How does one determine what type of transformers are used?

[DocRocz]

It's been a while since I've posted anything and since everyone here is so knowledgeable and nice I have a new question.

I was talking to a tech the other day and he said if I was going to build an amp I should decide on the type of tube I would go with and then I would pick the transformers based on that tube.

I found a tube KT66's and now I'm at a loss as to what information I need from the tube specs (or any tube for that matter) on how to
determine what transformer I should use.
I would like the output to be a 30watt amp.

Please be as specific as possible as I need all the input I can get.

Thank you for any help.

[Phil_S]

For starters, it will help a great deal if you have the data sheet for the tube:
http://tubedata.itchurch.org/sheets/086/k/KT66.pdf

Next, we need to know something about the circuit. 30W output doesn't tell us much. Is this PP or SE you are thinking about? According to the data sheet, KT66 is capable of 25W in SE operation, 30W in PP cathode bias, and 50W in PP fixed bias. Plate voltage is a big determining factor.

If you have a particular circuit in mind, or a variation on that circuit, it would help frame the question. 30W with a KT66 isn't much to go by.

[rsi]

The OT will be based on the load resistance KT66 wants to see in the configuration you choose (see previous reply).

The PT will be based on:

The HT voltage required for the configuration you choose (see previous reply again).

How much current draw for all the tubes you are using.

What type of rectification you will be using (if tube, you need another secondary to supply that tube).

There could be other considerations, such as multiple primary voltages if you want to use the amp in other countries.

[surfsup]

I'm learning too so this might be a good exercise for me:

For the PT:
How many KT66 tubes to you have? 1 or 2? or 4? Single Ended would be one tube. Push pull would be pairs of tubes 2 or 4. Since you say 30W I'll assume P-P

How many preamp tubes do you plan to use? (this might not even matter anyway in terms of current draw compared to the KT66s)

Since you say 30W amp and the KT66 puts out 14.5W each tube in P-P, we'll assume PushPull operation:

So lets assume two KT66s and three 12ax7 tubes
2-KT66s poweramp
2-12ax7 preamp
1-12ax7 PI

2 KT66s will pull 30W/Vrail (assume Vrail = 350ish) = 30/350 = 85mA draw for the plates
2 KT66s will pull an additional 20mA for the screens (I think)
3-12ax7s will be about 15mA (5 each)

That's 120mA total for the 5 tubes but you want a safety factor so 120*1.5 = 180mA. Say 200mA for a nice round number.

Now, do you want to rectify using a tube rectifier (another tube) or diodes? Diodes are easier and cheaper but the basis here is IF you are running the KT66s at 350V, you'll need a voltage at the secondaries of at least that amount, plus whatever will be lost from a dropping resistor, etc.

A full wave rectifier with capacitor will deliver (if memory serves since I'm typing from the hip) 1.4 times the secondary voltage of the PT. (a little unsure about this calculation)

http://www.hammondmfg.com/pdf/5c007.pdf

So if you want 350, you want to shoot a bit higher, lets say 375, so a PT with a secondary V of 270 would do.

Since all those tubes use 6.3V heaters, and the heaters will draw (2*1.3 for KT66 plus 0.3A for each 12ax7) 3.6A you need a PT that has 6.3V winding @ 4A

If you use a tube recto, these usually have 5V windings and most require 2A, but I think there are two recto tube types that need 3A, so a 5V winding with either 2A or 3A would be needed.

So, your PT would be 280V @ 200mA, 6.3V @ 4A, 5V @ 2A (last part optional if only using tube recto)

I'm sure I got some stuff wrong in there. Anyway, that somewhat explains the PT choice, but if you're designing your own amp, there's a TON more you'll need to know...take it from me, the resident ampgarage noob!

[Cliff Schecht]

Ss bridge rectifier is 1.414*RMS voltage and 0.62*rated current.

[surfsup]

Cliff, looking at that link I posted, one would actually use the average, not peak? so really its 0.9?

[rdjones]

A full wave rectifier with capacitor will deliver (if memory serves since I'm typing from the hip) 1.4 times the secondary voltage of the PT. (a little unsure about this calculation)

http://www.hammondmfg.com/pdf/5c007.pdf

Remember that a full wave (2 diodes) sees only half the secondary on each half cycle of AC, so it's 1.4x half the secondary voltage, or .7x the entire secondary.
The Hammond pdf shows this but I just wanted to call attention to it for clarity.

rd

[DocRocz]

To Phil_S, rsi, surfsup, Cliff Schecht, rdjones
I knew that the crew here was just amazing and would provide me with answers and even resources for me to pursue this information.

This is so amazing that I don't have the words to even come close to saying thank you for what you have shared with me.

If any of you have suggestions for reading material that I should look into for getting more information on this subject I would appreciate that very much.
What you have provided me is such a huge amount of information that I really want to get my hands on literature that will allow me to ask more concise questions but some day be a contributor.

Yes, I will be going over the links that have been provided.

Thank you so much.

[Kagliostro]

As a quick reference

may be you are also interested to read

this

http://www.trinityamps.com/phpbb/viewto ... =12&t=1457

and this

http://www.trinityamps.com/phpbb/viewto ... f=12&t=779

here a PT current calculator

http://www.dreamtone.org/Calculate_Current_Form.htm

Kagliostro

[surfsup]

this kinda covers the mA draw for the PT...

http://www.freewebs.com/valvewizard1/smoothing.html

His books are also great, and reasonably priced (and help cure insomnia as well! ha)

I also responded to another poster with some math here, but keep in mind I'm new, but this thread might help you get some of the basics down on voltage dropping, etc. I found that ALL the books were very good at explaining snippets from a circuit, but like you, it took me a while to figure out the simple basics like how many mA, V calcs, etc to get started:

https://tubeamparchive.com/viewtopic.ph ... ht=#154437

[David Root]

DocRocz, if you would be more comfortable with a proven circuit, take a look at the Marshall JTM45, an icon from the '60s, uses KT66 and the proper clone transformers are readily available. See Marstran, for example.

[DocRocz]

Again thank you so much and with what you have suggested I think it's going to be a while before I post another question.
Not that I don't have them, but I'd rather be more informed before I ask.

Thank you

[surfsup]

rdjones,

Remember that a full wave (2 diodes) sees only half the secondary on each half cycle of AC, so it's 1.4x half the secondary voltage, or .7x the entire secondary. The Hammond pdf shows this but I just wanted to call attention to it for clarity.

Sorry, I looked at the full wave BRIDGE rectifier. I see now on the lower left the full wave with two diodes not four.

Consequently I have the full wave (2 diode) in my HO. If I wanted to change out the PT, would you guys recommend going with a full wave bridge if you had the two extra diodes already?

Currently it is a 190-0-190 71mA 6.3V 2.5A (Hammond 260EX)
This puts (380*0.71)=265V after rectification...

So if I want 265V after rectification with full wave bridge, I'd need a 300-0-300, right?

Vavg (300*0.9) = 270V (close enough to 265V)

[Cliff Schecht]

With a FWB and a 300-0-300 transformer you'll end up with 600*1.414= ~850V! With a full wave bridge rectifier you don't use the center tap on the transformer, the diodes create an artificial centertap. If you used only half of the windings you would end up with 300*1.414 = ~425V. With your current transformer and a full-wave NON-BRIDGE rectifier you'll get 190*1.414=~270V (or you can look at it as 380*0.707= ~270V, same difference) and this requires you to use the centertap on the transformer, which is how it was designed to be used anyways.

[surfsup]

Cliff, so the 190-0-190 if I taped off the centertap and installed a FWB with two more UF4007 diodes, I would get 380*1.4 = 532.

I understand that, the bridge is using both sides of the cycle.

So you said "If you used only half of the windings"

I'm trying to figure this one out. To do this, would I use the center tap and one of the 190V secondary leads and tape off the other 190V secondary lead? That seems like it would be a FW rectifier not a FWB though, to me. how do you do that?