On Wed, May 02, 2001 at 04:49:27PM -0700, Alvin Oga wrote: > the diodes need to be power diodes... vs signal diodes > > given you cannot tie the power supplies at two diff voltages together... > you have to isolate it somehow... ( the power diode method )
I'm a bit hazy on how this actually works, but I would guess that each supply's output goes through a diode (forward biased) and the lower potential sides of the diodes would be connected. Thus not much current flows back the other way. > but more importantly, its primary purpose is to allow for the > two power supply at different voltages ( +5.25v and +4.75v ) to be tied > together > > at these extremes... the diodes wont help....and the dioes will simply > burn up due to the current it has to pass to get to that "voltage" > one side being a diode drop ( 0.7v ) across itself.. > - a power mosfet is better suited ... You will need some large diodes and you may well need cooling ie heatsinks. Hamish -- Hamish Moffatt VK3SB <[EMAIL PROTECTED]> <[EMAIL PROTECTED]>