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I'm a bit confused about reading current on a breadboard using a multimeter (I'm a complete beginner with electronics).

I'm using a self-powered breadboard which should deliver 5V / 1A. Multimeter is a Fluke 87V.

Voltage is OK, but when trying to read current on a simple circuit with no component (voltage out -> multimeter -> ground), the multimeter starts by giving a reading of ~2A, continuously decreasing (even below 1A).

When placing a single resistor on the circuit, the reading is perfectly correct, based on the resistor value, and stays stable.

So is this something normal (then why), or am I doing something wrong / missing something?

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    \$\begingroup\$ how are you measuring it "with no component"? Don't tell me you are shorting the multimeter in Ammeter mode to ground!? \$\endgroup\$ – KyranF Apr 24 '15 at 15:52
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    \$\begingroup\$ What is "simple circuit with no component"? There is no such a thing. Is it short-circuit? \$\endgroup\$ – Eugene Sh. Apr 24 '15 at 15:53
  • \$\begingroup\$ I think basically that's what I'm doing. Voltage out -> Multimeter -> Ground. I know I'm a complete newbie, but can you explain the reason why this is happening? \$\endgroup\$ – Macmade Apr 24 '15 at 15:59
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    \$\begingroup\$ You're a complete newbie, measuring with a Fluke 87V? Something doesn't add up here... \$\endgroup\$ – Mast Apr 24 '15 at 18:15
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    \$\begingroup\$ in addition to what's already been said: if the power source is rated 5V / 1A, never load it's output terminals with any resistance lower than Rload_min = Unominal / Imax = 5 Ohm - since that will most probably short the source itself by exceeding the allowed current rating. Instead of pumping the current into the load, the source will heat up its own internal resistance, effectively turning it from power source to a heater... not a good idea unless you want to let its en.wikipedia.org/wiki/Magic_smoke go out for a walk. \$\endgroup\$ – vaxquis Apr 24 '15 at 18:23
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You are using your multimeter's Current Shunt to short out your power supply.

I'm guessing that the power supply on your breadboard is going into thermal limit to protect itself. In other words, because there is a dead short across the supply, the supply is going into current limit.

Many (most?) supplies that are built into self-powered breadboards use linear regulators. One reason is cost, a better reason is that they have dramatically less noise on the output. Having a low-noise power supply is important when breadboarding sensitive analog circuits.

Most single-chip voltage regulators (78xx, 79xx) have thermal limiting internal to the chip. They will automatically reduce the current when the internal die temperature rises above some threshold. The hotter the regulator, the more the current drops.

These symptoms match what you are reporting.

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    \$\begingroup\$ Hopefully there's no such thing as a dumb question when you're a beginner, even if I realise now what I was trying to do was indeed stupid. Thanks a lot for your answer : ) \$\endgroup\$ – Macmade Apr 24 '15 at 16:16
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    \$\begingroup\$ Absolutely no such thing as a dumb question. In fact, it was downright smart for you to ask this particular question. You learned something valuable today. \$\endgroup\$ – Dwayne Reid Apr 24 '15 at 17:06
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    \$\begingroup\$ FWIW - even professionals will occasionally make a mistake and have their meter in Current mode when attempting to measure a voltage. One notable case (I can dig up the link later if requested) concerns a professional electrician having a major Arc-Flash accident because his inexpensive meter blew up when he connected that meter directly across a 480 Vac high-current feeder. Although he tried to sue everybody concerned, he lost because it was his mistake that caused the accident. \$\endgroup\$ – Dwayne Reid Apr 24 '15 at 17:09
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You DO NOT EVER connect an ammeter directly across a power supply - the ammeter is very low resistance - essentially a short circuit - so the power supply will deliver as much current as it can - possibly damaging the power supply or the meter.

You say the power supply is rated to provide up to 1 Amp, but it provides 2 Amp when the meter is first connected. That 2 Amp apparently overheats the voltage regulator, causing it to protect itself by reducing its output voltage and current.

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My recommendation is to get an amp clamp and remove the fuse on your voltmeter's amp circuit (disabling it). No more breaking circuits to install an amp meter for a couple minutes, easy to switch wires that you are monitoring without shutting down, safer on high voltage / high current applications. If your voltmeter does not have a fuse on the amp circuit, get a better meter.

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