IRFP240/9240 -Vbs,Drivers ??? - diyAudio
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Old 29th March 2005, 06:31 PM   #1
zox2003 is offline zox2003  United States
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Default IRFP240/9240 -Vbs,Drivers ???

OK..

I found a lot of different answers here on this question:

Do we need to have Vbs in amplifier with IRFP240/9240 or not(trim pot is just OK)?

Do we need to have drivers like in bipolar amps to drive these MOSFET's?

Can somebody give me answers from his working amplifier?
I have a plan to build amp with pair of these.How much power I can get?

Thanks
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Old 29th March 2005, 07:14 PM   #2
Did it Himself
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I think you mean a Vbe multiplier? The answer is yes, you do need one as IRF devices are positive temperature coefficient. A pot alone is not sufficient.

You can get away without drivers providing your VAS is running at high enough current. I like to try and use drivers.
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Old 29th March 2005, 08:49 PM   #3
ilimzn is offline ilimzn  Croatia
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You DEFINITELY need a Vbe multiplier for the IRF(P) parts. They have a positive temp. coefficient up to about 50-75% of maximum current - and I'm sure you don't want your idle current THAT high That being said, they have a lower temp. coefficient than BJTs so your Vbe multiplier needs to be 'degenerated' by inserting a small resistor in the emitter of the compensating BJT or it will overcompensate. You can also ose a resistor in series with the Vbe multiplier but that will make it difficult to adjust (effective temp coefficient will change with idle current adjusment). Finally, you could use a Vgs multiplier (MOSFET version of the Vbe multiplier).

As for drivers, depends on what you want to do. If you are not looking for extreme speeds, you may get away with driving them directly from a relatively high current VAS (10-20mA, more than that may get impractical due to power dissipation on the VAS which needs larger and usually slower transistors). For absolutely top speed, you need a driver stage. Also, a driver stage will sometimes reduce distortion as the MOSFET capacitances depend on voltages between it's electrodes, which means your VAS sees a nonlinear capacitance, which is a source of distortion. A driver stage will reduce the apparent capacitance seen by the VAS.
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Old 29th March 2005, 09:25 PM   #4
zox2003 is offline zox2003  United States
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Default Example please...

Can you post example of corect Vbe ( with resitor) please?
What kind of driver configuration is the best for driving MOSFET's?


Thanks
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Old 30th March 2005, 12:07 PM   #5
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Smile Hi buddy

Hi ZOX,

As you have access to complementary mosfets, you may use the following combination:

Input stage = dual differential 5mA per device equals to 10mA per pair with current sources.

VAS = cascoded loading for high speed with idle current around 40 to 60 mA depending upon the parallelling of mosfets.

no additional driver stage is necessary.

But the supply rails of the differential and VAS must be atleast 12 volts higher than output mosfet rails[better if regulated also].
use a symmetric VGS multiplier .

regards,
kanwar
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Old 30th March 2005, 12:16 PM   #6
Did it Himself
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kanwar,

What are your thoughts on using a MOSFET as the VAS so that you don't need to run the LTP at such high current?
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Old 30th March 2005, 12:37 PM   #7
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Default Hi buddy

Quote:
Originally posted by richie00boy
kanwar,

What are your thoughts on using a MOSFET as the VAS so that you don't need to run the LTP at such high current?

Hi ritchie rich,

I have once constructed VAS using mosfets. Driving the mosfets in VAS through gate relatively confers to increased high frequency distortion and phase-shift.
Either use a folded cascode or full cascoded loading when using mosfets in VAS.

regards,
kanwar
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Old 30th March 2005, 01:09 PM   #8
zox2003 is offline zox2003  United States
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Default HIgher voltage rails???

Mu power supply is +/- 42V .I do not have option to increase rails for input stage and Vbe.Is there other possible combination that will work with no problems?

Help,please.
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Old 30th March 2005, 01:47 PM   #9
Mr Evil is offline Mr Evil  United Kingdom
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Default Re: Hi buddy

Quote:
Originally posted by Workhorse
...use a symmetric VGS multiplier.
Why symmetrical? I've only used those in very specific circumstances because they tend to give lower performance than using a single transistor, the only advantage being the availability of a midpoint (and aesthetics, maybe).
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Old 30th March 2005, 02:47 PM   #10
zox2003 is offline zox2003  United States
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Default schematic?

Any good and simple but working schematic with 240 and 9240?
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