Regulated Voltage Supplies

[bob p]

So the sizing is murky.

Yes, a 100W per channel hifi setup playing typical music for repro will get by fine with a 500W tranformer. It will sag on prolonged peaks, which there may not be, depending on the size of the caps.

For monster, I-made-it-myself bulletproof applications, 2kva is about right and will will power both channels at full power indefinitely if the rest of the power supply is done right.

I have a general question about how amp manufacturers rate their regulation supplies. Audio Research lists an "Output Regulation" spec of 0.7 dB for one of their 100W amps on this page: http://www.audioresearch.com/Reference110.html

how do they arrive at that figure? is there a standard for this sort of specification, or is it subject to the nebulous ad hoc ratings of the HiFi industry?

[R.G.]

are those transformers actually potted? or are they just bolted into a metal box?

I'm not sure. I've seen the little ones (i.e. under 5kva) both ways.

Regarding the 1:10 crest factor rule, I'm not one of those guys who will try to pass off a 35W RMS amplifier as having 350 watts of "musical" power. To me, that kind of rating enters the realm of the HiFi snake oil salesmen. To me, 100W means 100W RMS continuous sine wave, and load testing a 100W amplifier would involve sweeping clean 100W RMS sine wave signals into a resistive load on the bench for a couple of days before it moves upstairs into my living room. So I guess my bias is to rate power with a continuous sine wave output, like the government would rate something for a military application.

That's kinda what I thought. You're defining yourself into the high end then.

I have a general question about how amp manufacturers rate their regulation supplies. Audio Research lists an "Output Regulation" spec of 0.7 dB for one of their 100W amps on this page: http://www.audioresearch.com/Reference110.html

how do they arrive at that figure? is there a standard for this sort of specification, or is it subject to the nebulous ad hoc ratings of the HiFi industry?

That's interesting. What they're quoting is not power supply regulation but "output voltage regulation", which is an odd way of stating what everyone else calls damping factor. What that spec tells you is that the output voltage will remain the same voltage with loads varying from open circuit down to 16 ohms within 0.7db. A solid state amp with a huge damping factor might have a spec for similar conditions of 0.001db.

But that's an interesting comparison. Your amp is going to come out much like that one in weight and size if you do a good assembly job. To a first approximation, there is nothing in there except the power supply, power tubes and output transformers. Everything else fits in the cracks.

[Steve Conner]

Hi guys,

Nice to hear of your compiler farm, Bob! I'm qualified as an EE, but I get sucked further towards being a programmer by the day. I just got a new job designing and programming an embedded system that uses a Windows CE front end and a bunch of DSPs. It pays well, but I find working with computers all the time just wears my brain out. Playing with tube amps instead makes a nice break, they are about as far from computers as could be.

I am kind of undecided on the power rating game. I agree that the wattage rating should be the maximum undistorted so-called "RMS" power that the amp can produce, but I'm not so sure on how long it needs to keep it up without melting. If you design it to deliver a full power sine wave for 5 minutes before popping a thermal fuse, it ends up a lot smaller and lighter than if you designed it to keep it up continuously.

Those figures I gave were just pure guesstimates. If you want your amp to run in pure Class-A all the time, then you have to bias it so that each side of the push-pull output idles at half the peak current you expect under full output power. That way, at each signal peak, one tube is passing twice the idle current, and the other one is exactly just turning off. It won't be exactly that because tubes aren't quite linear, but that is the general idea.

So, if you know the rail is 600V, say, and the power output is 100w, you can calculate a peak signal current of about 400mA. Therefore you have to bias at 200mA per side. The total draw would be 400mA @ 600V and the power dissipation at idle 240w. Since a KT88 can dissipate 40w, you need 6 of them, 3 per side.

For your regulator, I assumed the tubes would need a good few hundred volts across them to pass enough current. If you're dropping 300V @ 400mA you are dissipating 120w and hence need three KT88-sized tubes. You might get by with considerably less voltage drop if you used MOSFETs, or beam tubes with a separate screen supply instead of the usual triode connection. But you still need the headroom to cope with line voltage variations.

[bob p]

Steve, I think that everyone with a technical job ends up getting sucked into programming sooner or later. Its just the nature of the Beast, so to speak. I don't know how it affects you, but when I was doing the codehead thing, I found myself dreaming code at night. I took that as a bad sign.

I also like tubes as a means of escape from the over-siliconized world. But then I also like anything that's old and was built to last forever. Low quality disposable stuff just drives me nuts.

[bob p]

I am kind of undecided on the power rating game. I agree that the wattage rating should be the maximum undistorted so-called "RMS" power that the amp can produce, but I'm not so sure on how long it needs to keep it up without melting. If you design it to deliver a full power sine wave for 5 minutes before popping a thermal fuse, it ends up a lot smaller and lighter than if you designed it to keep it up continuously.

I guess that this is what the definitions of continuous commercial use vs. intermittent commercial / amateur / home audio use are all about. The duty cycle is very different, so its easy to "design down" for home audio without getting burned, and I think that's what most consumer electronics manufacturers do.

After all, what's the likelihood that anyone with a home audio system is going to be really pushing an amp for any extended period of time? Chances are if and when the consumer does unleash his amp, he's showing off for a few minutes while a friend covers his ears. For situations like that, a 5 minute limit is more than reasonable. The only problems occur when people take their home audio gear out and try to make it do a pro audio job. When that happens, you either trigger a thermal protection circuit or something gets melted.

[bob p]

Also if you can find a copy of "NBS Handbook of Preferred Circuits Navy Aeronautical Electronics Equipment," Vol. 1, Electron Tube Circuits, 1963 there are scads of regulated tube power supply circuits to pick and choose from.

Rob, I forgot to say thanks for mentioning this reference. I haven't found ityet, but I've got a librarian friend who's offered to try to find it on a loan.

[bob p]

Back to the subject of heat dissipation, I've been PM'ing with R.G. and I think that some of our discussion might be of interest to the community at large, so I thought I'd bring some of it back into this thread. I hope R.G. won't mind.

Oh, come on - I'm a techie. I'd prefer you built the chasses so they open with explosive bolts and the whole thing was powered by plutonium 8-) .

Explosive bolts. Hmmm.... I could use an extra triode and hold it at cutoff by connecting its grid to the bias supply voltage ... and then wire the triode in series with a relay ... so that the relay could trigger the explosive bolts in the event of a bias failure... HEY! It just might work!

Its too bad that I can't seem to find any explosive bolts on eBay. I've added them to my searches, so I'll get an email as soon as somebody offers them.


on a more serious note, you had mentioned that the regulation circuit would be a whole lot easier to design if it were supporting something more like Class A, where the current demands don't swing as widely. i've been thinking about this.

if the AB regulation circuit will require a lot of heat to be dissipated under the idle and typical music operating conditions (compared to the full load sine wave test), maybe the thermal downsides of the regulation circuit make it worthwhile to consider Class A as an alternative.

the original reason to go with Class AB was primarily based upon the need to get 100W of output power with as much efficiency as possible. (limited capabilities of conventional tube amp PTs, heat concerns, etc.). when the regulation idea came along, the idea of thermal efficiency kind of went out the window.

now that idea of using industrial control transformers has come along, we've effectively removed the iron constraint from the power supply, and a whole new array of options has become available. as you might expect, when a barrier like this is lifted, the revised outlook on what is feasible tends to revise my outlook on what is reasonable. yes, the target has begun creeping upwards.

now that iron is available to supply virtually any load, i guess its feasible to get 100W in more than one way: we could do it with a pair of 6550 in Class AB1, or with a gang of 4 or 6 of them running in Class A.

now i'm wondering how the MOSFET regulation heat dissipation and the total system heat dissipation would compare for a gang of tubes operating in Class A vs. a pair operating in Class AB. from a thermal perspective does it make more sense to design the amp one way or the other?

just so you know... i am trying to resist the temptation to buy a HUGE transformer and build the behemoth 300W KT88 Class AB1 stereo amp. ... MUST RESIST! ... Must Resist! ... must resist! ...

[R.G.]

you had mentioned that the regulation circuit would be a whole lot easier to design if it were supporting something more like Class A, where the current demands don't swing as widely. i've been thinking about this.

if the AB regulation circuit will require a lot of heat to be dissipated under the idle and typical music operating conditions (compared to the full load sine wave test), maybe the thermal downsides of the regulation circuit make it worthwhile to consider Class A as an alternative.

The Class A=easier comment was based on Class A current being essentially constant. You can regulate that down with resistors and a bypass cap. There is no design other than just finding power resistors that won't burn up.

the original reason to go with Class AB was primarily based upon the need to get 100W of output power with as much efficiency as possible. (limited capabilities of conventional tube amp PTs, heat concerns, etc.). when the regulation idea came along, the idea of thermal efficiency kind of went out the window.

now that idea of using industrial control transformers has come along, we've effectively removed the iron constraint from the power supply, and a whole new array of options has become available. as you might expect, when a barrier like this is lifted, the revised outlook on what is feasible tends to revise my outlook on what is reasonable. yes, the target has begun creeping upwards.

Of course. What good are features if they can't creep?

now that iron is available to supply virtually any load, i guess its feasible to get 100W in more than one way: we could do it with a pair of 6550 in Class AB1, or with a gang of 4 or 6 of them running in Class A.

now i'm wondering how the MOSFET regulation heat dissipation and the total system heat dissipation would compare for a gang of tubes operating in Class A vs. a pair operating in Class AB. from a thermal perspective does it make more sense to design the amp one way or the other?

just so you know... i am trying to resist the temptation to buy a HUGE transformer and build the behemoth 300W KT88 Class AB1 stereo amp. ... MUST RESIST! ... Must Resist! ... must resist! ...

As long as we don't use energy->matter conversion in there, all of the electricity going in comes out as heat. Since the transformer is putting out a reasonably constant voltage, the power that must be dissipated as heat is directly proportional to the DC current that comes out of the power supply.

We are restricted to choices which pick how much current comes out (that is, class A versus class AB) and where that power is realized, in the regulators or in the amplifiers, by how much of the raw voltage out of the first filter cap is taken up by the regulators versus the amps.

The transformer we're nominating will produce an output voltage of
480Vac*1.414 = 679V, and lose a couple of volts to rectifier losses, so call it 678Vdc nominally.

The power it can put out is 679W/A of DC, and that is maxxed out at 2000W/120Vac = 16.7A of primary current, and 4.17A of secondary current on the 480V winding. Using the RMS to average guesstimate of 1.8, that is 2.31A of DC available without overstressing the transformer wires, and that makes available a net of 2.31*679 = 1571W.

That's all coming out as heat. It's a bit more power than a hair dryer; these run about 1kW.

So you can have any amp or amps which use less than 679V at up to 2.31A. Less current means less power dumped into the room, and less voltage to the amplifiers has to be eaten by the regulators.

If you had two Class A amps that run 1A and 400W at 400V, then the total current fro the power supply is 2A (out of 2.31) and 400V (out of 679). The regulators have to lower 679V to 400, and do it at 2A. That's 279V and 2A in the regulators, or 558W to be dissipated on the heat sinks. It's generally unwise to dissipate more than 50W in a power device in free air, depending on the device, so you're talking at least 11 power devices to divide that power among. I'd probably use 24, spread over a flat backed heat sink of about 2 square feet of flat back area and 1" to 2" fins on the outside running vertically. You might still need a very quiet fan to keep air moving barely.

If you have two class AB amps that use 550Vdc and 0.75A on max signal, then the amps are dissipating 825W on peaks, and the regulators are doing (679-550)*1.5 = 193.5W.

That looks easier than the Class A setup, except that the Class A setup has essentially no current variation - it's always 2A. The "regulator" if you always wanted 400Vdc at the amps is just an array of resistors giving 279V dropped at 2A, or 139.5 ohms at 280W each. That's easy enough to do.

The variation is what makes the regulators have to go from low power dissipation (where the amps eat all the power) to high power (where there's a high current, low voltage load).

[bob p]

As long as we don't use energy->matter conversion in there, all of the electricity going in comes out as heat.

Since we won't be using the thermonuclear power source, I guess I won't need the explosive bolts.

There's one number in your calculations that I don't follow:

The power it can put out is 679W/A of DC, ...

i follow all of your calculations, with the exception that i don't understand how the power figure of 679W/A is derived. are the power units supposed to be watts per amp, or was that a typo???

The power it can put out is 679W/A of DC, and that is maxxed out at 2000W/120Vac = 16.7A of primary current, and 4.17A of secondary current on the 480V winding. Using the RMS to average guesstimate of 1.8, that is 2.31A of DC available without overstressing the transformer wires, and that makes available a net of 2.31*679 = 1571W.

i was thinking that with a transformer utilization factor of 0.812 (textbook number for a PT designed for a full wave bridge), a 2KVA transformer should provide 0.812 * 2000VA = 1624W of DC power. is this calculation valid, or am I doing something wrong?

[R.G.]

i follow all of your calculations, with the exception that i don't understand how the power figure of 679W/A is derived. are the power units supposed to be watts per amp, or was that a typo???

Not a typo, but admittedly cryptic. If a power supply produces 679 Vdc, then it puts out 679W for every ampere of current that it produces. It was a shorthand way of saying that the power supply voltage is constant and the power is determined by how much current flows.

i was thinking that with a transformer utilization factor of 0.812 (textbook number for a PT designed for a full wave bridge), a 2KVA transformer should provide 0.812 * 2000VA = 1624W of DC power. is this calculation valid, or am I doing something wrong?

Nothing wrong, just a different approximation than I was using. I calculated it from the other direction; the highly peaky current pulses into a full wave rectifier with a capacitor filter have a higher RMS current than a full wave rectified sine wave does. The capacitance forces all of the conduction to happen in a small fraction of the half wave. There are various rules of thumb on what the RMS factor is, and nonographs to calculate it. It peaks out at about 1.78...... times the DC average current out of the filter caps. Some people use 1.6, I use 1.8 because it's conservative. Doing that, I get an RMS current that the transformer windings see. The transformer voltages are fixed, so the 2kva rating really means such-and-such max amperes. When I do the calcs, I get a usable amount of power at 1.5KW DC out. Your "transformer utilization factor" was probably calculated a similar way, and comes up with 1.6+ kW. I consider that a sign that we're both kinda right.

If the utilization factor is more correct than my approximation, the transformer runs a few degrees cooler.

I guess I'm putting the lie to engineers calculating everything to three decimal places first. I like to know early whether it's near an edge or not. Later is many-decimal-places time.

[bob p]

thanks for clarifying that. i guess that there are many ways to skin the cat, so to speak, so i'm glad that i'm approaching the problem from a valid perspective, even if it isn't exactly the same way that a professional designer would approach it using conservative numbers.

i've got a couple of more questions though.

So you can have any amp or amps which use less than 679V at up to 2.31A. Less current means less power dumped into the room, and less voltage to the amplifiers has to be eaten by the regulators.

If you had two Class A amps that run 1A and 400W at 400V, then the total current fro the power supply is 2A (out of 2.31) and 400V (out of 679). The regulators have to lower 679V to 400, and do it at 2A. That's 279V and 2A in the regulators, or 558W to be dissipated on the heat sinks.

Hmm. My Class A numbers seem to have come out quite a bit lower than yours. Were your numbers just an example using round figures to demonstrate the math, or were they an estimate of the actual Class A operating conditions?

Here's where I got my numbers:

Looking at the GE 6550A data sheet, the figures for a pair of tubes running in Class A, cathode bias operation with a plate voltage of ~400 VDC are:

Zero-Signal DC Plate Current: 0.170A
Zero-Signal DC Screen Current: 0.0125A

Max-Signal DC Plate Current: 0.174A
Max-Signal DC Screen Current: 0.023A

Adding them up, at idle each pair of 6550A draw 0.1825A, and this increases to 0.197A under full load. The difference in current amounts to a whopping 14.5 mV. I guess that from a regulation perspective, that figure is negligible, and everyone just says that the current demand stays constant.

The cathode biased P-P pair under these conditions will put out 34W of useable audio power, so it would take three parallel push-pull pairs to reach the desired output of 100W in Class A. (I would probably use fixed bias Class A, but the data sheet does not have those figures.) Multiplying the sum of the Max-Signal currents for the pair by 3 pairs of tubes, it looks like the output section for each channel in a stereo amp would require 3 x 0.197A = 0.591A. Multiply that by 2 for Stereo operation: 2 * 0.591 = 1.18A. At ~400V, that comes out to 1.18A * 400V = 473W.

to be consistent we should add maybe 40 mA per channel for the preamp tubes (that amount is baked into the .375A per channel Class AB figure we've been using all along). doing that brings us up to a conservative estimate of 0.640A per channel, call it 0.650A to make the numbers round. Double that for stereo, and you've got 0.650*2 = 1.3A needed at 400VDC, for a total power used by the tubes of about 1.3A * 400V = 520W.

The regulators would have to lower the power by 679 - 400 = 279VDC, and by my numbers they'd have to do it at only 1.3A, for a total power dissipation in the regulators/heatsinks of 279*1.3 = 363W.

The DC circuit then, would have to supply 679V of power at 1.3A, or a total of about 883W of power. 363W of this gets dissipated into the regulator, and the remaining 520 is dissipated by the tubes. Does this sound right, or am I way off?

---

for the Class AB example, my numbers are also a bit different. My guesstimate for the 100W Class AB amp is that each channel would use up to 0.375A under full load, so that two channels combined would use 0.75A under full load. In your example, it looks like you used 0.75A per channel, which would be on-target for a 200W amp that used 2 pair of 6550.

If you have two class AB amps that use 550Vdc and 0.75A on max signal, then the amps are dissipating 825W on peaks, and the regulators are doing (679-550)*1.5 = 193.5W.

With a plate voltage of 550 VDC at 0.75A of current for TWO channels, the amps would dissipate 550V * 0.75A = 412.5W on peaks, or half of your figure.

Similarly, the regulators would be doing (679-550)*0.75 = 96.75W. That is, if it is actually correct to base my numbers on figures that are half of your numbers. By any chance are your numbers twice as big as mine because you've multiplied by a safety factor of 2, or was it just an oversight?

---

That looks easier than the Class A setup, except that the Class A setup has essentially no current variation - it's always 2A. The "regulator" if you always wanted 400Vdc at the amps is just an array of resistors giving 279V dropped at 2A, or 139.5 ohms at 280W each. That's easy enough to do.

Completing the crunch of my numbers then, The Class A "regulator" has to dissipate 279V at 1.3A, or 363W. I'm not sure how you reached a value of 280W, so I can't provide a comparison figure.



Making the Big Comparison then:

Class          Power Out        Dissipation       Dissipation         Total
                                in Tubes          In Regulator
A              100W              520W               363W              883W      
AB             100W              412W                97W              509W

does this seem right to you? if the numbers are correct, they suggest that:

a. buildling a Class A amp doesn't make sense unless we could get a PT that has lower voltage secondary taps. its just too thermally inconvenient to throw away that many volts.

b. Based upon a textbook number for a transformer utilization factor of 0.812 for a FWB, building a Class A amp at the listed figures would require a transformer bigger than 1KVA. A 1 KVA transformer with a utilization factor of 0.812 would only provide 812W of power and we need almost 900. A 1.5 KVA transformer should do it.

c. If the 100W Class AB application only requires 509W total, a 1 KVA transformer should provide 812W DC, which should be plenty.

d. if we could find a PT that would give us a B+ closer to 400 VDC, that would compare favorably (from a thermal perspective) to using our power distribution transformer for the Class AB amp; the Class AB setup uses about 509W total, and a Class A setup that had the right B+ would need no voltage dropping "regulator" and would be close to 520W.

or are my numbers just way off because I've made a fundamental mistake somewhere along the way?

[kg]

just so you know... i am trying to resist the temptation to buy a HUGE transformer and build the behemoth 300W KT88 Class AB1 stereo amp. ... MUST RESIST! ... Must Resist! ... must resist! ...

come to the dark side.



seriously i've got to get off my duff and get soldering again. i've had all the parts for a rather large, rather heavy stereo tube amp for years in my basement, just mocking me:

couple of 1k5va power trannies--one big EI bastard and a wind it yourself toroid
two hammond 1650w output trannies
twelve el509 beam tetrode power tubes with magnoval sockets
four "widowmaker" 470uF@500vdc caps

the fact that i have a two year old running around is somewhat of a concern with plate caps and all. it would have to be very well caged and protected, obviously.

i think the other thing holding me back is the fact that i don't yet have suitable, appropriate speakers for such a beastly amp! someday, though.

ken

[Steve Conner]

Yikes, feeping creaturism! Looking at what you guys have written, I like the idea of the Class-A setup. You would use a tranny with a lower voltage to give a 400V DC rail (how about a 208 to 480 industrial tranny, but feed 120V into the 208 tap? would reduce buzz and stray flux too) and just ditch the regulator entirely. The current draw of Class-A is so near to constant that you don't have to worry about sag. You can always use a C-L-C filter to remove hum.

I think the subjective sound quality of a Class-A amp would be better, too. If you push a single pair of 6550s to 100W in Class-AB, that 100W comes with a LOT of harmonic distortion, and being odd harmonics, it's not necessarily euphonic either. With the Tung-Sol RI 6550s I tried, the distortion manifested itself as a funny expander effect when biased at 40mA each in my guitar amp. They had so much more gain for large signals than small ones, that it was audibly obvious. I found myself wanting to crank the bias as hot as possible to minimize it, and I think three pairs cooking along in Class-A would sound so much nicer, especially at the conversation volume you often listen to a hi-fi at.

Long time no see, KG! My dreams of a BAGA clone are getting closer too, I now have a set of iron from a Fender 300PS and a ham transmitter, a big 19" chassis, and an assload of assorted 6550s and KT88s that yielded a sort-of-matched sextet. I don't have any explosive bolts, but the weight of the iron should disintegrate the chassis just fine without them...

[R.G.]

Hmm. My Class A numbers seem to have come out quite a bit lower than yours. Were your numbers just an example using round figures to demonstrate the math, or were they an estimate of the actual Class A operating conditions?
...
or are my numbers just way off because I've made a fundamental mistake somewhere along the way?

Oh, the numbers were pure fabrications for math illustration. To a certain extent, with a power supply designer hat on, I don't care what the amps pull. You tell me what they pull, I tell you the bad news about what it will cost.

I'll look at your numbers and see if i disagree. Probably I won't.

[bob p]

come to the dark side.

Ken, i'm getting the feeling that Resistance is Futile.