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Electronics savvy folks: Question.


L. Ron Hoover

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I haven't heard it used with respect to amps, just motors. :idk:

 

Since we commonly use "load" to mean resistance and/or reactance with respect to amps, it paints a strange picture. I guess it could be used to test voltage in / voltage out and look for losses in the amp. Or it could be distinguishing between electrical load and mechanical load.

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No speakers. But who knows what "no load" means to some laymen.

 

And how does one fully test an amp without a load? :idk: I'm guessing that's where you're going with this.

 

Maybe they tested the line level in/outs? But if they cracked it open and tested voltage points (like Kindness mentioned)... I'd be leery about it.

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I haven't heard it used with respect to amps, just motors.
:idk:

Since we commonly use "load" to mean resistance and/or reactance with respect to amps, it paints a strange picture. I guess it could be used to test voltage in / voltage out and look for losses in the amp. Or it could be distinguishing between electrical load and mechanical load.

 

 

Let me rephrase a little:

 

What does the general term "no load" mean to you with respect to electrical circuits?

 

Does it refer to an open circuit or a short circuit?

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Let me rephrase a little:


What does the general term "no load" mean to you with respect to electrical circuits?


Does it refer to an open circuit or a short circuit?

 

 

In this case, they mean open circuit at the output.

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"no load" is quite misleading here. The "load" on an amplifier without a speaker plugged in is extremely high, equal to the resistance of the air gap between the two contact points in the output plug.
:D

 

...which is also misleading since "no load" denotes an open in a circuit... and an open in a circuit means a resistance of infinity... and air's resistance is accepted as infinity.

 

:p

 

But yeah... I know what you're getting at. It's all theoretical at that point. :D

 

Does it refer to an open circuit or a short circuit?

 

Open.

 

If you try to measure the resistance of an open, you get infinite resistance - it's immeasurable. But a short has a measurable resistance. Theoretically it's 0... realistically it's in the fractions of a milliohm... but 0 is still measurable.

 

Now consider what the term "load" means (in electrical terms). A load is placing a "weight" on a circuit to induce current flow.

 

Theoretically, an open tells us almost nothing because there is (virtually) no path to ground - infinite resistance. There is nothing to induce a current flow in the amp.

 

Any measurable resistance... as well as a short, will induce current flow.

 

So the load could be 0 ohms. It's still a load of some sort.

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Let me rephrase a little:


What does the general term "no load" mean to you with respect to electrical circuits?


Does it refer to an open circuit or a short circuit?

 

 

I've heard it used to describe an open circuit on a transformer secondary.

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You can't measure power without a load. The open circuit voltage is useless.

 

 

Crank a power amp with no device connected and stick a multimeter across the outputs. You get a zero volts reading?

 

EDIT: Trick question... "Power" is a measurement of the output and the typical unit is watts, no source means no output.

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Wow. I'm borderline confused now. :confused:

 

OK...

 

Volts is the electrical potential. It is NOT the actual flow of electrons (from negative to positve). That is amperage. And electron flow (amps) cannot take place in an open circuit.

 

(Keep in mind, this is discrete circuit theory. We all know that lightning flows through an "open". But we're not talking about 1.21 gigawatts of electricity here, are we? :D).

 

Wattage is total power and is calculated by Volts * Amps.

 

So, in an open circuit, there is no amperage flowing... which means there is no power being dissipatated anywhere. Usually... power dissipation (wattage) will occur in the load (although a lot of it gets dissipated as heat too... but we won't get into that).

 

If the load just happens to be 0 ohms... electrons will flow and power will be dissipated. In an amplifier, it WILL be dissipated completely as heat since amplifiers are obviously not designed to run 0 ohm loads. In this case, the power dissipation (heat) will most likley affect the weakest electrical part in the circuit... and fry it.

 

This is exactly what fuses are designed for. To be the weakest part of the circuit in the unlikely event of a short (or other over-power consumption).

 

 

 

I know all that wasn't asked for. But it seemed like it might help clear some things up that were not elaborated upon in this thread.

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