Most likely you have the newer, worse model Virtual Ground power rail, cheapy meters. Make sure you don't allow the (-)/(+) power lead to come in contact with the (-)/(+) probe of the meter.
Also, put a cap and a small load across the wallwart, most likely have one hell of a ripple on it and over voltage.
My guess would be that the battery eliminator is a switching type that needs some sort of a load to start up. Does it weigh less than a few ounces? If so, it's almost certainly a switcher.
Use Ohm's Law to determine a proper resistive load, if it's less than
100 ma. Put an appropriate resistor across the output and then measure it. A light bulb of the proper size (which you must determine by the output capacity) would be easier to obtain if the output is more than
100 ma.
Post the milliampere rating of the supply here and someone will help you figure the proper load.
Measure the output of the eliminator when it's connected to the HF meter. My guess is there's no output from it due to the load being too small which is common with SMPS types. A non SMPS one should work ok - that will be heavier than a similar output SMPS one. Or add a load large enough to switch the SMPS on - an LED and resistor selected to draw 20mA should be enough. 390 ohms in series with a standard red LED should be ok. Make sure you get the polarity of the LED correct.
--
*Rehab is for quitters
Dave Plowman dave@davenoise.co.uk London SW
To e-mail, change noise into sound.
To clarify (I think): Many digital meters require that the power supply be isolated from the circuit being measured.
And, make sure the battery eliminator actually measures 9 V without a load, since the meters won't be much of load.
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Yes, for a single meter AND with neither power supply connection also attached to one of the meter inputs. And assuming he didn't try to measure the wall wart's voltage with the meter! :)
-- sam | Sci.Electronics.Repair FAQ:
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