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I have this cheap DC power supply.

  • Tekpower DC Variable Power Supply, 1.5-15 V @ 2A, HY152A

enter image description here

It says it is limited by 15 V or 2 A. My question is, if I short it out, can it damage the unit? Is there anyway I can hurt myself with this?

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    \$\begingroup\$ @the power supply link - "Ideal for tattoo use" Amusing! \$\endgroup\$ Feb 27 '12 at 11:12
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From the information given it is likely but not certain that it is short circuit protected.
A site referenced below says that it is.
Caveat Emptor (and look inside - see below)

It is quite likely, based on cost and common practice that it uses a version of the LM317 regulator, which is short circuit protected. Your supply voltage output has a lower limit of 1.5 volts, whereas the LM317 allows output from 1.25 Volts and up. (Given the analog meter output and the worst case output minmum of 1.3V the claimed 1.5V minimum matches the LM317). There is a "big brother" LM350 regulator available but current is above 2A.

SO - if you are comfortable with looking inside your supply * and there is an LM317 visible then the supply is protected against short circuiting. * - Probably 4 screws underneath the case - with that style of case they are probably near the rubber feet and possibly also holding the feet on.

The supply is advertised on some sites as suitable for use wih tatooing machines - probably because the voltage and currrent suits the motors used. The way they are used probably operates the supply in current limited mode, also suggesting safe short circuiting.

If it does not contain an LM317 it still MAY be short circuit safe - and the site below says it is.

This site says that the HY152A power supply is short circuit protected. They claim -

  • Protection: Reverse polarity protection, short-circuit protection.

Load / voltage / Current:

Fairly obviously (hopefully) if a supply limits current to 2A when the load gets excessive, then the voltage supplied MUST drop. Failure to do this would violate Ohms law and break most of the laws of physics coincidentally. ie

  • I = V / R

For a given current, if you reduce R (= heavier load) then Voltage must reduce proportionately. If you increase load (decrease R) and voltage and current stays the same then Scotty gets upset ("Y' canna break the laws a' Physics!)

If you want current limiting and your load cannot tolerate low voltage without damage for some reason then you need an alawrm or a complete cutoff, regardless of how good your supply is


LM317: The LM317 naturally current limits first just by "running out of steam" - there is only so much current that it is able to pass. Then in most cases, the voltage drop across the regulator x the current will cause heating and temperature rise. If the heatsinking is not able to limit temperature rise to an acceptable level the IC has internal circuitry that further limits current so that current will fall to maintain temperature at an acceptable limit. This will usually result in a much lower than 2A current until the short is removed.

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Most benchtop power supplies have a current limiter built-in, but there's only one way to know (apart from trying and possibly damaging the device): read the specifications. If it doesn't have any, don't buy it.
There are two different ways of current-limiting. The simple method just drops the voltage but keeps supplying the full current, which may cause damage to your circuit. A more advanced method is foldback current limiting, where shorting causes to drop the current as well. The power supply needs to be reset to go back to normal operation.
More expensive benchtop power supplies have adjustable current limiting.

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  • \$\begingroup\$ The fact that a supply is overload-protected or short-circuited means the supply should not be damaged by over-current conditions, but it does not imply that output current will be limited except to the extent necessary to prevent damage to the supply itself. A manufacturer may make a 12V 5A supply and a 5V 12A supply which are identical except for the voltage setpoint. If the 12V supply is connected to a 0.42-ohm resistor, it may perfectly happily feed about 12 amps to it. The 5A rating means the voltage won't sag under a current draw less than that, but does not mean... \$\endgroup\$
    – supercat
    Nov 12 '13 at 18:41
  • \$\begingroup\$ ...the supply won't put out more than 5 amps. The supply won't be able to output unlimited current, but the highest-voltage supply in a product line (with the lowest rated current) may be able to output just as much current at lower voltages as could the units with the lower rated voltage and higher rated current. \$\endgroup\$
    – supercat
    Nov 12 '13 at 18:46

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