As you can see in the picture, the switch is set to the off position.
When I test the continuity, there is no beep.
However, when I test the outgoing voltage, it is 18 Volts D.c.
You did not specify the input voltage or if you had a load on the ouput.
Let me make a wild assumption. YOu are using a digital voltmeter and
there is no load on the output. The digital voltmeters draw such little
current you are probably picking up a leakage voltage due to the
insulation being damp, an arc over that is leaving a carbon trace, or
just weak insulation. If you put a very small load on it,the voltage
should go away.
People run into this all the time with the 120 volts AC in houses. You
can almst sitck one end of the meter in the wall socket and hold the
other lead in your hand and show a voltage.
Without some kind of load on it, you can't trust voltage readings on a
digital meter. For reference, even a "good" analog meter imposed about
20,000 ohms per volt on the scale, (10v scale was 200,000 ohms).
I worked in a very large plant and there was so much wire in the
conduits the digitals were almost worthless for electrical power trouble
shooting. Used the Simpson most of the time. I think Fluke has came
out with some meters that will place a load on the circuits now to help
eliminate the problem.
I knew I had the breaker pulled on a circuit but was getting a shock.
Even the simpson was showing about 70 volts . That was even with the
wires disconnected from the breaker. I discovered if I started off on
the highest Simpson range and stepped down the meter would stay in the
same relative position if it was the induced voltage.
Rig up an adapter that lets you put about 10 meg ohm across the leads
and it will work fine. For 99.99% of the users, it should be inside
the meter anyway. It is only when you are probing CMOS or similar
circuit that inserting a load that small into the circuit will give
you a bad answer. Even then all CMOS pins are supposed to be driven or
Some companies have adapters like that.
However 10 megs is not nearly low enough. That is the impedance of most
better quality meters. The adapter should be much lower, probably a few
thousand ohms at the most.
Where I worked we had so much induced voltage in the wiring that I made
a tester out of a 7 watt small light bulb as we had thousands of them on
the equiipment as indicators , so had lots of spares.
On Saturday, April 6, 2019 at 9:39:40 AM UTC-5, Ralph Mowery wrote:
I found this.
This is an old-style unregulated "wall wart" adapter. The internal circuit
is probably this, and nothing more:
simulate this circuit – Schematic created using CircuitLab
The transformer is designed to fail in a safe manner (no external flames, n
ot too much smoke and isolation between mains and output maintained) if it
develops a shorted turn or is fatally overloaded.
The reverse leakage of the diodes in BR1 is quite low at room temperature a
nd the electrolytic capacitor (perhaps 10,000uF rated at 16VDC) itself typi
cally has not much leakage, so the charge can be maintained for minutes or
more with nothing connected.
More modern switching supplies (as well as linear regulated adapters) typic
ally drain down much faster.
If you're connecting such an adapter to a circuit of your own making, it's
worth remembering this as the adapter can easily hold enough charge to dest
roy something if it's connected incorrectly. With it pulled out of the wall
socket, you can short the output with a wire and most of the charge will b
e removed, but a bit of voltage will creep back due to the way capacitors b
ehave (dielectric absorption effect).
On Saturday, April 6, 2019 at 10:08:41 AM UTC-4, A K wrote:
So to summarize from your last two posts:
The switch is on the primary side.
You're measuring the 18V at a power supply output, with the switch
turned off and it's apparently the charge left inside the power
If you had included that important info initially, the responses would
have likely been different.
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