NDSU
January 3, 2013
Maximum Power Transfer
Problem: Determine R to deliver maximum power to the load.
Solution: Since you can reduce any circuit to its Thevenin equivalent, let's use that.
Rth
Vth
+
-
RL
Case 1: Rth is fixed. Find RL to maximize the power to the load.
Note that there is a maximum point:
If RL = 0, the power to the load, P = I 2 R L = 0
If RL = infinity, I = 0 and the power to the load is again zero.
Somewhere between RL = 0 and RL = infinity is a maximum power transfer.
You can compute this as
V th 
I =  R th +R
L
P = I2RL
RL
 V th
P = 
(R th +R L ) 2 
The maximum occures when the derivative is equal to zero:
RL
d 

dR L  (R th +R L ) 2 
=0
Recall the derivative of a fraction is
 ho de hi - hi de ho 
ho ho


 (R L +R th ) 2 −2R L (R L +R th )  = 0


(R th +R L ) 4
(R L + R th )(R th − R L ) = 0
The maximum power transfer happens at RL = Rth
The maximum power delivery happens at RL = - Rth. You can ignore this solution since negative resistance doesn't
normally exist.
In MATLAB, you can verify this. Assume for instance that Vth = 12V and Rth = 2 Ohms:
-->Vin = 12;
NDSU
January 3, 2013
-->Rth = 2;
-->RL = [0:0.01:10]';
-->I = Vin ./ (Rth + RL);
-->P = (I .^ 2) .* RL;
-->plot(RL, P);
-->xgrid(5)
-->xlabel('RL (Ohms)');
-->ylabel('Power Transfer (Watts)')
Note that at maximum power transfer, you are 50% efficient: 18W is dissipared in the load, and 18W is dissipared
in the source resistance (Rth).
Case 2: RL is fixed. Find Rth to maximize the power to the load.
The solution in this case is not Rth = RL. If you think about it, you want Rth = 0 or as close to zero as possible.
The maximum is at Rth = -RL, at which point the loop resistance is zero and the current is infinity. If you can't have
negative resistance, make it as close to -RL as possible (i.e. as small as possible.)
You can check this with MATLAB. Let Vth = 12V and RL = 2 Ohms. Plot the power to the load for Rth = 0 to 10
Ohms:
-->Vth = 12;
-->RL = 2;
-->Rth = [0:0.01:10]';
-->I = Vth ./ (RL + Rth);
-->P = ( I .^ 2 ) .* RL;
-->plot(Rth, P);
-->xlabel('Rth (Ohms)');
-->ylabel('Power Transfer (Watts)')
NDSU
January 3, 2013
As you would expect, Rth is bad: it makes the source less efficient at delivering current. Ideally, you want Rth as
small as possible.
Case 3: Find RL to maximize the efficiency of the system.
Efficiency is the power to the load divided by the total power dissipated.
P

eff =  PLoad
total 
or
eff =
I2RL
I 2 (R th +R L )
R
eff =  R L +RL th 
This again has a nonsense solution. The maximum efficiency is when RL = infinity. At that point, 100% of the
power goes to the load. Of course, it's 100% of zero, but it's still 100%.
That's one of the problems of delivering power to a load
For high efficiency, you want RL >> Rth
For maximum power, you want RL = Rth