Non-Ideal Power Sources, Maximum Power Transfer

Non-Ideal Power Sources

When modeling circuits, ideal power sources are often used for simplicity. However, power sources actually have an internal source resistance which can be measured. In this experiment, we use a method to show that an internal resistance of a power sources exist.

We use the diagram above to model our circuit. The yellow portion indicates the voltage source V_s with its internal source resistance R_s. A resistor, R = 22 Ω, is wired in series so that we can measure the output voltage V_out across it. A source voltage of V_s = 1 V is used. The source current, I_s, and power dissipated by resistor R can be calculated.

First, we consider an ideal power source V_s (in red). This means we consider the circuit without R_s. It is apparent that the output voltage is simply the source voltage, 1 V. Using Ohm's Law, the source current can be calculated, I_s = 45.5 mA. The power can now be easily obtained by multiplying the current and voltage, P = 45.5 mW.

Next, we consider the non-ideal power source (in blue). Using a voltage divider, we find the output voltage to be V_out = R(1 V) / (R + R_s), which for the resistor chosen yields 22 / (22 + R_s).

The source current is then calculated as I_s = 1 / (R + R_s). The power then can easily be calculated, giving us P = R / (R + R_s)^2. All of these are functions of R_s and R (which we will vary 3 times in this experiment).

Measuring V_out (22.4 Ω) = .976 V

The source is provided by Analog Discovery interface using WaveForms. We apply a source voltage of 997 mV which gives us a reading of 1.000 V on the multimeter. We then apply this voltage to a resistor R which is varied.

The chart shows the output voltages (V_out) measured for the three measured resistor (R) values. The three calculated columns show the source resistance (R_s), source current (I_s) and power through resistor R (P_R). All values are in standard units (V, Ω, A, and W). There is a source resistance of about 0.43 Ω (average). The power dissipated by the resistor R then an average of 0.0296 W.

Maximum Power Transfer

The maximum power that can be delivered to a load is investigated in this experiment. A derivation can be a bit lengthy but can be summed up here: Maximum Power Transfer. It turns out that maximum power is achieved when source resistance is equal to load resistance, R_s = R_l.

A voltage source of 5 V is given and the two resistors are 2.2 kΩ. The theoretical output voltage (2.5 V) is calculated as half of the source voltage (5 V) since the resistors are of equal value. The power is then calculated as shown, yielding 2.2 mW.

The circuit is then wired with 5 V voltage source from the Analog Discovery and two resistors measuring 2.16 kΩ. 

The output voltage of 2.49 V measured is then used to calculate power over the load resistor. The power obtained is 2.87 W (33% error).

The error is too large to ignore, so the theoretical value is recalculated using the actual resistance of the resistors. P = (V_out)^2 / R = (2.5)^2 / (2160) = 2.9 mW. This shows that the high error was due to the difference in theoretical resistor (2.2 kΩ) and actual resistor (2.16 kΩ) used. This would now give a percent error of only 1.0%.

The maximum power then is about 2.9 mW. If we used a potentiometer to change the resistance of one of the resistors, we would see that for any other resistance value, the power would decrease.