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Old 8th January 2007, 07:27 PM   #1
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Default Gate-source voltage needed to switch a FET

According to most datasheets of powerFETS, the suggested gate-source voltage to switch a FET into full "ON" mode is at least 10V, while the max gate-source voltage is 20V (as listed under absolute maximum ratings).

==> Is there much of a difference between switching a FET using a gate-source voltage of 12V, as opposed to 15V, as opposed to 18V? I know the higher the Vgs, the faster the FET switches (and thus switch losses can be reduced), but is there a "law of diminishing return" here? I don't have much practical experience with switching FETs; I wanted to ask before I try anything dangerous...
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Old 8th January 2007, 07:35 PM   #2
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I would say a good Gate-Source voltage would be 15V. This way, no problems.
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Old 8th January 2007, 07:36 PM   #3
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Default Re: Gate-source voltage needed to switch a FET

Quote:
Originally posted by rtarbell
According to most datasheets of powerFETS, the suggested gate-source voltage to switch a FET into full "ON" mode is at least 10V, while the max gate-source voltage is 20V (as listed under absolute maximum ratings).

==> Is there much of a difference between switching a FET using a gate-source voltage of 12V, as opposed to 15V, as opposed to 18V? I know the higher the Vgs, the faster the FET switches (and thus switch losses can be reduced), but is there a "law of diminishing return" here? I don't have much practical experience with switching FETs; I wanted to ask before I try anything dangerous...

The data sheet also shows you characteristics of Id vs. Vgs for a given Vds. You can see that the Vgs actually controls the Id. So, you really have to be sure you have enough gate drive for the expected Id you want to switch. Err on the plus side: too high Vgs doesn't harm (as long as you stay below the max; better use a zener shunt here!).
If the 10V Vgs is enough to turn it hard on, there's not much point in going higher. Going higher does NOT turn it on faster: it turns on faster if you can sink more current into the gate capacitance, so use a driver with (relatively) high currrent capacity, at 10 V.

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Old 8th January 2007, 08:21 PM   #4
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Yeah, that's what I meant!
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Old 8th January 2007, 09:21 PM   #5
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Quote:
Originally posted by N-Channel
Yeah, that's what I meant!

... but it's not what you said

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Old 8th January 2007, 09:43 PM   #6
batee is offline batee  United States
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Default High Side FET drive...

Here's what I did for a FET switching in a motor control project, but it's pretty generic. Works for 0-100% PWM duty cycle.

http://www.diyaudio.com/forums/showt...81#post1088581
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Old 8th January 2007, 10:28 PM   #7
batee is offline batee  United States
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Schematic and board here:

http://www.diyaudio.com/forums/showt...90#post1088590
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Old 8th January 2007, 10:48 PM   #8
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A Vgs higher than 10 volts will turn the MOSFET on faster (with a given gate resistance) because the effective gate capacitance will be charged faster. However, it will take longer to discharge this same capacitance, so the total turn off time will be longer. In addition, the power to drive the gate will increase. All in all, it's best to use the vendors spec. which is typically 10 volts.
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Old 9th January 2007, 12:25 AM   #9
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A higher Vgs usually produces lower Rds (or lower Vce for IGBT) and thus lower conduction losses for any given current, however, the improvement is not always worth the effort or the risk of going near the 20V limit (datasheet has the last word on that). I routinely use 19V gate driver supplies in most of my circuits.
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Old 9th January 2007, 11:27 AM   #10
VEC7OR is offline VEC7OR  Lithuania
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Yes, and reduces useful life of fets, also you have to suck more charge out of gate.
IMO anything over 15V is overkill, but again depends on fets.
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