Internal Resistance and the Effects of Loading

Posted by Jeff Schuler on Jul 27, 2009 in Uncategorized |

purely-resistive-circuits

Loading is a fairly difficult concept to grasp.  Consider the two circuits in Figure 1.  Which one presents the larger load to the source?  Though it may seem that the bigger resistor is the bigger load, the smaller resistor actually places the larger load on the source.  This is because the smaller resistor draws more current from the source, requiring the source to deliver more power.  Applying Ohm’s law shows that a 100 Ω resistor will draw twice the current and twice the power as a 200 Ω resistor.

Internal Resistance and the Effects of Loading

In the example above, the difference between the two resistors didn’t really matter.  Yes, the 100 Ω resistor drew twice as much power, but other than a higher energy bill, the smaller resistance had no effect on the source.  The source in Figure 1 is an ideal voltage source, meaning it has no internal resistance and can supply an unlimited amount of current.  An ideal voltage source is a theoretical phenomenon not found in real life.  In theory, you could short circuit an ideal voltage source, draw an infinite amount of current, and everything would be fine.  Real life voltage sources, however, have an internal resistance.  Circuit theory models a real voltage source by placing a resistor in series with an ideal voltage source.

Ideal voltage sources vs. real voltage sources

ideal-and-real-voltage-sources

Figure 2 contrasts an ideal voltage source with a real voltage source.  The open circuit voltage is the voltage measured when nothing is connected to the source.  Because nothing is connected, no current is drawn and there is no voltage drop across the internal resistance (V = IR).  Therefore, the open circuit voltage will be the same for both ideal and real voltage sources because there is no current drawn from the source.

While the internal resistance has no affect on the open circuit voltage, this small amount of resistance begins to matter once a load (a resistor or circuit) is connected.  Consider the two circuits in Figure 3 to see how internal resistance affects real voltage sources.  Note that Vsupply has been replaced with Vopen circuit.  These two terms are interchangeable.

internal-resistance

Both sources are set to 15 V, but the real source has 50 Ω of internal resistance (the same as the function generator found in the lab).  Once the 100 Ω load is connected, current runs through the 50 Ω resistor, creating a 5 V difference between the open circuit voltage and the terminal voltage (the terminal voltage is the voltage measured across the load).  The difference between the two can be found using Ohm’s law.

V_{oc} - V_{T} = I R_{i}

Where Voc is the open circuit voltage, VT is the terminal voltage, and Ri is the internal resistance.

Minimizing the Effects of Loading

The example in Figure 3 reveals a problem often encountered in electronics: how can the effects of loading be avoided?  The answer is that while loading is always present, its effects can be minimized.  The best supplies have a terminal voltage that is virtually the same as the open circuit voltage.  When a large portion of a supply’s voltage is lost across its internal resistance, less voltage is available for the load.

internal-resistance-current-flow

For the supply in Figure 3, the easy way to reclaim that “lost” voltage is to reduce the load on the supply.  Let’s replace the 100 Ω load with a 20 kΩ load (a common resistance in the audio world).  Figure 4 shows that increasing the load resistance decreases the load placed on the supply.  The new load draws a much smaller current than before and, as a result, the terminal voltage is virtually the same as the open circuit voltage.

This example demonstrates an important method for avoiding the negative effects of loading: when hooking up two circuits, the circuit that supplies the signal should have a low internal resistance, or “output” resistance.  Meanwhile, the circuit that receives the signal should have a high “input” resistance. The terms “output” and “input” are used because the signal comes out of the first circuit and into the second circuit.

Note: Keep in mind that the load resistors used in these examples could possibly represent a much larger circuit.  The first circuit placed a heavy 100 Ω load on the supply while the second was a much lighter load of 20 kΩ.

Input and output resistance and loading will come up over and over throughout your electronics courses.  Don’t worry if these concepts mean little or nothing to you now.  It’s important that you are exposed to loading now so that you’ll recognize this concept in a couple of months when you’re working on your final project and loading suddenly become important.

Finding Internal Resistance Using Ohm’s Law

Problem: A power supply is connected to a circuit.  The terminal voltage is measured to be

19.2 V when the circuit draws 80 mA of current and drops to 18.8 V when the circuit draws

120 mA.  Find the open circuit voltage and internal resistance of the power supply.

Solution: First, you should understand what the question is asking.  You should understand that Ri and Voc are the same in both instances.  The only difference between the two is that the circuit draws more current in one instance.  Next, draw the two circuits.  You know that the power supply must be real.  Otherwise, the terminal voltage would be the same no matter what current the circuit draws.  If the supply is real, then the circuit must look similar to the circuit on the right in Figure 3.  Figure 5 shows the two circuits used to solve the problem with all known variables labeled.

calculating-internal-resistanceEquations for the voltage drop across R_{i} can be written for each circuit:

General Equation: V_{oc} - V_{T} = I R_{i}

Circuit 1: V_{oc} - 19.2 = 0.08 R_{i}

V_{oc} - 0.08 R_{i} = 19.2

Circuit 2: V_{oc} - 18.8 = 0.12 R_{i}

V_{oc} - 0.12 R_{i} = 18.8

Now we have two equations and two unknowns.  Solve the system of equations to find Ri and Voc.  The easiest way is to eliminate one variable by subtracting one equation from the other.

V_{oc} - 0.08 R_{i} = 19.2

-V_{oc} - 0.12 R_{i} = 18.8

______________________________

0 + 0.04 R_{i} = 0.4

Solve for R_{i} to find the internal resistance of the power supply.

R_{i} = 10 \Omega

Finally, plug the answer for R_{i} back into one of the equations to find the open circuit voltage.

V_{oc} - 19.2 = 0.08 R_{i}

V_{oc} = 0.08 R_{i} + 19.2

V_{oc} = 0.08 (10) + 19.2

V_{oc} = 20V

Thanks to Ryan Eatinger, Kansas State University (reatinge@ksu.edu) for material contribution.

 

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