As was suggested earlier. A 12 V car battery.
Remember this thing is an AC clipper or should drop about 6V or -6V depending on polarity. It's a shunt regulator, so there HAS to be a load.
If you use a 12 V bulb and a 12 V battery, then if the regulator failed shorted, the bulb would be at full brightness. If it opened, it would not be lit.
If it works correctly, the12 V bulb would act as if 6V was applied to it. i.e. 6V across the bulb and 6V across the regulator.
You cant really break anything using the 12V bulb and a 12 V supply capable of powering the bulb.
That's what I did (but no bulb) and all I got was battery voltage.
I could ask the same questions, what battery voltage and the bulb voltage? If it was 6V, then the behavior could be OK, but it doesn't tell you if it's actually working.
The regulator won't drop exactly 6V. So, a if the regulator was working at 7V and the battery was 6 V the regulator would be off.
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Tests could be performed with a 9V battery or a 12 V wall wart but you would have to use a resistor for the load and size it appropriately.