why did the wires melt when i hooked up my amplifier?

I had someone replace an LED indicator in my Rockford Fosgate car amplifier the other day. When that was replace we hooked up a little 12vdc wire and ground to its connectors and remote switch to see if the LED would come on which it did. But when they were connected, the wires got very hot very fast and went very limp b/c of the heat.

Why did that happen with no speakers attached? The amplifier’s max output is around 900 if that matters.

It sounds like there’s a short at the 12v power input somewhere.

Consider Ohm’s Law: V=IR. Voltage equals current times resistance.

The battery wants to be V=12 volts between terminals. But the resistance, R, of just the wires and their contact point is so small, that the current, I, has to be absolutely effin’ huge.

Now, consider Watt’s Law, P=IV. Power equals current times voltage.

Since V is… about… the same as it usually is, and I is “absolutely effin’ huge,” then their product, Power, is also going to be “absolutely effin’ huge.”

Since Power is defined as rate of Energy consumption or supply, this means that the amount of electrical energy being deposited in the resistance, i.e., the wires, is “absolutely effin’ huge.”

And finally, since electrical energy deposited in resistors gets converted to thermal energy, aka Heat, then the temperature of the wire is gonna get “absolutely effin’ huge.”

And that’s when the insulation melts.

I would check to make sure you didnt short the power wires at the input, and that there’s no short inside the unit. Since you just had it serviced, the latter is entirely possible.

There isn’t a short b/c i hooked it up to the 4ga wire that runs off the battery in my car and it was fine, no heat at all really. It’s just with those skinny wires and no load at all that they couldn’t take the amount of amperes flowing thru? Does an amplifier literally suck more amperes thru a wire to charge the capacitors faster?

lil’ bump

The small wire you originally used… what gage was it?

What are you saying: You merely replaced the LED and made a temporary hookup with small gage wire to see if the LED was lit?
What is the size of wire the amp. is normally connected to battery with?
How may amps does the amp. draw at no load?

You just answered your question. The smaller wire couldn’t handle the current that was flowing through them.

As for no load, you are just assuming that the amp doesn’t draw any current if there is no speaker attached. You seem to be wrong in that assumption.

There are a lot of different types of amplifiers. Home stereos will typically use what’s called a push-pull system, where you have one type of transistor pushing in one direction, and when the signal goes negative, you have another transistor pulling in the opposite direction. The benefit of this is that when there is no sound, then there’s (almost) no power used either. The problem with this type of amplifier is that it requires a dual polarity power supply, one to drive the transistor on the positive side and another to drive the transistor on the negative side. Cars don’t have a dual polarity power supply. They just have a single 12 volt supply. If you are going to drive an amplifier off of a single supply, then you make your output 6 volts and let it swing up to 12 or down to 0 (so you essentially have +/- 6 volts). In this arrangement, with no sound, the amplifier is “half” on, which means that it’s going to draw a fair amount of power when there is no sound.

The second problem you are going to run into is that 12 volts (+/- 6 volts) is only going to make so much sound out of 8 ohm speakers. If you want a lot of sound, you need more than 12 volts. What higher power car amplifiers do is use DC to DC converters, so that they can make a power supply for the amplifier that has a lot more than 12 volts. Since these aren’t 100 percent efficient, the DC to DC converter will also tend to draw power, even when the amplifier isn’t making noise. If you are going to use DC to DC converters anyway, it’s possible to make both a single supply output and a dual supply output. The latter simply requires two converters.

Since your amp is fairly high power (which should be obvious from the 4 ga wire required to drive it) we can be certain that it has DC to DC converters in it. Beyond that, I can’t say for certain whether it’s a class A amplifier (single transistor, on all the time) or a class B (push-pull), but I’m guessing that due to the rather high amount of current it drew that you have a class A design.

In the range of say, 18–ya know, computer power supply wire size :smiley:

Yes. 4 gauge. I don’t know, but if someone tells me the setting i should use on my multimeter i’ll go measure.

[qoute=Dag Otto]As for no load, you are just assuming that the amp doesn’t draw any current if there is no speaker attached. You seem to be wrong in that assumption.

Well i assume it draws *some * juice since i can watch the filament in the power wire bend as soon as it is connected (fuser holder is coming right off the positive on the battery). There are 8 plum-sized capacitors inside that need charging ya know? I just didn’t think it would actually suck so much amperage thru a wire to actually make it melt within seconds of continuance.

The subwoofers are 4 ohm each. I run the amplifier in stereo mode so each speaker output goes to each speaker. I’ve heard rumors that bridging the speakers would lower resistance to 2 ohm each. … . not sure how that works.
Thanks for the info!

When speakers (or any other simple impedence like light bulbs) are in parallel, the admittance (which is the inverse of the impedence) is the sum of the admittance of each speaker.

Or, in other words… 1/R = 1/R1 + 1/R2

So, 1/R = 1/8 + 1/8 = 1/4, so R = 4.

A web search on “ohms law” should give you some nice examples with diagrams and such to make it more clear.