Electronics > Projects, Designs, and Technical Stuff
Audiophile help please - Ohms and power
ejeffrey:
--- Quote from: DW1961 on July 31, 2020, 06:54:29 am ---
--- Quote from: ejeffrey on July 31, 2020, 05:27:14 am ---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.
--- End quote ---
OK I get that. So is the voltage drop constant or depends on the amp chip? It looks like they are losing 16% from an RMS of 17V.
It looks like they lose 16% in the conversion?
24V @ 8 Ohms = 72 watts / 2 =36 watts
36 -30 = 6 watts lost.
16.6% loss? Is that linear for each voltage?
--- End quote ---
No, it would typically be approximately constant voltage drop for the same topology and is limited by the Vbe of the transistors (or equivalent turn-on voltage of MOSFETs if used). With a Darlington output stage you are limited to about 1.2 volts away from each rail (2x Vbe) for a Sziklai output stage about 0.6 V. MOSFETs have considerably higher Vgs than a BJT Vbe, so unless used in high power (and high voltage) amplifiers where this dropout is negligible they would usually employ a small auxillary supply to allow the gate drive to go over the main supply rail. In this case the output could get to nearly the supply voltage. These factors are basically independent of supply voltage, so they become less significant as the supply voltage increases.
DW1961:
--- Quote from: dzseki on July 31, 2020, 11:11:08 am ---At different loudness levels the human ear perceives the bass and treble notes differently, consult with the Fletcher-Munson curves. Some amplifiers until the 90's had a "loudness" button to mimic the human hearing.
--- End quote ---
It increases the top and bottom end when levels are low. I sure do remember those buttons! Always sounded better with it on. It was like a automatic equalizer boost at lower levels that would go away the higher you got it.
DW1961:
--- Quote from: ejeffrey on July 31, 2020, 03:57:13 pm ---
--- Quote from: DW1961 on July 31, 2020, 06:54:29 am ---
--- Quote from: ejeffrey on July 31, 2020, 05:27:14 am ---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.
--- End quote ---
OK I get that. So is the voltage drop constant or depends on the amp chip? It looks like they are losing 16% from an RMS of 17V.
It looks like they lose 16% in the conversion?
24V @ 8 Ohms = 72 watts / 2 =36 watts
36 -30 = 6 watts lost.
16.6% loss? Is that linear for each voltage?
--- End quote ---
No, it would typically be approximately constant voltage drop for the same topology and is limited by the Vbe of the transistors (or equivalent turn-on voltage of MOSFETs if used). With a Darlington output stage you are limited to about 1.2 volts away from each rail (2x Vbe) for a Sziklai output stage about 0.6 V. MOSFETs have considerably higher Vgs than a BJT Vbe, so unless used in high power (and high voltage) amplifiers where this dropout is negligible they would usually employ a small auxillary supply to allow the gate drive to go over the main supply rail. In this case the output could get to nearly the supply voltage. These factors are basically independent of supply voltage, so they become less significant as the supply voltage increases.
--- End quote ---
So are my calculations correct given a 16% decrease in RMS voltage due to the DC loss, and I could calculate power using the formula above, mainly getting the new RMS and subtracting 16.6%
gman76:
The loudspeaker typ has a spec (individual drivers in a loudspeaker certainly do) called ‘sensitivity’ that relates a given voltage to an SPL acoustic output. When an amplifier drives a loadspeaker, it is driving a voltage. Current and therefore power is simply a consequence of the voltage applied to an impedance. 8 ohm vs 4 ohm ratings are approximations. The impedance varies a lot with freq and the industry provides these as ‘standard’ values that are roughly the average over the audio freq range. Back to sensitivity - it’s standard practice to quote it as 2.83Vrms into 8 ohms which is 1 watt. Whether the speaker is 4 or 6 or 8 ohms, generally it doesn’t matter. They almost always quote sensitivity as 1W into 8ohms. When you apply 2.83V rms, you get some acoustic output that is measured at 1meter, and that’s the sensitivity in dB.
So, you could call a speaker that delivers 90dB more efficient than one at 88dB since both are driven by the same voltage. But is a 90dB 4ohm speaker more or less efficient than the 88dB 8ohm speaker? This is a hypothetical. The 4ohm speaker draws more power but produces more output, so maybe it’s a toss up.
The audiophile world generally isn’t concerned about efficiency or power consumed or power required. After all, Class A amplifiers or vacuum tube amps are fairly common in that space. Space heaters.
But I have seen virtually the same drivers, 4 vs 8 ohm, that show better sensitivity for the 4 ohm model. More voice coil winding in the 8 ohm. The 8 ohm draws less current, lower power. Seems like a basic trade off. How much output current capability does the amplifier have? The flip side is that your amp needs more voltage headroom to deliver the same SPL (at high volume levels) for the 8 ohm model.
DW1961:
--- Quote from: gman76 on August 01, 2020, 07:09:00 pm ---The loudspeaker typ has a spec (individual drivers in a loudspeaker certainly do) called ‘sensitivity’ that relates a given voltage to an SPL acoustic output. When an amplifier drives a loadspeaker, it is driving a voltage. Current and therefore power is simply a consequence of the voltage applied to an impedance. 8 ohm vs 4 ohm ratings are approximations. The impedance varies a lot with freq and the industry provides these as ‘standard’ values that are roughly the average over the audio freq range. Back to sensitivity - it’s standard practice to quote it as 2.83Vrms into 8 ohms which is 1 watt. Whether the speaker is 4 or 6 or 8 ohms, generally it doesn’t matter. They almost always quote sensitivity as 1W into 8ohms. When you apply 2.83V rms, you get some acoustic output that is measured at 1meter, and that’s the sensitivity in dB.
So, you could call a speaker that delivers 90dB more efficient than one at 88dB since both are driven by the same voltage. But is a 90dB 4ohm speaker more or less efficient than the 88dB 8ohm speaker? This is a hypothetical. The 4ohm speaker draws more power but produces more output, so maybe it’s a toss up.
The audiophile world generally isn’t concerned about efficiency or power consumed or power required. After all, Class A amplifiers or vacuum tube amps are fairly common in that space. Space heaters.
But I have seen virtually the same drivers, 4 vs 8 ohm, that show better sensitivity for the 4 ohm model. More voice coil winding in the 8 ohm. The 8 ohm draws less current, lower power. Seems like a basic trade off. How much output current capability does the amplifier have? The flip side is that your amp needs more voltage headroom to deliver the same SPL (at high volume levels) for the 8 ohm model.
--- End quote ---
That's roughly how my head was processing it. However, My 8 Ohms speakers are 93 dB @ 2.83V/1M, so there's that aspect too. So you can both of world's and less power that way I guess.
All fascinating, but back to my question above about calculating power output: for the above amp calculating for power using a 19V driver instead of 24.
I'm still not clear on the voltage % drop due to DC to AC and back conversion. Otherwise, it's DC voltage calculated for RMS to AC, then then the % drop from conversion itself in the amp.
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