Maximum power transfer vs Max power efficiency

• fisico30
For example, if you have an amplifier with an output of 10 million ohms and you connect a loudspeaker with an impedance of 50 ohms, the amplifier will deliver 10 million ohms of power to the loudspeaker. However, if you connect a loudspeaker with an impedance of 100 ohms to the amplifier, the amplifier will only deliver 1 million ohms of power to the loudspeaker.In cases where the power being delivered is not important, but the speed at which the power is delivered is, max power transfer is the better option. For example, if you have a battery that has an emf of 10 volts and an internal resistance of 5 ohms, and you connectf

fisico30

Hello forum,

what is the main difference between max power transfer and max power efficiency?

In the case of max power transfer, the goal is to transfer the most energy to the load per unit time (impedance matching). Power is I*V and is about delivering energy very fast...
High power means receiving lots of energy very fast or little energy at an even faster rate...

10\$/day for a month represents higher power than receiving 10 millions over 10 centuries. the total final amount is less but the per unit time amount is larger...

In the case of power efficiency, it seems to be about the load having the highest voltage. I am not clear why would be a form of power efficiency.How do I need to interpret that in terms of energy?

Some devices need a large voltage and don't care about high power...any example?

thanks
fisico30

Are you familiar with the fact that max power is transferred to a load,R, when R is equal to the internal resistance,r, of the power supply?
If you have a battery with emf = E and internal resistance = r connected to a load resistor = R
Then the current in the circuit I = E/(R+r) and the voltage across the load (the terminal pd)
V= IxR. For example if E =10V, r=5Ω and we make R a large value, let's say 95 then I = 10/100 =0.1A. The power dissipated in the load R = I^2 x R = 0.95W.
If we now make R small, say 1Ω then I =10/6 = 1.67 and the power dissipated in R = 2.8W
Now make R = r = 5Ω Now I = 10/10 =1A and the power dissipated = 1x1x5 =5W.
If you put more values into the equation you will find that 5W is the max power that can be transferred. Ironically it also means that max power is being wasted in the internal resistance which means that the efficiency when max power is being transferred is 50% !
The efficiency is given by (power in load)/(power from battery)
power in load = V x I, power from battery = E x I so efficiencey = VxI/ExI = V/E
this means that the efficiency can only be 100% when V = E but this means something ridiculous !
V can only be equal to E when there is NO CURRENT flowing so the efficiency is 100% when it is doing nothing !

In the limit case where r=0 and R-> infinity the battery voltage is all across the load R.

there is no power dissipated at all. It is strange to talk about "power" efficiency, since neither r or R are receiving any power...

Of course, if r<<<<<R but both are still finite, very little power is output from the battery (the less power as R/r gets bigger) but most of that power is dissipated across R and almost no power dissipated across r...

In the impedance matching case, 50% to r and 50% to R. But that could be 50% of 10000 W.
When there is no impedance matching, R receives almost 100 % of the power, but that power is super small...
That's what we get when we are greedy: you want it all but it is not worth much.

I think I get it now...

But not all devices need a lot of power. Some need a high voltage instead. Any example? All devices need energy though...

Thanks!
fisico

The importance of max power transfer is in cases such as loudspeaker connected to an amplifier. To get max power transferred the resistance (impedance) of the loudspeaker should me 'matched' to the output resistance (impedance) of the amplifier.
Not surprisingly this is called 'impedance matching'