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Can you put two diodes in parallel in order to obtain the double load?
Wiki User
∙ 14y ago
As a rule, NO!
The main reason is thermal runaway:
When you warm up a diode, its "conductance" at any particular voltage increases.
(The Shockley diode equation gives the exact equation).
Even with two identical diodes in parallel, if the temperature of one diode was even slightly hotter than the other diode, more of the current would go through the hotter one. The one with more current would heat up more rapidly, lowering its conductance even more, and after a short time the hotter diode is hogging most of the current.
The final effect is that a diode will carry almost all the current, while the other stays almost unused.
There are a variety of things that cause two diodes to *not* be identical, which only speeds up the thermal runaway.
*the unavoidable imbalance between the two diodes voltage drops.
*(with an AC signal): the diode with slightly faster turn-on time will absorb more turn-on loss
*(with an AC signal): the diode with slightly slower turn-off time will absorb more turn-off loss
There is an obvious exception: use identical diodes, with exactly the same electrical and themal chracteristics. Individually selected diodes with voltage drop matched to less that one mV, and carefully assembled for thermal matching, may carry the same current, so effectively double the rating.
It is a very unstable condition, like keeping a coin vertical on a table.
Another arrangement is to add current sharing resistors in series to each diode, at least 0.3/0.4 V of additional voltage drop is required, so is a very unefficient solution.
Considering high frequency signals (as in switching power supply) adds even more problems, so the answer is always: use just one diode, leave the task of paralleling only to very skilled people dealing with such high currents that no single diode in the market can withstand.
Wiki User
∙ 14y ago
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