This I have thought about. We know that shorting the supply creates zero resistance path between the vdd and gnd. We also know that current will pass through the least resistance path and since there is a zero resistance path, all current will pass through that path based on Ohm's Law. It follows then (if I am not mistaken), all connections to the supply will be considered as open circuit since no current will flow through them (because all current flows through the zero resistance path) and only the supply should be harmed?

  • \$\begingroup\$ Never zero resistance (unless you're dealing with superconductors) - just a very very low (but finite) value \$\endgroup\$ – JIm Dearden Sep 26 '13 at 10:03

Many power supply have built in short circuit protection, therefore it is not always the case that shorting the output of the power supply will harm it. Though it is not advisable to do this even when the short circuit protection is in place.

Your conclusion about the current being steered into the zero resistance path is correct, but you should not conclude that all other connections are open circuits, or that this can't harm devices connected in parallel to short.

Simple example:


simulate this circuit – Schematic created using CircuitLab

We are charging some very big capacitor with a power supply having \$50 \Omega\$ internal output impedance. This impedance limits the current which can be supplied to the cap and the charging process completed fine.

Now you are closing the switch, shorting both the power supply and the capacitor. Lets assume that the supply is fine - it has SC protection in place. However, due to very low resistance of the switch, the discharge current of our big capacitor is huge. The capacitor has some low Equivalent Series Resistance and gets very hot due to high discharge current. This heat causes the capacitor to be destroyed.

A single capacitor is the simplest example I could think of, but there are many more.


Shorting the output of power supply to the ground can damage both the supply and the equipment connected in parallel to the short. The potential damage to other equipment depends on the equipment's internal implementation.

  • \$\begingroup\$ this is a good point. That's why most devices have protection diode. \$\endgroup\$ – Standard Sandun Sep 26 '13 at 10:22

Unfortunately things are not simple as that. Below are a few scenarios that components also get damaged.

  1. Assuming that your power supply is a SMPS supply, then its driver will try to increase its power output and compensate voltage. However with time [milliseconds here] the short circuit will melt and be an open circuit. A transient high voltage spike would transfer to your components (the current should find a path to flow). This is true with high response time SMPS. But no designer could consider on all the parameters.

  2. Short circuits will take take higher inrush current, that would generate EMI that would harm nearby components.

  3. Short circuit will lead power MOSFET/Transistor to thermally run away from its operating point and suddenly breakdown. Such a situation will generate reverse spikes (Don't know why, but I read this somewhere.) Such a reverse spike is bad for LED's and this is a known issue with LED drivers. Common with the SMPS designs without a transformer.

  • \$\begingroup\$ I lately had an accident in a project that consists of multiple separate PCBs and parts fead from the same Power supply. The PS shorted during maintenance and 64 HALL sensors suddenly died. I'm still wondering what exactly happened there and the last thing I could think of is EMI. Now I wonder, could simple decoupling caps (bulk and fast non polarised) somewhat protect from such cases? I'm thinking of what I'll change to protect the new sensors before I buy all of them new... \$\endgroup\$ – Julian F. Weinert May 10 at 21:28

Not true. First, it is never Zero Ohms. That is only in perfect conditions. Every piece of copper adds a little resistance. If it was Zero Ohms, no energy would be wasted into heat.

Some electrical components are very sensitive to both current and heat. Essentially, most components can be compared to fuses. They can handle only a certain amount of current going through them before they pop. The tiny wires and circuit paths inside an IC, depending on the IC, might only handle a few tens or hundreds of mA. Digital Potentiometers often have maximum wiper current in the 5mA or less range. As will LEDs, transistors, or even wires and pcb traces.

Spot Welding is essentially a high current short circuit created on purpose. Now imagine that happening inside an IC you just short circuited.

But, shorting to ground might not cause the Power Supply to die, if the power supply has protection, or if the part where the short occurred dies first. A short circuit in a microcontroller with a max current of 200mA and a power supply of 10A could easily blow the microcontroller away fast enough that the power supply would never exceed a few Amps.


You need to consider the whole current loop.

enter image description here

The short circuit is just across the output terminals but there are other SERIES circuit elements (most at risk from damage) connected in the circuit path such as resistances, transistors, diodes etc. These will produce voltage drops within the loop and so it would not be true to say that the rest of the circuit can be just "considered as an open circuit".


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