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Old 20th May 2006, 01:39 AM   #1
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Default Class G and switch mechanism

Okay, for a class G rail switching amp, a friend of mine is considering to use a power mosfet as a switch in order to switch the high voltage rail on/off. The Vgs voltage will be raised to about 10V via an optocoupler mechanism.

==> What kind of optocoupler output should he use? Photovoltaic seems appropriate here because it provides a floating output voltage.

==> When the transistor switches from on to off or off to on, how time critical is the driving mechanism (to ensure that the FET doesn't hang around in active mode too long and burn power)? Will an optocoupler switch fast enough for this?
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Old 20th May 2006, 02:02 AM   #2
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Hi.
The thing you are talking about is Class-H not Class-G...
The Photvoltaic optocoupler couldnot be used because the gate of the Switching Mosfet requires current magnitude much larger than supplied by the photovoltaic coupler....

For more info read My thread "The Class-H Amplifier" carefully...


The Class - H Amplifier

Cheers,

K a n w a r
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Old 20th May 2006, 04:11 AM   #3
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Thank you Mr.Workhorse,

Yes, I see there is a bit of a nomenclature difference (I was taught that class G is a rail switcher and class H is a rail tracker). The technique I am referring to is the rail switching amp.

I saw in the thread your schematic with the rail switch. Here are my questions:

You use the TLP250 to drive the N-channel FET switch. I understand that pins 5 and 8 of the TLP250 are the power pins and determine the maximum and minimum voltage that the output of the optocoupler (pin 6) assumes.

==> How are pins 5 and 8 receiving power by the upper-most power supply? If I am looking correctly, pin 8 (the positive voltage reference of the optocoupler) is getting coltage from the top source through the resistor and diode. However, in the return path when current flows out of pin 5, current is going against the diode that is in parallel with the RC snubber network.

==> Concerning the n-channel mosfet switch itself, how can you be sure that when the optocoupler is turned on (and the Vgs of the FET increases) that the FET will act as a switch and not be in active mode? I see that the TLP250 will put a positive Vgs on the transistor, but how can you be sure that Vgs is greater than Vds? If Vgs is less than Vds, isn't the FET in active mode?

Thank you for your time, and I am truly interested with your work!
-RT
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Old 20th May 2006, 05:26 AM   #4
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Hi RT

==> How are pins 5 and 8 receiving power by the upper-most power supply? If I am looking correctly, pin 8 (the positive voltage reference of the optocoupler) is getting coltage from the top source through the resistor and diode. However, in the return path when current flows out of pin 5, current is going against the diode that is in parallel with the RC snubber network.

The Pins 5 & 8 are connected to Bootstraped power supply consists of Series Resistor->Fast Recovery Diode->16V Zener paralleled with 220uFD cap draws current from Upper Tier positive Rail...
During Down Swing from Peak positive value of output voltage of amplifier[Assuming Tier Transition from Higher to lower] , it starts decreasing and inturns creates a return or discharge path for the PIN 5 because the Switched Bus is connected to the Power output devices of amplifier itself...Also when the amp is in idle posistion the idling current through output devices automatically defines the return path for the optocoupler supply and inturn it gets charged to 16V peak....


==> Concerning the n-channel mosfet switch itself, how can you be sure that when the optocoupler is turned on (and the Vgs of the FET increases) that the FET will act as a switch and not be in active mode? I see that the TLP250 will put a positive Vgs on the transistor, but how can you be sure that Vgs is greater than Vds? If Vgs is less than Vds, isn't the FET in active mode?

Again this actions corresponds to the voltage provided by bootstrapped supply which provide approximately 15-16Volts to the photocupler IC to deliver it as instantaneous Vgs to the gate of FET and it acts as a Switch saturated to its limits...
In my case the upper tier voltage was +-150VDC and lower Tier was +-75VDC and bootstrapped voltage derived from it was 16VDC, which rides Floating with respect to Source of FET and always keeps its elevation higher than Vds when FET is Turned ON....

Hope this helps......

BTW: Do you understand what is Bootsraping or not.....I think you dont understand it.....

Cheers,
K a n w a r
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Old 20th May 2006, 05:01 PM   #5
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I think I understand your circuit now:

When the circuit is using the low supply rails, the bootstrap capacitors are charged.

When the output voltage of the amp increases and requires the high voltage supply rails, the bootstrap capacitors act as voltage sources to keep the output of the optocoupler (and thus the gate of the FET switch) above the drain voltage.

Another question if you would:

For the bottom switch on the negative supply side, I would assume that it works in the same way as the top switch.

==> When the bottom switch is turned on for the larger supply rail, I understand that the gate-source voltage of the bottom switch can still be held at ~16V by bootstrapping. Now the drain voltage of the bottom switch is floating (it is not tied to the supply rail like the top switch is), so how can you be guaranteed that the bottom switch won't be in active mode?

Thank you for your answers, I am learning a lot!
-RT
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Old 21st May 2006, 03:54 AM   #6
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Hi,

It seems to me that you lack certain basics in electronics....

The Drain Voltage is not Floating , When FET is OFF then the drain voltage is held at the lower negative rail voltage but when the FET is switched ON and the Drain voltage is switched to larger negative voltage, The FET is Turned on as "Switch" because its source is held at constant voltage and gate to source Turn-ON VGS is referenced to its source, Drain has nothing to do with it....

BTW: I think you have more of the interest in this project than your friend has....

Cheers,
K a n w a r
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