Thevinin Equivalents

PROBLEM
 Suppose you have two power supplies, and multiple loads. Given a minimum acceptable voltage across a load, determine the smallest equivalent load resistance that can be successfully used, and what is the power consumed. What voltage will exist across the terminals of the load if we remove it (open circuit/highest voltage/Voc) and what short-circuit current will flow if we replace the load by a short (isc/highest current)?

METHOD
 We model a system similar to Figure 1:

Figure 1

We want to reduce the circuit to it's Thevenin Equivalent consisting of a voltage source and a resistor -- we want to "thevenize" the left-hand side of the circuit. We remove RL2 from the circuit to find Voc=Vx=Vth.

Vs1=Vs2= 9 V

We use nodal analysis at node x to find Vx, which results in the following nodal equation:
Vx/680 + (Vx-9)/100 + (Vx-9)/39 = 0  >> Vx = 8.659 V = Voc = Vth

We find Rth two ways:

We first find Rth using equivalent resistances: Rth = Rc3 + 1/(1/Rc1 +1/Rc2) = 66 ohms

Additionally, we can find Vy using nodal analysis. Next we find isc = Vy/Rc3; finally, we can find Rth = Voc/isc= 66ohms

The nodal equation for Vy:

(Vy-9)/39 + (Vy-9)/100 + Vy/680 + Vy/39 = 0 >> Vy = 5.11 V >> isc = 5.11/39 = 0.131 A
Thus, Rth = 8.659/0.131 = 66 ohms

Finally, we have our Thevenin equivalent circuit:

Vth = 8.659 V      Rth = 66 ohms

Given that the minimum load across Vload is 8 V, we can find the smallest permissible value for RL2 using a voltage divider. We also find the isc using Ohm's Law, and the open circuit voltage by inspection.

Vload=Vth*RL2/(RL2+Rth) >> RL2 = Rth*Vload/(Vth-Vload) = 66*8/(8.659-8) = 801 ohms

isc = vload/RL2 = 8/801= 9.99 mA

Voc = Vth = 8.659 V

DATA
 Next, we build the Thevenin equivalent circuit to find the measured voltage across the load, Vload and the open current voltage Voc. Percent error is calculated.
Components Data Table

Component	Nominal Value	Measured Value	Power/Current Rating
Rth	66 ohms	66.5 ohms	0.3 W
RL2,min	825 ohms	826 ohms	0.3 W
Vth	8.659	8.71	10.0 A

Thevenin Equivalent Experimental Data Table

Configuration	Theoretical Value	Measured Value	Percent Error
RL2=RL2,min	8.00 V	8.04 V	0.50%
RL2=infinity	8.569 V	8.54 V	0.34%

Next, we build the system in Figure 1 on the breadboard and again measure the voltage across the load, Vload and the open current voltage Voc. We also calculate the power supplied to/absorbed by RL2 when RL2=Rth. Percent error is calculated.
Components Data Table

Component	Nominal Value	Measured Value	Power/Current Rating
Rc1	100 ohms	100. 3 ohms	1/8 W
Rc2	39 ohms	38.1 ohms	1/8 W
Rc3	39 ohms	39.1 ohms	1/8 W
RL1	680 ohms	679 ohms	1/4 W
Vs1	9 V	8.96 V	10.0 A
Vs2	9 V	9.05 V	10.0 A

Figure 1 Circuit Experimental Data Table

Configuration	Theoretical Value	Measured Value	Percent Error
RL2=RL2,min	8.00 V	7.94 V	0.75%
RL2=infinity	8.569 V	8.64 V	0.83%

Maximum power to RL2 when RL2=Rth:
P=((Vth)^2)/4Rth >> (8.659)^2/4*66 = .284 W
Thus, Pmax = 0.284

Finally, we verify the maximum power by adjusting values of RL2 and measuring the load voltage each time and calculating power, where Pload = ((Vload^2))/R

Power Data Table

Configuration	Vload	Pload
RL2=0.5Rth	2.75 V	0.229 W
RL2=Rth	4.14 V	0.259 W
RL2=2Rth	5.57 V	0.235 W