Electrical experts, educate me: Gain and voltage

[Supergumby5000 5,000+ posts]

I was just doing some thinking today, and im kind of confused. I'm no electrical engineer or anything, so this may be way more in depth than i think, or on the other hand it could be much simpler, but im curious nonetheless.

Setting gain on an amp.

Okay, i understand how to set gains on an amp, but what confuses me is the voltage.

for example, 4000RMS @ 1ohm requires an output voltage of 63.2volts

4000RMS @ 2 ohms requires an output voltage of 89.2 volts, which is quite a bit more voltage than a 1 ohm load.

theoretically, the 1 ohm load will have much more strain on an electrical system than a 2 ohm load, but why does it require less voltage?

Also, when i look at this, i feel like an amp running 4krms at 2 ohms would need to be a lot more beefy than an amp at a 1 ohm load, when regarding voltage output, however my amplifier is capable of 2000RMS x 2 @ 1 ohm or 4000RMS @ 2 ohms bridged. technically from the above information, wouldnt this mean one of these ratings is under/overrrated?

Is the strain on the amp due to a reaction between the amp and the drivers/wiring? I guess what im saying is i dont understand how the whole voltage thing works in theory to thermal and mechanical limits of an amplifier.

Any input on this would be awesome, im just looking to understand this a bit more.

Click to read more...

 

[RidnClean 5,000+ posts]

my brain hurts from reading that //content.invisioncic.com/y282845/emoticons/crap.gif.7f4dd41e3e9b23fbd170a1ee6f65cecc.gif

I'm staying tuned to learn somethin...

 

[Skip01 5,000+ posts]

Might have more too do with the amperage

Like 89volts and 45amps for the 4000 @ 2ohm

And 63volts and 64amps for 4000 @ 1ohm

Just a guess though, i really have no clue

And i didnt read whole thing haha

 

[IDSkoT 5,000+ posts]

souce: http://www.thesuicidaleggroll.com/gain.htm

Probably the two most basic equations in electrical analysis are Ohm's law (V=IR, V is voltage, I is current, R is resistance) and the power equation (P=IV, P is power). When you combine the two, you get P=V^2/R. Rearranging this gives us P*R=V^2, or V=sqrt(P*R). This means that if your amp does 100rms at 4ohm, we have that V=20. In order for this amp to actually do 100rms @ 4ohm, it would have to put out 20Vrms on the speaker terminals. Unlike methods 1 and 2 above, this method depends on the power rating by the manufacturer to be exactly right in order for your gain to be right. If the amp is overrated, then by following this method you might actually be clipping the amp at full volume, and if the amp is underrated, then by following this method you might not be able to reach the full potential of the amp. Another disadvantage is you have to rely on the accuracy of your dmm, and many can be pretty innacurate, especially with AC signals. However, it's easy and fast, and some people like that.

As far as "bulky"... are you referring to the heat sink or the board? A bulkier heat sink just allows for better heat dissipation. As far as "strain", the amp is less efficient at lower resistances, thus it requires more current.

 

[Supergumby5000 5,000+ posts]

Might have more too do with the amperage
Like 89volts and 45amps for the 4000 @ 2ohm

And 63volts and 64amps for 4000 @ 1ohm

Just a guess though, i really have no clue

And i didnt read whole thing haha

hmm interesting, maybe thats where my confusing is arising.

and Idskot, i understand all that. Maybe you misunderstood my original post. Its worded kind of weird, but it was the best i could do haha.

 

[Rashaddd 5,000+ posts]

Power = Voltage x Current

Dropping voltage doesn't mean dropping power, so long as amperage increases. This is also what can cause amps to blow.

For example, let's say you have an amp (we'll give it 100% efficiency), putting out 1kw. At 14v power input, this means the amps internals have to be able to handle about 70 amps of current. What happens when voltage drops to 10v? Well, either you lose power, and are now only getting about 700 watts max out of it, or you increase current to compensate, which means the amp will try to pull 100 amps. If you have proper fusing, your amp should be fine, and hopefully the fuse blows. If you don't, there's a good chance you'll see smoke out of it (even with proper fusing you may blow it)

 

[Supergumby5000 5,000+ posts]

Power = Voltage x Current
Dropping voltage doesn't mean dropping power, so long as amperage increases. This is also what can cause amps to blow.

For example, let's say you have an amp (we'll give it 100% efficiency), putting out 1kw. At 14v power input, this means the amps internals have to be able to handle about 70 amps of current. What happens when voltage drops to 10v? Well, either you lose power, and are now only getting about 700 watts max out of it, or you increase current to compensate, which means the amp will try to pull 100 amps. If you have proper fusing, your amp should be fine. If you don't, there's a good chance you'll see smoke out of it (even with proper fusing you may blow it)

this makes a lot more sense, thanks.

 

[IDSkoT 5,000+ posts]

Power = Voltage x Current
Dropping voltage doesn't mean dropping power, so long as amperage increases. This is also what can cause amps to blow.

For example, let's say you have an amp (we'll give it 100% efficiency), putting out 1kw. At 14v power input, this means the amps internals have to be able to handle about 70 amps of current. What happens when voltage drops to 10v? Well, either you lose power, and are now only getting about 700 watts max out of it, or you increase current to compensate, which means the amp will try to pull 100 amps. If you have proper fusing, your amp should be fine. If you don't, there's a good chance you'll see smoke out of it (even with proper fusing you may blow it)

I wish there was a 1kw amp with 100% efficiency //content.invisioncic.com/y282845/emoticons/drool.gif.b5e863e893038027711d4402f340dad0.gif

The fusing thing is a problem because most fuses will allow a higher amperage than what they're rated for a certain amount of time (depending on the fuse, of course.) It's not like a 250 amp fuse will blow once Amperes == 250. Same goes for the fuses inside of the amp. They only work with high alterations in amperes.

 

[IDSkoT 5,000+ posts]

this makes a lot more sense, thanks.

So you shit on my post, filled with information, and thank him?! Dick!

Haha.

 

[Supergumby5000 5,000+ posts]

So you shit on my post, filled with information, and thank him?! Dick!
Haha.

//content.invisioncic.com/y282845/emoticons/wink.gif.608e3ea05f1a9f98611af0861652f8fb.gif

I understood what you said. The whole amperage relative to voltage thing was what i was missing.

 

[ngsm13 5,000+ posts]

http://www.bcae1.com

nG

 

[chrisw91 10+ year member]

this is something that has been bothering me as well. i saw someone on these forums selling an amp that is claimed to put out 1000 watts @ 1,2, and 4 ohms. how is that possible when most other amps have the power cut in half from 1 to 2 ohms and 2 to 4 ohms?

 

[Rashaddd 5,000+ posts]

I wish there was a 1kw amp with 100% efficiency //content.invisioncic.com/y282845/emoticons/drool.gif.b5e863e893038027711d4402f340dad0.gif
The fusing thing is a problem because most fuses will allow a higher amperage than what they're rated for a certain amount of time (depending on the fuse, of course.) It's not like a 250 amp fuse will blow once Amperes == 250. Same goes for the fuses inside of the amp. They only work with high alterations in amperes.

Doesn't matter though, as for the same reason the fuse won't blow instantly, no electrical component will. They all have added tolerance over the amount of power they can handle "all day", and no amp that should be fused at 250 amps will blow from a momentary 260...

This does create a problem most people don't realize though. What happens when the fuse blows? That means the fuse finally was put under such stress that it was enough to blow it (was likely taking over its rating for several minutes if not more, depends on how much over its rating we're talking about). So your fuse blows before your amp is, but your amp is also experiencing this stress and has heated up and is closer to blowing until it can cool/etc back to it's "resting" state. Many people will go ahead and pop in a fresh fuse right away if they have one on hand.

The problem with this is, lets say each of these fuses can handle 300 amps for 5 mins before it blows, and your amp can handle that for 6 mins. So, the fuse blows first, the amp is close to blowing (for simplicity, it only will take one more minute to blow), you pop in that fresh fuse which can handle 300 amps for 5 mins again, while your amp now can only handle 300amps for 1 min until it gets a good rest (because its already taken the other 5), and now you have the equivalent of using a fuse too big, resulting in a blown amp instead of a blown fuse.

 

[Skip01 5,000+ posts]

this is something that has been bothering me as well. i saw someone on these forums selling an amp that is claimed to put out 1000 watts @ 1,2, and 4 ohms. how is that possible when most other amps have the power cut in half from 1 to 2 ohms and 2 to 4 ohms?

regulated power supply

Like most JL amps

http://mobile.jlaudio.com/products_amps_pages.php?page_id=228

 

[scoob8000 10+ year member]

Power = Voltage x Current
Dropping voltage doesn't mean dropping power, so long as amperage increases. This is also what can cause amps to blow.

For example, let's say you have an amp (we'll give it 100% efficiency), putting out 1kw. At 14v power input, this means the amps internals have to be able to handle about 70 amps of current. What happens when voltage drops to 10v? Well, either you lose power, and are now only getting about 700 watts max out of it, or you increase current to compensate, which means the amp will try to pull 100 amps. If you have proper fusing, your amp should be fine, and hopefully the fuse blows. If you don't, there's a good chance you'll see smoke out of it (even with proper fusing you may blow it)

Good explanation here. It's all Ohms law. Google it for more in depth reading. //content.invisioncic.com/y282845/emoticons/smile.gif.1ebc41e1811405b213edfc4622c41e27.gif

With any two values, you can figure out the related values.

Keep in mind, depending on how you calculat the watts you are assuming the amp is 100% efficient.

Watts = Voltage*Voltage / Resistance

Watts = Amps*Amps * Resistance

Current = Voltage / Resistance

Current = Watts / Voltage