Slight OT - Calculating Transformer valve heaters (DC)

[simon111]

I need some help guys.

I have built up a PSU, standard variable type using a 5Amp Variable regulator (LM350T) - also using a 4amp bridge rectifier, to power the heaters for an amp I am building.

I have a 30VA 9V Transformer.

And 2x 12AX7 (setup for 6.3v) and 2x EL84 valves.

Everything works except that I can't get more than 5.4volts out of the voltage regulator. As i remove each valves the voltage goes back upto the original 6.3v with no valves in.

I have measured the AC current for each valve on it's own and the 12AX7's are drawing 0.55amps once warm, and the EL84 1.25amps once warm @ 6.3v DC. But when all the valves are in the overall current is 2.2amps (this figure seems fine if you add up all the valves datasheet current specs) and the voltage drops.

My transformer is 30VA , therefore 3.34amps @ the outputed 9V.

When I connect up a 12v car battery I can get the desired 6.3v at the heaters.

Is my transformer too small?
If so would the next size up of 50VA suffice?
I don't want to buy another transformer only to find that that isn't big enough as they are rather costly.

another small question. I can set the heaters to the desired 0.3amps for the 12AX7 but the voltage (DC) is really low. Is this ok? I.e. Do you have to have 6.3v or 0.3A? - and does DC make any difference to AC?

Sorry for the slight OT, but hopefully the theory will come in good.

Thanks.
Simon

[brett]

Hi
check the voltage out of the rectifier (before the regulator).  As you insert each tube, I suspect that the voltage will fall, indicating that the output impedance of the transformer is too high for the job.  Weight (kg) is a good indicator of the "true" VA rating in most EI power transformers.

I notice that the rated output is at 9V AC, so there's going to be a lot of heat emitted by the regulator (the regulator is reducing the voltage from about 11 V DC to 6.3 V DC, right?).  Is the regulator fitted with a suitable heatsink?
cheers

[simon111]

Yes the IC has a big heatsink.

What size transfomer do you recommend I get?

[R.G.]

It's probably only Mother Nature telling you stuff again.

Both the LM350 and the transformer are imperfect. The LM350 specification says that it can't produce the desired output with less than about 2.2V across it when the output current is 2.2A. And the transformer has some resistance in it.

To get a better mathematical fit for what the transformer is doing, you need to get me some measurements on it.
1. Measure the open circuit secondary voltage and the primary voltage that causes it. To do this, measure your AC power line voltage (only if you already know how to do this safely; if you don't already know this, you should not be wiring up power line AC) and also the secondary voltage on the transformer when there is NO load on it. These two numbers tell us the voltage the transformer is trying to do and sidesteps the resistances by putting almost no current through them.
2. Measure the DC resistances of the primary and secondary windings with the transformer unplugged from everything.

It will go something like this.
Assume that the transformer voltages you measure are 120Vac and 9.0Vac on the secondary with no load; and also that the transformer primary resistance is 43 ohms and the secondary is 1 ohm, flickering - as low as you can measure. The voltage ratio is 120V/9V = 13.333 in the primary direction and 9V/120V in the secondary direction. The current ratios are the inverse.

What that means is that anything that happens on the secondary looks like it had 13.3times as much voltage and 1/13.3 times as much current when you look at the primary side. So 2.2A on the secondary is 2.2/13.3 = 165ma of primary current. That current flowing through the 43 ohms of the primary wire resistance drops V= 0.165*43 = 7.1V away from the 120 you're feeding the transformer. The secondary voltage then sags to (120-7.1)/13.3 = 8.49V just because of the resistance. But that's only at AC. Rectifying to DC makes it worse.

It's worse because a full wave rectifier both subtracts two diode drops (1.4V) from the peak voltage to make the bridge work, and because the filter caps make the diodes only conduct in very short, sharp pulses. All of the DC current coming out has to be provided to the cap in those sharp pulses once each peak of the AC line. How big are those pulses? They're big. Each half-cycle of AC power is 8.6ms long. It's very common for the time the current is turned on to be less than 1ms, sometimes a lot less. Making the filter capacitor bigger makes that time shorter. Since the time to charge the cap is very short, the current has to be high to get enough current in there.

The math is fairly involved and also involves the loading and the capacitor size. But it's easily 5-10 times the DC current in some power supplies. And that pulse of current flows through that primary side resistance too, lowered by the secondary->primary ratio. So if our power supply wants 5x pulses to charge the filter cap, the charging current is not 2.2A/13.3, but five times that, 11A/13.3. AGH! That would drop over 35.4 V from the AC power line at the primary and make the best secondary voltage look like 120-35.4/13.3 = 6.36Vac.

That 6.36AC gives a peak AC value of 1.414*6.36V = 9Vpeak, and we lose 1.4V to the diodes in the bridge rectifier, for 7.6V at the filter caps. The LM350 has to have 2.2 of those so the biggest voltage the 350T can put out is 5.4V. It can't go any higher because any more DC load lowers its input voltage.

Obviously, I cheated in doing that. I took your output voltage and worked it backwards to get the primary resistance. But that's not an unreasonable primary+secondary resistance in small transformers. I also left out the effect of ripple current, which might also be severe and would make things worse.

It does depend on your transformer. If it's really 3.34A at 9Vac, the open circuit voltage will be higher. If it's 9Vac open circuit, you have a problem with the transformer. Small transformers sag - it's hard to make them efficient. Do the measurements and we can tell better what's happening.

Tube heaters want to see rated voltage. You can't independently set them to 0.3A and not have the voltage across them rise to 6.3V. If you have to choose, put slightly less voltage across them. They'll live a bit longer. AC and DC make no difference to tube heaters for the heater operation. AC will cause some hum, but the heaters are at the same temperature. In fact, the term "RMS" was thought up to express an AC current that did the same heating value as a DC current. 1ADC heats exactly the same amount as 1A AC RMS. 10Vdc heats a resistor exactly the same as 10Vac RMS. By definition.

[simon111]

Thanks for your very detailed response. (don't know what many people would do without your help)


I got a bit lost reading through it the first time so I will read through it again, but here are my measurements as requested. (maths was never my strongest subject!)

Primary AC = 237.15v (UK)
Secondary AC = 10.54v

Primary Resistance = 83.77R
Secondary Resistance = 0.52R

Hope that helps.
Thanks
Simon

[R.G.]

That is a useful set of measurements. Let's calculate.

We have the transformer ratio as 237.15v/10.54v = 22.5.

We can choose to consider that the wiring resistances are in either side of the transformer. I did it in the primary side last time. Let's do it in the secondary side this time. It'll let me talk about impedance transformation.

Transformers convert voltages and currents; they also convert impedances. That's how a tube amp output transformer works. It transforms the impedance of the speaker to match the tubes. The impedance transformation ratio is the square of the voltage ratio. In this case, it's 22.5^2 or 506.25. So the 83.77R in the looks exactly the same as 83.77/506.25 = 0.16547. This adds to the secondary resistance of 0.52R for a total of 0.685R.

We can now look at the voltages. 10.54Vrms *1.414 = 14.9V peak.
Subtracting the diode voltage we get 13.5V after the rectifiers.
For a 2.2A draw and a 5:1 pulse ratio, we get 11A in the secondary on charging pulses. That drops 11*0.685R or7.5V, gives a net of 5.9Vdc on the filter caps! Less 2V for your regulator, that would be only 3.9V to the output of the regulator.

That's even worse than you're seeing in reality! What gives?

It's probably that I just guessed at the 5:1 current pulse ratio. As you can see, the calculation is extremely sensitive to how much we guess that to be. Your current pulse ratio is probably less. Also measuring low resistances is full of errors, so your resistor measurements may be a bit high.

But that's how you do the math. And the answer is - yes, you have too small a transformer. The high pulse ratio of the full wave bridge has tripped up a lot of people - including me! How do you think I got so familiar with the way it gets to you?   

[simon111]

Excellent R.G. Thanks so much for this.

Could I ask your advice on what transformer you think would be suitable?

Ideally I would like a toroidal. Any help on what suitable model/make, would be greatfully received.

Or... how to reverse the maths round to work out what I do need in terms of specifications for the transformer.

Cheers.
Simon.

[GibsonGM]

Thanks a lot, R.G.! That response goes a long way towards helping me understand my dilemma with that "big old transformer" I've been playing with.   7 ohm primary, 50 ohm secondary. 120Vac in, 220vac out.  With a 1K load resistor, my voltage drops from 220VAC (when primary is open) to 12vDC after the bridge, LOL!   I'm still not sure if my power supply circuit is limiting current, but I am highly suspect of the transformer's ability to deliver current after reading this post. 

Is there a quick and dirty test of a transformer's VA? I'm the hands-on type.  I connected a 60W light bulb to my bridge...current draw was a mere 165mA, leading me to conclude that, since I expected at least 100mA more to be drawn, there is not enough VA here...the darn thing Does say "KVA .075", so you would think!!!  It could be damaged, I suppose. 
Frustrating...

[R.G.]

how to reverse the maths round to work out what I do need in terms of specifications for the transformer.

That's a smart guy! Let's do that. It will be a little frustrating because there are in infinity of answers. It will be like a line in the sand, not a particular place: anything over this line will work, anything short of this line will not.

We take the required output voltage - 6.3Vdc. Add to it the regulator overhead, 2.2V. Add to that the ripple voltage on the DC power supply. I like to use no more than 5% ripple, so we're up to 8.5V*1.05 = 8.925V. Add the diode drops, and be generous, so add 1.8V to get 10.725V peak out of the transformer. That's what comes out after the losses in the transformer windings. Now we have to start guessing.

I used a factor of 5 for the pulse current to DC average current. It could be worse than that. Let's play safe and assume a factor of 8. So we expect pulses of 2.2A*8 = 17.6A. The transformer peak secondary voltage must produce at least 10.725V out when its secondary is loaded with 17.6A through the equivalent resistance of its windings.

Side road: referring resistance from primary to secondary to primary. There are two winding resistances, primary and secondary. We can use the tranformer ratio to refer one of these to the other side to make a single combined resistor to make calculation easy. It works out the same however you do it, and one resistor is easier to handle than the other. It's easier for us to get it all into the secondary right now, so we'll take the measured primary wire resistance Rp and refer it to the secondary. The factor to use is the square of the voltage ratio. So the primary resistance Rp when referred to the secondary (which we'll call Rp') is Rp' = Rp* (Vs/Vp)*(Vs/Vp)

For a 240:10V transformer, that's Vs/Vp = 10/240 = 1/24 = .042, and squared that's .001736. So 100 ohms in the primary is the same electrical effect as 0.1736 ohms in the secondary.

Back on the main track: We have to calculate how to get 10.725V after the peak secondary voltage provides 17.6A through the resistance Rs + Rp'.  And here we hit the infinity of solutions. There are an infinite number of voltages and resistances that do that. So we have to start saying "bigger than..." and "smaller than".

We have to have Vs-pk> 17.6A*(Rs+Rp')+10.725V
Vs-pk is 1.414 times Vsrms, or Vsrms> (17.6A*(Rs+Rp')+10.725)/1.414   

The problem here is how do we figure out Rs+Rp'? The transformer makers do tell you this sometimes. They just hide it. It's in that "regulation" number. Transformer makers usually specify their transformers with a certain % regulation. They will say something like "240Vac to 15Vac with 10% regulation no load to full load". You just have to know how to interpret it.

It can mean "This transformer will put out at least 15Vac (for example) when powered with 240Vac and its full rated load current pulled by a resistive load; the open circuit voltage will be 10% higher than that." Or it could mean "This transformer will put out 15Vac unloaded when powered by 240Vac, and it will sag no more than 10% when it's at its full rated current in a resistive load." Some few makers will tell you exactly what they mean, and a very few will specify the winding resistance for you, that being what you really want. But the important thing is, they tell you what the sag amount is, and that lets you calculate the combined primary and secondary resistances.

But even if it's ambiguous, you can use it. All you need to know is whether you get enough voltage. So you take the most pessimistic description if you have to. If we have the last " for example" transformer, the pessimistic case is it puts out 15Vac and has 10% regulation down from that. The transformer  - if new - will have either a secondary current rating or a VA rating. Let's pretend that the transformer in question is a 45VA transformer.

That means that it is supposed to be able to put out a voltage-current product of 45. The voltage is nominally 15, so the amp rating is 3A, which will be in RMS (it always is). So the transformer sags 10% in voltage when it's driving a resistive load of 3A. 10% of 15V is 1.5V, and ohm's law gives us that the combined secondary resistance of (Rs+Rp') must be 1.5V/3A = 0.5 ohm, which is the magic number we needed to know. We can take that and use it with the peak voltage and current numbers to give a good estimate of the DC voltage at no load and full load on a rectifier/filter power supply.

Is there a quick and dirty test of a transformer's VA? I'm the hands-on type.  I connected a 60W light bulb to my bridge...current draw was a mere 165mA, leading me to conclude that, since I expected at least 100mA more to be drawn, there is not enough VA here...the darn thing Does say "KVA .075", so you would think!!!  It could be damaged, I suppose.

Yes, there is a quick and dirty test, but you won't like it. You load the transformer up and let it get hot. You keep applying load and letting it heat until the temperature stabilizes. When the outside of the transformer gets so hot that you cannot keep your index finger on the core, it's at its max VA.

The VA rating of a transformer is a thermal limit. You can pull power out until it gets to the temperature limit of the internal insulation. Exceed that and it self-destructs eventually. Transformers are as tough as nails. You can overtemp them for short periods of time and the insulation doesn't break down instantly, but takes some time. There's some overload in them. But over long periods of time, cooking the insulation at over its rating will kill it.

All commercial transformers are made from at least insulation class A, a temperature of 105C in the hot spot inside the transformer. That equates to an external temperature of about 130F, which happens to be the highest temperature a normal human being will voluntarily keep their fingers on. That's a bit of luck, as it's a quick test.

Class A is the lowest recognized insulation temperature system, so it's a conservative measure. There are insulation systems which are much higher, and some transformers will happily run at temperatures which will give you an instant skin burn, like a clothes iron.

Notice that just the VA rating is not useful for our purposes. By the time the tranformer is running at its VA rating, it's sagged so badly that most circuits are not in regulation. You have to design to the regulation numbers, not the power rating.

[GibsonGM]

Ah, I see ;o)  Moving from 9V circuits into the tube thing is a lot like being a total noob looking at a Dist + and feeling overwhelmed by its complexity, lol.  It sounds like my problem is regulation then, rather than the ability of this thing to deliver current.

It really does seem like the $75, delivered, charge for a nice and properly designed Hammond transformer is worth it.  By the time someone at my level is able to mickey mouse an unknown transformer into tube-powering specs, so much time has been wasted.
Thanks for going thru the #'s for us, R.G., as always, an eye opener!