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dboomer 
Member - Posts: 11
Member spacespace
Joined: August 21, 2007
Location: United States
Posted: August 24, 2007 at 8:24 AM / IP Logged  
I would suspect the PS couldn't handle the 2ohm load on the output. Your pictures didn't show the output transistors. How many pairs are there? There is generally 2 PS fets for every output pair. With 8 PS fets you should have a maximum of 4 pairs of output transistors per channel. Also considering its made in China, the PS transistors and output transistors may not be properly matched...they may have stuck whatever was cheapest and readily available.
csound 
Member - Posts: 18
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Joined: July 27, 2006
Location: South Africa
Posted: August 28, 2007 at 8:26 AM / IP Logged  
The one channel amp has six output transistors on the "P" side and six on the "N" side (class AB). With eight FETS per power supply (Dual power supply).
The amp is rated 1ohm stable by the manufacturers. How do they "match" power supply and output transistors? Do you mean that the IRFZ44 FETS can't handle the current that the output transistors can = 2SC5200 and 2SA1943.
Marius Minnie
dboomer 
Member - Posts: 11
Member spacespace
Joined: August 21, 2007
Location: United States
Posted: August 28, 2007 at 10:06 AM / IP Logged  
That would mean a 2.4:1 ratio (2.4 PS fets to each output pair (1 NPN and 1 PNP)) which would be good for a 4ohm load. I am guessing that the output is probably working off 50 to 60 volt supplies (pos and neg). That means the transformers would have a 4.5 to 5 turn ratio putting tremendous load on the Z44’s. Z48’s would have been a better match. I believe they over rated the capabilities of this amp (as with most amps). If you wanted to repair this amp (if the board is not charred too bad) and run a 2 ohm load, I would install Z48’s. They handle more current and I have had success in interchanging them without altering the circuit. Although I guarantee nothing…lol.
dboomer 
Member - Posts: 11
Member spacespace
Joined: August 21, 2007
Location: United States
Posted: August 29, 2007 at 9:20 AM / IP Logged  

I did some calculations on that amp. It produces 750 w/rms at 2 ohms so take the square root of 750x2 which equals 38.7, then multiply by 1.4 to get peak voltage 54.2, then multiply by 1.25 to compensate for efficiency loss(output stage only, which can even be higher depending on the circuit) now your around 68v or higher that the PS has to produce. Now divide 68 by 14.4 (battery voltage which could be from 12 to the 14.4 that I generally use) and now you have the transformer turns ratio of 4.7 @14.4v or 5.6 @12v. Now we need to figure the amperage of the output stage. Take you 750w / 68v and your at 11 amps. Multiply 11a x 4.7 and then compensate for efficiency loos of the PS of 1.2 and your at 62 amps of 14.4 volts. Now divide that by 16 (the number of z44's) and you have each one having to produce 3.9 amps. Multiply 3.9 by 14.4 and your at 56 watts per fet. The data sheet shows that these fet will dissapate 94w cold and only 50w warm. And considering that there are 16, the chances of one being weaker than the rest is quite high. I would say this is/was a great 4 ohm amp but not a 2 ohm. Also take into consideration that these figures are at 14.4v and very high effifiency, actual load on the fets are probably much higher. Hope this helps.

csound 
Member - Posts: 18
Member spacespace
Joined: July 27, 2006
Location: South Africa
Posted: August 30, 2007 at 3:53 AM / IP Logged  
The best posts so far on this topic, thanx. It will take me some time to figure out your calculations and formulas, but it will be worth my while to study and remember it. Thank you for the help, appreciate it.
Marius Minnie
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