Single Ended 6V6/EL34/KT66/6550/KT88 Amp

[ThermionicScott]

Looks very nice, and although the microphone clipped a lot, you can hear that it's a good-sounding amp.

I couldn't help but notice the 1M grid resistor on the output tube. While cathode bias is more forgiving than fixed bias, that's a lot higher than the 220k recommended for KT88, or the 500k recommended for 6L6/KT66. It might be a good idea to reduce that value in case you encounter an unstable or gassy new-production tube.

- Scott

[pdf64]

For anyone in the EU who's inspired to try a similar project, Ampmaker has a tempting kit that's along the same lines, see
Double Six 1W/6W/12W amp

[priesty]

Gents,

Thanks for your comments, nice to get some feedback.

If I can clear up my rationale for the Fuses and Mains Switching:
Fuses: To save space on the build, I used an IEC mains socket with integral fuses on live and Neutral. These are located in a fuser "Drawer" included in the socket body. SCHURTER 4301.0501 from Farnell.
Switch: I don't know about other countries, but in the UK supply feed Neutral is tied to Earth and is always identifiable from Live. Morgan Jones book "Building Valve Amplifiers" he does mention that two pole mains switch could be fitted across Live and Neutral, but draws atttention that in such an arrangement, the shorting of the live contact due to mechanical failure would appear to be OK, as item would be still switched on and off by the Neutral half of the switch, but in both "Off" and "On" states the Live supply is still present. Morgan Jones reccomended switching the live circuit only as this was the safest and most reliable if the live/neutral polarity is known.

[R.G.]

Ahmmmmmm, no.. It's actually a great source of entertainment for my colleges at work ; when they hear things like ; when the short circuit current is higher on the primary winding than the secondary winding, yet some people choose to put fuses on the secondary winding..

I think that says more about you and possibly your colleagues (look it up: colleague not colleges) than it does about the electronics.

OK. Shall we delve into this technically again? I did the math and pulled up the tech references for you. But I can do it again.

Did you not get that primary fuses are for protecting (a) the box and electrical system against fires and electrical safety issues and (b) only secondarily protecting the power transformer?. And did you not get that fuses **also** on the secondaries are to protect the power transformer? These are two different purposes, as I explained to you at some length.

The size of the short circuit current on one winding versus the other is not an issue. The objectives in the use of the fuses is an issue. You really do need to be able to embrace more than one idea at a time, Gary.

Did your colleges [sic] think about both uses of fuses? I'm guessing that if your colleges [sic] are as good as you think they are, they'd be able to follow the two different uses. And how to size them. If you presented the issue that way, and not in a way that begged the question.

Do you need me to copy the write up from the earlier set of posts that you just decide not to talk about any more. I can do that if that would help you understand.

Or would you rather not talk about this issue at all?

[R.G.]

If I can clear up my rationale for the Fuses and Mains Switching:
Fuses: To save space on the build, I used an IEC mains socket with integral fuses on live and Neutral. These are located in a fuser "Drawer" included in the socket body. SCHURTER 4301.0501 from Farnell.

That's cool. If you're using hardware that has a dual-fuse setup, there's no reason to go trying to defeat it. However, the fuse in the line/hot side is the one that's critical. I have no issue with that at all.

Switch: I don't know about other countries, but in the UK supply feed Neutral is tied to Earth and is always identifiable from Live.

Yes, it's that way in the USA as well. However, we also have to worry about the electricians that wired the buildings getting live and neutral mixed up in the sockets, and so I always treat live and neutral as though they may be flipflopped in power switching. I recommend breaking both of them with the power switch. I think your decision to do so is good practice.

Morgan Jones book "Building Valve Amplifiers" he does mention that two pole mains switch could be fitted across Live and Neutral, but draws atttention that in such an arrangement, the shorting of the live contact due to mechanical failure would appear to be OK, as item would be still switched on and off by the Neutral half of the switch, but in both "Off" and "On" states the Live supply is still present. Morgan Jones reccomended switching the live circuit only as this was the safest and most reliable if the live/neutral polarity is known.

Yeah, I'd agree with that - with the caveat that "if the live/neutral polarity is known" is a big issue. I have come to not trust electrical power wiring. In my experience, about 5-10% of new construction outlets are wired with line and neutral switched. I carry one of the plug-in outlet testers with me for just this issue when I'm setting up. I highly recommend it for everyone.

Also, the two-pole switch failing with one side shorted is possible, although I think that one pole fail open is more likely. That is just my opinion and experience and is not by any means definitive.

[mooreamps]

OK. On the high voltage secondary, you calculate the short circuit current on the secondary winding and on the primary winding..

-g

[R.G.]

OK. On the high voltage secondary, you calculate the short circuit current on the secondary winding and on the primary winding..

Just in the interest of technical accuracy, I'll elaborate. You may be getting it, but I can't quite tell from your post.

1. All AC powered devices need some means of limiting the power line current in case of an internal fault.This can be done a couple of ways, but the only common one is a fuse in the primary AC power line into the circuits. This is necessary for fire and electrical safety whether or not there is even a transformer in the device.
IT'S NOT TO PROTECT THE EQUIPMENT, IT'S TO KEEP FROM BURNING THE BUILDING DOWN.

A primary fuse may have the side effect of protecting the power transformer or the secondary circuits in a tube amp. That's not what it's there for, though. If it does protect them, it's coincidental.

2. To protect the power transformer from overloads on its secondaries, you need current limits on the secondaries. This is because:
(a) You can't protect a transformer from an internal flaw it has already developed; the war is already over if the transformer melts down on its own. That's what the primary fuse is for; it protects the rest of the world from the transformer or secondary circuits starting fires, and it does not discriminate between transformer internal faults or secondary overcurrents.
(b) Each secondary has a current limit for its own wiring that is unique to the secondary. It is not in general possible to limit the current in one winding to less than the current that burns out that winding by fusing the primary in a multi-secondary transformer.

A multisecondary transformer is designed to parcel out the power which comes in the primary winding. In a hypothetical tube amplifier, you may have a high voltage winding at 350-0-350 and 400ma DC load, for a VA rating on the winding of 196W and a secondary of 6.3Vac at 8A, or 50.4VA. The transformer would reflect a load to the primary of 250VA, roughly. That's about 2A of line current. It would probably be fused at 3A-4A to allow for operating overloads and startup surges. A short on the heater winding would force the primary to supply 76A of current in the heater winding before the primary fuse could think of blowing, and 152A to make it blow in under a second, using typical I-squared-T charts for slow blow fuses.

I can weld a 3/8" wide, 1/4" deep bead of molten steel with a welding rod at 150A. It is very much a horse race whether the primary fuse opens or the heater secondary melts into a puddle. Even if the primary fuse opens, the secondary insulation is often decomposed and a permanent short happens.

If instead the 6.3V secondary is fused at, say, 10A, a short on the heater clears the fuse in a fraction of a second on its way up to 76-152A. The secondary does not have time to overheat and either decompose its insulation and/or melt.

The closer a secondary's power usage approaches 100% of the primary input power, the better a primary fuse can protect the secondary. If the 6.3V heater power from the example was provided by a separate 6.3V/8A power transformer, the primary current at normal operation would be 6.3*8/120 = 0.42A and the primary could be fused with a 0.5A slow blow fuse. A short on the secondary would then cause the primary current to hit 1A with only 16A on the secondary and clear the primary fuse in a second or so. Notice that even here, a secondary fuse could be closer to the secondary rating and protect it better than a primary fuse.

This is a factor any time the power in a secondary is vastly different from the total primary power. A high voltage winding using only 10% of the input primary power is also at risk because the primary can happily power 100% of the power into the winding designed for 10% of the rated power. The high voltage winding can then burn out easily.

Do you understand the differences?

[mooreamps]

Here is what I understand..
...

power primary = power secondary....

120 volts X 0.300 amps = 36 watts ; 360 volts X 0.100 amps = 36 watts
120 volts X 0.600 amps = 72 watts ; 350 volts X 0.200 amps = 72 watts
120 volts X 0.900 amps = 108 watts ; 360 volts X 0.300 amps = 108 watts
120 volts X 1.2 amps = 144 watts ; 360 volts X 0.400 amps = 144 watts
120 volts X 1.5 amps = 180 watts ; 360 volts X 0.500 amps = 180 watts.

So, small changes in the secondary current give you large changes in the primary current, Thus, line fuses should go on the primary side of the power transformer...

Also, you did not answer my question... What is the short circuit current for the secondary winding, and then for the primary winding...



-g

[R.G.]

Here is what I understand..
power primary = power secondary....

I think I begin to understand the difficulty you're having now. That's an OK beginner's approximation to understanding the normal operation of transformers. It is inadequate for understanding either multiple secondaries or fault conditions.

The way it actually goes is
Primary input power = primary winding resistive losses + core losses + secondary resistive losses + (sum of secondary output powers). It is a tribute to transformers in general that they can be designed so that the output power is so close to the input power under normal operation.

And it's important to remember that a shorted winding is not normal operation.

120 volts X 0.300 amps = 36 watts ; 360 volts X 0.100 amps = 36 watts
120 volts X 0.600 amps = 72 watts ; 350 volts X 0.200 amps = 72 watts
120 volts X 0.900 amps = 108 watts ; 360 volts X 0.300 amps = 108 watts
120 volts X 1.2 amps = 144 watts ; 360 volts X 0.400 amps = 144 watts
120 volts X 1.5 amps = 180 watts ; 360 volts X 0.500 amps = 180 watts.

Hey, you've got pushing the buttons on the calculator down to a fine art. Good work. I fully believe that for A*B = (a constant) that any value of A will produce a corresponding value of B that makes it true. Of course, that's 10th grade algebra, and not all that helpful in designing transformers, because this statement:

So, small changes in the secondary current give you large changes in the primary current,

... is only true for a step up transformer. For a step down transformer, like a 120:6.3V transformer, the opposite is true by the same reasoning. Big changes in secondary current make small changes in primary current. I'll, um, let you do the table of mulitplication. I'm pretty sure you can handle that. Maybe it would be help for you to punch the buttons. But the fact that most tube amp power transformers have both a step up secondary for the primary and a step down secondary for the heaters mean you have to consider both cases if you're trying to figure out what to do when doing a competent design for what happens when you have a secondary fault. That means that this:

Thus, line fuses should go on the primary side of the power transformer...

... is another non sequitur - which, I'll remind you from our earlier discussions, is Latin (that's another language that sometimes gets used) for "it does not follow", and is often used. [URL="http://en.wikipedia.org/wiki/Non_sequitur"]quoting wikipedia[/URL

Also, you did not answer my question... What is the short circuit current for the secondary winding, and then for the primary winding...

Actually, you did not ask a question. I'd have been happy to answer it if you had. Here's what you actually posted:

Ahmmmmmm, no.. It's actually a great source of entertainment for my colleges at work ; when they hear things like ; when the short circuit current is higher on the primary winding than the secondary winding, yet some people choose to put fuses on the secondary winding..

and

OK. On the high voltage secondary, you calculate the short circuit current on the secondary winding and on the primary winding..

I'll simplify for you - so you see any question marks in those posts of yours?

But since you think you did finally ask about the short circuit current in the primary and secondaries, I'll be happy to answer that for you now that I know you want to know.

The quick answer is that short circuit currents can't be calculated without knowing more about the transformer than the voltages. That's because a theoretical voltage source can supply any current whatsoever. What limits the current is either a resistance or a current limiting device, which could be an electronic limiter, thermal or electromagnetic breaker, or fuse.

To estimate the fault currents in a transformer, you need to know not only the primary and secondary voltages and currents, but also the resistances of the windings. You also really should know the leakage inductances to estimate the inductive limitation of currents and time response, but we'll ignore that for the moment, not to complicate things too much. You also need to know the impedance of the AC power line into the transformer, by the way. We'll call that 1 ohm, because almost all power outlets can pass 120A or so for a few cycles until the breaker trips. I've measured 200A peaks on the startup surge of large toroidal transformers and variacs before which did not trip the breaker. So we'll just approximate it with the 120A value; we know it's actually lower, and as we'll see, it doesn't make much difference to the actual transformer operation.

I did a little research and came up with one actual resistor measurement of the wire resistances in a Bassman-style clone. The PT winding measured 1.7 ohms, the 6.3V and 5V windings measured 0.4 ohms, (although I question that because it is notoriously hard to measure low ohms accurately with a standard meter), and the 650Vac CT high voltage winding measured 28 ohms. That may not be perfect, but as we'll see, it doesn't matter much as long as the inaccuracies are not huge.

With a 120VAC input, the ratio to the windings are:
Half of high voltage: 325/120= 2.78 - so the current ratio to the primary is 0.369
whole high voltage: 650/120= 5.42: current ratio of primary to secondary is 0.1846
6.3Vac heater: 6.3/120 = 0.0525; current ratio to primary = 19
5V heater: 5/120 = 0.042; current ratio to primary = 24

That means that if you pull 1A of current through the 6.3V winding, for instance, that the primary has to supply 1/19 th as much, or 51.6ma. If you pull 100ma through a 325V half-secondary, the primary current will be 271ma.

Now we have to look at what the voltages and currents do when the resistances are taken into account, and when we short a winding.

Let's short the high voltage winding. The AC power is supplying 120Vac into a 1.7 ohm resistor (primary resistance) supplying a 650Vac secondary with a resistance of 28 ohms. To get the primary current, we will refer the secondary resistance to the primary. The 28 ohms becomes 0.954 ohms, and the total resistance is then 2.6 ohms. The fault current is then 120Vac/2.6 ohms or 45A. Yep, that will blow a fuse.

Near as I can tell, that's what you wanted me to say, or is as far as you can get in the analysis. Don't know which. But it's by no means a complete answer. It ignores what happens when you have a short on one of the other secondaries.

Let's pretend to short the 6.3V winding. The 0.4 ohms reflects to the primary as 0.4*19*19 = 144.4 ohms. We add the 1.7 ohms of the primary and get 146.1 ohms. The fault current for just a shorted 6.3Vac winding is then 120Vac/146.1ohms = 0.821A. Notice that this will *not* all by itself, blow a reasonable sized primary fuse which ought to be about 3A for this PT.

So what will happen is that the primary fuse will let that condition go on forever in the secondary. The secondary is meanwhile dissipating a power equal to the current squared times the secondary resistance. On the secondary side, the current is 19 times bigger, or 0.821*19 = 15.6A. The power in the secondary wires is the current squared times the resistance, or 15.6*15.6*0.4 = 97.4W. So the heater wires themselves are sitting there dissipating nearly 100W, and the primary fuse will not blow. If the 100W of fault current heating can't get out, the secondary wires will heat until the insulation decomposes and the winding shorts. Then the short becomes internal, and permanent, and the transformer is permanently disabled, since it can now no longer provide any external power to heat the tube heaters. It's dead.

Now, let's examine - will this blow the primary fuse with the other secondaries loaded up? After all, they are pulling current too, and that might add up to enough to pop the fuse. The answer is a solid "maybe". The magnetic workings of a transformer are such that the volts per turn of the winding is forced to be the same for all windings. The high voltage winding has 650/6.3 = 103 times as many turns as the 6.3V winding. But the 6.3V winding is shorted. There is still some small voltage that appears across the internal magnetic circuit, but it is vastly reduced by the shorting, and heavily dependent on the actual transformer construction. There will be some small voltage that appears on the 5V winding and high voltage winding, but it is very much reduced. Hence their normal load currents are dramatically reduced. It depends on the transformer and how it's wound, but it is entirely possible that the primary fuse will not clear because the heater winding short reduces the loading on the other secondaries by clamping their volts per turn to a dramatically lower value.

Notice that the 5V heater winding for the rectifier tube can do the same thing, but more so because its voltage and current ratios to the primary are even bigger.

So that's the answer to the question you finally asked (I think). I've computed the primary current for a couple of situations, a short on the high voltage winding and a short on the heater winding. A short on a high voltage winding can blow the primary fuse, which is as far as you got, but there are situations where a shorted heater winding can burn itself out without popping the primary fuse, which is what I've been trying to tell you. The fact that the shorted heater winding will not pop the primary fuse lets the overcurrent go on forever until something burns out.

The short circuit current in the heater winding which is limited by the winding resistances of the primary and secondary can and in some circumstances will limit the primary current to a value that will leave the primary fuse intact. The actual resistances of the windings will vary from transformer to transformer, but these are representative values. The same math process can be used to evaluate any specimen of power transformer.

And it illustrates what I've been trying to help you understand - fusing only the primary side of a transformer may not protect the power transformer. You still have to fuse the primary side for fire and electrical safety reasons, but if you're trying to protect the power transformer, which is the single most expensive part in most amps, you have to fuse at least the low voltage heater windings, because if they're shorted, they can burn themselves out without popping a reasonable value primary fuse. And by clamping the volts per turn, they divert the otherwise normal power that would go into the other secondaries and reduce the otherwise normal primary current to let them do this. The moral of the story is that you can't protect a tube amp power transformer from a heater winding short by fusing only the primary. To protect the power transformer, you have to fuse the heater windings, too. In the illustration, a 10A fuse in the 6.3V heater winding would clear before the winding can overheat.

Here's the thing, Gary. I've been admonished ever so gently that you're just trolling. I like to think you're not malicious, just naive and stubborn, and that there is some hope of educating you. So this is a test. If you can't follow, or at least ask intelligent questions about the parts of the above that you don't understand, I give up educating you and conclude that you either can't follow the reasoning or are determined not to understand for reasons I can't fathom.

[mooreamps]

Not the point... But if you want to fuse the HV secondary, it's no skin off my nose..


-respectfully

-g

[Steve Conner]

Sorry but RG is right. A fuse in the primary isn't guaranteed to protect every winding of a multi-winding transformer. If the VA rating of the winding is small compared to the total VA of the transformer, such as a heater winding or bias tap, it'll just burn without blowing the primary fuse.

However, Gary is right in a sense too. If I only had one fuse, I'd put it in the primary and hope for the best. You can't sell a product like that in Canada, though. The CSA want the secondaries fused too.

That's also moot, because if some turns short in the transformer, it's quite capable of catching fire without blowing any of the fuses. (the VA rating of one shorted turn is quite small)

[mooreamps]

I've said once before I had no issue with fusing the heater windings , because the secondary current would be higher than the primary current... .but not much more.. Given if you have a 4 amp load on a 6.3 volt secondary, the load impedance is already 1.6 ohms.. You put a short on that winding, and it will just sink it's current to ground and run hot. Sure it would eventually fail at some point. You would have to fuse this example right at 4 amps or so, and hope the surge current doesn't take the fuse out.




But yes, it can be calculated. Secondly, it does not protect the HV winding if there a short. If there is a short in the HV winding, the entire field will collapse producing "no voltage", pulling "no current" through the secondary fuse, thus reflecting the short back on the primary.

-g

[R.G.]

I've said once before I had no issue with fusing the heater windings , because the secondary current would be higher than the primary current...

You may have said that,

.but not much more..

But this is factually incorrect. I just did the analysis for you. What part of the math showing that the secondary current would be high enough to burn out the secondary did you not understand?

IF it's the following, I'll go through what is wrong with it.

Given if you have a 4 amp load on a 6.3 volt secondary, the load impedance is already 1.6 ohms..

Or the equivalent. That is correct.

You put a short on that winding, and it will just sink it's current to ground and run hot.

It's not clear to me that you understand this yet. The current that will flow is limited only by the winding resistances of the primary and secondary wire windings, and may be very much higher than normal operation. It will not "just sink its current to ground, with the implication you've made that it will be a little higher than the ordinary load. It will be many times the ordinary load current because the wire resistances in primary and secondary will be much, much smaller than the 1.6 ohms equivalent of the normal heater load. It will get MUCH hotter.

Sure it would eventually fail at some point.

Good. A teachable moment. Yes, it will fail.

You would have to fuse this example right at 4 amps or so, and hope the surge current doesn't take the fuse out.

No, I wouldn't. If you had been taking notes during the fuse discussion (remember that one?) You'd have noticed that the clearing time for a fuse is dependent on the amount of overcurrent beyond the guaranteed carrying current. The heater is a NICE load to protect with a fuse because (a) its fault current is many times the rated full load current, and (b) it can withstand some seconds worth of this overload before being permanently damaged. This lets you fuse for, say, 200% of normal load and do the math to be sure that the power on surge will not cause nuisance trips. So you would not have to fuse at 4A and hope; you could fuse above the normal load, allow for short overloads, including power on surge, and be relatively certain it will not have nuisance trips. If you understand transformers and fuses, that is. Trying to help you here.

But yes, it can be calculated.

And we have another winner!

Secondly, it does not protect the HV winding if there a short.

That was never an objective of a heater winding fuse. It's there to protect the heater winding. If you want to protect the high voltage winding, you have to do the analysis to find out what its current does under fault conditions, and what the primary current does under fault conditions. You may or may not need a fuse in the HV winding. But without doing the math, you can never tell. You MAY be able to do without a HV winding fuse (see the math in my posting on figuring this out) or you MAY need a HV fuse. Or you may try to cheap out when you realize "Oh, MAN! You mean I need more than one fuse to do it RIGHT! Sheesh!" and get by with only the safety-required primary fuse. It's up to you as the designer of your amps how much you think protecting your customers from meltdowns is worth. If you think you can get by without taking the time and effort to protect the power transformer, it's, as you say, no skin off my nose.

If there is a short in the HV winding, the entire field will collapse producing "no voltage", pulling "no current" through the secondary fuse, thus reflecting the short back on the primary.

I think you're confused here. If you mean by that that a short on the HV winding will not cause an overcurrent on the heater fuse, yep, you're right. But that was never the point.

And as nearly as I can translate this statement, it's also factually incorrect. If there is a short on the HV winding, it does not "collapse the entire field." What it does is to cause the primary voltage to be lowered by the voltage dropped in the primary and reflected secondary (HV secondary in this case) resistances, but there is still a voltage, if small, on the primary, and still a field. If there were no field in the core, there would be coupled energy to the HV winding, no secondary (HV secondary, let's keep that straight) current and no short. So this statement as is is physically impossible.

This does reduce the voltage on the heater secondary, which is not experiencing the fault, and not participating in a HV winding fault. If you're confusing a fuse for the heater secondary with protecting the HV winding, you're going to need to think about it a bit.

Is this how you and your colleges [sic] design rockets and space equipment?

[mooreamps]

1. We are are not building spacecraft here.
2. The entire field does collapse.
3. Why don't you short out the heater winding through a current meter and find out ? I did..
4. Take a pill.... :|

-g

[R.G.]

1. We are are not building spacecraft here.

Do you mean "not building spacecraft in this forum" or "not building spacecraft where I work"? I kind of think that knowing the technical basics is necessary for either one, but what did you mean by "here"?

Also, if one does not understand something simple like power transformers, does that mean that it's OK for them to do design in some more demanding field, like maybe the proverbial rocket science, without knowing one of the simpler things? How does that work?

2. The entire field does collapse.

Prove it. Or perhaps define what you mean by "collapse".

3. Why don't you short out the heater winding through a current meter and find out ? I did..

(a) I have.
(b) the results vary by transformer, as you'd know if you were paying attention
(c ) what did you find out from your experiment?

4. Take a pill.... :|

Whatever for? Just trying to help you understand how to calculate transformer action.