why do two diodes in parallel not share the current load?
They may, but the current sharing depends on:
The forward voltage of the diodes
The "dynamic resistance" of the diodes (dVf/dI)
They don't? The shares may not be equal but how could they not share? You mean the current goes all through only one of them?
Yes, 1 device tends to hog most of the current. Diodes have a highly non=linear I-V curve, i.e. Id = Is*exp((Vd/Vt) - 1). Important to know is that "Is" is a strong function of temperature, and varies among parts. Due to mismatches in parameters, as soon as 1 diode draws more current, it gets hotter. Then Is goes up with the higher temperature, which increases diode forward current Id further. Then temp goes higher, Is increases, etc. It is a slim chance that 2 paralleled diodes share current 50%/50%. Possible, but very unlikely.
If, however, each diode is connected to a small valued series resistor, then the diode-resistor pairs are connected in parallel, the resistors force current sharing. Does this help at all?
"Sharing" does not imply 50-50. As long as one portion is not zero, they share.
and since transistors are also nonlinear this accounts for their observed behavior as well?
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