6KW class D

K

kiakiaka

New member
I like this chip and the concept,thats why i puted it here for discussion.They r offering solutions with external frequency clock or with internal clock which is not constant.I would like to hear what u guys think which solution is better and why.PCB and everything else is not a problem to make but some discussion can be very usefull before that.
 
S

stewin

Member
Dimonis said:
What are you talking about? 2092 can withstand 200V single or +/-100V supply , but due to pumping and other factors , the safe operating is +/-85V .
So with and ideal PSU of +/-85V you can get not more than 1800W on 2Ohm load.:D

P=U*U/2/2Ohm =85*85/2/2= 1806Watts
Click to expand...

if you bridge it will work o.k like 4kwts at 4ohms at +/-85v
 
Dimonis

Dimonis

New member
stewin said:
if you bridge it will work o.k like 4kwts at 4ohms at +/-85v
Click to expand...

Again P=U*U*2/4Ohm = 85*85*2/4= 3612 Watts and this calculation doesn't include losses on mosfets , output inductors . And the supply must be an ideal voltage source of +/- 85V.

So in reality for this BTL 2500W 4Ohm is a good result.
 
I

ILTON

New member
Hello Friends
Someone already built the amplifier 6k?
he would have the same good results?
Would like your opinions
I'm waiting for your opinions
and when respondelas'll do in Proteus PCB and post here
thank you very much
 
I

ILTON

New member
Ok Dimones you have any suggestions of a Class D amplifier with good power so I can build? since I've been taking a look with the DIYs, amplifiers and other Argentine friends. further documentation incopletas have or are not trustworthy. I say this only remembering the large power amplifiers class d. I know that DIY is even more so just wanted to post something more complete and with good knowledge here in Brazil with parts really easy and very good thank you friends
 
M

Malmir

Member
Dimonis said:
Again P=U*U*2/4Ohm = 85*85*2/4= 3612 Watts and this calculation doesn't include losses on mosfets , output inductors . And the supply must be an ideal voltage source of +/- 85V.

So in reality for this BTL 2500W 4Ohm is a good result.
Click to expand...

Did i miss something here ?
P = U*U/R = 85*85/4. In Bridge You have double supply voltage =>
P = U*2*U*2/4 = U*U = 7225
From where do You get Your "2" and why only 1 time ???
 
Dimonis

Dimonis

New member
I'm calculating RMS power.
85V - is peek (amplitude) voltage . RMS - 85/1.41=60.3V.
In bridge mode the supply doubles , so we have 120.6 RMS voltage on load.
120×120/4=3600W
 
M

Malmir

Member
Then lets feet the FETs with +/- 255V :)or near that. That should be possible for the 4kW only +255 if bridged :).
 
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