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In this answer @stevenvh points out that \$V_{BE}\$ for a TIP110 darlington may be as high as 2.8V. This looks extremely high to me, even for a double junction.
In normal use you don't need to drive a darlington with more than a few mA, and I expect \$V_{BE}\$ for the first transistor to be less than 800mV. Then the voltage drop for the second transistor would be 2V! Even when the 2.8V is equally shared that's 1.4V per junction. A 1N4001 needs 20A to get this high voltage drop! I doubt that a transistor will survive 20A of base current.
How can this high \$V_{BE}\$ be explained?

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  • \$\begingroup\$ Good question! I was asking myself the same when I read the value in the datasheet. \$\endgroup\$ – stevenvh Jun 20 '11 at 10:55
  • \$\begingroup\$ Seems to be common in darlingtons, and strongly dependent on collector current. For currents several amperes you often see \$V_{BE}\$ values higher than 2.5V \$\endgroup\$ – stevenvh Jun 20 '11 at 12:34
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They must be just leaving too much of safety margin for themselves.

In the worst possible conditions one may get 1.1V drop on each transistor (-55C, high current), so instead of writing 2.2-2.3 in max they just did 2.8.

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