Re: Calculating the drain I - SSTC

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Original poster: "Steven Ward" <srward16-at-hotmail-dot-com> 

Sue,

There is a small trick to help reduce that cross conduction.  Use a 10 ohm 
resistor on the gates of the fets, but, in parallel with that resistor 
place a 1n5918 (schottkey, same ones in your 442X gate driver 
protection).  Put the Anode towards the gate.  In this manner, there will 
be 10 ohms of resistance for the RISE times, but the schottky will short 
out for 0 ohms of resistance on the FALL times.  This should help with any 
cross conduction at the expense of adding the 10R to slow down the rise 
times.  Im using this method for switching my big 80N50 mosfets because 
they turn ON faster than OFF, so there is a chance of a bit of cross 
conduction if one fet turns on before the others turn off.  Also, there 
will be more ringing on the falling edge of the wave because the 10R is not 
dampening it, so make sure to employ some 15V back to back zeners to clip 
any negative (or possibly positive) voltage spikes.  Good luck!

Steve


>From: "Tesla list" <tesla-at-pupman-dot-com>
>To: tesla-at-pupman-dot-com
>Subject: Re: Calculating the drain I - SSTC
>Date: Sat, 27 Dec 2003 20:27:42 -0700
>
>Original poster: "S.Gaeta" <sgtporky-at-prodigy-dot-net>
>
>Hi Terry,
>
>Thanks again for the calculations. I am building the circuit on vector
>board, and last night I thought it was rediculous that I used 16 AWG
>stranded wire to go from the H bridge to my terminal block which will
>connect to the 10 AWG primary (it's about a 3" run). I was going to double
>up on the wire, but I guess with 4 amps, it's not really going to be a
>problem the way it is.
>
>A kind of bizzare thing happened this afternoon. I built the 4420/4429 part
>of the driver, and tested it out by feeding a 200 KHz, 5V square wave from a
>function generator into it. I was very pleased to see that I phased all the
>driver transformer windings correctly, and the MOSFETs will switch when they
>are supposed to. There was no voltage applied to the drains of the MOSFETs.
>I noticed that the drivers were only running very slighty warm. I was
>noticing that I was getting a slight overlap on the two wavefoms because the
>rising, and falling edges were slightly sloped on the output of the driver
>transformer, but not on the input of the driver ICs. This really troubled me
>as this slight cross conduction will cause big problems later when I power
>the MOSFETs. I was trying to think about how I would be able to alter the
>duty cyle but, my thoughts were interrupted when my waveforms suddenly
>turned to crap because one of the drivers stopped working. I was powering
>the 4 chips with 15 volts DC, and they were only drawing a total of about
>200mA. Now that shouldn't have happened! The only thing I can figure is that
>maybe they didn't like impedance of my signal generator. Wait a minute....
>At one point I was varying the frequency all over the place just to see what
>would happen, and if the cross conduction would go away (it didn't).
>Something in the circuit would whistle when I hit 11 KHz (Yea, who switces
>at that frequency!). Sometimes I get a little too brazen, and foolish when I
>play with the little low power stuff.
>
>All the best,
>Sue
>
>----- Original Message -----
>From: "Tesla list" <tesla-at-pupman-dot-com>
>To: <tesla-at-pupman-dot-com>
>Sent: Saturday, December 27, 2003 3:02 PM
>Subject: Re: Calculating the drain I - SSTC
>
>
> > Original poster: Terry Fritz <teslalist-at-twfpowerelectronics-dot-com>
> >
> > Hi Sue,
> >
> > The #10 wire is 0.1019 inches in diameter.  The skin depth is 0.00553
> > inches.  So, we can find the conductive area and the resistance.
> >
> > area = pi x r^2
> > outside = 3.14159 x (0.1019 / 2)^2 = 0.008155 square inches
> > inside = 3.14159 x ((0.1019 / 2) - 0.00553)^2 = 0.006481 square inches
> >
> > 0.008155 - 0.006481 = 0.0016739 square inches of conductive area.  That is
> > the same area as #17 wire (.0016089) which has a resistance of  0.00516
> > ohms per foot.  At 4 amps, every foot of wire will dissipate
> >
> > 0.00516 x 4^2 = 0.08256 watts
> >
> > It is really not that exact, but it should be sort of close.
> >
> > So if I have not messed up anywhere, your wire should stay cold and have
> > only tiny loss.
> >
> > Cheers,
> >
> >          Terry
>