If you short a bench supply, where the power go? - Page 1
Forum section: electronics / beginners

author: alank2
author_rank: Super Contributor
timestamp: July 17, 2020, 01:48:19 pm
content: I've seen some that recommend you do this to set the maximum current. I connected the positive/negative leads of my DP832 through my DMM to calibrate it. I can visual what happens if you have a load on it like a huge resistor, but without that, is there anything consuming the power? Assuming leads are shorted and you have the current set for 500mA, 500mA is moving from the positive to the negative, but with nothing to hinder or resist it, is the power unrealized?

author: greenpossum
author_rank: Frequent Contributor
timestamp: July 17, 2020, 01:54:38 pm
content: I assume you are setting it to constant current mode. Everything has a resistance however small. The power dissipated in that element (wire, connector, meter) is given by I^2*R. Now if the PSU is a linear type, then the power will be dissipated in the series regulator. If the PSU has means to throttle the generated voltage, say it's fed from a switching power supply, then the power supply will generate enough voltage to overcome the loop resistance for the given current.

author: VooDust
author_rank: Regular Contributor
timestamp: July 17, 2020, 10:25:47 pm
content: What actually happens when you set it to 500mA max is that it reduces the voltage as much such that only 500mA will flow.E.g. when you simply connect the leads and the leads have 0.5 Ohms resistance, the bench supply will set itself to 0.25 Volts (watch the display!). There is always some resistance somewhere, even in a piece of wire.Thus the power is always "realised" or dissipated, as we call it. Some of it is dissipated in the internals of the bench supply itself (look up "source resistance" if you want to know more).

author: alank2
author_rank: Super Contributor
timestamp: July 17, 2020, 10:43:02 pm
content: Thanks guys - that helps me see it!

author: golden_labels
author_rank: Super Contributor
timestamp: July 18, 2020, 01:25:33 am
content: Look at the schematics of even the simplest power supply. Put a short across the output. Think what happens.If it’s a linear power supply and you short it, the power gets dissipated mostly in the transistor that controls voltage.There is always some minimal resistance in the connectors and wires, but it is minimal and — taking VooDust’s example to draw the picture — the voltage drop will be minuscule. For 0.25V and 0.5A, that will be 0.125W dissipated. This is negligible. But the supply is powered by much higher voltage. If 0.25V is dropped on connector/wires and it’s a 30V power supply, 30V - 0.25V = 29.75V is the voltage drop across the transistor(1)and the power lost on it is 14.875W. If the current is 5A, the power lost is nearly ~148W.Keep in mind, however, that many power supplies have a multitap transformer. So the actual voltage it uses may be be much lower, down to a few volts. That includes even the cheap Chinese linear supplies. In such a case the power lost on the transistor is usually between a few two dozen watts.____(1)A bit more, because the actual supply voltage must be a bit higher than the maximum output voltage. There are also at least sense resitors, so a bit of power is lost on them. But this is not important for the big picture.

author: tautech
author_rank: Super Contributor
timestamp: July 19, 2020, 09:38:43 am
content: Google "PSU foldback current limiting".

author: max_torque
author_rank: Super Contributor
timestamp: July 19, 2020, 10:12:22 am
content: This is a question fundamental to our modern, electrically powered world!Power is, simply speaking, Voltage x Current at any given moment.If you have purely resistive elements, called a Linear circuit, then the current is constant at all points in the circuit and the voltage drops across those elements in proportion to their resistance.Imagine a 'magic' 10v battery, that has no resistance itself, and always supplys exactly 10 volts (a so called "perfect" voltage source). Put two 5 ohm resistors in series across that battery.The current flowing in series through every part of that circuit is 1 amp , because 10volts is being driven into 10 ohm total.The battery therefore is supplying 10 watts of power (it's lifting 0v to 10v, a 10 v change, and has 1 amp flowing, so thats 10v x 1 amp = 10 wattsEach one of the 5 ohm resistors, again, with 1 amp flowing throug it, has 5 watts of power/heat being created, because each resistor has 5v across it, and the total power in the two resistors is 10w, the same as the battery power, so our circuit is balanced (we are obviously also coupling our circuit up with 'magic' wires here that have a zero resistance themselves!)So, lets look at a PURELY linear PSU. Lets assume it has 240v coming in from its AC mains supply, and you set it to have 1 amp constant current output. You short the ouput together with a 'magic' zero ohm bit of wire as the 'load'. That zero resistance bit of wire, has zero volts across it, so disipates zero power, and therefore the voltage dropping across the PSU is 240v (we have 240v in, and zero volts out, so that's a 240v drop) and being totally linear, if 1amp flows in our 'load' then we also have 1amp from our supply. So the PSU is now having to disipate 240 watts!!! yikes! that's 1/3rd of a horsepower!This, clearly a problem, and the reason that in our modern world, very very few circuits are purely resistive any more! The answer to the problem was to use magnetism to convert between two different voltage levels without very much resistance. Using either an AC transformer, or a high frequency type switching circuit (called a DC/DC converter) we can now drop voltage in a NON linear way, because we are not using pure resistance to impeed the voltage.A typical bench PSU used to have a large AC transformer in it, that took in the 240V input, and had a range of different windings on it, meaning depending which winding was connected you could magnetically reduce the input voltage to a say 5, 10, 20 or 30 volts. A series of relays was used to select the most appropiate winding for the output voltage at any given moment, and a much smaller linear voltage controller was used to do the final regulation on the output. This is why old transformer PSUs would "click" when the output voltage changed, as the PSU tracked that voltage and selected the most appropriate transformer tapping to supply it.If you take our original case, our magic zero ohm load and our 1 amp constant current setting, the psu would 'click down' to its lowest 5v transformer range, and now the linear stage instead of disipating 240watts, would have to get rid of just 5 watts! (plus in reality, a bit from the loss of that large transformer) That's a BIG improvement. If we ignore the transformer losses for the moment, the current being pulled from the AC supply is no longer the same 1amp as the load is being delivered, thanks to that transformer, it is now just 20 mA! (5 watts is 0.02 amps at 240v)These days, DCDC converters have become more controllable and are a cheaper alternative, and a lot of PSU's may only use a switching type circuit, with no AC transformer, and not have any purely resistive elements in the chain at all. These trade high efficiency for output noise and ripple (anything that switches introduces a frequency onto the output)

author: Zero999
author_rank: Super Contributor
timestamp: July 19, 2020, 10:34:58 am
content: In a nutshell, the power supply doesn't regulate both the voltage and current simultaneously, which is impossible. It has two modes: constant voltage and constant current and flips between the two, depending on the resistance of the load. Set it to 1V, 1A and it will be in constant voltage mode for load resistances above 1Ohm and constant current mode for load resistances below an Ohm.

author: JustMeHere
author_rank: Frequent Contributor
timestamp: July 19, 2020, 07:35:30 pm
content: Into magic smoke if you're unlucky.

author: AG6QR
author_rank: Frequent Contributor
timestamp: July 19, 2020, 08:26:14 pm
content: An ideal Constant Voltage power supply loves an open circuit, because it dissipates no power when delivering zero current. It hates a short circuit, because it requires infinite power to maintain voltage across zero resistance.An ideal Constant Current power supply loves a short circuit, because it dissipates no power when delivering zero voltage. It hates an open circuit, because it requires infinite power to maintain current across infinite resistance.Most power supplies we deal with behave more like constant voltage sources than constant current sources, but not all are that way.A typical bench power supply behaves (almost) like an ideal voltage supply when driving a high-resistance load, and (almost) like a constant current supply when driving a low-resistance load. So it's never called on to provide infinite power. It loafs along, providing zero power, when faced with an open circuit or a short circuit. It has to work at its hardest when the load resistance is such that it's called on to deliver its maximum current and its maximum voltage.

author: paul8f
author_rank: Regular Contributor
timestamp: July 25, 2020, 08:12:16 pm
content: Everything has some resistance. Short the output, wait a few minutes, and see what gets hot!

author: Brumby
author_rank: Unknown
timestamp: July 26, 2020, 07:08:20 am
content: Quote from: paul8f on July 25, 2020, 08:12:16 pm** Sometimes that could be a few seconds.

author: paul8f
author_rank: Regular Contributor
timestamp: July 28, 2020, 08:07:44 pm
content: Quote from: Brumby on July 26, 2020, 07:08:20 am.... or even milliseconds!

author: TimFox
author_rank: Super Contributor
timestamp: July 28, 2020, 09:39:29 pm
content: I personally prefer CV/CC laboratory power supplies where a pair of lights indicates as the supply switches between CV and CC modes. As mentioned above, you can't demand both voltage and current values simultaneously with a given load, and the power supply control chooses which of the two program values (max V or max C) is the more stringent limit on the output and adjusts the supply voltage accordingly. Absent such indicators, you need to look at the voltage and current meters to determine the mode.

author: tautech
author_rank: Super Contributor
timestamp: July 28, 2020, 09:53:58 pm
content: Quote from: TimFox on July 28, 2020, 09:39:29 pmOr have a PSU that displays Set and Out valuesandhas CV/CC LED indication.They do exist.