Within the field of electronics, a delicate balance between energy transfer and efficiency must be maintained; however, in some circumstances, maximum power ( P ) must be delivered to a load without regard to how inefficient the process may be. Herein lies the concept of maximum power transfer. The mathematical derivations relevant to this process involve calculus, and thus, they are beyond the scope of an entry-level electronics course. Nonetheless, Thevenin equivalent circuits can indeed be used to give an introductory glimpse into the underlying concepts at hand. We will begin our brief introduction to this topic by viewing a diagram of a Thevenin equivalent circuit. It is customary, however, to replace the usual VTH and RTH designations with Vs and Rs for the sake of simplicity:
For review purposes, recall that three power equations are of significance to us:
- P = IV
V = IR, so…
- P = ( I )( IR ) = I2R
I = ( V / R ), so…
- P = ( V / R )( V ) = V2 / R
The power equation of primary interest to us in this instance is equation 2, because altering the value of the load resistance ( RL ) has measurable effects on the current ( I ) delivered to the load:
I = [ Vs / ( Rs + RL ) ]
As it turns out, the maximum amount of power that can be delivered to a load is delivered when Rs and RL are equal in value. Furthermore, it is of crucial importance to understand that Rs is the internal resistance of the source at hand. Thus far, the source resistance has been treated as if it is non-existent, but when paired with a load resistor of equal value, this resistance value enables power delivery that is necessary to operate various types of audio systems in particular. As far as efficiency is concerned, the ratio of power delivered to a load vs. the power that is actually converted to useful work is summed up as follows:
Efficiency = ( Output / Input ) x 100%
Suffice it to say that the power transferred from the source is consequential to the Rs and RL values of the circuit. To the contrary, the power utilized by the load is calculated using RL by itself. The relationship between power input and output is simplified by setting Rs and RL equal to one another within the input power derivation:
Power Transferred From the Source
Ps = ( I2 )( Rs + RL )
Rs = RL
Ps = ( I2 )( RL + RL )
Ps = ( I2 )( 2 RL )
Power Utilized by the Load
PL = ( I2 )( RL )
Efficiency = ( PL / Ps ) x 100%
Efficiency = [ I2RL / ( I2 )( 2 RL ) ]
Efficiency = 50%
What happens when the load resistance is either greater or less than the internal resistance of the source? If the load resistance is less than the source resistance, almost no current is being captured and used by the load; thus, power utilization by the load is minimal at best, and heat losses occur within the source in the form of a shorted circuit ( ouch!!! ). If the load resistance is greater than the source’s internal resistance, the system becomes more efficient, but less power is generated by the load. As the load resistance approaches infinity, current cannot pass through the load resistor, and the power utilized by the load has a value of zero:
Maximum Power Delivery
Rs = RL
Maximum Efficiency with Lower Power Delivery
RL > Rs
Minimal Power Delivery
Rs > RL
or,
RL → ∞
As is always the case, the best proof of the validity of any theorem matches theory and mathematical observation. By measuring power output as a function of load resistance, we can plot several power output values onto a graph that visually matches and satisfies the aforementioned claims:
Q: The source in the diagram below has an internal resistance of 75 Ω:
What is the load power when the load has the following resistance values?
- 0 Ω
- 25 Ω
- 50 Ω
- 75 Ω
- 100 Ω
- 125 Ω
A:
a. PL = ( I2 )( RL )
I = [ Vs / ( Rs + RL ) ]
I = [ 10 V / ( 75 Ω + 0 Ω ) ]
I = 0.133 A = 133 mA
PL = ( 133 mA )2( 0 Ω ) = 0 mA
b. I = [ 10 V / ( 75 Ω + 25 Ω ) ]
I = 0.1 A = 100 mA
PL = ( 100 mA )2( 25 Ω ) = 0.25 W = 250 mW
c. I = [ 10 V / ( 75 Ω + 50 Ω ) ]
I = 0.08 A = 80 mA
PL = ( 80 mA )2( 50 Ω ) = 0.32 W = 320 mW
d. I = [ 10 V / ( 75 Ω + 75 Ω ) ]
I = 0.0667 A = 66.7 mA
PL = ( 66.7 mA )2( 75 Ω ) = 0.333 W = 333 mW
e. I = [ 10 V / ( 75 Ω + 100 Ω ) ]
I = 0.0571 A = 57.1 mA
PL = ( 57.1 mA )2( 100 Ω ) = 0.327 W = 327 mW
f. I = [ 10 V / ( 75 Ω + 125 Ω ) ]
I = 0.05 A = 50 mA
PL = ( 50 mA )2( 125 Ω ) = 0.313 W = 313 mW
RENAISSANCE PHYSICS 