Define amplifier power rating from DC rails + speaker impedance?

[stevehertz]

How do you define or calculate the power rating of a solid state amplifier when you know its main DC rail voltage of, say 50V and that the speakers are a nominal 8 ohms say? I'm not bothered about all the stuff that can put a spanner in the works; "it depends on this, on that" etc etc. What is a basic, rule of thumb equation to arrive at an approximate power level based on an amp's DC rails and the nominal speaker impedance? Assume the amp has a good current capability to deliver easily into 8 ohms. Yes I know speakers have impedance dips below this. Thanks.

[kalee20]

If it's 50V supply, transformerless output stage (usual complementary NPN/PNP pair), with a chunky capacitor coupling to the speaker, then the output drivers will sit at half the supply, or 25V.

Then assume that the output transistors lose a volt, the speaker then can see 24V swing in each direction. So the peak undistirted AC sine wave will be 24V.

RMS value will be 16.8V.

Power in speaker = 16.8 squared over 8 watts, or 35 watts, near as dammit.

How's that for you?

[trsomian]

Assuming a single supply rail of, say 50V, then the peak to peak amplitude of the output is, assuming perfect devices, 50V, so the single peak voltage is 25V. For a sine wave the RMS voltage is 25/SQRT(2) = 17.6V

Power = V * V / R where R is the load resistance. So for an 8 ohm loudspeaker power = 17.6 * 17.6 / 8 = 39W

That is for a sine wave, and assuming no losses in the output devices etc, etc, so the actual undistorted output would be somewhat less than that

[Jimmyhaflinger]

Record player amplifiers with ad161-162 output transistors sound much louder than amplifiers with ac187-188 transistors even though both run off 12v supplies and drive 4 ohm speakers.

[Radio Wrangler]

And that is it trspmian's calculation gives you the absolute maximum power that could be achieved even if everything was perfect. Things can only go down from here.

No 'peak music power' 'IHF' and such twaddle. Honest to goodness Watts. Force times distance travelled Watts. Watts that can be measured in a calorimeter.

No real world amplifier will get terribly close to this limit due to voltage drop in transistors, wiring etc. Most amplifiers run darlington emitter followers as output stages, so there goes about 1.4v times two off of the peak to peak swing. Also the voltage amplifier driving said darlingtons won't make it all the way to the rails so you can lose a couple more even if the designer was careful.

One way to avoid some of this is to have a higher voltage supply for the driving circuits than for the final power devices. This really makes a difference if you use Mosfet power devices because their gates have to swing several volts further than their outputs. At least the added supply doesn't have to source much current. My power amp in the lounge runs +55 and -55v supplies for the output stage, with two added supplies regulated at 15v more than these, so the driving stages get +70 an -70v. Only the 55v supplies have to do heavy lifting.

Another thing the basic calculation does no do is give any information about how long an amplifier can sustain that level of output power. That depends on the facilities for removing heat from the bits that get hot.

It's quite nice that when there are arguments going on about power levels and how many watts a so-and-so can do with people claiming shock and awe figures, a simple calculation cuts through all the hype and bull. There can only be more power if the voltage goes up or the load resistance comes down.

If the load impedance falls, then you can do a similar sum about available current and use I squared times R to get power.

So my amplifier, limited by its voltage swing (110v) can do no more than 189W into 8 Ohms. Its current swing is limited to +/- 15A So the load that would take most power would be 15A at 55v which is 3.666.. Ohms Theoretically it could give 15 over root two =10.606 amps into that load which is 412.5W per channel when you do I squared R.

The mains transformers are 400VA on each channel so that fits, the bridge rectifiers will do 15A if they have to. The reservoir capacitor banks are going to have a hard time, but the big beast should hang on until the heatsink temperatures trip the safety cutout switches.

Letting some reality into the theoretical max figures, it will do 200W into 4 Ohms each channel, both running at once, indefinitely, with comfort margin. I tested it for a full afternoon but I did have a fan blowing at it.

Gross overkill? Yup!
Very silly? Oh, yes. Very very.
Fun? Darn tootin'

Still, if Deep Purple happen to drop in some afternoon, I wouldn't want them to feel lacking in power. And it fits my tag-line, below.

David

[Skywave]

Originally Posted by stevehertz

How do you define or calculate the power rating of a solid state amplifier when you know its main DC rail voltage of, say 50V and that the speakers are a nominal 8 ohms say?

With many commercially-built hi-fi power amplifiers, the PSU voltages that power the output devices tend to drop at maximum audio power out. This, obviously, has an effect of such a calculation. If the PSU is adequately designed and regulated, then that consideration obviously does not apply. The simple alternative to an electronically-regulated PSU is to use a massively over-rated mains transformer, large rectifiers and big capacitors, but for me, that seems a rather clumsy, inelegant approach.

I didn't read a mention of this 'PSU loading effect' above, so I thought I'd just mention it.

As for a quick rule of thumb calculation, if you can measure the peak-peak voltage swing (say Vp-p) across the load just before the onset of clipping and you know the load resistance (say RL), then the output power is (Vp-p)²/8*RL watts. Obviously that ignores any considerations of distortion.

Al.

[stevehertz]

That's good enough for me guys - many thanks!

So - being equally simplistic - you can work backwards to find what voltage rails you need to give you a certain RMS power output/ohms.

[Radio Wrangler]

That's how you do it.

Decide what power you want into what range of impedances. Pick the highest impedance and highest power.... calculate the rails you need to get the voltage swing.
Look at the lowest impedance and the highest power, and you similarly find the current swing you have to rate things for.

David

[dominicbeesley]

Can that not all be (more or less) doubled with a H output configuration?

(edit: effective volts that is i.e. *4 power)

[kalee20]

Yes, of course it can.

And it can also be altered to anything you want, if you have a transformer coupling to the loudspeaker. So if you are stuck with, say, 12V supply rail but you want 50 watts into an 8 ohm load, you use a step-up transformer.

Hence, the assumptions in mine and trsomian's posts above!

[mhennessy]

Now you've had the theory, I'd offer these typical "rules of thumb", based on real-world designs:

100V DC: 100W
70V DC: 50W
50V DC: 20W

All into 8 ohms.

These are off-load voltages, measured when the amp is idle. The voltages obviously drop under load - hardly any amplifiers use regulated power supplies, and there are compelling reasons to stick to the simple approach.

For example, the Quad 405 is rated at 100 watts, and its rails are +/-50V (100V total) when idle.

Note the very round numbers here - it's surprisingly difficult to be more precise, frankly. It's actually normal to build several prototype mains transformers until you settle on one that gives the required nominal output power in practice (at the required price point).

Don't forget also that you have to build in some margin for when the mains supply is low. As a result, the "official" rating given by the manufacturer is conservative. Most hi-fi amplifiers I've tested give more than 10% over their rated output - sometimes as much as 25%. I think the most I've seen was the Musical Fidelity B200, rated at 60 watts per channel, but gives something like 130 watts per channel (from memory - I'll dig out my notes if anyone wants this verified, but it's certainly in that order of magnitude). This, BTW, explains why the 120VA mains transformers used in that model are incredibly unreliable!

These ratings are "continuous average sine wave power", or "Honest to goodness Watts" to use David's phrase. What some Marketeers incorrectly call "Watts RMS", which is nonsense, but established nonsense, for better or worse. Now, a sine wave is a tough signal for an amplifier (and loudspeakers!); real music is much easier. As a result, when playing music the rails sag less compared to the sine wave testing, and so you get a bit more "dynamic power". There is a standard for measuring dynamic power, but that's probably straying beyond the original question, I'd guess. However, most power amplifiers with conventional unregulated supplies can usually supply 1 or 2dB extra with music - figures higher than that suggest a "saggy" supply with a marginal transformer; lower figures suggest a "stiff" (or perhaps "over-engineered") power supply. Or to put that another way, it's why you can get away with under-engineered power supplies (though the B200 is not an example to follow here!).

[G6Tanuki]

I've never really got my head round "directly coupled" transistor audio-stages where the 'speaker is connected direct to the amp output; to me there should be a transformer in there to re-combine the two half-wave class-B waveforms and match them to the load [which could be a loudspeaker, a "100Volt Line" audio-distribution network, 600-Ohm telephony circuits, or the impedance of an anode-modulated RF output-stage]

One of my first 'real' semiconductor audio-amplifiers was a 100-Watt-continuously-rated-sinewave job which ran off 12V (or more likely 10.mumbleV because of discharged lead-acid batteries and supply-wiring-resistance) - it had two 'parallel-quads' of Newmarket NKT404 transistors [think of them as Newmarket's equivalent to Mullard/Philips OC28/OC35 offerings, but at 2/3 the price] - it had an autotransformer as its output-load and drove an interesting string of 15-Ohm cabs wired in series/parallel.

When you're paid to deliver "Loud" you don't fret too much about low-level distortion.

[dominicbeesley]

Originally Posted by kalee20

Yes, of course it can.

And it can also be altered to anything you want, if you have a transformer coupling to the loudspeaker. So if you are stuck with, say, 12V supply rail but you want 50 watts into an 8 ohm load, you use a step-up transformer.

Hence, the assumptions in mine and trsomian's posts above!

I wasn't being narky. A genuine question....

[mhennessy]

It's a good question. In reality, you rarely get 4 times the power though. Each amplifier is "seeing" half the impedance, so driving twice the current, hence conductive/resistive losses double. Compounding this, the PSU sags more...

A good rule-of-thumb is to assume you'll get 3 times the single-ended output, and regard more than that as a bonus

[Skywave]

Originally Posted by mhennessy

. . . hardly any amplifiers use regulated power supplies and there are compelling reasons to stick to the simple approach.

Where, presumably, the "simple approach" entails not using a regulated PSU. So, just out of interest, what are those "compelling reasons" for not using a regulated PSU - apart from cost?

Al.

[kalee20]

In response to mhennessy and Dominic's posts: Absolutely! Though you can uprate the cables, output devices, etc, to cope with the increased current.

There are also amplifiers that have + and - supply rails. Here, the output is centred at zero, there's no output capacitor, and the output swings to within a volt or so of each rail. But the same basic calculation applies.

Dom's point about bridge amplifiers is good, and it's done a lot in practice. Such an amplifier, of course, doesn't have a 'cold' output terminal: both terminals of the speaker have audio on them.

[G8HQP Dave]

Originally Posted by G6Tanuki

I've never really got my head round "directly coupled" transistor audio-stages where the 'speaker is connected direct to the amp output; to me there should be a transformer in there to re-combine the two half-wave class-B waveforms and match them to the load

Fortunately a simple wire junction is sufficient to recombine the two Class B signal currents. Of course, a piece of wire cannot do load matching but in most cases that is not needed as it has been embedded in the design.

[G6Tanuki]

Originally Posted by Skywave

Where, presumably, the "simple approach" entails not using a regulated PSU. So, just out of interest, what are those "compelling reasons" for not using a regulated PSU - apart from cost?

The problem is that to power an amplifier the regulated PSU has to be built with a slew-rate that can follow the rapidly-fluctuating current demand of the amp. Your regulator becomes, in effect, another audio-amplifier!

Rather than regulating the main power-supply rails quite a few amps use what's best described as an "active decoupler" between the main PSU rails and the low-signal-voltage stages - sometimes as simple as a transistor, a resistor and a capacitor. This is a great way to keep the audio-frequency fluctuations out of the front-end.

[Radio Wrangler]

You can never get enough bandwidth out of a power supply regulator, and so you have to engineer a change-over zone where feedback ceases to keep the effective source impedance low nd decoupling capacitors take over for higher frequencies. This drops you right in the classic problem of trying to drive a capacitive load while maintaining stability. Wherever you choose to pitch the changeover, you wind up with the capacitive load getting serious at the point where the feedback system is trying to withdraw gracefully. This has to be so, because it's what you're trying to do.

If you're designing an amplifier, one with unregulated supplies amounts to a series pass device between a ripply voltage on a reservoir capacitor and your loudspeaker. If you choose to use a regulated supply, then you have two series pass devices in series. This isn't necessarily better than just one. You can have a bit of decoupling between them and it does guarantee the supply to the second series pass device.

But if you have a single device and arrange good enough power supply rejection the regulator wasn't needed. What's more you get the overhead voltage that the regulator used to consume to play with and so you can have a bit more power for les clipping on transients.

In short, a good enough amplifier design doesn't need regulation on the main power rails.

The Quad 303 was regulated. It seems they did this because they were going close to the voltage ratings of the 2N3055 family, the beefiest devices of the period it was designed in, and the regulator saved them having to lower the power rail voltage and reduce the amplifier's power spec to allow for mains voltage variations. Clearly the right decision at the time.

David

[kalee20]

Brilliantly put, David!