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Old 16th December 2010, 07:35 PM   #721
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The purpose of overload protection, is to protect itself from high currents due to short circuit or too low impedance on output. If you can take care of never short-circuiting or never using lower impedance load, then you can neglect it of course if you want to.
The clipping indicator won't show any of these, and won't protect from any of thes. But its useful, because driving an amp to clipping can even take it to high temperature (if you design it to take only normal audio signal), and continous output power can burn your speakers.
If you do a calculation and the price and time to make an overload protection is more than you spend for substituting the bad FETs (if no other thing fail, exept a fuse), then you can even say its better to leave it - as the errors are rare. (i dont use any overload protection for my class b amps, the last failed 3 years ago because a drunk guy cut the speaker wires with wire-cutter...)
I am working on overload protection too, but dont want to use the original shunt method.
My idea is now to use a current transformer. A simple T37-2 can take 50 amps with 1 turn without saturation, and its secondary with enough thin turns could directli connected to SD pin. But its only an quick idea now...
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Old 17th December 2010, 02:43 PM   #722
opor is offline opor  Thailand
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Hi all,

I have try T150-60 from Micrometal and EI33 from my PC but core are still very hot. It seem that we are facing the same problem.
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Old 17th December 2010, 05:20 PM   #723
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Quote:
Originally Posted by opor View Post
Hi all,

I have try T150-60 from Micrometal and EI33 from my PC but core are still very hot. It seem that we are facing the same problem.
Material -60 is not a good choice, you must use cores with much less permeability, like material -2 or material -6. Are you sure its an EI33 core? Isn't it EE33? How much air gap did you used? What are the wire resistances of you inductors? How much is the switching freq? How much very hot is?(approx in degrees, for me very hot is the boiling vacuu-oil in the lap, i can easily get out things from boiling water)
With the original switching freq of 140kHz and 50kHz low pass, almost any core will be hot, just calculate it... if there is no sufficient air gap, or bad core material, it will saturate...
You can calculate the induction in the core with the following:
B=mű0*műrel*n*Imax/(l_core+műrel*l_gap) (where mű is the greek letter for permeabilities, mű0 for vacuu, and műrel for core, n is the turns, Imax is the current, l_core is the path lengt in the core, l_gap is the length of gap ) put everything in in SI, so you will get it in Tesla. It should be lower than 0,3T.
Then calculate the copper losses in the resistance fo the inductor (best is to simulate).
If you dont do any calculation and design, just try everything without mind, how do you think you will have the luck to make a good amp?
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Old 17th December 2010, 05:23 PM   #724
opor is offline opor  Thailand
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The more turns the less heat. Why? 100uH is colder than 30uH??
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Old 17th December 2010, 05:34 PM   #725
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Quote:
Originally Posted by opor View Post
The more turns the less heat. Why? 100uH is colder than 30uH??
Its more complicated than you say.

1. More turn on the same corn gives higher induction (from currents) in the core, so it will increase core losses, and it can even saturate. (you must see datasheet, but B<0,3T is a good start)

2. But bigger inductance gives better filtering (decreases corner frequency), so decreases ripple current, so if the resistance is the same, it will have less copper losses

3. But more turns with the same wire thickness gives more resistance, so more copper losses.

4. more turn decreases induction from voltage ripple on the core.

You must calculate all the losses, because you should minimise the sum of them, usually this involves finding which is significant. . Usualy the increase in the one gives decrease in the other...
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Old 17th December 2010, 06:04 PM   #726
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Lorylaci:

I know few people who use current transformers and its good idea especially for big power but for AB clas with Si transformers with Ei core

What about cliping inidicator from my previous post? Or do you have something else that can show me when is maximum from amplifier?

And this, how much power transformer I need for 400W amp (this amp) for class D, i ask because for AB class transformer must to be double of amp power, class A 5x of amp power...understand me in any case, what I ask you

Thank you
Regards
Nikola
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Old 17th December 2010, 07:51 PM   #727
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Quote:
Originally Posted by Dzony988 View Post
Lorylaci:

I know few people who use current transformers and its good idea especially for big power but for AB clas with Si transformers with Ei core

What about cliping inidicator from my previous post? Or do you have something else that can show me when is maximum from amplifier?

And this, how much power transformer I need for 400W amp (this amp) for class D, i ask because for AB class transformer must to be double of amp power, class A 5x of amp power...understand me in any case, what I ask you

Thank you
Regards
Nikola
The current transformer thing is only an idea now, im doing calculatons, simulations, then test it on breadboard, so i will post updates whether its a bad or good idea...

The best is to use a led (or déprez if you prefer) vu or peak power meter, and calibrate its 0dB to max undistorted output. It will tell you enough
(attention its only measures output voltage, it doesnt care its 2 ohm or 200 ohm)

The transformer depends on your philosophy (and purse):
- All the power all the time: the bigger the better
- Lightestcheapest: the one which produces acceptable heating while operating
- Economic design for normal operating conditions: 6dB crest factor is the minimum that feels like only hard but not really distorted (some clipping). So if you plan for example 2x400W sinus undistorted, that means 2x800W peak power, and for 6dB crest factor 2x200W "normal max average output power". The transformer you need will be 2x200/nű (greek letter for efficiency, >0,8 is easy to get), so 500W should be good. If you use enough caps, and a good 500W transformer, that should handle the musical peaks. If you drive the amp harder it would be audibly distorted, so otherwise care should be taken not to drive harder, nowadays every big transformer has a bimetal switch, so its protected from overheating. The calculations are the same for class-b but with nű of 0,5 about... (class-a is other, cause it drives a lot of current all the time)

If you have money, and dont care about more weight (spare more on fitness-room ticket), and especially if you cant look over the vu (or peak power) meter, you can always use bigger transformer.
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Old 17th December 2010, 09:56 PM   #728
Pafi is offline Pafi  Hungary
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Quote:
Originally Posted by lorylaci View Post
Its more complicated than you say.

1. More turn on the same corn gives higher induction (from currents) in the core, so it will increase core losses...
Laci!

Core loss is caused by fast induction change, not just induction, and higher inductance means lower induction change (at switching freq, which is the relevant).

Quote:
2. But bigger inductance gives better filtering (decreases corner frequency), so decreases ripple current, so if the resistance is the same, it will have less copper losses

3. But more turns with the same wire thickness gives more resistance, so more copper losses.
You can continue: with same thickness the resistance increasas with N, but inductance increases with N^2, so ripple current copper loss will be less (1/N^3), but the same wire may doesn't fit, must choose A2=A1/N, so this loss will be 1/N^2.

Quote:
4. more turn decreases induction from voltage ripple on the core.

You must calculate all the losses, because you should minimise the sum of them, usually this involves finding which is significant. . Usualy the increase in the one gives decrease in the other...
Yes, there are two contradictional form of losses, but they are the switching freq. and audio freq. losses. It's simply because at fsw there is a constant voltage, and at audio freq the current is constant (I mean not dependant of inductance).

Back to opor's question:
Quote:
The more turns the less heat. Why? 100uH is colder than 30uH?
Basically yes, at idle. Because (dB/dt)^2 is proportional to core loss. But at full load always take care of saturation! Higher RDC can increase audio freq loss, but this is generally insignificant because of high crest factor of audio signal.
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Old 17th December 2010, 10:00 PM   #729
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I found increasing the switching frequency reduced the core temperature.

I also found adding extra capacitance lowered it a bit too.

I can only guess that this moved the switching frequency further away from the LC resonant frequency at which the LC is a short circuit.

Moving the switching frequency from 140KHz to 220KHz reduced the core temperature by 40 degrees C.
The resonant frequency of my LC was 40KHz.
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Old 17th December 2010, 10:38 PM   #730
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1. Pafi, you are right! I knew Ploss~dB, but I knew it wrong then.
2. 3. Hmm. I thought it would be N/N^2*N, so I was surprised, but I calculated it and really, Ploss~1/N^2. But corner freq moves - so it would be same to recalculate output LC to a lower corner freq.

But its more complicated than just "more turns less heat". The last few cores noted here are surely saturated! And you even should calculate idle losses, and load conditions. For me idle temp should be lower than +10°C to ambient, and even a normal clipping (the 6dB crest factor) should be max +50°C to ambient

nigel... at 140kHz of switching freq everybody should lower CORNER FREQ (not resonant freq, if real resonant freq from the parasitic capactiances of the inductor would be 40kHz that would be bloody hell) of filter. For a 140 kHz switching freq the LC corner freq should be lower, which means larger inductors and capacitors... byebye tiny full range class-d amp.
I did increase switching freq over 200kHz, like nigel, so more filtering. If you mean you increase LC filter capacitance you move corner freq too.
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