equations. (i was bored)

[Sniper 0]

So... I got bored today and decided to write up equations for amplifiers, like loss, power with efficiency, how to find how efficient it is. basically, a variation of ohms law.

as we all know, (V)olts * (A)mps equals (P)ower (V*A=P)

but, in amplifier circuitry, this cant be used because amps have their own loss.

so lets say our imaginary amplifier is 160A, running off 10.65 Vs, and has an effieciency of 77%.

it'll push out around 1312.08 Watts.

now to find each number, we use these numbers to find them.

Basically, 1312.08W / 10.65V = 123.2a - 160a = 36.8 amps of current lost in the amplifier, 36.8 / 160 = .23, subtract that from 1 and we get .77.

simple.

I'm quite sure all amp builders already know this, but it was interesting to me, I was trying to find what voltage that the Sundown 1500.1 would have to be at in order to put out EXACTLY 1500 watts.

the amp is 80% effiecent and runs off 160A.

so we would need exactly 11.71875 Volts behind it (11.7) in order to put out 1500 exactly if its 80% efficient.

because no amp is 100% effiecient.

(V)x(A)x(E in 0.X form) = (POWER)

(P) / (V) = (amps after loss)

(amperage left after loss) / (total amperage) = (E)ffciency

(amps left after loss) - (amps total before loss) = (amps that were lost in the line)

Well, thats about all I know with numbers for now, might throw more info on here, maybe even some info from one of my acoustics books (yea, i'll get some t-line equations on here)

[shizzzon 461]

while this may be true, it can not be used as set in concrete.

Efficiency changes with resistance.

The higher the resistance, the higher the efficiency.

I used to own a Cadence ZRS10 and i measured 94% efficiency at 4.2 ohms. I'm sure it's a lot lower than that at 1 ohm.

Since resistance changes at every frequency, it is impossible to say that someone will achieve 1,500w of real power at 11.71v because of this.

Due to efficiency changes, it may only be 72% efficient or 90% efficient but at such a high load it would be putting out no where near expected power.

This is why it sucks for trying to quickly assume or quickly test power output...

It requires multiple tools at once to find power in\power out specs.

[sadistic 13]

that a good point and here is another are you using rms or peak watts peak is vxa= va now to converter it to a rms value you take the va and multiply it buy .707 so if it was 14.4volts times 160 amps =2,304watts right?? now multiply this buy .707 would equal 1,628.928 rms watts assuming there 100% eff but of course they are not. it's close already to what it should be but of course it mite be like 96.9% eff but there is no telling unless unless the chip they used the transistor chip is perfect and does not use that much energy as heat it could be 96.9 % eff but you will never know. there performance aspects and how they act in that circuit is the difficult part . this is why they make big charts on this stuff. plot it etc... and there is no telling if there is step up or down transformers in the circuit with the transistor chips. which could change the power to the chip.

[Sniper 0]

that a good point and here is another are you using rms or peak watts peak is vxa= va now to converter it to a rms value you take the va and multiply it buy .707 so if it was 14.4volts times 160 amps =2,304watts right?? now multiply this buy .707 would equal 1,628.928 rms watts assuming there 100% eff but of course they are not. it's close already to what it should be but of course it mite be like 96.9% eff but there is no telling unless unless the chip they used the transistor chip is perfect and does not use that much energy as heat it could be 96.9 % eff but you will never know. there performance aspects and how they act in that circuit is the difficult part . this is why they make big charts on this stuff. plot it etc... and there is no telling if there is step up or down transformers in the circuit with the transistor chips. which could change the power to the chip.

using fuse ratings to get amps, and only using RMS feed coming through the amp.

this is why you need to figure out what the loss from the amp is, saying its a 160A thats getting 140 through, thats 14/16 = 7/8 which equals .875, or 87.5%

thats how I find how efficient my amps are, I just take whats coming through unclipped, then compare it to how many amps the amplifier is pulling.

i dunno, just a thought.

[Sniper 0]

while this may be true, it can not be used as set in concrete.

Efficiency changes with resistance.

The higher the resistance, the higher the efficiency.

this is true, I think if I play around with other variables, I might be able to plug in resistance, that will make these equations usable.

[sadistic 13]

This is why it sucks for trying to quickly assume or quickly test power output...

It requires multiple tools at once to find power in\power out specs.

very true

this is why formulas should not be used they are inaccurate most of the time. a real reading should only be used it the most accurate way.

[///M5 2,833]

I should just delete this garbage list of misinformation wtf is your point of confusing yourself in public???/

[Sniper 0]

I should just delete this garbage list of misinformation wtf is your point of confusing yourself in public???/

it isnt wrong, but it cant be used in the regular enviroment.

its the right information to find certain numbers, but you need numbers to use it, essentially its a version of ohms law that allows efficiency to be found.

and im not confused, i understand the equation.

the only way to properly use it is to use POWER as it comes off with an ohms-load on it.

meaning my house amp does :

153 watts @ 4ohm

175 watts @ 2ohm

215 watts @ 1ohm (gets really hot at 1ohm, no wonder though )

to plug this in you need to know that ; house current is 115VAC and the amplifier has a 4A fuse rating, since all the other outputs do about 34 watts rms, this means

since its a 5 channel amp we cant properly find efficency because the subwoofer line uses its own ammount of the voltage's amperage. this means the only way to find it out

is to add up total output then find what 115x4 equals. this amp does 311 watts rms total (Saying the subline is wired to 1ohm) and 4x115 equals 460.

this means my amplifier is exactly 67.6% efficient.

(my highs are all wired to 2ohm & in this scenario my sub is wired to 2ohm)

Now tell me, is this misinformation? I think not.

for this equation to work you have to know what the system output's exactly as power. meaning this requires a good multimeter and the RMS chart. (or a true RMS Clamp if you've got the money for one)

[topgun 262]

to plug this in you need to know that ; house current is 115VAC

Question. Have you measured the voltage at your wall outlet and gotten 115V? It varies depending on time of the day, load, line sag and what not.

Question 2. Wtf is 'voltage's amperage'??

[Tirefryr 1,742]

OH boy.

Just a little teaser here; are you testing the amp with a resistive or reactive load?

[Sniper 0]

to plug this in you need to know that ; house current is 115VAC

Question. Have you measured the voltage at your wall outlet and gotten 115V? It varies depending on time of the day, load, line sag and what not.

Question 2. Wtf is 'voltage's amperage'??

Measured with a multimeter before and during testing.

the amp is rated at 4 amp usage.

[Sniper 0]

OH boy.

Just a little teaser here; are you testing the amp with a resistive or reactive load?

I was using a subwoofer, so it would be both resistive and reactive, because the coil impedance is 2ohm+2ohm, allowing me to get 1ohm, 2ohm, and 4ohm, Le is 2.50mH.

[topgun 262]

*chuckles*