The negative volts on feedback into LTP is due to output mosfet driver circuit not being symmetrical.
It only need 1.4 volts to turn on lower mosfet but about 5 volts to turn on upper mosfet.
I use a pot in LTP current mirror to adjust DC offset.
This results in negative voltage on LTP at idle.
It only need 1.4 volts to turn on lower mosfet but about 5 volts to turn on upper mosfet.
I use a pot in LTP current mirror to adjust DC offset.
This results in negative voltage on LTP at idle.
Because Gm of the mosfet increases significantly at higher currents, the effective Ft also increases at higher currents. (not the case with BJT of course
) This may lead to a clean output signal with no instabilities until the right conditions occur. Then under load a RF oscillation burst may occur and cause all sorts of havoc, particularly device failure and magic smoke dissipation.
A gate to drain Zobel will prevent this from happening. Small, like 150R in series with 47pf or so, mounted right at the device pins.
Small, like 150R in series with 47pf or so, mounted right at the device pins.
I have empty spaces on the pcb for snubbers close to mosfets.
It was originally suggested 33R and 220pf.
Yes, it would be relative to the size of the internal capacitances'. I'm just so used to using smaller sized mosfets.
I find to get it optimized, it takes playing with the values a bit and that of the gate stopper as well. But as I learned from Bob, the gate drain Zobel works really well for taming mosfets.
Hi Nigel. You make no mention of load testing yet. I'm not suggesting you test an amplifier to destruction but your OP poses the question "300 watts RMS out of 2 mosfets ?" so why not consider a test to at least 200W and post the results as something that logically needs doing to answer your own OP (I know from a previous thread that you have at least one massive heatsink lying around there somewhere) .
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Nigel, there is no such "thing" as WATTS RMS.
Wattage cannot be a root mean square number...Period!
The sq root of 2 multiplied by the sq root of 2 is NOT root 2 but simply the digit 2.
You go no about modelling with a sinewave and you semm to forget that most people, maybe not yourself listen through real world speakers of which the majority are moving coil.
Now I guess that you are not wanting to use the 300w amp to drive tweeters and so I presume woofers would be the load.
Think about phase angle and how it typically doubles the heat dissipation in any amplifier (Class A excluded).
A woofer typically has a phase angle of about 45 degrees and this causes the doubling of heat dissipation.
Nigel, there is no such "thing" as WATTS RMS.
Wattage cannot be a root mean square number...Period!
The sq root of 2 multiplied by the sq root of 2 is NOT root 2 but simply the digit 2.
You go no about modelling with a sinewave and you semm to forget that most people, maybe not yourself listen through real world speakers of which the majority are moving coil.
Now I guess that you are not wanting to use the 300w amp to drive tweeters and so I presume woofers would be the load.
Think about phase angle and how it typically doubles the heat dissipation in any amplifier (Class A excluded).
A woofer typically has a phase angle of about 45 degrees and this causes the doubling of heat dissipation.
Your answering to a 3 year old thread.
I worked all that out years ago !
Moved on now, on to USB PC scopes.
True if you are literal but you arrived late to the party, everybody and his brother has known that for decades.
Yet "watts RMS" has been used for decades (and continues being used) by all kinds of Companies, even "serious" ones run (or advised) by Engineers who do know their Math.
So? ...... why does it still happen?
Because the implicit idea is that "Watts RMS" is shorthand for "Watts calculated using RMS Volts"
I write this only because *regularly* somebody "rediscovers the wheel" and posts it again.
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Actually im working on similar project
Can this Onsemi / Fairchild Mosfet FDL100N50F 100A - 500V 2500W Power Dissipation
12000PF input Capacitance - Total Power dissipation 5000W 1 pair
work in quasi configuration with snubber for 500 - 1000W Classs AB @ 4 Ohm
Rail 90V DC
heavy Mosfet low price 10 EUR pcs.
Can this Onsemi / Fairchild Mosfet FDL100N50F 100A - 500V 2500W Power Dissipation
12000PF input Capacitance - Total Power dissipation 5000W 1 pair
work in quasi configuration with snubber for 500 - 1000W Classs AB @ 4 Ohm
Rail 90V DC
heavy Mosfet low price 10 EUR pcs.
You will NEVER be able to dissipate any more than 250-300W/device, even if You keep Your heatsink 25C! This FETs are developed for switching, not for linear application. The case is the limiting factor.
Sajti
I find it helps because it implies that the manufacturer is not using PMPO or some other made up measure to claim higher power output than the amp is actually capable of.
ok
have one more question, how about when using Mosfets developed for Linear operation ? 1 pair IXTK90N25L2 or IXTK11020L2 for 1000W Class AB
IXYS have Linear Mosfets for heavy Linear Use
IXTK90N25L2 or IXTK11020L2
250V - 90A - 960W Power Dissipation
N-Channel Enhancement Mode, Extended FBSOA
Applications
Solid State Circuit Breakers
Soft Start Controls
Linear Amplifiers
Programmable Loads
Current Regulators
Same. TO-247, TO-264 package can dissipate maximum 250-300W if You mount them isolated on heatsink. But this is still theoretical value, as I calculate it with 25C heatsink temperature. If the heatsink is 50C the maximum dissipation is 200W.
If no isolation to the heatsink You can dissipate 50-70% more. Using more smaller output devices will reduce the heatsink size, and give better reliability.
Sajti
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