If one diode has a forward voltage drop of 0.7 volts, and you put a second diode same type parallel, will it change the forward voltage drop to half 0.7V (0.35V)?

This need not be a long discussion, just a curiosity. I COULD go downstairs and test such a device, but I'd rather make jello for the grandchildren.

 

No.
The diode with the lower forward voltage conducts.

 

The forward voltage drop is a band-gap energy level thing, not a resistance function. Thus the voltage will stay close to the same. BUT there is also a small resistance componant in the forward drop and the voltage due to that will change a bit. But you may not see it on your meter, because it is SMALL!

 

Thanks @MisterBill2 and @MrChips.

Just like LED's the lower Vf will conduct. Like I said - it was a dumb question. But the answer makes perfect sense when I think about it. WHEN I think. Isn't often sometimes.

 

If one diode has a forward voltage drop of 0.7 volts, and you put a second diode same type parallel, will it change the forward voltage drop to half 0.7V (0.35V)?

Nope. If the two diodes are absolutely identical, resulting in them sharing the current equally, the voltage drop will decrease somewhat, perhaps by a few dozen millivolts. But if they aren't identical (and they never are), the reduction in voltage drop may be more or less than that.

Putting diodes in parallel is such an unpredictable crapshoot that it's rarely done.

Found this graph of some experimental data in my files. It shows forward voltage (vertical axis) for an assortment of different diodes as a function of forward current (horizontal axis), from 100 pA to 10 mA.

 

Even with a difference in forward voltage one will conduct more than the other but not take all the current.
Simulation below shows the current sharing between two 1N4148 small signal diode for a total current of 10mA, with the simulated forward drop difference going from 0 to 100mV.

Of course the higher current diode will start to hog more of the current if the current is high enough to cause it to significantly warm it, which lowers its forward voltage.