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28th October 2005, 12:42 PM  #11 
diyAudio Member

Um... I see nothing wrong on hyperphysics. In fact they are a valuable resource and should be referenced more often.
What you are misunderstanding is a basic concept of maximum power transfer vs. output impedance. Power transfer is maximized when the load impedance is equal to the Thevenin source impedance of the signal source. Let me put it this way: Connect an ideal voltage source V in series with a resistor Ri. Now connect a load resistor RL to complete the loop. The current can be expressed as voltage V through the sum resistance, I = V / (Ri+RL). The voltage developed across RL is proportional to that current: VL = RL * [V / (Ri + RL)] = V*RL / (Ri+RL). Power is voltage times amperage, or PL = VL * I = [V / (Ri+RL)]*[V*RL / Ri+RL)], or P = V^2*RL / (Ri+RL)^2 . This says power goes up with the square of voltage, but the rest of the equation isn't easy to figure out by inspection. You could graph the equation with RL as the variable and see how it works, but there's a better way to find what we want. I'll take the derivative ("rate of change") of P with respect to RL. When dP/dRL = 0, we know we are at either a peak or valley in the above equation, because for a small change in RL, there's basically no change in P. Nice, eh? This is a quotient rule problem so I won't write the steps; suffice it to say, dP/dRL = V^2 * [(RiRL) / (Ri+RL)^3] (yes, the denominator is cubed.) That has even less meaning, but if we take dP/dRL = 0 and solve for RL, we can find all the peaks and valleys in the original equation. Now this part gets real fun  I mean it, the answer just plops out in all its elegance: First take distributive, seperate the +Ri and RL terms (they still have a common denominator): 0 = V^2*Ri / (Ri+RL)^3  V^2*RL / (Ri+RL)^3 Add thee negative term to both sides (move it to the left): V^2*RL / (Ri+RL)^3 = V^2*Ri / (Ri+RL)^3 Now cancel some things. What can we cancel? Heck, almost the entire equation! V^2 drops, and the denominator too. We are left with: RL = Ri. This was found by taking the derivative, which means we used the same RL, Ri, etc. as in the original equation. Nothing's changed, so we can go back to it: P = V^2*RL / (Ri+RL)^2 Since RL = Ri for dP/dRL = 0, we know when RL = Ri is either a maxima or minima. We don't know which! There are two ways to tell: 1. take the second derivative and determine if it is positive or negative; 2. look at values of P at, less than, and greater than, RL = Ri. (Remember RL is the variable and P is a function of RL, i.e., P(RL).) I don't feel like taking the derivative again so let's try plugging values into P. Not real numbers, we don't need a specific case.. let's just set RL to some logical values. First of all, we need to know P at RL = Ri. This comes to P = V^2 / 4Ri. First of all, power and resistance cannot be negative because such would be meaningless. Second, even if voltage were negative, V^2 is always positive. This tells us P(Ri) (i.e., RL = Ri) is positive, so to prove it is a minima, we need to find values greater than V^2 / 4R. A logical starting point is taking extremes: RL = 0 and infinity (i.e., short and open circuit). We know from electronics that both of these situations have zero power dissipated, but let's check just to make sure. P(0) = V^2*(0) / (Ri + 0)^2 Anything times zero is zero, and zero divided by anything is zero. Zero is clearly less than V^2 / 4Ri, so this end checks out. For large values of RL, we are tempted to take a specific case and say RL is some large value, say 1Gohm, but how do we know Ri isn't a teraohm? There are two ways to handle this: we can let RL equal to some arbitrary multiple of Ri, say two times, or let RL approach some insane value, like infinity. Both work: the first, taking RL = 2Ri, results in a coefficient of 2/9ths, which is less than the 1/4th coefficient we got for RL = Ri. If we take the limit as RL approaches infinity (you can't actually let RL = infinity because infinity isn't a number ), P goes to zero. (Think of it this way: as RL becomes much bigger than Ri, the denominator looks like RL squared. This cancels with the RL in the numerator giving V^2/RL [for large RL]. For even larger RL, P becomes a very small fraction, and eventually, zero. Since zero at both ends is less than 1/4 V^2/R, we can conclude that power is at a maximum at RL = Ri. There. I did all the math, algebra, and even some calculus for 'ya to prove that Pmax is at RL = Ri. What's that you say, so why do power amps have Zo != RL? Yep, I just spent half an hour talking about ONLY Thevenin equivalent circuits. A power amplifier does not necessarily have an equivalent representative circuit inside it. In fact they rarely do (which does at least make Hyperphysics' diagram erroneous). So what then? You have to understand the difference between maximum power and output impedance. So what is output impedance? It's the exact electrical equivalent of me putting on a policeman's clothes. Just because I look the part doesn't mean I have any legal power over you... In the same way, you can adjust any *active* electrical circuit to appear as as miliohm output impedance, but that doesn't mean you'll be pulling kiloamperes from it when you short the output! Amplifiers also have to worry about efficiency. Take this example: say you generate a square wave with a pair of MOSFETs. MOSFETs are essentially electrical switches with a welldefined ON resistance. Say the squarewave is 10V tall and the FETs are one ohm each. If you designed the circuit for 90% efficiency, then you intended that the load be about 9 ohms  10V aross 9+1 ohms is 1 ampere, which is 9V across the output and 1V across whichever FET is turned on, and likewise, 9W output, 1W dissipated in the FET, hence 90% efficiency. This current situation has damping factor = 9, 9W power output and 90% efficiency. If you reduce load to 1 ohm, it matches the Thevenin resistance of the generator and the 10V supply is shared evenly across both: 5V in the FETs, 5V on the load. This makes 5 amps, which is 25W  far more power than before  but also puts 25W into the FETs, astronomically more than the *1W* they were chosen at design time! In a linear amplifier, you can account for output impedance effects by detecting the loss and applying more signal in proportion. This is negative feedback. You still have the same old amplifier, so Zo can go stone dead zero perfection, but you're still left with the same limits on voltage and current. Understand now? Tim (*saves post for adding to website*)
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28th October 2005, 12:57 PM  #12 
diyAudio Member
Join Date: Sep 2002
Location: Sweden

No Tim, it is wrong. As I said previously, this is only interesting when the source impedance is fixed and you can choose the load impedance freely in order to maximize the power transfer. In this case, the load impedance, ie. the speaker, is fixed and this guy claims you should choose an amplifier with the same impedance as the speaker. This would meant that you have to burn away half the power in the amplifier to no purpose. You could use the theory correctly, though, so if you have for instance an amplifier with 0.1 Ohm output impedance, you should choose a speaker with 0.1 Ohm output impedance in order to maximize the power transfer. However, that is only relevant in theory, I am afraid, since no audio amplifier would be able to deliver the currents required for such low load impedances.

28th October 2005, 12:57 PM  #13 
diyAudio Member
Join Date: Jun 2004
Location: Warsaw

That's a great example how math with wrong assumpions leads to wrong answers.
Math stayed pure here, elegant... dP/dRl makes not much sense, because it is very hard to change Rl, do you imagine dynamic speakers of 1mOhm or 1 MOhm impedance? I don't. You should count dP/dPi because electronic circuit can have different Zout, like mOhms, MOhms, negative... If you are able to count derivatives... well? I if one can use a hammer everything looks like a thumb. BTW I mailed to the author and didn't get any answer so far. 
28th October 2005, 01:43 PM  #14  
diyAudio Member

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In my later FET based example, I mention what would happen if you attempted to match to an (uncompensated) amplifier, and noted that such conditions most often bring the amplifier out of design ratings. THAT is why doing it in general is a bad idea. Quote:
[quote]dP/dRl makes not much sense, because it is very hard to change Rl, do you imagine dynamic speakers of 1mOhm or 1 MOhm impedance? I don't.[/b] Er, this applies to any sourceload circuit. From spot welders to lightning bolts and everything inbetween, if you can establish a Thevenin equivalent circuit for the source, you can apply the power transfer theorem (which I proved above). It's quite easy to change RL, either by selecting resistors and graphing data points or by using a continuously variable resistor. Taking it to exactly zero and near infinity is just one of the mathematical tools that's more powerful than a single experiment can be. Quote:
Tim
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28th October 2005, 01:53 PM  #15  
diyAudio Member
Join Date: Sep 2002
Location: Sweden

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28th October 2005, 02:20 PM  #16  
diyAudio Member
Join Date: Jun 2002
Location: Left Coast

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28th October 2005, 08:46 PM  #17 
diyAudio Member
Join Date: Jun 2004
Location: Warsaw

I've once read about matching output impedance vs. freq to the speaker impedance vs. freq as an experiment. The book was from early '50s I think and author claimed that output impedance makes little audible difference in HiFi, but bigger difference in LoFi.

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