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Old 26th October 2007, 02:24 PM   #1
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Default p-channel fets

How would an output stage made with p-channel together with n-channel fets work massive shoot through ?

I bought same quantity of IRF9540 as i did of IRF540 just to try this.
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Old 26th October 2007, 06:26 PM   #2
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Shoot through largely depends on the timing of the control signals regardless of what type of fets are used for the output stage.
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Old 28th October 2007, 06:29 AM   #3
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Think about this:

Shoot-through happens when the positive rail FET should be turned off and the negative rail FET is turned on before the gate charge is sucked out of the positive rail FET.

Linear Technology has an app note that compares the decaying voltage from the gate charge of the FET turning off and inhibits the other FET from turning on until the first gate drops below conduction threshold (about 2V).

What if your modulator circuit could swing 10V? You could tie the gates of an N-channel FET and a P-channel FET together. The voltage driving the P-channel FET could not rise enough to turn on until the gate charge on the N-channel device dropped below it's conduction threshold.

No shoot through!

How could we swing any real voltage into the load with only 10V on the gates?

Ground the sources of the FETs and hook the speaker hot to the transformer center tap.

A good opamp with a pair of class B followers would have enough muscle to drive the gate capacitance at high frequencies.

Fairchild has a nice 200V P-channel FET with only 0.17 ohm RDSon, or the IRF parts would work with rail voltages below 50V
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Old 29th October 2007, 11:22 AM   #4
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Hi djk, the concept is workable but will result in the PSU having to pass all the switching frequency through it.

I've attached a drawing of a TINA spice simulation I did of a version where the inductor is moved to between the PSU and the FETs, but needs to be wound bifilar. A large value capacitor ac-couples the drains of the FETs.
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File Type: jpg class-d simple4.jpg (36.0 KB, 158 views)
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