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Old 17th February 2005, 12:38 AM   #1
B.I.G is offline B.I.G  Romania
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Default mos-fet amplifier question ...

would this circuit work ? i see no thermal compesation ..... so wouldnt it blow the output devices ? i see the author recomending irfp240/9240 that are not lateral devices.....
http://users.swing.be/edwinpaij/ampli_mosfet_simple.htm
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Old 17th February 2005, 01:00 AM   #2
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That is basically one of the most common Mosfet configurations around, and does work quite well. Thermal compensation isn't required as Mosfets have a negative temperature coefficient, meaning the current reduces as heat increases thereby reducing the heat.

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Old 17th February 2005, 01:08 AM   #3
B.I.G is offline B.I.G  Romania
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i thought only lateral mos-fet`s have negative temperature coefficient
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Old 17th February 2005, 03:11 AM   #4
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According to IRFP240 datasheet, "on" resistance increases with increasing temperature, so therefore more heat less current.

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Old 17th February 2005, 03:46 AM   #5
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But that's resistance when fully turned on. Under quiescent conditions they exhibit a positive temperature coefficient. You can see this in the graph of Vgs vs Id, which shows zero tempco at about 15A, positive below, negative above.

You can get away with no thermal compensation if the devices are very well cooled and the source resistors are large, but I wouldn't be happy doing it myself, especially with only a resistor to set the bias.
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Old 17th February 2005, 07:17 AM   #6
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The output stage works quite well, but if I were using thr IR devices I would replace the trimpot wuth a VGS multiplier.

Jam
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Old 17th February 2005, 07:24 AM   #7
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It will be work well only with lateral devices .
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Old 17th February 2005, 08:11 AM   #8
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Hi everybody,

I think there is need for something little more on switching vertical mosfets.

These mosfets do need thermal biasing compensation only when there is a high bias [Id~100-200mA]is implemented , but when a low biasing [Id~10-20mA]is used there is no need for any thermal compensation.
Similarly when the mosfets are subjected to heavy loads i.e. large current drawing loads, their RDS starts increasing due to the phenomena of positive temp. coeff. and the current starts decreasing which ease in paralleling and current sharing.


Secondly , the effect of applied rail voltage is very evident in terms of Vgs~Ids criteria.

regards,
kanwar
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Old 17th February 2005, 10:33 AM   #9
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workhorse, I think it's the other way around - temperature coeficient is negative at small values of Ids, but once a certain threshold is reached, it becomes positive. Lightly biased MOSFET stages need thermal compensation, but once the bias gets high enough they so not.

From this thread: http://www.diyaudio.com/forums/showt...108#post576108

Quote:
Originally posted by Nelson Pass
[snip]

In Class A with a good sized bias, the temperature coefficient
gets less, and as the current increases it swings the other
way. This point occurs at a value somewhat higher than where
Class A amplifiers are usually biased, but you can use much
smaller Source resistance as the current goes up, not only
from less temperature coefficient, but also because the resistive
voltage drop of the resistor becomes a larger part of the bias
voltage and is linear.

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Old 17th February 2005, 10:43 AM   #10
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To any newcomer who want to build this circuitry,be careful with the layout,unless u want to build a power oscillator
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