How Power is Drawn from an Amplifier

[Luke0011]

I had an interesting question from a customer that i would like to put to this forum to get an idea how you would answer it.

He has bought a 1000w pdx 1.1000

He is intending to wire up a type x but in the meantime he wired a jvc or jbl sub rated at 150 rms and had it on full gain and was happy with the output but asked me that his concern was that if the gain was at full and this sub sounded ok then he was worried the amp would not be able to run a 1000w type x very well.

How would you answer this question???

[VZ-Rob]

tell him he's dreaming
would have sounded crappy and distorted
the thing is with bass many people think distortion is o.k.
IT'S NOT will kill the amp / speaker eventually
would need to know the settings as well
1000 watts peak MAX on 12.5 volt? is say 80% efficient Hmmmm !
more like 60% efficient so putting out 750 watts peak MAX
should still PUMP real hard

[StoneX]

Ask him if the gain was at max and the volume on the head unit was near max... I'm sure it wasn't.

I'm sure you know what the gain does, so even with the gain at max the amplifier still makes it's power from 0-1000W as you vary the volume control on the source. His volume control could have been set so it was making 300W, 500W, who knows.

Also, speakers will generally take more than their rated RMS power if the power is clean and free of distortion.

I'd just tell him, not to worry and wait till he hears it


VZ-Rob... Amp power ratings are already at the output so efficiency ratings don't come into play.

[holdenboy]

^^^^ Exactly, im doubtful he had the head-unit volume on max.

[Luke0011]

Interesting StoneX but considering different source units output AC voltage from anywhere between 1 volt and 8 volts at maximun volume throw that theory out the window.. I would suggest that the device it self would determine power drawn from the amplifier but not limiting exculsively. Theres also the possibility that the digital amplifier works somewhat differently in its output capability. Some amps even output a certain amount irrespective of there impedance

[StoneX]

Throw which theory out the window?

Amplifiers have an a gain exactly for the reason you mentioned, because source units have different maximum RCA output voltages.

The impedence of a speaker determines what power it receives, so two 4 ohm speakers will receive about the same power if hooked up to the same system. Speakers do have an impedence curve thought which means the impedence changes at differring frequencies, so you can end up with slightly different power at different frequencies.

I assume you're referring to class-D amplifiers when you say "digital". They interact with speakers the same way as a normal amp. The only difference is how the transistors turn on and off to produce the sound wave.

The amplifiers that output the same power at diferent impedences have regulated outputs.

How does this relate to your original question though?

[theSeekerr]

Oh, bloody hell, here we go again....

OK. Typical MOSFET designs are current output devices - that is, they look, to the sub, like a Voltage Controlled Current Source. The controlling voltage is the input signal.

We can model this, roughly, as I(out) = g*V(in), where g is the gain of the amplifier. Note that for 99% of amplifiers, that's actually a fixed value, and what we consider the gain control is actually an attenuator on the input signal.

OK, let's take a hypothetical amplifier with g = 5. We apply a 1V RMS signal on the input. Since g = 5, the amplifier attempts to force 5A through the speaker.

If the speaker is, say, a 4 ohm sub, this results in a voltage of 20V across the output terminals, for a total output power of 100W.

OK, what if we swap in a 2 ohm sub? Now we see a voltage of 10V across the output, for a total output power of 50W.

So why can an amp deliver more power to a 2 ohm sub than a 4 ohm one? Well, that's down to the power supply. If the switchmode PSU in the amp runs rails of, say, +/- 50V, it can swing an output signal of (nearly) 100V peak to peak. That's an RMS voltage of about 35V. Since this is the largest voltage it's possible for this amp to swing, it has a peak theoretical output power of only about 300W into 4 ohms, but 600W into 2 ohms.

Upshot is, if your amp is rated to 1000 watts into 4 ohms, but with the "gain" at full you aren't swinging 178V peak to peak (63V RMS) across the outputs, your amp still has more left in it....you just need to crank up the signal source more.

[StoneX]

I don't know why you bother getting so technical... That will go over 99% of the readers heads in here

[theSeekerr]

You're probably right, but at least I know I've supplied the correct answer if anyone wants it.

(EDIT: Also because I'm studying MOSFET amps at the moment in Fundamentals of Electrical Engineering. Trying to explain parts of it to people helps me keep it straight in my head, too)

[StoneX]

No worries

I think your last line is the most important part in relation to his question.

[Tasmaniak]

wow, I sometimes forget how boring things can be at component level...Whilst I understood it. My eyes just glazed over and my mind over to much more fruitful topics. Like Sex!

[Luke0011]

Throw which theory out the window?

Amplifiers have an a gain exactly for the reason you mentioned, because source units have different maximum RCA output voltages.

The impedence of a speaker determines what power it receives, so two 4 ohm speakers will receive about the same power if hooked up to the same system. Speakers do have an impedence curve thought which means the impedence changes at differring frequencies, so you can end up with slightly different power at different frequencies.

I assume you're referring to class-D amplifiers when you say "digital". They interact with speakers the same way as a normal amp. The only difference is how the transistors turn on and off to produce the sound wave.

The amplifiers that output the same power at diferent impedences have regulated outputs.

How does this relate to your original question though?

Sorry stonex miss read your quote as max gain as max volume.

Oh, bloody hell, here we go again....

OK. Typical MOSFET designs are current output devices - that is, they look, to the sub, like a Voltage Controlled Current Source. The controlling voltage is the input signal.

We can model this, roughly, as I(out) = g*V(in), where g is the gain of the amplifier. Note that for 99% of amplifiers, that's actually a fixed value, and what we consider the gain control is actually an attenuator on the input signal.

OK, let's take a hypothetical amplifier with g = 5. We apply a 1V RMS signal on the input. Since g = 5, the amplifier attempts to force 5A through the speaker.

If the speaker is, say, a 4 ohm sub, this results in a voltage of 20V across the output terminals, for a total output power of 100W.

OK, what if we swap in a 2 ohm sub? Now we see a voltage of 10V across the output, for a total output power of 50W.

So why can an amp deliver more power to a 2 ohm sub than a 4 ohm one? Well, that's down to the power supply. If the switchmode PSU in the amp runs rails of, say, +/- 50V, it can swing an output signal of (nearly) 100V peak to peak. That's an RMS voltage of about 35V. Since this is the largest voltage it's possible for this amp to swing, it has a peak theoretical output power of only about 300W into 4 ohms, but 600W into 2 ohms.

Upshot is, if your amp is rated to 1000 watts into 4 ohms, but with the "gain" at full you aren't swinging 178V peak to peak (63V RMS) across the outputs, your amp still has more left in it....you just need to crank up the signal source more.

Thanks for the technical response, i understand this partially so i have a few questions.
1. if the 2 ohm sub produces 50w is to be incorrect then 4 ohm at 100 watts would also be incorect also. what if you started with the 2 ohm example first and state the 4 ohm example as being incorrect???
Confusing.
2. Is it safe to assume that if the 2 ohm example is used the current increases and the voltage drops. As in the sub draws 20 amps thus producing 200 watts

Is it a similar principle to a car battery droppping voltage to increase current while cranking as is the input voltage in respect of the gain control setting aproxximating a set current and voltage differenciating output wattage.

[Tasmaniak]

can we get back to the topic at hand here please? .....sex.


-Edit-....I mean both hands

[theSeekerr]

Sorry stonex miss read your quote as max gain as max volume.
Thanks for the technical response, i understand this partially so i have a few questions.
1. if the 2 ohm sub produces 50w is to be incorrect then 4 ohm at 100 watts would also be incorect also. what if you started with the 2 ohm example first and state the 4 ohm example as being incorrect???

I didn't say that anything was incorrect in my original post....so now I"m the one who's confused.

2. Is it safe to assume that if the 2 ohm example is used the current increases and the voltage drops. As in the sub draws 20 amps thus producing 200 watts

Is it a similar principle to a car battery droppping voltage to increase current while cranking as is the input voltage in respect of the gain control setting aproxximating a set current and voltage differenciating output wattage.

No. Nothing like that at all, actually.

Sorry, my original examples were a bit poorly worded. In reality, your input signal would be an AC source which is continuously variable in amplitude (that is, voltage) and frequency. As a result, your output current is similarly variable in amplitude. The voltage is related by Ohms Law, that is, V = IR, where I is the output current and R is the speaker impedance, in this case.

Since Power = VI, you can substitute Ohms law and derive the expression Power = Current Squared * Impedance, which is also equiavalent to Power = Voltage Squared / Impedance

So yes, 200 watts into a 2 ohm sub takes more current and less voltage than 200 watts into an 4 ohm sub. However, that has nothing to do with the way a battery's terminal voltage sags when delivering high current, which is a purely chemical effect.

If you'd like anything more explained, just quote the relevant parts of my posts and ask away.