Electronics > Projects, Designs, and Technical Stuff
Audiophile help please - Ohms and power
DW1961:
--- Quote from: magic on July 30, 2020, 08:10:40 pm ---
--- Quote from: DW1961 on July 30, 2020, 07:43:28 pm --- 2 × 50 W Into a 4-Ω BTL Load at 21 V (TPA3116D2)
2 × 30 W Into a 8-Ω BTL Load at 24 V (TPA3118D2)
2 × 15 W Into a 8-Ω BTL Load at 15 V (TPA3130D2)
--- End quote ---
For supply voltage of ±21/24/15V maximum sinewave output is approximately 14/16/10V RMS, respectively. Do the math (P=V²/R) and you get the numbers they give.
--- End quote ---
24V RMS 2*/8 ohms = 72 total Then /2 = 36 watts per channel.
Why do you say that 24 volts they use is maximum and not RMS? If it's a 24V power supply, then that's RMS, isn't it? I mean you don't rate power supplies by peak voltage. That's what Where in the link is the voltage being lost? If I can get this, then were done with the Amp discussion. I'll get it.
I'm reading this and trying to figure it out:
"Now let’s look at the RMS values. In our example, we know the RMS voltage is 10 volts. We saw earlier that 10 squared is 100, and 100 divided by 5 [ohms] gives us a calculated power of 20 watts."
https://geoffthegreygeek.com/understanding-amplifier-power/
ejeffrey:
24 volts is the DC supply voltage to the amplifier. It's a constant input voltage, so peak, RMS, average are all the same. When the amplifier has to produce a varying waveform from the 24 V input source, it can produce any voltage up to (slightly less than) 24 V. The voltage is dropped across the amplifier's power transistors. So if you want to produce a sine wave with maximum power output, it would have a 24 V peak. In that case the RMS voltage would be 17 V. (17 V)^2/(8 ohm) is 36 watts into an 8 ohm speaker. In reality it won't be able to go all the way to 24 V, there will always be some voltage drop on the power transistors, hence the 30 W rating.
DW1961:
--- Quote from: ejeffrey on July 30, 2020, 10:43:30 pm ---
--- Quote ---Also, back to speaker Ohms, if you had that amp and had two identical speakers, but one with 4 Ohms and one with 8 Ohms with the above amp specs, how is that going to effect speaker output?
--- End quote ---
It won't. The difference between an 8 ohm and 4 ohm speaker is the voice coil. An 8 ohm speaker will have more turns of thinner wire than a 4 ohm speaker. When driven with the same power (higher voltage but lower current) the coil will produce the same magnetic field, and thus the same sound.
--- Quote from: DW1961 on July 30, 2020, 07:43:28 pm ---Supports Multiple Output Configurations
2 × 50 W Into a 4-Ω BTL Load at 21 V (TPA3116D2)
2 × 30 W Into a 8-Ω BTL Load at 24 V (TPA3118D2)
2 × 15 W Into a 8-Ω BTL Load at 15 V (TPA3130D2)
--- End quote ---
--- End quote ---
"It won't. The difference between an 8 ohm and 4 ohm speaker is the voice coil. . . . When driven with the same power (higher voltage but lower current) the coil will produce the same magnetic field, and thus the same sound."
But they aren't being driven with the same power for that 3116D amp. One gets 50 watts per channel and one is getting 30. They do have diff. voltage and current, but the power output is different.
I really apologize if I am just hacking this thing to death. I really am trying. I know you can change voltage and current and get the same power rating. But you did say "same power" so I'm going with that :)
Siwastaja:
--- Quote from: Dundarave on July 30, 2020, 11:02:51 pm ---Another, simpler (I think) way to look at this is from an impedance-matching perspective:
For maximum power transfer between any amplifier circuit and its load, the output impedance of the amp needs to match the speaker.
--- End quote ---
Noooooo!!!11
This stupid thing, taught in university classrooms, goes around and around misunderstood from decade to decade.
Well obviously, from a very limited perspective, it is true. If you have an arbitrary Thevenin equivalent source, say an amplifier which output transistors define output impedance of 10 ohms, then if you do want to get the maximum possible power from it, you match the output load of 10 ohms.
But then you make the logical mistake and turn it around - if you have a 10 ohm load, to get maximum power possible, do you use a 10 ohm output impedance amplifier? Obviously, no! Use 1 ohm, and you'll get even more power. Run the math, it's trivial!
The key you didn't think about in school is this:
Often:
* the power used by the load does actual work (is converted to sound in a speaker, is converted to mechanical energy in a motor, is converted to light in a lightbulb...)
* the power used by the amplifier is wasted power (converted to heat in the amplifier)
From this perspective, matching the impedances so that power use is shared 50%-50% is a ridiculous idea. Instead, if you choose Zamplifier = 0 and Zload = whatever to bring the voltage & current relationship to the most optimal point for practical design, you have just the amount of power you need, with perfect load voltage regulation, and best possible amplifier efficiency.
The imaginary amplifier with 10 ohm output impedance was probably never meant to drive a 10 ohm load, but maybe a 1000 ohm load. Why?
Because amplifiers are often designed to be voltage sources that can drive the output to whatever value within its rails. If it had the equal output and load impedance, then the output voltage would change by factor of 50%. Try to output 10 volts - only 5 volts available to the load. 5 volts wasted as heat in the amplifier (this obviously won't matter in a linear amplifier which wastes power as heat anyway).
This is why amplifiers have the feedback, which is designed to bring the output impedance close to zero. The transistors and PCB traces themselves still have some on-resistance, so the feedback needs some voltage leeway to work with. Want to output exact 10V, to a load which hogs 1A? Say, transistors drop 1V because of their internal on-resistance. Is the output dropped to just 9V, and is the output impedance 1V/1A = 1 ohm? No, and no. The feedback measures the output and adjusts the transistors to be "more on" until the output is at 10V, and the output impedance is hence 0V/1A = 0 ohms.
So whenever you want to drive a load with a voltage source, the correct impedance is not the load impedance, but zero. In case of linear amplifiers driving intermediate levels, it's about voltage regulation, giving the load the correct voltage. In case of switching amplifiers, it's all about heat as well. If you match the source and load impedance, the efficiency of the amplifier is limited to just 50%, which would be very crappy for such circuit.
Siwastaja:
--- Quote from: DW1961 on July 31, 2020, 05:38:17 am ---"It won't. The difference between an 8 ohm and 4 ohm speaker is the voice coil. . . . When driven with the same power (higher voltage but lower current) the coil will produce the same magnetic field, and thus the same sound."
But they aren't being driven with the same power for that 3116D amp. One gets 50 watts per channel and one is getting 30. They do have diff. voltage and current, but the power output is different.
--- End quote ---
To drive them with the same power, you need to up the voltage. This may mean changing to another amplifier chip designed to handle higher voltage, but now this chip doesn't need to deal with as much current.
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