First Lm3886

[PJPro]

OK. Thanks for the advice.

[akis]

Originally Posted by Atilla

Akis, I've got a laminated core transformer like that, we estimated it at 15-17VA, 18V. It gets hot as well, very slowly, no matter what the load on it is. I was wondering the same thing but it's been 1/2 a year now and it doesn't seem to be troubled at all by this. It's powering my headphone amp, which is running almost all the time

Oh, and there is a very good line filter before the transformer. Apparently it makes no difference.

I was told by Farnell that these are encapsulated, and they trap the heat in, and as a result reach 70 C easily.

My "mistake" was that I placed it too close to a relay (4mm) and now the relay gets warm unnecessarily.

[PJPro]

I've been doing a little testing this evening. Interestingly, it was more of an experiment to support some statements I had made elsewhere on a forum I frequent.

I'd be really grateful if someone could take a look at the thread and reassure me that my argument is sound. If it isn't I need to make my apologies and eat some humble pie!

The debate is in my build thread here. Skip to the middle of page 6 where I describe the working of the power supply module. The debate builds from that point and covers theory, first principals and practical experimentation to attempt to arrive at the truth. However, my arguments do not sit well with a electronics man of 30 years experience. If I'm talking crap, tell me.

[akis]

First, your explanation is incorrect (but I sped-read it). You achieve the negative -35 Volt DC rail line, not by using the negative cycles of the

wave, but rather by using a separate winding, another 25 V AC secondary which in conjuction with the first secondary gives you a 50 V AC - or expressed

in peak-to peak as it should be, +/- 70 V or 140 Volts peak to peak. By rectifying this voltage you then get +/- 70 V DC (excluding losses) and having

a centre tap (the two leads of the windings together) you achieve a 0 point.

The AC voltage is a sinusoidal and it reaches a peak, positive and negative. It can also be described as "voltage peak to peak" which is much easier to

read off the oscilloscope. Eg your 25 V AC secondary is a +/- 35 V shape and you say "this a 70 V peak to peak". I prefer peak to peak because it saves

making conversions all the time when I read things off the scope. My signal tone generator also displays peak to peak, not RMS.

RMS is about energy - the area under the curve. Use RMS when you want to do power calculations for components. A square wave's RMS would be its peak as

the area under the curve has no gaps, a triangular shape would be half the peak, because the area of a trinagle is half of its enclosing rectangle, and

the area under the sinusoidal is 0.7 times the area of the square (it has been decades now, but I bet the integral of the sinusoidal function must be

a/sqrt(2)).

When you FULLY rectify an AC you convert the negatives to positives, you flip whatever is under 0 to the top. You have posted a picture that shows that

process in your other thread. The peak is the same on the positive side, it has not changed, but now you are getting twice as many peaks on the

positive as you did before, and nothing at all on the negative.

This new shape is not AC: it is a heavily modulated DC. The energy it carries is still the same (the RMS value).

You then feed that voltage into the filter capacitors, which charge to the peak and then discharge into the load. If there was no load, then one cycle

would be enough to charge the caps to the peak value. You would then measure 35 volts even after disconnecting power. 35 volts is the peak reached.

When there is a load however the capacitors discharge into that load, so they reach the 35 volts momentarily and then fall gently as the load is eating

up their charge, until the next cycle where they would charge pretty rapidly. That seen on the scope is a sawtooth pattern, I have posted a picture

hereabouts of what it looks like. The peak-to-peak extend of that ripple depends on the current being drawn and the size of the capacitors. Unless you

have really smoothed down and eliminated that ripple, you cannot really call that "DC" neighter can you give it a value, it is all approximate.

The above explanation ommits various effects, eg the voltage drop across the bridge rectifier and the energy spent on it.

Another way to think about this is that "25 V AC RMS" does not really exist, it is a shorthand notation to describe "70 volts peak to peak in a 50Hz

sinusoidal shape".

[Ted205]

the dc voltage value after the rectifier is the Peak AC voltage value

so if you're starting with 25V ac rms the peak will be something like 25VACrms x (root2) = 35VDC so peak to peak will be x2 = 70

[akis]

With a 25V AC secondary I got 77 Watts RMS into 7.5 Ohm before clipping. It does get hot and I have added a fan (one fan to 3 chips).

[wintermute]

Originally Posted by PJPro

wintermute

I took a rather different approach to my LM3886 project I did a P2P. The thread of my (well most of it) journey is here --> [sorry I've lost the damn URL...it's was truncated when I saved it]. There were a few other threads where I asked specific questions before and after starting the build but that pretty much documents the build process

Stuey

PJ, here's a pretty good PDF about ground loops etc. if you have the time to read it!

[Again, the damn url is truncated!]

----------------------------

Don't suppose wintermute and Stuey still have those URLs available?

Hi PJ Yep here it is
Wintermute's Gainclone

also here is a link to your upgrade island post First LM3886......... They got merged into the relevant forums a week or so ago, but were a few pages back... you need to dig to find them but the date range was between I think about 15th to 22nd Aug

Tony.

[Redshift187]

I think you are mistaken about doubling the voltage for + and -. That comes from using two secondaries and two rectifiers. DC voltage is approximately AC peak voltage for a bridge rectifier, which is RMS * sqrt(2) = RMS * 1.41. Minus 1.4V for diode drop of course.

[PJPro]

Thanks guys. Some interesting replies. I not sure I meantioned doubling of the voltage for + / - but two of you highlight it....so that impression must have come from somewhere.

Anyway, I feel a lot more confident that my statements are indeed correct.

[PJPro]

A few pictures from my other thread showing some preliminary testing...


The basic rig.


Checking voltage from the power inlet


Checking voltage from the transformer


Checking voltage from the power supply board

I did the use the light bulb tester. But nothing happened i.e. I didn't see it light up. I am assuming that this is good.