Superposition Theorem: Independent Sources and Circuit Design

February 13, 2020 Cadence PCB Solutions

Comic-text pop that says Super!

 

One could argue that the word “super” has become less-than-super with all the various uses in our language. Let’s think about this for a moment. We have superheroes, super-sized burgers and fries, supercarriers, and—of course—one-of-a-kind super deals. Given a little time, I’m sure that you can think of even more super things.

Linear Circuits are Supercalifragilisticexpialidocious

Of all the super things that I know about, the superposition theorem has always been a favorite. The theorem shows as:

“The current in any circuit element or voltage across any element of a linear, bilateral network is the algebraic sum of the currents or voltages separately produced by each source of energy.”

In its very basic sense, the superposition theorem tells us how to solve a circuit that has more than one independent source.  Using this method, we can consider the effects of the sources one by one. A circuit that has n independent sources requires solving n separate circuits. 

While the superposition theorem seems to-good-be-true for avoiding simultaneous equations when solving a complex network, it only applies to linear circuits and linear circuit responses. When we think of Ohm’s law, we’re working with linear components such as resistance, inductance, and capacitance that remain constant with respect to current and voltage. Any linear circuit—whether resistive, inductive, or capacitive—has a proportional output for any given input. Linear responses represent the current flowing through any branch and the voltage drop across any component.

The Superposition Theorem Has a Super Duper Rating

Superposition occurs through voltage division, current division, series-parallel combinations, and Ohm’s law. To analyze the effect of each source on the circuit, we must set all the values of the other sources to zero. We can accomplish this by replacing every voltage source with a short circuit and every current source with an open circuit. After considering each source, we can discover the total current or voltage for an element by superimposing each of the component currents or voltages.

We can use the superposition theorem to find the currents flowing in the circuit shown in figure one. 

 

Circuit schematic showing current flow and resistor voltages

 Figure One

Starting with the currents caused by the left-hand source, we can obtain several currents. The total resistance (RT) in the circuit equals:

RT = 2 + (3 x 6) / 9 = 4Ω and

I4 = 4v / 4Ω = 1A

As we move to figure two, we can use current division to continue two additional current values:

I5 = (6/9) x I4 = 0.667A and I6 as

I6 = 0.333 A

 

 

Circuit schematic diagram

 Figure Two

After solving the left-hand source, we can determine currents for the right-hand source.  With this analysis shown in figure three, RT =

6 + (2 x 3) / 5 = 6 + 1.2 = 7.2Ω

while

I7 = 6 v / 7.2Ω = 0.833A

 

Again using current division, we can determine the values of the I8 and I9 currents.

I8 = (2/5) x I7 = 0.333A

I9 = 0.5A

From there, we can determine the current flowing through the 2Ω resistor.

I1 = I4 + I9 = 1.0 + 0.5 = 1.5A

And the values for I2 and I3

I2 = -I6 – I7 = -0.333 – 0.833 = -1.166A

I3 = -I5 +I8 = -0.667 + 0.333 = -0.334A

  

Circuit diagram with resistors and battery sources

 Figure Three

 

Even Superman Sometimes Loses His Powers

Kryptonite for the superposition theorem is that it only applies when a linear relationship exists between voltage, current, and resistance. Since the power loss in a resistor results from a squared relationship, we can calculate the power by superimposing power losses.  Instead, we must calculate the power from a total voltage or current value. Using the 2Ω resistor from figure one as an example, the power for that resistor does not equal:

P = [(I2 + I4) x 2] + [I2 + I9 x (1)2 x 2] + 0.52 x 2 = 2 + 0.5 = 2.5 watts     

Instead, we must use the following calculation to determine the power at the 2Ω resistor.

P = (I2 x I1) x 2 = 1.52 x 2 = 4.5 watts

Verification is important for any circuit design. As a design engineer and as a design team working towards getting your electronic out for production with as few mistakes or vulnerabilities as possible, it is key to maintain a close relationship with your circuit simulation tools. Knowing about superposition theorem helps; however, having tools that can accurately and reliably simulate these results for you throughout your design process is invaluable.

No matter the circuit you’re working on, you can rest assured that the suite of tools available from Cadence for PCB design and analysis will get your job done. The simulation offerings from PSpice can augment any design, whether you’re working on switch-mode power supplies, or working on improving the yield and reliability of your circuit, rest assured that the easily integrated modeling results will optimize your design.

If you’re looking to learn more about how Cadence has the solution for you, talk to us and our team of experts

About the Author

Cadence PCB solutions is a complete front to back design tool to enable fast and efficient product creation. Cadence enables users accurately shorten design cycles to hand off to manufacturing through modern, IPC-2581 industry standard.

Follow on Linkedin Visit Website More Content by Cadence PCB Solutions
Previous Article
Steady State vs. Transient State in System Design and Stability Analysis
Steady State vs. Transient State in System Design and Stability Analysis

The importance of determining the steady state and the transient state is critical to proper circuit design...

Next Article
Determining Annular Rings for Via Management in PCB Design
Determining Annular Rings for Via Management in PCB Design

Annular rings are particularly necessary to get right for adequate via placement within PCB designs.

OrCAD Work From Home Program

Request Access