We can use the concept of impedance matching that we just discussed. To talk a little bit about how to connect an amplifier and a loudspeaker to get maximum power transfer, and therefore maximum volume and efficiency out of your system. So, we're going to look at this circuit. It's basically the same circuit that we were just looking at. But no I m going to, instead of saying I have a DC source, let's say that the amplifier is some kind of a time varying signal coming out of it. And so I just used the AC, voltage source to, represent that. Now let's say you, pick up a loud speaker with an 8 ohm impedence. Now a real loud speaker is not, does not just look like a simple 8 ohm resistor at all frequencies, it's a much more complex, circuit element than that. And we'll talk a little bit about that in the future. But, for now, let's just say it's 8 ohms. And so, we just decided from the previous analysis, that to get the maximum power transfer from the amplifier to the loud speaker, we ought to build an amplifier that has an internal output impedance of 8 ohms. And then if the loud speaker is 8 ohms, I'll get maximum power transfer from the amp to the loud speaker, and I'll be making most efficient use of the power that the amplifier is drawing from the the wall socket. Now, just as of a little aside. Let's say your trying to build a speaker cabinet, like say a base cabinet with four speakers in it. And how would you connect those? And let's say then that you have an amplifier with an internal impedance of 8 ohms, well how would I connect these 4 8 ohm speakers to get maximum power transfer? Okay, so the question is, how do we connect 4 8 ohm speakers, so I can get maximum efficiency out of this amplifier. So, I've got lots of choices on how I can do that. I could take and connect all of them in series, so I could connect this one to that one and then like this. If I did that, though, this would be a total of 32 ohms. And so if this, is a 32-ohm impedance this is going to not be the most efficient power transfer, so that's not a good solution. the solution to this is going to be to take two of these speakers, connect them in series, like this, and that gives you 16 ohms. And then take the other pair, connect them in series. That's another 16-ohm impedance. And then if I connect these two pairs of speakers in parallel, I have 16 in parallel with 16, that's going to give me 8 ohms. So that's the, the way to connect these speakers together, so I get maximum power transfer from my 8 ohm output impedance amplifier, to my array of 4 8 ohm loudspeakers. Okay, now before we go on and talk about [UNKNOWN] amps, just to complete the set of sources we're going to be dealing with, I need to talk a little bit about current sources. Now, let's start by talking about the ideal current source. Now, it's just what it says. a cur, a source of current is, is just something that if I connect the load resister to it, it's going to force its current through that resister regardless of the value of the load resister. And so, that means that the ideal current source can generate any voltage that you want. So, if I take, let's say I had a 1 amp current source here, and I connected a 100 ohm load resistor to it. That means that this current source would have to generate 100 volts to force, to continue forcing its 1 amp through this 100 ohm resistor. Now just like voltage sources current sources, real current sources are non ideal, that means they could produce only so much voltage. There's a limit to how much voltage they can produce. So, the way to represent a real current source is we add an internal resistance in parallel with the ideal source. And so, this pair of terminals is looking into the real current source. There's this internal resistance that we can't separate from the current source. It's internal to it, it's part of it, it's, it's there you're stuck with it. So now, if I take a look at at this and I ask what is the maximum voltage I can generate across a load. Don't forget now I've got this parallel internal resistance with the load, and so the voltage is limited to I times R internal. Now to look at the limiting case, let's say that RL you make it bigger and bigger. So, it goes tends off to infinity. So, it's a mega ohm, 10 mega ohms, 100 mega ohms, you can make it bigger and bigger. the parallel combination of RL with R internal is going to just approach R internal. remember the expression for parallel combination of two resistors is going to be R internal times RL times RL over R internal plus RL. So when RL gets really large, you can ignore the R internal downstairs, and I have RL over RL, that's a 1. So, this whole thing in limit of RL going to infinity, this whole thing goes to R internal. So, if I put a really large load resistor, the parallel combination is just R internal, and then the voltage all of the current is going to go through R internal, and our voltage is limited to i times R internal. So, an ideal voltage source cannot supply. I'm sorry. An ideal voltage source can supply any amount of current. An ideal current source can generate any voltage. A real voltage source has a limited current generating capacity, can only supply so much current. A real current source can only provide so much voltage, a limited voltage. So, there's a nice symmetry there, and so keep that in mind.