What's the difference between a dead short and a high current draw?

[skillz21]

Let's just say I have a power supply that is rated for 10v, 5 amps. What is the difference between me connecting a load that will draw 5 amps from it, and just shorting the output?

[ataradov]

The obvious answer - the dead short is a load that consumes infinite amps.

If your power supply does not have constant current mode or overcurrent protection, then it will overload and probably die.

[alsetalokin4017]

What's the difference between a dead short and a high current draw?

The color of the smoke?

[alsetalokin4017]

 the dead short is a load that consumes infinite amps.

The amp-eater?            

[skillz21]

That makes sense, Thanks!

[Brumby]

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.

The obvious answer - the dead short is a load that consumes infinite amps.

This is the direction my points were leading to, but infinite amps requires zero resistance and that would require a superconductive current path.  However, for a theoretical discussion of an ideal system, this is correct.

[Mechatrommer]

Let's just say I have a power supply that is rated for 10v, 5 amps. What is the difference between me connecting a load that will draw 5 amps from it, and just shorting the output?

just to add to the clarification... load that consumes 5 amps from 10V is 2 ohm connection (V = I x R), dead short is theoritically 0 ohm. V / R = 10 / 0 = Inf or NaN.

[alsetalokin4017]

Amps are not consumed!

Don't believe me? Then connect an (infinite range of course) ammeter (not an amp-eater!) on one side of your "infinite amp consuming" direct short, or any other load, and another meter on the other side of it. If the load is consuming amps... why do both meters read the same?

[ataradov]

Amps are not consumed!

You must be fun at parties. It is a colloquial term.

[GeoffreyF]

Amps are not consumed!

You must be fun at parties. It is a colloquial term.

Colloquial engineering / physics is a thing?  The only time I've considered infinite amps is when I was thinking about hypothetically achieving time travel, but I digress.   Current is limited by the design of the power supply a fuse, or the current it takes to produce the heat when it ends in fire and/or melting.   I think this answer would be more fun at a party.  Remember this for safe parties: Take the lamp shade off the lamp, at least unplug your prop, before you put it on your head.

[aheid]

The obvious answer - the dead short is a load that consumes infinite amps.

I feel that's the wrong way to look at it. Instead you should think of the actual load as having nearly zero resistance. In that case, the internal resistance of the PSU (PCB traces, capacitor ESR, cabling etc etc), is the effective load.

As mentioned, what happens to the power supply depends on the design.

[ArthurDent]

The obvious answer - the dead short is a load that consumes infinite amps.

The obvious answer - the dead short is a load that TRIES TO consumes infinite amps. (fixed it) 

The original post might need some further explanation. My take on the question is based on my having several lab type bench supplies that are adjustable voltage and current and the way to set the current is to short the output and adjust the current limit. I could also set the voltage to 10 VDC unloaded, connect a 2 ohm, and slowly adjust the current knob up until the current read 5 A and the voltage returned to 10 VDC.

If you had the typical SMPS that is built into equipment rated 10 VDC 5 A, you probably wouldn't be asking the question because the current limit, if it has one, would be fixed just slightly over the maximum current of 5 A.

[macboy]

The short answer is: the difference is the voltage across the load, and hence, the power delivered to it.

Your ideal maximum load of 2 ohms will have the output voltage at 10 V and 5 A of current delivered, consuming 50 Watt. The power supply is operating normally and is in voltage-controlled mode (not current limiting).

Assuming it's a current limited supply, if you apply a dead short (let's say a length of wire having 0.01 ohm resistance), then the supply goes into current-limiting (current-controlled) mode, and the current delivered will still be around 5 A, but the voltage across your 0.01 ohm "load" will be 0.05 V. The power delivered will be only 0.25 Watt. Note also the maximum output current will be higher than the rating, usually 5% to 10% higher. Also, remember this assumes that the power supply has active current limiting of some kind, but if it is not current limited then all bets are off. A traditional power transformer can deliver several times its rated current into a short, for example. 

Also note that power supplies usually have relatively large capacitors on their output, and in the case of suddenly applying the "dead short", the spike of current delivered from the capacitor alone will be significant, and can be high enough to damage the connectors/contacts which completed the connection. It can also cause small wires or thin PCB traces to open.

Another consideration, especially for a linear supply, is power dissipated in the supply itself. A 10 V linear supply may have a transformer & rectifier that makes about 15 VDC unregulated (at full load), to generate the 10 V regulated output. At full load then, about 5 V is dropped by the regulator, so at 5 A output, that's 25 W of heat dissipated in the regulator pass transistors. With a (current limited) short circuit, all 15 V are dropped across the regulator, so about 75 W is dissipated; that's 3 x as much. This could damage the supply if it is not designed for continuous operation into a short circuit. For this reason, some supplies have "foldback" current limiting, which will allow the full rated current when the output voltage is normal, but limits the current to a lower value when the output voltage drops (due to a low resistance load like a short).

This is also why a traditional lab or bench supply with a "current limit" control should generally not be used as a current source or current-controlled power supply. They are designed for low output impedance, which is the opposite of what's required for a current source.

[pelule]

I am a bit confused - why asking same item in several differents posts?
    https://www.eevblog.com/forum/beginners/whats-the-difference-between-a-dead-short-and-a-high-current-draw/msg1936558/#msg1936558
   https://www.eevblog.com/forum/beginners/safe-to-test-stability-of-power-supply-board-by-shorting-output/msg1941466/#msg1941466
   https://www.eevblog.com/forum/beginners/how-to-measure-power-in-watts-in-a-power-supply-with-current-limiting/msg1941697/#msg1941697
/PeLuLe

[KL27x]

^ Despite all the answers already posted, it is possible the OP still doesn't have the answer to his question. 

[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]

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.

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.