Transformer choices?

[ToneMerc]

The ones we got are a little bit different from TM version.

Yep, David's were the default wiring colors, but as far as I know the voltages and current ratings were the same as Jasons.

TM

[Mark]

That sounds like the transformers I was referring to.

Good idea too, I never saw the point of the extra transformer when you could buy one with the extra winding.

I dare say HAD was an over the counter kind of guy.

[Mikka]

If you don't use rectifier valve, why do you continue to use CT High Voltage secondary ?

This is a nonsense !

Cooper and iron are so expensive to day, why don't you try to use less metal ?

[Mark]

How much extra wire is involved in a centre tap?

I wouldn't think any really. An extra piece of wire is used for the external centre tap wire.

What exactly were you proposing?

[Mikka]

How much extra wire is involved in a centre tap?

Twice !

What exactly were you proposing?

It's very simple, when manufacturing a transformer CT is used twice as much copper as a single coil. Finally a CT transformer for a given power is designed as twice rating power than a non CT transfomer scheduled for the same power.

In addressing the problem in reverse, we can double the secondary current capability of a CT Power Tranny simply bypassing the CT.

For example, I operated a push-pull EL84 with a transformer CT originally intended to power a single EL84 in class A.
The transformer in question is a hammond 369GX. I just opened it to find the junction point of the CT and then separated the two windings to be released in two separate wires and then the two coils wired in parallel.

I have developed my approach here.

So if you're not attached to the diode valves, you can make great savings on the power transformer. On the high voltage is not necessary to use a transformer center tap if you put silicon diodes bridge (full wave), which will save you the half power to the secondary HV.

Transformer manufacturers are fully aware of this but also know that they were losing money in disclosing such information.

[ToneMerc]

How much extra wire is involved in a centre tap?

Twice !

What exactly were you proposing?

It's very simple, when manufacturing a transformer CT is used twice as much copper as a single coil. Finally a CT transformer for a given power is designed as twice rating power than a non CT transfomer scheduled for the same power.

I believe you have it backwards, in a non-center tapped PT design you have double the HT current rating. Thus, in order to increase the current rating of the HT winding you have to increase the diameter of the wire used. So, if you have to increase the diameter of the winding are you terribly decreasing the amount of copper? Most likely not.


Neverthess, carry on.

[martin manning]

Thinking this through:

A CT transformer has twice as many turns of wire in the secondary since there are two secondary windings that produce the HT voltage. The wire size can be smaller than a non-CT (FWB) design because each half of the secondary only conducts on half of the cycle.

Roughly, the wire diameter for the non-CT secondary would have to be increased by the square root of two to double the section area, so the net result for a non-CT secondary is 0.5 times the length of wire at twice the section area... about the same amount of copper. The core size would go with the total winding cross section area, which would also be about the same. I'm not seeing much difference except the need to wind 2x the turns and add the connection and the lead-out for the CT.

[Mikka]

You probably don't take time to read my thread that I put the link.

You think wrong because you approch the problem by the bad side.

The tranny power is first influence by its impedance because U = ZI what is the element that sets the rate regulation expressed as a percentage.
The CT doubles the HV impedance.

Wiring the two widing in // of my hammond tranny halves the impedance because in the same time doubles de wire diameter twice by making the two widings become one.

Z = (Z1 x Z2) / (Z1 + Z2)

Take the hammond datashett of the 369GX :
-50VA
-HV CT 225V-0-225V 65mA DC or (225+225) x 0.065 = 32.5VA
-Heater 6.3V - 2.5A DC or 6.3 x 2.5 = 15.75VA

32.5 + 15.75 = 48.25VA

We found something close to the total power gives by the datasheet.

But with 32.5VA everybody knows that it 's not possible to operate a Push Pull of EL84 in classe A. This tranny is given for a SE of 6V6 or El84.

By wiring the two coils in parallel, I doubled the diameter of copper and halved impedance. This has created a virtually new coil without using more materials. Thus, from 65mA DC available I switched to 130mA DC for a nominal voltage of 225V DC. On the high voltage winding for 32.5VA given in CT, I could get 65VA now.

However with the same transformer, and therefore the same amount of iron and copper, I show that we can get, by releasing CT, twice as current of the HV winding and thus ultimately double the HV power.

Believe me or not but that's working and as a person who has been trained in the manufacture of transformers I know my job.

When building a high-voltage winding with center tape, we consider this and we know the way to design this winding considering twice the power required which means the carcass of a sheet as a result.

Take time to compare to two transformers given the same power but with a center tap and the other not. You will see immediately that the one with the secondary winding in CT is two times bigger than the other.

[Mark]

Mikka, I had a look at the link and sorry but I have no idea what you are talking about.

Martin, I agree entirely.

[martin manning]

I'd like to understand this more completely. I have seen discussion about CT transformers requiring a bit more wire and iron, but I have also seen the 0.62 factor for the current capability of a FWB with capacitor input vs. 1.0 for a FW (Hammond data sheet, e.g.). That would say that the wire cross section area would have to be 1/0.62 times as big, I think. That is very close to the increase needed to double current capability one sees in wire tables, nut that would put even more copper in the FWB (no CT) arrangement. I'm no transformer expert, though. Vacuum Voodo (Aleks) has looked into this, maybe he can explain?

[rob@tele]

There is one factor that is ignored here and that is the duty cycle of the two sides of the CT.

If you make a Power supply using the CT and diodes per side, you will have a RMS value of 0.5 per side.

Doing a full wave, bridge rectifier with both sides of the HT secondaries in parallel, you get a RMS value per side of 0.7071. The power for each side is then 150% in stead of the 100%. So the transformer will get hotter!

[Mikka]

Anyway my amp with 13W EL84 push-pull works very well since the beginning of November 2010 with a transformer Hammond 369GX given by the manufacturer to 50VA with a secondary high voltage center tap given to 32.5 VA.

I mean, I do not increase the power of the transformer in itself but I double the available current.

Nothing is more fair and accurate than a scientific approach and explanation :

The datasheet says it's a 50VA tranny with center tap 225/225V 65mADC HT.
The heater is given for 6,3V / 2,5AAC so maths say around 15VA ... and by the way I know the HT have around 35VA.

[img:600:530]http://images4.hiboox.com/images/3810/6 ... f946d0.jpg[/img]

Ok, it's very simple until now.

It's easy to see that CT 35VA HV winding are not enough for a 6V6 or EL84 13W push-pull.
Moreover, the manufacturer advises this transformer for SE EL84 or 6V6.
However the center tap don't help to use the total power habilities hide in power trannies.
Center tap was used because of the rectifier design but today if we are not against to use silicium diodes we better have to use a full wave design.

I tested and took few mesurements on the 369GX PT.
HT coils R => 175/165 ohms, U=240V (no charge, no consumption) for a mains of 230V with a 240V primary wiring.
(240 / 230 ) x 240 = 250V
250 - 225 = 25V => 25 / 225 = 11% of regulation ratio.
175 + 165 = 335 Ohms
25 / 335 = 74.6mA
0.0746 x (225 + 225)* = 33.5VA

It's closed to what I assumed above.

So I can say, if I can separate the two secondary coils, and wire them in // then the new HT caracteristics will be :
335 / 2 = 167.5 Ohms (becaus of the // coils)
25 / 167.5 = 150mA
0.150 x 225* = 33.5VA.

* -> This is the detail that makes all the difference !

Yeah !
At the first vue it's seem to be a good value for a EL84 / 6V6 PP amp.

I can assume that :
UA+ = 225 x 1,4 = 315V with C / R filtering stage. (approximative value)

Cool, that seems to be a good voltage value for a kind of AC15 power stage !

So I started to work on my 369GX PT modifications.

I opened it and do that things.

[img:620:530]http://images4.hiboox.com/images/3810/4 ... 299894.jpg[/img]

Nota : Don't take care about the current and power values on this picture because they are wrong.

That was not easy.
Cut the insolating cartboard was not the more delicate things.
I found very quickly de center tap. I desolder The red/yellow wire and the two distinct coil wires were able to be separated and at this point it was very difficult to resolder the red/yellow wire on one of both of them and the other one with a orange wire for differencing them. But it was done without all break. After that I insulated and fixed them.

I tested my mod and it seems to work well.

[img:150:113]http://images4.hiboox.com/vignettes/3910/2730c0c2471b9e2696633e7c5692891d.jpg[/img]

I give you the result of the tests.

You remember that the primary is wiring for a mains voltage of 240V
The secondary is in // with a nominal voltage of 225V

I would like to precise that the ambient Temperature is around 30°C, because I live in a tropical erea.


The first test was done with a 25W bulb.
Mains Voltage : 235VAC
HV Voltage : 228VAC

The theorical Iac = 25 / 228 = 110mA

Iac measurement was done with my oscilloscope and a 1 Ohm 1/2W 1% resistor.
I read on the screen a voltage of 319mV peak to peak.
Irms = (319 / 2) x ( V^2 / 2 ) = 113mA
Understan V^ by square root.

No signifient temperature elevation on the tranny.

The second test was done with a 60W bulb.

Mains voltage : 236V

-At the begining : HV=210V

-After 15 mn : HV=209V / Tranny T° = 40°C / I can touch and keep my hand on the tranny

-After 20mn : HV=208V / T° = 45°C / I can touche and keep my hand on the tranny.

-After 25mn : HV=206V / T° = 50°C / I can touch and keep my hand on the tranny but I feel it's pretty Hot.

-After 30mn : HV=205V / T° = 55°C / I can touch the tranny but it's hard for me to keep my hand on it more than 15s !

The theorical rms ac current was around 280mA.

Peak current is :
Ip = 280 x 1,4142135 ... = 396mA

here is the load line of the transformer I have drawn from the measurements I made.

[img:838:764]http://images4.hiboox.com/images/4010/0fd32c1857e1919826bebcc6ba736453.jpg[/img]

Now I will discuss the various information that I used for the study of the power stage.

Basing myself on this datasheet and the output transformer, Hammond 125E, I deduce the different possible load values as follows:
6.8k / 8.2k / 11.6k / 12.8k (plate to plate of course)

I want a power ranging from 10 to 15W, which seems quite feasible.

To study the power supply, I use PSU Designer II.

Helping me with my experience and software, I went by intuition and step by step to finally reach this compromise.

[img:977:684]http://images4.hiboox.com/images/4010/03e35ccbc6f34e0205f540710f13ed5f.jpg[/img]

[img610]http://images4.hiboox.com/images/4010/abef042809a27811ea2656c69656a1e1.jpg[/img]

[img:565:544]http://images4.hiboox.com/images/4010/3305fd5499d4db1cc856d9a860331ec4.jpg[/img]

This chart is extremely important in understanding what happens in the transformer and allow us to link with the load line of the transformer described above.
We now need to read this graph by calculating the RMS current.

Irms = (Ix0.7)/ V^2 = (0.360x0.7) / 1.414 = 0.178A

[img:838:764]http://images4.hiboox.com/images/4010/ed27c1da33c057f7b3845cd24e292313.jpg[/img]

221 x 0.178 = 39W

The transformer is given for 35VA on the HV winding.
It should be able to keep the 39W required especially as we are in class A.

RMS power of the amp :

To do it, we have to come back on our loadline.

[img:977:684]http://images4.hiboox.com/images/4010/03e35ccbc6f34e0205f540710f13ed5f.jpg[/img]

Prms

Irms = (0.085 - 0.0375) / 1.4142 = 0.0336A
Urms = (288 - 15) / 1.4142 = 193V
Prms = Urms x Irms = 193 x 0.0336 = 6.5W

We are in class A so this value is for one tube, because we have two tubes

=> Prms = 6.5 x 2 = 13W

It is therefore possible to operate a push-pull EL84 with a high voltage non CT winding of 35VA around when using a full wave rectifier silicon diodes bridge.

In the case of a high voltage winding with center tape, we would have had an available capacity of about 60 to 70VA.

[Mark]

Mikka, I think we will have to break this down bit by bit as I don't see the point of some of your calculations.

Code: Select all

It's easy to see that CT 35VA HV winding are not enough for a 6V6 or EL84 13W push-pull. 
Moreover, the manufacturer advises this transformer for SE EL84 or 6V6. 
However the center tap don't help to use the total power habilities hide in power trannies. 
Center tap was used because of the rectifier design but today if we are not against to use silicium diodes we better have to use a full wave design. 

I tested and took few mesurements on the 369GX PT. 
HT coils R => 175/165 ohms, U=240V (no charge, no consumption) for a mains of 230V with a 240V primary wiring. 
(240 / 230 ) x 240 = 250V 
250 - 225 = 25V => 25 / 225 = 11% of regulation ratio. 
175 + 165 = 335 Ohms 
25 / 335 = 74.6mA 
0.0746 x (225 + 225)* = 33.5VA 

It's closed to what I assumed above. 

So I can say, if I can separate the two secondary coils, and wire them in // then the new HT caracteristics will be : 
335 / 2 = 167.5 Ohms (becaus of the // coils) 
25 / 167.5 = 150mA 
0.150 x 225* = 33.5VA. 

In the final result of both equations is the VA rating of the transformer. It proves that you have gotten the spec wrong. It is 225-0-225vac @0.075mA. Which gives a VA rating of 33.75

Could you explain the equations and their importance?

[martin manning]

I must admit I have a hard time following too. My main interest here is in the relative merits of the FW vs. FWB when the DC power requirements are fixed.

In my simple analysis above, I assumed that the wire cross section of the non-CT FWB would have to be doubled with respect to the CT FW to permit the single secondary winding to handle the current on both halves of the charging cycle. In that case, however, the series resistance of the non-CT secondary would be half that of the CT version, and the reduced losses would produce less I^2*R heating and a lower voltage drop in the winding, making the FWB more efficient.

I’m not sure about the relative inductive and capacitive losses (power factor), but the winding resistance for a given voltage ratio is, I believe, where the principal advantage lies for the FWB. This is essentially what Mikka has demonstrated by paralleling the secondaries of a CT transformer, and it's a useful data point because it directly compares the two rectifier configurations using the same core size and the same mass of copper. In reality the optimum design for a FWB transformer would take advantage of the improved efficiency, and a smaller, lighter transformer would be the result for the same delivered DC voltage and current.

Since it is the losses that are being reduced, I’m not convinced that it is as much as a factor of two in material costs. This is a very complex question due to the number of variables involved, including things like the effects on ripple voltage and any effect on the other components that would be used in the two designs.

Incidentally, before silicon rectifiers were available, consider that a FWB arrangement would have required two additional and costly vacuum rectifiers, the additional volume, the associated hardware, wiring, and heat rejection, plus the additional 4-6A of heater current from the power transformer to run them. In addition to all of that, the FWB incurs two diode voltage drops of 20-50V instead of just one for the FW. All of this would make a FWB power supply using vacuum rectifiers prohibitively expensive and inefficient as compared to a FW supply.

[Mikka]

Incidentally, before silicon rectifiers were available, consider that a FWB arrangement would have required two additional and costly vacuum rectifiers, the additional volume, the associated hardware, wiring, and heat rejection, plus the additional 4-6A of heater current from the power transformer to run them. In addition to all of that, the FWB incurs two diode voltage drops of 20-50V instead of just one for the FW. All of this would make a FWB power supply using vacuum rectifiers prohibitively expensive and inefficient as compared to a FW supply.

I totaly agree with that !

In the final result of both equations is the VA rating of the transformer. It proves that you have gotten the spec wrong. It is 225-0-225vac @0.075mA. Which gives a VA rating of 33.75

Maybe you're a little too fussy.


The manufacturer gives as 32.5VA nominal power.
In my first deduction is approximately 35VA I then finally after measurements and calculations I find 33.5VA.
In the first case I am to 9 percent difference in the second case in less than 3 percent.

Note that in my area the mains frequency is 50Hz while the Hammond transformer specifications from 60Hz. This has an impact on results.

Could you explain the equations and their importance?

Ok, I can try !

The first is to calculate the rated voltage unloaded of high-voltage secondary.
knowing that the primary is wired for 240V, I measured a mains input voltage of 230V and I read a secondary voltage of 240V, I calculate the nominal voltage of the secondary winding as follows:
(Up / U) x Us = Un => (240 / 230) x 240 = 250V

I am fortunate to know the rated voltage is 225V secondary. From the voltage unloaded I deduce the voltage drop regulation ratio which will serve me for the simulation in SPICE.
250 - 225 = 25V => 25 / 225 = 11% of regulation ratio.

After measuring the impedance of the two secondary windings and knowing the voltage drop over rated, I can calculate the rated Alternative current.

U = ZI => I = U / Z = 25 / 335 = 0.075 AC

Knowing the rated current and rated voltage, it is easy to calculate the power rating of the secondary. Il faut pour cela tenir compte du CT.

0.0735 x (225 + 225) = 33VA

Then if I put the two windings in parallel I change two things.
First the impedance of the secondary.

335 / 2 = 167

Second, the secondary voltage because of the only one effective winding.
Which ultimately gives:

25 / 167 = 0.150A AC
0.150 x 225 = 33VA

The accuracy is plus or minus 5% because of the tolerance of the measuring instruments.