Voltage & current sent to speakers...

[Fisherdude]

This may be a dumb question, but I'm really curious, and I don't think I've ever read anything about this.

What is the typical voltage & current of an amp's output that's actually sent to the speakers? I know this isn't a simple P=IE thing, since we're dealing with impedances and speakers certainly aren't a pure resistive load. At a solid, but not thunderous, listening level.

1. For a typical PP tube amp, say 15-30 RMS w/ch?
2. Would SS be different, due to different output impedances?
3. Finally, a kick-ass SS amp in the 300 w/ch range.

Thanks! I feel like I need some education today! :stupid:
Clay

 

[Ed in SoDak]

Not dumb at all, but my answer may be!

Even though many peaks are transient, the loads are real and you need the current handling to deal with them.

Many SS supply voltages can be 60, 80 volts or higher. Lower volts would mean more amps are needed to supply the same "watts" to the speaker, so higher volts generally means you can get by on less current.

If the output is 65 watts for example, divide that by the 80 volts to get .8 amp of current. A 40 volt supply would have to be able to provide double that amount of current. Just means heavier wire is needed and a beefier supply, so in practical terms, a 40 volt supply won't be asked to pump out 65 watts.

Next up is transformer or power supply sag, where, when high current is asked for, the supply can't deliver it, so voltage drops. Ohms Law now requires yet more amps of current, as watts delivered must balance out. At some point, something gives. I guess we'd call it distortion of some kind or another to keep this simple.

At the other end, can the transistors take the heat without going into thermal runaway? Doubled or tripled outputs, beefy transformers, huge capacitors in the power supply, big wiring, hefty connectors and circuit board traces, bigger heatsinks... We follow the path upwards in greater power handling, so yes, it all matters very much indeed.

For a real-world example, though not recommended, I put a 60-watt bulb on the speaker outputs of my old SWTPC Universal Tiger 90wpc amp. No problem making it glow almost like it was plugged into the wall, though it was pulsing at the beat of the music. Remember those "light boxes" of the 70's? Sorta like that. ;-)

-Ed

 

[ilimzn]

You are correct, though P=IE, if you take phase into account. Phase difference between I and E is introduced because of the impedance nature of the speaker - this varies both in phase and magnitude with frequency.

The important thing to know is that in the above equation, only E is a constant. This is what a regular (voltage drive) power amp does, it tries to be the perfect voltage source - maintaining the same desired voltage in each instant of time, regardless of the load. So, when an amp is rated P into Z ohms, it really means that it outputs square root of (2 * Z * P) Volts peak sine wave with rated distortion.

Let's take 100WPC into 8 ohms as an example - sqrt(2 * 8 * 100) = sqrt(1600) = 40V. Peak voltage of the sine wave means that the highest/lowest points of the sine are 40V from zero - the positive peak is at +40V the negative at -40V. If there were no losses and the power supply was perfectly regulated, this voltage would also be the minimum power supply voltage for a directly coupled SS amp. Keep this in mind for later on...

The formula above holds true regardless of SS or tube, with some footnotes to add. You just have to remember that the impedance is not a constant so neither is the power if we assume the amp is capable of maintaining it's output voltage - i..e it has an infinite dampling factor.

With SS, it's relatively easy to get a high damping factor. A DF of 100 only introduces a 1% error in the above calculation. This is important because it tells you that if the impedance is lower (either in general or at some frequency or band of frequencies), because E is constant, and I=E/Z, more current (and since P=IE more power) will be demanded, if it is higher, the oposite is the case.
Typically, the case where less power is demanded is no problem, but the case where more is demanded can lead to overload - after all, the power supply as well as power transistor current ratings or protection trip points are finite.
Normally, it is considered sound to design an amp rated for Z=8 ohm operation, to provide enough current for 4 ohm operation (not necessairly on a long term basis). In theory, this would mean that the amp is capable of 2x the rated power. In practise, since an 8 ohm nominal load does not dip to 4 ohms that often, and if it does, it's not for the full audio spectrum, AND due to the nature of an audio signal full power is not demanded long term, the current capability of the output is dimensioned for 4 ohm operation (often less), but not the cooling (heatsinks) or the power supply. The latter is normally overdimensioned by 50% and a hefty short term reserve over that is provided by correct choice of the filter caps in the power supply. For the former, there is usually an overheat protection circuit. This is also the reason that most amps rated at 100W into 8 ohms will not drive the theoretical double that into 4 ohms.

So, now we have one half of your answer - for a, say 150W SS power amp rated for 8 ohm load, the output voltage would be +-49V, and current CAPACITY (at least short term) would be 49/4 = 12.25A. If you would look at the output transistors, you would probably notice that their voltage and current ratings are signifficantly higher, but this is because the typical bipolar output transistors are not capable of withstanding BOTH high current and high voltage. The full datasheet gives something called a SOA curve (Safe Operating Area) which will tell you what the particular part can do in a particular application. Note that rated power increases with the square of the rated maximum output voltage (easily grasped intuitively: 2x voltage on the same load means 2x current, and since power=current x voltage, 2x voltage with 2x current means 4x the power)

Now, a typical PP tube amp will be somewhat different in this respect because it would typically be trtansformer couplled. The same is true for the rare transformer coupled SS amp.

A transformer is used to present the output of the amp with a particular favorable impedance it works best with (obviously a big factor when choosing the output devices). Therefore, the purpose of multiple output taps on the output transformer is to connect the load to the best match. An 8 ohm tap on a 100W amp will be capable of giving you 40V with at least the required 40/8=5A current, while a 4 ohm tap will give you 28.3V (sqrt(2) less than for 8 ohms) V but at least 7A current capacity (sqrt(2) more than for 8 ohms). If you use the P=(E^2)/(2xZ) rule, you will see that the idea is to keep the power output constant. Even for transformer coupled amps, be that tube or SS, the current capacity is igher than the minimum required in order to cater for dips in the load impedance. That said, in general the headroom is less than with direct coupled - this is because a load that tends to be largely lower than it's declared value would then be connected to the lower impedance tap on the transformer. There may be a small maximum power penalty but the amp will be able to better drive the load, without nasty problems like overheating and transformer core saturation.

Now lets get back to SS for a moment - some SS amps, in the interest of keeping the outputs safer and the available power better utilized, do the same thing by changeing the taps on the power transformer, so that the power rails are set for the particular impedance.

I should also add that difficult loads (not necesairly low impedance, but for instance very capacitive or inductive) will see the same overal figures of E and I. The problem is that due to phase difference between E and I, the actual power transmitted to the load will be lower, and the rest of the intended power will be reflected back into the amp, where it will have to be dissipated. So, this is one more reason why, looking at the parts in the amp, you may see apparenty grossly overdimensioned parts. At least if it is a good amp, that is ;-)

 

[Fisherdude]

Thanks, guys!! :thmbsp:

 

[cabinover]

Let me go back to the beginning again, and again, and again,and............I know there is at least one point I can grasp here besides P=IE

 

[ilimzn]

Ed, you are confusing two things (power and current) and also using a misleading load (light bulb) although your basic premise is correct. Let me address this one at a time:

Even though many peaks are transient, the loads are real and you need the current handling to deal with them.

Correct. Even though the maximum power is only seldomly required, what we view as seldomly (fractions of a second), is sufficiently long for the speed of the gain elements in an amp, that in most cases it can almost be viewed as long term. For one thing, the power stage of an amplifier must be designed to withstand transients even though on a very long term basis they may be very rare. The power supply is not as problematic as the filter caps act as a sort of a 'flywheel', it is easyer to provide longer term transient power. Finally, heatsinks are the slowest element on this list, the mass of metal involved provides thermal inertia, so again a sort of a filter, or flywheel efect.

Many SS supply voltages can be 60, 80 volts or higher. Lower volts would mean more amps are needed to supply the same "watts" to the speaker, so higher volts generally means you can get by on less current.

In general this is true but for one flaw - the amp itself only determines the MAXIMUM power that is available without damage or protective measures. The load is the real determining factor as to how much power gets delivered, as the load is really what makes the demand. The thing that ties the power rail voltage and the maximum power obtainable is the load impedance (or if we simplify things, resistance), as stated in two laws: Ohms and Joule's (see below).

If the output is 65 watts for example, divide that by the 80 volts to get .8 amp of current. A 40 volt supply would have to be able to provide double that amount of current. Just means heavier wire is needed and a beefier supply, so in practical terms, a 40 volt supply won't be asked to pump out 65 watts.

OK, here you are using the wrong law.
If you wanted 64W out of a 80V power supply, you need a resistor of the appropriate resistance to draw those 0.8A when connected to 80V. This resistor, according to Ohm's law has a value of 80V/0.8A = 100 ohm. Definitely not even close to the more usual 8 ohms

Look at it this way:
Suppose you wanted a maximum of 40V out of your amp. Obviously, i it's not transformer coupled on the output, you would need at least a 40V from the power supply since the amp works by 'wasting' the excess off the power rail voltage when it doesn't need as much. At best it can waste nothing.

Now, since we are talking AC, let's assume we are using what the testing lab is using, i.e. a sine wave (actual audio is a fair bit more complicated). Since AC changes direction, for our purposes, we can do that by providing +40 or -40V out of the amp. Therefore we use two power rails, one for the positive part and one for the negative part of the output - each 40V, one + one -.
An aside: this is not one 80V power supply. Even amps that are capacitor coupled and have only one power rail (in this case 80V), actually divide this into a positive and negative rail by means of an output capacitor, which is charged to half of the 80V, i.e. 40V. So, from the standpoint of the output, we again get + and - 40V.

Now, since we are using a sine wave, we have to remember that even though the peak of the sine wave is 40V, the sine wave is at this maximum level only for a very short time. If a DC 40V supply was connected to an 8 ohm resistor, we would get (via Ohms law) a current of 5A through it, and a power of P=IE = 5 * 40 = 200W. If you use P=EI and figure ohms law in it as I=E/R, you get that P=(E^2)/R, or E squared over R.
Since we are talking about a sine wave, there are two important things to note:
1) because positive voltage = positive current, negative voltage = negative current, and anything squared is always positive, the fact that it is AC does not change the P=EI or the P=(E^2)/R formula.
2) A sine wave varies in amplitude, and is only at it's maximum for a short time. For all other times it is lower, and for two short times it is zero or near. The point is, a sine wave with a peak at 40V will produce less power across a given resistance compared to DC which is always 40V. To be able to compare the power effectiveness of a sinewave AC to DC, the term 'effective voltage/current' is used. 40V peak sine AC has the same effect power-wise as 28.3V DC (40V / square root of two), or, as is commonly said, the effective voltage of a 40V peak sine is 28.3V. All other formulae stays the same.

Now, that we have this settled, how much power do you get out of a 40V peak sine connected to an 8 ohm load? Simply use the effective voltage in the calculation: P=(E^2)/R = (28.3 * 28.3)/8 = 100W.
that being said, all the elements of the amp itself DO have to be designed for the peak voltage and currents. Typically if you did put a square wave into the amp (effective voltage of a square wave is equal to it's peak voltage DC), it would come quite close to providing the full 200W, at very high efficiency. For a sinewave, the amp has to waste the excess voltage from the power rails for when the output sine wave is not at peak value, so efficiency is reduced. At maximum power (peak of sinewave as close to power rail as it can get) it's about 70% in the ideal case.

Next up is transformer or power supply sag, where, when high current is asked for, the supply can't deliver it, so voltage drops. Ohms Law now requires yet more amps of current, as watts delivered must balance out. At some point, something gives. I guess we'd call it distortion of some kind or another to keep this simple.

The flaw here is that 'power must balance out'. This is not the case. Power is not constant, output voltage is (or would be if the amp were ideal, but for a SS amp it comes quite close). If the voltage rails sag (and they do), the maximum voltage the amp can deliver is determined by the lowest voltage the power rail sags to (this is the worst case, when the output signal maximum coincides with the rail minimum). The difference between the unloadd and loaded rail is wasted in the resistance of the tranmsformer windings and rectifiers, as heat. Even though the amp proper sees less voltage and thus, has to waste less for the times the output waveform is below the rail voltage, so therefore it heats up less, the transformer, seeing extra losses, heats up more, and so does the rectifier. How much power one can actually safely get from an amp (as a component) is a balance - there are big differences as to how often and how long the maximum power is demanded (i.e. how long the transient lasts in teh audio signal), how much distortion is tolerated (*) and how long it takes the components in the amp and power supply to reach unsafe levels. At some point, if there is an overdemand, and there is a protection circuit, it will limit the power or shut off the amp entirely to protect it from damage.

The 'something's got to give' mostly falls into the cathegory of the input signal demanding more output voltage from the amp thjan it's power rails. At this point, the output waveform is clipped at the level of the power rail - and this is how you get clipping distortion.
Now, people usually figure the clip is a nice straight line cutting off the top portion of a sine wave. in reality, this can be very distorted, especially at low frequencies as the power supply ripple is superimposed on the top of the waveform, and even entire portions of the waveform, right down to zro, can be missing if there is a protection circuit involved.

For a real-world example, though not recommended, I put a 60-watt bulb on the speaker outputs of my old SWTPC Universal Tiger 90wpc amp. No problem making it glow almost like it was plugged into the wall, though it was pulsing at the beat of the music. Remember those "light boxes" of the 70's? Sorta like that. ;-)

This is not even close to a real world example. Even though a speaker has a varying impedance with frequency and also with amplitude (but to a lesser extent), it does not vary nearly as much as the resistance of an incandescent bulb. A resistance of a bulb increases very rapidy as the filament heats up, often by 10 times. Also, because the heat dissipation from teh filament is slow, there is no real frequency dependency unless we are talking very low frequency, well below 20 Hz. Also, at all times the bulb's resistance remains almost entirely an ideal resistance, there is only a tiny inductive component. A speaker, in turn, will have dips as low as half rated impedance, and peaks as high as several times rated impedance, depending on frequency to the greatest extent, and will, depending on frequency behave heavily inductive as well as heavily capacitive. That being said, a dummy load resistor is not a real world example either, since it has constant and pure resistance.

 

[waylyn5945]

Whew, having just read all of that, its going to take a while to dissect and absorb it. I've been an electrical engineer for many, many years but always in the industrial side using 120 - 600 volts AC and sometimes as much as 34,500 VAC. Believe me, this audio stuff is totally different - I've got a lot to learn. But many thanks to guys like EchoWars and ilimzn who share their wisdom. :thmbsp:

 

[Fisherdude]

Ain't that the truth!

Although I made it to my senior year as a EE, after three years in the army (and three years away from calculus) I ended up taking my degree in something else. It's almost painful trying to drag this out of my long-term memory banks!

I won't mention HOW long ago, but I did my Fortran programming on punch cards!

Just don't bring up wye & delta phase diagrams!! Aaaaiiiiiiieeeeee!!!

Clay