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WastefulMiser:
You need to study the characteristics of semiconductor diodes. There is not a constant forward voltage drop. There are two major factors for a given part number. These are forward current and the junction temperature.
A quick illustration on a small sample of 1N4148s. Two diodes from same batch. Ambient temperature about 70 F. As the current thru the diode is changed its junction temperature changes. Thus, one waits a short time after changing the current to measure the voltage.
1N4148 is a small signal diode.
Current . Diode 1 Diode 2
0001 MA 0.614 V 0.615 V
10.4 MA 0.728 V 0.727 V
106. MA 1.024 V 1.014 V
233. MA 1.069 V 1.053 V
An MR752 is a larger diode, higher current rating
235 MA 0.72 V then added some heat to one lead 0.63 V.
Really the voltage drop is related to current density per unit area, the junction temperature, the material, and processing. Total current is current density times chip area.
You need a characteristic curve for a particular diode to determine current vs voltage, or at least one point and general information of the curve shape.
Some sites to look at:
http://en.wikipedia.org/wiki/Diode_modelling
http://www.physics.csbsju.edu/trace/CC.html
Your next problem is load sharing between the power supplies. You need to know what the peak current load on the supplies will be. Can one supply safely and reliably support this peak load. If so, then you really don't care about how the load is distributed. Otherwise there needs to be a load balancing method that includes the diodes.
How well regulated does the voltage need to be after the diodes? This will greatly affect how you do load sharing.
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