Electronics > Beginners
What's the difference between a dead short and a high current draw?
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rx8pilot:
A load "tells" the power supply how much current it will draw by creating a path to ground with a given equivalent resistance.

In this case the load is behaving like a 2 Ohm resistor. The resistor limits the amount of current that can flow. Even if the power supply could supply 10,000 amps - when connected to a 2 Ohm load, only 5A will flow.

In a short, the resistance is effectively 0 Ohms so the load has nothing to limit the current. At that point, it will rapidly exceed the capability of the power supply of 5A.

The reaction of the power supply will be to activate a safety mechanism to control the current or the weakest component in the supply will overheat and stop working.
David Hess:
The difference is that a dead short produces a lower output current because your well designed power supply includes foldback current limiting.
Zero999:

--- Quote from: Brumby on November 03, 2018, 05:08:44 am ---If the supply is current limited, the answer is simple.  Same if you have a fuse.  However ... but I assume your question is directed at supplies that do not, in which case things become a little more interesting....

5 amps is not a huge current in itself.  Yes, it is at the high range of the sort of thing you might encounter on your average bench, but it is by no means scary.

A 10V supply delivering 5A means the load will be equivalent to 2 ohms.  If the supply is rated at these specifications, then Ideally) it should be able to deliver that current indefinitely.  Expect some parts to heat up, but they should all operate within their capabilities.  (In practice, this is not always the case - but let's not dwell on that right now.)

Now, increase the load by putting, say, 1 ohm across the output.  You are now trying to draw 10A.  This is a (significant) overload.

In this situation, the ratings of the output transistors may be exceeded and/or their heatsinking may be inadequate.  Failure is on the cards.  In the case of a linear supply (big, heavy transformer) the voltages within the unit will almost certainly sag as the transformer is incapable of supplying the demand.  In the case of a switchmode unit, you could get shutdown or hiccuping.  In either case, power and current ratings of components within a unit that is trying to work will be at risk of being exceeded.  Things will generally get hot and you are very likely to have something "give".

Now let's move onto the "short".

Let's say you stick a screwdriver across the output terminals, presenting an effective load of, say, 0.1 ohms.  This would have the supply try to deliver 100A ... which just isn't going to happen - and something is going to fail.  It's a similar process as the overload mentioned above, but is usually much quicker.

--- End quote ---
Another thing to note is that because P = I2R, doubling the current results in four times the power dissipation in any resistance, so by overloading a power supply by a factor of two, the power dissipation in many of the components i.e. transformer, cabling, PCB traces etc. will increase by a factor of four.
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