RE: Full-Bridge -vs- Half-Bridge

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Tom Luttrell PWRCOM" <tom-at-pwrcom-dot-com.au> 

Assuming identical switching devices, the only advantage a full bridge
has is that it can reverse the polarity across the load (the switching
devices are in series in an H bridge so the maximum current limit
remains the same).

This would mean double the peak AC voltage is available at the same
current (i.e. you can drive the same amount of current through a higher
impedance load with the same input voltage).

For a square wave V peak = Vrms therefore two times the peak voltage at
the same current will give 2 times the power out (P=IV) for an H bridge.

For a sine wave V peak = SQRT(2).V and thus the power advantage of a
full H bridge in this case would be 1.4 times.


 > -----Original Message-----
 > From: Tesla list [mailto:tesla-at-pupman-dot-com]
 > Sent: Friday, 25 June 2004 10:49
 > To: tesla-at-pupman-dot-com
 > Subject: Re: Full-Bridge -vs- Half-Bridge
 >
 > Original poster: humanb-at-chaoticuniverse-dot-com
 >
 > I posted this last week but didn't get any answers, so
 > I'll try one last time ;-)
 >
 > On Fri, 18 Jun 2004 12:52:22 -0700 (PDT),
 > humanb-at-chaoticuniverse-dot-com wrote:
 >
 >  >
 >  > Hi, can someone tell me just how much more power a
 >  > H-bridge can process over a Half-Bridge? It's not just
 >  > 2X as much is it? It isn't that simple, is it? Also,
 >  > what would be the best way to calculate the optimum
 >  > Capacitance in the voltage divider for a Half-Bridge?
 >  > Currently I am using two 1uF caps, but when running a
 >  > resonant tank one may want to adjust this, so I would
 >  > like to find a way to calculate the optimum ballance.
 >  >
 >  > Thanks,
 >  >
 >  > David Trimmell
 >
 >
 >
 >