Electrical – Resistance between VCC and GND drops to 3k Ohms

When I've heard about power supply "isolation testing" before, the context has been a test done on the mains inputs to a system or device. I've also heard these tests called "insulation resistance" tests or "hipot" tests, though those could be slightly different tests. The test confirms that voltage applied to the mains input won't connect to the case of the UUT and cause a shock hazard.

The test involves applying a fairly high voltage (1 to 5 kV or so) to verify there's no spark gaps or insulation breakdown paths between the mains and user-accessible surfaces.

According to this, isolation testing tests

The maximimum voltage (ac or dc) that can be continuously applied from the input to output or input to case of an isolated power supply. ... Testing power supply isolation requires specialized equipment (Hipot tester) and can be destructive.

The main point is that this is a test you would do on the mains input of an isolated power supply. It is not a test you should be doing on the inputs where your operating circuit expects to receive its local regulated input power.

I nearly always put such a 'fools diode' in my circuit, even when I am breadboarding. Reverse power will kill most of the chips I use. Over-current protection is a function for the power supply, not for the powered circuit. This makes more sense:

• most power supplies are current limited in the first place, think of basic batteries, and 7805-like regulated power supplies (but NOT NiMh accu packs!)

• Having the protection in the attached circuit will not protect against an error in the circuit, or the wring to the circuit, so it must still be complement by a protection in the supply

• I build much more powered circuits than I build power supplies, so it makes more economic sense to put the current limiting in the supply.

When I use a battery (NiMh etc) that does not inherently limit its current I add a fuse, polyfuse or the like (often the battery supplier has already done this). A series resistor is often not practical because it makes the voltage the circuit sees dependent on the current it consumes.

You circuit is fed by 9V, most likely it is intended to be fed by a 9V battery, which is current limited by itself (unless maybe when it is an NiMh 9V pack? or are those protected in any way?) If you are really worried I suggest you either

• put a fuse or polyfuse between the battery and this circuit, or

• as your circuit probably draws very little current and probably can live with a small voltage drop, put an schottky diode in series (an 1N5819 will drop ~ 0.4V at 100mA). This circuit was probably designed long before schottky diodes became common place.