Transformer choices?

[Mark]

Mikka, lets break it down a bit further.

Why is it necessary to make this calculation and what is its purpose?

(Up / U) x Us = Un => (240 / 230) x 240 = 250V

I understand Vp is the intended voltage applied to the primary winding, I understand V is the main voltage at your GPO (the wall), where does Vs come from?

What is the significance of Vn?

I assume it is the unloaded secondary voltage, but the following quote is a little confusing.

I read a secondary voltage of 240V

I understand this as saying you've measured the secondary voltage.

Another factor I noticed was the efficiency of the bridge. AC current doesn't equal DC current with a bridge rectifier.

Sorry to ask so many questions, but I figure, what is the pointing of posting, if no one knows what you're talking about?

[greekie]

Mikka, I am sorry to burst your bubble on this mass conspiracy of copper waste, but your theory doesn't hold true I am afraid.

It is the matter of two different ways of rectifying the even and odd cycles of the alternating current of the transformer. Not to be derogative or anything, but this is explained in very basic electronics 101's like this:

http://www.electronics-tutorials.ws/diode/diode_6.html

The biggest difference between a full wave bridge and a full wave centertap is that the voltage and power lost in the diodes is less (about half) with the full wave center tap. The voltage is half and the current is double the full wave rectifier, but it requires a center tap on the transformer secondary.

You are not unleashing hidden power by modifying your transformers. You are just switching what we could call "topology", which could also be done by purchasing a different product. What youøve found is that your transformers are more beefy than the minimum requirements, which is always nice.

Cheers

Jake

[Mark]

http://www.altronics.com.au/download/Da ... ormers.PDF

This site states DC current = 0.62 X AC current.

I vaguely recall from college that it was all much as muchness, but I'd still like to hear Mikka's explanation. If I'm incorrect, I'd like to be corrected, and if Mikka is incorrect, I'm sure he'd feel the same way.

Jake, your reference doesn't mention the current limitations associated with a bridge rectifier. In fact most sites only make reference to voltage.

[Mikka]

Mikka, I am sorry to burst your bubble on this mass conspiracy of copper waste, but your theory doesn't hold true I am afraid.

Ok, but then explain me why my tranny does'nt toast while it supplies two EL84 since November 2010 and I use this amp three times by week in rehearsal and on stage ?

You are not unleashing hidden power by modifying your transformers. You are just switching what we could call "topology", which could also be done by purchasing a different product. What youøve found is that your transformers are more beefy than the minimum requirements, which is always nice.

When moving from CT to mount a single coil assembly I have not altered the available power of the transformer. I've never said anything to that effect! What must be understood is that the secondary voltage is increased from twice 225V to 225V at once. Thus by dividing the voltage by two, for the same power available, I double the current available since I use the two windings. It is mathematically simple and logical.

http://www.altronics.com.au/download/Da ... ormers.PDF

This site states DC current = 0.62 X AC current.

I vaguely recall from college that it was all much as muchness, but I'd still like to hear Mikka's explanation. If I'm incorrect, I'd like to be corrected, and if Mikka is incorrect, I'm sure he'd feel the same way.

I don't totaly agree with that factor because it doesn't take care all power supply component. Only one factor is not enough even it can give a global ratio. Be aware that the current peaks generated in the transformer is strongly influenced by the value of capacitors, especially the first. It is therefore impossible to deduce a valid factor for all head-mount capacitor power supply.

We can look at it by taking my simulation in Duncan PSU Designer :
http://www.hiboox.fr/go/images-100/fell ... 1.jpg.html

If you do the total consumption of the amplifier are:
0.075 + 0.0075 + 0.0013 + 0.005 = 0.0888A DC or 88.8mA DC
88.8 / 0.62 = 143mA AC

In the end I find this factor a little high and therefore in a certain way risky.

Now resume what I calculated on the basis of this simulation:

Irms = (Ix0.7)/ (Square root of 2) = (0.360x0.7) / 1.414 = 0.178A or 178mA AC

If I make the connection with the consumption of rectified current I find the following factor.

88.8mA DC / 178mA AC = 0.498

I have a relatively low factor because I have rather large current peaks due to my filter.

By selecting a factor of 0.62, the manufacturer is based on a very positive case which is surprising and risky.

My method is much more demanding and thus allows me to design power supplies more stable and more reliable.

where does Vs come from?

Measurements

What is the significance of Vn?

Unloaded nominal voltage

Sorry to ask so many questions, but I figure, what is the pointing of posting, if no one knows what you're talking about?

I thought I was on a technical forum and I had to do to connoisseurs.

[Mark]

I thought I was on a technical forum and I had to do to connoisseurs.

"and I had to do to connoisseurs" This illustrates my point entirely. I understand english is probably not your first lanuage, and thats cool.

This comment makes no sense at all (what did you have to do to connoisseurs?), but that isn't the problem, the problem is that you assume everyone knows what you are talking about.

Thus I asked you to explain reason for the equations and the values in the equations.

As is evident, I don't know everything about electronics, but I'm opened minded enough to learn.

To be convinced I think I would need further explanation, a whole lot of calculations doesn't do it for me, after all you might have made a mistake and not picked up on it. I would have to prove this for myself.

If this is too much hassle, then lets drop it now while we are all still friends.

[Mikka]

Hi Mark,
I must apologize to you.
Rereading my post I realized how insensitive I was.

You're welcome, and I will try as much as possible to answer your questions.

But this case is that it is possible for some and in some cases to save on the price of the transformer. It seemed that it might interest many people.

I wish to remain as your friend.

[Mark]

Sounds great Mikka.

If you have time could you explain the reason a certain calculation is required and what does it prove. It is necessary to state the components in the equation and where these figures came from.

In short a text book explanation.

Only if you have time of course.

[Mikka]

To define the technical data of a transformer must take few measures and make some fairly simple calculations.

In our case, we already know the nominal output voltage. By far, I noted two other elements, the open circuit voltage (no consumption), and the impedance of the windings. It allows me to calculate the voltage drop in charge, the regulation ratio, and the nominal current.

Knowing the voltage drop and impedance of the windings, it is easy to find the current using Ohm's law.

V = ZI => I = V / Z

Initially I have two CT winding each measure about 170 ohms.
Z = 170 + 170 = 340 Ohm

The voltage drop at rated current charge is as follows.
250 - 225 = 25V

Knowing Z and V so I can now calculate the current rating.
25 / 340 = 0.074A (AC)

I have at this level all the elements to calculate the power rating of the CT secondary high voltage winding.

P = VI = 0.074 x (225 + 225)* = 33.3VA

*Because CT secondary is two secondary windings having a potential difference of 225V twice from one end to another.

Hope can help you !

[Mark]

Mikka, sorry I still haven't looked at this as I want to devote the time to that it deserves.

[Mikka]

No problem Mark.