are designed to draw a substantial amount of current when they're "doing nothing" or idle. That mode is called "AB", where the tubes are basically half-biased, and never entirely cut off on either positive or negative input signals swings. Some audiophile fops consider it to be the ultimate (of seventy nine thousand possible) amplifier configuration.
The middle case (filament is "on the spot") gets the highest vote for either of two positions: the previous paragraph's conjecture or the transformer is whack, and its Hi-V secondary is miswound. If it is miswound, then there are only 3 things you can do to "tame the shrew". [1]: replace it, [2] drop the output voltage a bit or [3] drink imprudently. While I can't recommend the third approach for its effectiveness, it sure beats the heck out of chucking the cat, then the amplifier out the window.
If [4] improbably, the filament is slightly low … then you really do have a miswound transformer. If the difference, the shortcoming, is pretty small (not more than 5% low) then all the solutions in the previous paragraph apply. Tubes won't care if they get 5% too low filament/heater voltage. They are designed for such consumer travails.
If entirely improbably, the filament voltage is way low, then you really need to get a different transformer. Chalk it up to crâppy Chinese suppliers. It happens. Remember, in China, a deeply seated and long-held ideal is the parable of a clever Han that makes a money-printing machine (which of course no one else has). The key is "printing".
Your Chinese made circuit board almost perfectly fits that vision. It is the "magic" that allows the supplier to box up a bunch of other stuff using employees who don't know the difference between transformers other than they have the right number (and colors) of wires hanging out. Call it "ignorance" or "naïve", it would account for #4 and #5 cases, and it is not an uncommon complaint.
However, being a betting goat, I'd place my bet on "the filament windings are also 25+ % high", which means you just have a mains-spec to transformer-input mismatch. Get a Variac. When you finally figure out (with the variac) what the right primary (lines-side) voltage should be, then using the voltage-drop and the inductor's version of ohm's law, you might just consider putting a series-inductor on the primary side of the transformer.
It will act like a low (almost no) loss series resistor, which will do its job at reducing your lines voltage, whether there is a load on the main transformer or not. Inductors are serious voodoo in the A/C world. But they ain't cheap.
Anyway, I wrote too much.
The coffee clearly has kicked in.
Good luck.
Buy a VARIAC, like this one:
MASTECH 2KW Transformer Variac 2000VA 0 250V 220V Input | eBay
2 kilowatts. 200 to 250 volts input. 0 to 250 volts output, variable. Less than $100 including shipping. And I would take a bet … it won't gather dust on YOUR bench.
GoatGuy
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If you are using the 320v winding without a load the output should be about what you found 465v ( well 320X1.4= 451V ) but again without a load I expect that 320v winding is probably closer to 330v. But when you put a load on that rectified supply the math changes a lot it now works like 320 X .9 = 288v and you actualy lose 2 few volts in the tube rectifier so it should be about 280v when under a load. ( perhaps that was 320 X .88 not 320 X .9 I forget but you see the point)
hi guys, cheers for the input. there are lots of valuable points made and starting to get a bit clearer now. i will check all the other voltages/heaters etc tomorrow as having a beer tonight. i was following the start up procedure to the letter so i might just throw it all together and hope for the best as it seems until under load i am not going to find out if its correct voltages. i am 99.99% confident that its built correct and resistors and caps are correct as being a little out of my depth i check recheck and check again and then once more for luck.
i would dearly love to have a variac as such above but they dont turn up at that price or power rating in the uk sadly.
i would dearly love to have a variac as such above but they dont turn up at that price or power rating in the uk sadly.
Get a 100W light bulb and socket.
Wire the light bulb in series with the hot lead to the unit under test (UUT). The light bulb will light dimly when you power it up if all is well. If all is not well, the light bulb will light brightly, but will protect the UUT.
Steven
Wire the light bulb in series with the hot lead to the unit under test (UUT). The light bulb will light dimly when you power it up if all is well. If all is not well, the light bulb will light brightly, but will protect the UUT.
Steven
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