adding fuses to secondaries

[gsr]

I don't know the power ratings of your first builds PT. Regarding the primary fuse, it's value depends on the whole power the PT is able to provide. 350V x 0.07mA = 24.5VA (voltampere), 5V x 3A = 15VA, 6.3V x 3A = 18.9VA. Whole power = 58.4VA. 58.4VA / 120V (line voltage) = 0.48A. I usually double the value and take the closest match of fuse (in this case 1A). BTW I use slo blo too.

My first build was 330-0-330 @ 100ma. Similar on the filaments. I built a 5e3 type amp. I think I'll switch its fuse to a 1 amp slo blo and see if it pops. RG, in my new build, if I use a virtual center tap for the 6.3v, does it matter which winding wire I put the fuse in? Same with the 5v and the high voltage?

[mooreamps]

'The primary line fuse would either be 3/4 amp or 1 amp ; fast blow.'

What's your rationale for advising a fast blow type?

Slow blow for windings on motors and synco-servos where in-rush circuits are significantly higher than for other electrical circuits.

-g

[mooreamps]

My first build was 330-0-330 @ 100ma. Similar on the filaments. I built a 5e3 type amp. I think I'll switch its fuse to a 1 amp slo blo and see if it pops. RG, in my new build, if I use a virtual center tap for the 6.3v, does it matter which winding wire I put the fuse in? Same with the 5v and the high voltage?

Just wire it up like this.

http://www.schematicheaven.com/fende...a764_schem.pdf


-g

[R.G.]

I don't know. I think my job is OK with space systems company, at least for now. {keeping my fingers crossed here.....}

Cool. Hope the job remains secure. I spent the last few years at IBM ducking bullets while they played layoff roulette. However, back at the techie stuff:

But, since we all know now what the short circuit current is on the primary ; for the high voltage winding ; eddy currents and back flows and such as mentioned ; either part would be the most correct choice.

That is kind of the source of my squeamishness about random fuse blows. I spent many hours in front of a test rig looking at the inrush currents on transformers, just on the primaries. It is very difficult to know that the primary currents will be.

They can be surprisingly big. The primary fuse has to carry that inrush surge and not pop.

I don't have the proper instruments to set up a quick test rig and take pictures, so I went hunting for pics in google. Here's as good an illustration as I could find without a lot of searching. It's bigger than a typical guitar amp, but it illustrates the points I've been making.

Some gems gleaned out of this that agree with my experiences:
- transformer inrush, all by itself, is ten to twelve times the rated (i.e. maximum) primary current, and the article mentions that the difference is on how much current the AC line can provide, not the transformer. The wall-socket matters. That, all by itself, is a good argument for slow blow. However, this is worse for bigger transformers as I noted.
- filaments have a cold resistance about 1/10th their hot resistance. It's the change in resistance that provides the local feedback that keeps them from burning out every time. But cold filament current will be up to ten times the hot filament current if the transformer allows it, in an inrush that lasts for one or more seconds. This is up there in the "pops fast blow fuses, maybe" time scale.
- peak inrush currents through the diodes to the uncharged caps is big. This is a high-ratio transformer in the article, and it puts 26A into the caps on the first cycle.
One interesting thing to note is that every startup is a shorted-winding transient, it just happens to be one that runs backwards from worst towards normal operation. And it depends partly on the characteristics of the AC line you plug it into, and the exact instant in the AC line phase when the power switch contacts make.

Precision in predicting the AC line inrush current is quite difficult to do. Fortunately, transformers are robust enough to withstand big surges for a few seconds. That few seconds lets us pick a fuse which is slow enough to not pop on the 1-2 second inrush transient, but still open before the transformer goes up in flames. Mostly.

True, they can be expressed in ratio's. But consider a 6.3 volt winding capable of sourceing 6 amps. This implies it is capable of driving a 1.05 ohm load. 1 ohm is not that far away from zero.

It depends on your perspective. To a guy who only does signal electronics, 1 ohm may as well be zero. To a guy who does speaker design, 1 ohm is a normal working resistance, far away from zero. To a guy who designs high current power supplies, 1 ohm may be a nice safe, 5% minimum load to keep your power supply from having zero-load funnies. To a mathematician, 1 is an infinity away from zero. So the exact wiring resistance, and just as importantly, the leakage inductance in the transformer windings matter.

It might run hot, but given a few hundred milliohms of wiring resistance, it might survive driving a dead short for a short time.

The problem with the heater winding surviving for a while in the face of a dead short is not that it blows out the heater winding. The problem is that it won't necessarily pop the primary fuse in any reasonable time, maybe forever.

If the heater winding is a few hundred milliohms - let's call that 0.3 ohm for something to calculate with, makes not much difference either way - the current which flows is 6.3V/0.3 = 21A. The normal current is 6A, so the heater current has increased by 15A from normal.

15A more in the heater winding increases the primary current by 15*6.3/120 = 0.63A. If the normal current in the transformer primary is .6A, then you have only doubled the normal current by shorting the 6.3V heater winding. The news is twice as bad for the primary if you happen to have a grounded heater CT because the same current will flow in the heater winding that's shorted, but will increase the primary current only half as much.

With a 1A fuse in the primary, how long will it take 1.2A to blow it? Let's go back to the fuse datasheets to get the optimistic case that the fuse makers want you to use. Here's a representative fuse, the Cooper-Bussman AGC*fast-acting* series. Give a look at "electrical characteristics", upper left hand corner. The blow time for 100% of rated current is literally forever. For 135% of rating, it's sixty minutes, maximum. Note that they won't tell you any closer than that. And for 200% of rating it's two minutes. Again, this is a fast acting fuse.

By the maker's own admission, the 1A version of this fast acting fuse may take one hour to blow at 135% of rating. Meanwhile, back at the ranch... er, transformer ( )
We're trying to figure out how or if the transformer dies. If we have a short on the 6.3V winding, and the transformer is protected by the recommended 1A fast blow fuse, we've calculated that the primary current goes from the 0.6A normal current up to 0.6A plus 0.63A from the shorted heater winding in excess current. That's 1.23A, and we could expect the fuse to blow in - how long??

We know that all the maker will tell us is that a bigger current, 1.35A, this fast blow fuse will blow in less than one hour. At only 125% of rated current, presumably this will be longer, because the fuse is designed to never blow at 100% of rating. There's a current-time curve there that is headed for forever in this direction. But let's assume the best case we have any info on without actually testing fuses. Let's say it blows in an hour.
(NB: yes, I see the time-current curves. Those show the 1A fuse going at 100s at 1.???A. The curve is nearly vertical there. Tiny differences in the fuse, wiring, whatever, make huge differences in blow time.)

So our transformer sits and cooks the heater winding for an hour at 21A. The only real question to be answered now is - what dies first? The power transformer, the wiring to the tube sockets, the tube socket to the (possibly) defective tube, what?

In many cases, it's the heater winding. If that happens, you're out $100 to $300 if you buy retail.

Let's play the problem the other way. If it were my amp, I would put a fuse in the heater winding. I would make it a 7A or 8A slow blow. Cooper-Bussman (my friends!!) make a suitable device, the MDL-7R. These things will let through the 6A normal current forever, will let through the few-cycles inrush without blowing, but will open in less than 120 seconds at 200% of rating. I picked the 7A version just to be sure that a marginal fuse won't give nuisance opens.

The fault current of 21A is three times the rated fuse current, and should clear the heater fuse in well under the specified 120 seconds @ 200% of rating. So the heater winding endures 21A for under (probably well under) two minutes max instead of an hour.

It's just me, but I'd rather pop a heater fuse than hope the primary fuse clears within two hours. I'm a belt and suspenders kind of guy when it comes to stopping electrical faults.

in my new build, if I use a virtual center tap for the 6.3v, does it matter which winding wire I put the fuse in? Same with the 5v and the high voltage?

If you make %@$#$ certain the winding CT can't contact ground, it makes no difference. The resistor/virtual CT has so much resistance in series with it, it does not cause fault currents like the winding by itself. When the fuse opens, the winding is floated, and so the voltage can't cause any currents. If the winding CT is hard-grounded, then currents can flow in the non-opened side. A CT winding is actually TWO windings, connected together, so you have to either fuse both sides, or open the CT so the two windings are both opened by the same fuse. Virtual CT makes it possible to use one fuse.

[mooreamps]

sigh ; ..... which is why we put them on the primary side where the Isc is higher than the secondary.

-g

[R.G.]

sigh ; ..... which is why we put them on the primary side where the Isc is higher than the secondary.

I'm confused. Maybe you could explain that a bit more. I'll kind of ramble through some more analysis and you tell me what I get wrong.

Perhaps I missed something in my analysis. Here are the things I based the math on:
1. Inrush currents in the primary can be big, but last for only a few AC cycles.
2. Predicting primary inrush current is hard because it depends on many factors.
3. Secondary short currents can be large as well. However, especially for the heater, the reflected current in the primary is not a huge increase. It may easily be near 100% of the normal current in the primary.
4. From actual fuse specifications, fuse blow time on currents of 135% and 200% of rated current can be long. Sometimes very long, up to a couple of hours.
5. From actual fuse specifications, fuse blow time on currents of 300% and up of rated current get down into the range of seconds to a minute.
6. Trying to protect secondary loads with primary fuses with high step-down ratios is quite difficult to analyze.
7. The primary current change from normal to faulted-secondary is small, on the order of 100% of normal primary current for the special case of 6.3V heater windings, even though the heater windings are running 300% or more of their rated currents.
8. From actual fuse specifications, the change from 200% to 300% of rated current makes a dramatic difference in clearing time.

#2 is taken directly from my personal experience in front of a workbench and a 'scope. The rest are taken from documents that are publicly available.

For the issue of Isc from the AC power line: This is a big deal for trying to figure out whether the interrupting current rating of your primary fuse will actually open the AC power line in the case of a hard fault on the AC power side. If the Isc is, say, over 10,000A, the common cartridge fuse may not be able to interrupt it. You'd need a higher rated interruption current fuse.

However, interruption current resulting from Isc is not the same as rated fuse for primary protection. Rated current for a fuse is "it's got to carry that much current.". Interruption current is "it's got to interrupt that much current without forming a permanent arc, somehow.) They're different, as the fuse datasheets note. Cartridge fuses are wholly inadequate for interrupting lines with many kiloamps of available Isc.

However, in a transformer with a secondary fault, the impedance of the transformer itself prevents Isc from being significant. Let's say we had an AC supply with Isc = 1Mega-Amp, just to put it way beyond what can really be had. Most wall breakers will not interrupt that, but we'll posit it. We have here a transformer rated for 120VA (120V times 1.0A primary current at max load.)

The number I'm trying to get to is the equivalent impedance inside the trannie. With a power line able to produce an essentially infinite current, the transformer insides themselves are what will limit the current into a secondary fault. We can get at this from the transformer regulation spec. 5% "regulation" (output voltage sag, actually) is a good number for a 120W transformer.

The impedance that makes the output voltage sag is the internal wiring resistance, mostly, with some contribution to leakage inductance in some cases at power line frequencies. Let's pretend it's all resistance. Let's further refer all the resistance to the primary, and pretend that the actual transformer is ideal, perfect transformation at a 1:1 voltage ratio.

That puts all the sag on the primary side, and we can then calculate what the series resistance is. For the example trannie, the output voltage of the ideal backend trannie sags 5% because its primary voltage sagged 5%. It did that because the referred primary and secondary resistance lowered its input voltage by 5%, or six volts. The primary plus referred secondary wiring resistances must then be 6V/1A = 6 ohms.

Notice that this is true for a 5% regulation on the secondary no matter what the secondary voltage is.

So if we short the secondary of a transformer with similar specs, the AC line voltage is driving a six ohm resistor into a short circuit. We can now compute the short circuit current the transformer will let flow no matter what Isc the AC line can provide even if that's a mega-Amp, or even infinite.

The current that flows in the primary is about 120V/6 = 20A. It can't get bigger than that no matter what current the AC power line could do all by itself. Isc being bigger than, say, 100A, is immaterial. The transformer can't suck in any more than that on its own for a shorted secondary fault. Note that a directly-shorted primary side on the transformer is a different matter. That is a short before the transformer is even participating. I think this means that Isc on the primary side is irrelevant to a secondary-side short.

It's not irrelevant to a primary-side short. On the primary side, whatever the transformer does is largely irrelevant. If the transformer suddenly becomes a perfect primary short, the Isc of the primary line is available to the short, and the primary fuse now must be able to clear the Isc and voltage available. If it can't, the energy available goes into melting and vaporizing conductor metal into droplets and gasses hot enough to start fires.

The fact that there is an internal set of resistances in the trannie that limit the primary current on secondary shorts, and the transformation ratio of secondary current seems like it argues that to stop overcurrents, you need to protect two places, for different reasons. The primary fuse stops hard faults on the AC wiring from starting fires. The transformation ratio to a low voltage/high current secondary means that a fault in such a winding does not increase the primary current enough to make clearing a fuse timely - based just on the transformer and fuse ratings, no matter how much AC line current is available - so you need a secondary fuse where the much bigger change from normal to faulted conditions makes selecting a suitable fuse possible.

I'm always willing to learn a new trick or two, so please tell me which parts I missed.

[mooreamps]

The A/C side of the high voltage secondary windings is a higher impedance than the primary side.
For example ; lets say a secondary winding is rated for 300 volts at 100 milliamperes. Primary current equals Zout/Zin ; or 250 mils. Now, lets say the secondary short circuit current is approx 20 times the rated current ; as a rule of thumb ; or 2 amperes. The primary current jumps to 5 amperes.


-g

[R.G.]

The A/C side of the high voltage secondary windings is a higher impedance than the primary side.

Oh, OK. You were referring only to the high voltage winding. That's why I asked the questions - your statements were confusing me, as they appeared to not apply.

So you're OK with my analysis for filament windings, right? You're only worried about whether a fuse is a good idea in a high voltage secondary? OK, let's go there.

For example ; lets say a secondary winding is rated for 300 volts at 100 milliamperes. Primary current equals Zout/Zin ; or 250 mils.

Hmmm. I'd calculate it this way - the secondary is rated for 300V at 100ma. The primary current is equal to the secondary current times the inverse of the voltage ratio. The secondary voltage is 300, the primary voltage is 120, so the current in the primary is 300/120 times the secondary current or 250ma.

Transformers have no impedance as such, only voltage and current transformation ratios. See Transformer - Wikipedia, the free encyclopedia for a good beginner sendup.

I believe what you may have done is calculate an equivalent impedance for the secondary as 300/100ma, then reflected that load into the primary as the square of the turns ratio (120/300)^2 or 0.16 times the 3000 from the secondary loading calculation, then divided 120Vac by the resulting 480 ohms to get 0.25A. It works, I guess. It seems to me that multiplying the secondary current times the inverse of the voltage ratio is less roundabout, but whatever for you is fine as long as the numbers come out the same. Mother Nature doesn't care how you calculate the numbers, She just insists that they come out according to the laws of physics.

Now, lets say the secondary short circuit current is approx 20 times the rated current ; as a rule of thumb ; or 2 amperes. The primary current jumps to 5 amperes.

Let's do some quick calculations. If you'll go with a 5% regulation transformer like I posited in my last post, then you can model it as an ideal transformer with a 120:300 voltage ratio following a series 6 ohm resistor. This represents the referred winding resistances, and is part of the transformer. So with a shorted secondary, the available primary current is 20A max. The shorted 300V winding will sustain 50A, not two amps - or whatever the real numbers are for the transformer in question.

Remember the high voltage winding resistance is already factored into the overall transformer losses in the regulation number.

Obviously, a 100ma winding which is trying to support 50A is going to have some problems. What likely happens is that one part of either primary or high voltage secondary gets into thermal runaway almost instantly.That might hold the fault current down a bit.

Like a 10% regulation transformer? OK, the internal impedance is a 10% voltage sag, 12V at 1A of the full primary current, so the internal resistance can be modeled as 12 ohms, and the primary current simply can't go beyond 10A. Your rule of thumb of the shorted secondary current going to 20 times normal current amounts to picking a number for the transformer's internal resistances. Those vary, depending most heavily on the VA rating, and what that VA rating forces the practical windings to be. But OK, in the range of 10A to 50A - at least until things start to come apart.

Remember back on the fuse datasheets? Remember how the fusing time dropped dramatically as you got over 200-300% of rated? We're now faced with something of an imponderable. The primary in our now well-worn transformer is carrying 1000 times the rated current (if we used a 1A fuse there.) The secondary, if we fused it at, say, 200% of rated, or 200ma, is trying to conduct 50A. It's trying to go to 25,000 times the rated current(NB - with such a situation, the parasitics start weighing in and whether it will get to 5000, 10000, or 25000 times rated gets really hard to track down by modelling).

Which fuse clears first?

I'd bet on the statistics favoring the secondary fuse. First, it's a smaller rating, so it's not dealing with as much current to interrupt, and its being whacked with massively more of an overload than the primary, which is "only" being asked to clear 10x its rated current. That'll go pretty fast. But the secondary fuse will be hitting 10X overload when the primary is still down in the single digits.

In any case, I would want a high voltage winding which is headed for a massive overload, many times more than its normal current, to have a fuse which cleared as little over the rated max as possible.

Again, I would want both fuses - the primary-side, actually AC power line, fuse to prevent fires, and the secondary fuses which can be matched closer to the overload on that one winding, to protect each winding. It is *because* you can match the fuse rating and overload characteristics to the winding better than more accurate fuse selection is possible. Putting the "protection" fuse on the primary side muddies things up.

So - you didn't say anything about that Isc stuff from the last note, whether I had something wrong or what it was if I did, nor where I may have made a mistake in calculating.

Please - tell me clearly if I missed something.
- Was my analysis on Isc being important to the primary AC line fuse but not the transformer incorrect?
- Does Isc have anything to to with secondary currents in the face of the transformer's internal resistance limiting current flow?

And as always, if my analysis in this note was incorrect, please tell me plainly and succinctly "This was wrong, here's the right way to calculate it." I really would like to learn the right way if I'm missing something.

By the way, I'm having some problems with reconciling how pressure thrust, momentum thrust and specific impulse are related. Wonder if you could help with that?

[mooreamps]

Well, I'm not going to do this any more. If you put a short on the secondary, it's going to reflect on the primary. The primary current is going to shoot up higher than the secondary current.

-g

[R.G.]

Well, I'm not going to do this any more.

OK, if you like. I've been trying to tell you some things where either I'm misunderstanding what you're saying, or I have things completely backwards, or you have the concepts a little fuzzy. I'm trying to back up what I say with publicly available references so you don't just have to take my word for it, and I've invited you to tell me what I'm getting wrong. If you don't want to talk about it, that's OK too.

If you put a short on the secondary, it's going to reflect on the primary. The primary current is going to shoot up higher than the secondary current.

This is true in certain cases. Those are when you have a secondary with a higher voltage than the primary. The situation is reversed when you have a low voltage winding. In this case, the primary current increases less than the secondary current does. It so happens that most tube amps have at least one of each kind of these windings.

Details matter, and they especially matter when there is a fault that may damage things. The Devil is always in the details.

[Enzo]

Well, I never worked for IBM or the space program. Plenty of space cadet musicians walk through my door though.

SO let me consider a Fender something or other - a Twin reverb maybe - 3A mains fuse. Now if I put that dead short on the 6v winding and it draws the 20 some amps as discussed earlier, to me that represents about 120 watts drawn through the transformer. That translates to 1 amp through the primary. (assuming 100% efficiency and 120v mains) That 3A primary fuse will never blow.

Or maybe there is a 360VAc HV secondary, and a VERY leaky filter cap in the B+ circuit is dragging 1 amp from it. In my mind that is 360 watts coming through the iron, and that means 3A drawn through the mains fuse. That 3A primary fuse won't blow.

Aren't those faults worth protecting against?

Any number of times someone has asked me, "That little 1/2 watt resistor there burnt to a crisp. How come the main fuse never blew?" Maybe i am wrong, but my response was, "That 3A mains fuse won;t blow until more than 360 watts is being drawn from th wall. It doesn;t take nearly 360 watts to burn up a 1/2 watt resistor."

[soundmasterg]

Is there a reason why the modern blade type fuses aren't used in amps and the cylindrical ones are still always used?

Greg

[Enzo]

I don't know, I've never used them, but thinking about it I'd wonder:

1. Do those come in 250v ratings?

2. SLow blow?

3. You may have noted that the old bayonette fuse caps have changed, the fuse is now much farther inside, so you cannot touch the end of the fuse. I suspect those blade fuses don;t offer enough protection against touching - the fuse is not inside a closed holder. Can they be UL approved, and do they meet the Canadian and European standards? few companies would use two completely different types of fuses for domestic versus export models.

4. Do they come in the range of ratings we have in the cylinder fuses? Like the T500ma for a Marshall 50W? I know for auto use they come in 10-15-20 and up ratings, but do they also come in small values like 1/2, 1,2,3, etc.?

5. What is the relative cost of those compared to glass? When your factory buys a million fuses a year, even 1/10 of a penny adds up. I have no idea what they cost.

And it may be none of that.

[pdf64]

Slow blow for windings on motors and synco-servos where in-rush circuits are significantly higher than for other electrical circuits.

-g

I'm not familiar with sync-servos, but to my understanding the start up in rush current of motors and transformers (as used in tube guitar amps) are comparible. In-rush mitigation devices are advertised as being suitable for motors and transformers. I can't comprehend how a quick blow fuse could be used to protect a tube guitar amp.
Certainly in the UK they wouldn't last many switch on cycles, pretty likely none.
Maybe the supply impedance is lower here (thereby possibly increasing the in-rush current)?
Could the higher line voltage/lower frequency affect this?
But even so, I can't think of any guitar amps that use a quick blow line fuse (that don't have additional in-rush mitigation) - can anyone else?

[R.G.]

Is there a reason why the modern blade type fuses aren't used in amps and the cylindrical ones are still always used?

It may be safety issues. Do you mean the kind of blade fuse like in autos? Got a picture link?