In the last post concerning circuit analysis (Simple Circuit Analysis Techniques You Should Know) I went over some very basic but important techniques.
First, we had a quick over view of resistors in series, parallel, and series-parallel circuits. Then, we talked about Ohms Law. Finally, Kirchhoff’s Laws were introduced.
This time, I’m going to show you a few simple tricks for combining parallel resistors before I go into circuit analysis using some other very important (and slick) methods.
Circuit Analysis 201: Slick Ninja Techniques
Supersleuth Shortcuts for Combining Parallel Resistors
Last time I promised to reveal a few tricks that work on certain parallel circuits. I’ll start by saying that the shortcuts that follow not only apply to resistors, but also apply to other components.
Recall from the first post in this series that parallel resistors add up with the formula below. We also know that inductors follow a similar pattern.
Of course, this can extend beyond 3 resistors to any number by adding more 1/R terms on the bottom.
The problem is that doing calculations like this is tedious and boring. Good thing there are common instances where it’s totally unnecessary!
Consider the example of two resistors in parallel (a common situation). I’ll spare the derivation (though it’s not hard), but equation 1 is a quick, easy to remember formula for this situation.
Things get even better if there are multiple resistors of the same value in parallel.
For example, most of you can probably see that two 100 Ω resistors in parallel have a total resistance of 50 Ω. This suggest that if all the resistors in a parallel network are the same value, the total resistance is the value of one resistor divided by the number of resistors.
So, if we have ten 100 Ω resistors in parallel, the total resistance is 10 Ω.
Feast your eyes upon figure 1 below.
Figure 1: 3 parallel resistors of the same value.
You may recognize this picture from the first post in this series (with slight modifications). Assuming all three resistors are the same value, we can write:
In general, we can write:
Where n is the total number of resistors in the network.
Remember, equation 2 only works when all the resistors are the same value and equation 1 only works for two parallel resistors (even if they’re not equal).
Enough about combining parallel resistors — Let get to the good stuff.
Voltage Division: Ninja Circuit Analysis Trick 1
The need to find the voltage drop across a resistor is a need you’ll often encounter in the world of electronics.
Consider the circuit in figure 2.
Figure 2: Look familiar? This is the series circuit from the last post in this series.
If we want to calculate the voltage drop across each resistor, we need to perform several steps:
- Calculate total resistance by adding up all the resistors.
- Once we know the total resistance and the source voltage, we can calculate the current.
- Using the current, we can then calculate the voltage drop across each resistor using ohm’s Law.
Now this is all fine and good, but there is an easier way with less steps to do this.
Voltage division allows you to calculate the voltage drop across any resistor without having to first calculate the current. This allows you to bypass the last two steps given above. That makes your life as an electronics enthusiast easier.
Deriving the formula for voltage division from Ohm’s Law isn’t that difficult, but I’ll leave it up to you to do so if you choose. Equation 3 gives the formula for voltage division:
Where VR is the voltage drop across the resistor of interest
VS is the source voltage
R is the value of the resistor of interest
And Rtotal is the total series resistance
Although voltage division can give you the voltage drop across a particular resistor in a series circuit with any number of resistances greater than one, you’ll often apply it to a circuit (or part of a circuit) with two resistors. In this case equation 3 becomes:
Note that we can also use voltage division to find the drop across two or more series resistors. For example, if we want to know the voltage drop across the series combination of R2 and R3 from figure 2, we can use voltage division to find that simply by treating R2 and R3 as one, larger resistor.
Let’s do a couple examples using voltage division.
Find the voltage drop across R3 in the figure below.
All we need to do is apply equation 3:
VR3 = 12[(7.5k)/(5k + 10k + 7.5k)]
Doing the arithmetic, we get:
VR3 = 4 V
The second example is a circuit (or more likely part of a circuit) that shows up a lot in the world of electronics.
The circuit in the figure below is a voltage divider. A voltage divider is a simple circuit which turns a large voltage into a smaller one. They are very common and are one of the most fundamental circuits in electronics.
If Vin is 5 V, what is Vout?
This problem is a piece of cake, yet a very important and practical depiction of reality.
Vout = Vin(R2/(R1 + R2)) = 5(10k/(10k + 20k)) = 5(10k/30k)
Vout = 1.67 V
Voltage dividers and voltage division in general show us a basic yet important concept in electronics. They show us that the source voltage in a series circuit is divided between the resistors in direct proportion to their resistance. The bigger the resistor compared to the others, the larger the voltage drop.
This, of course, is in accordance with Ohm’s Law which tells us that voltage equals current multiplied by resistance, or V = I x R.
Note that potentiometers (and other things like light dependent resistors) often find use in voltage dividers as they’re being adjustable makes the divider more versatile and easy to change if need be.
Some that are new to electronics may be wondering why we should even bother with voltage dividers.
Why not just start with the voltage that we need in the first place?
The problem with this is that many circuits need more than one voltage and building a power supply to deliver each one is impractical and expensive.
The Ninja Gets Loaded: A Voltage Divider Caveat
Consider the voltage divider from example 2 above. By itself, it can do little more than act as an aid for the study of electronics.
In other words, the reason we build voltage dividers in the first place is so that we can attach something to them, and this is the norm.
So, when we attach a load to a voltage divider, the circuit in example 2 ends up looking like the circuit in figure 3.
Figure 3: voltage divider with load attached.
The load in figure 3 could be anything, it doesn’t matter – loads are modeled as a resistance attached to the circuit. This is because every circuit or piece of equipment offers a certain amount of resistance.
But what happens to Vout when we attach this load?
The short answer is that Vout changes because we’re “loading” the circuit.
How much it changes depends on the load. Let’s take a closer look at this.
Let’s assume that RLoad in figure 3 is 30 k and that Vin is still 5 V. When the circuit was unloaded, Vout was 1.67 V, what is it now?
The series circuit we once had has now become a series-parallel circuit with R2 || RLoad .
By applying the supersleuth shortcut for two parallel resistors, we see that their combination offers 7,500 Ω. Also, since they’re in parallel, the same voltage appears across both.
Now, we use voltage division again to see what this new voltage is:
5((7.5k)/(7.5k + 20k)) = 1.36 V.
We “lost” some of the output voltage.
Let’s do another example, only this time we’ll make RLoad much bigger than 10 k.
Assume Vin is still 5 V and that RLoad is now 1 MΩ. Find Vout.
The parallel combo of 10 k and 1 M is 9900.99 Ω (using our supersleuth formula, of course).
Again, using voltage division, we get a Vout of about 1.66 V.
This value is a lot closer to the value we originally wanted, which was 1.67 V.
This suggests that when using a simple voltage divider such as this, you’ll want to use a load with a high input resistance if possible to avoid loading the circuit.
If not possible, you’ll need to calculate how much current the load will pull and then alter the voltage divider resistor values to offset the loading effect.
Just for kicks, let’s see what happens to the output voltage if the load resistance is 500 Ω.
The parallel combination of 500 and 10 k yields about 476 Ω.
Using voltage division, we get an output voltage of only about 0.15 V when we thought we’d have 1.67 V!
Now we know why having a load with a high input impedance (or resistance) is often desirable.
One More Thing About Voltage Dividers
Don’t use a voltage divider as a voltage supply for any load that requires even a modest amount of power. Not only would this be extremely inefficient, you’re likely to burn up the resistors and destroy it.
Current Division: Ninja Circuit Analysis Trick 2
Though not as often used as voltage division, current division is another trick which can come in handy. Unlike voltage division, which works on series resistors, current division works on parallel combinations of resistors.
Current division tells us that the total current in a parallel network is shared by the resistors in inverse proportion to their resistance. You’ll see what I mean in a minute.
It comes in handy when you know the input current but not the input voltage.
Consider the circuit in figure 4.
Figure 4: current division.
It’s easy to see that the two parallel resistors divide the current provided by the source amongst themselves, but how much current goes through each resistor?
The answer is easy if we use the current division formula:
Notice the striking similarity to voltage division. But watch out – the numerator in the fraction is the opposite of that for voltage division!
As is similar to voltage division, current division can extend to a parallel network with any number of resistors great than one.
Do we really need an example problem?
I don’t think so.
Superposition: Ninja Circuit Analysis Trick 3
Sorry — we’re not talking about anyone’s sexual escapades here. What we are talking about is perhaps even better than that.
The superposition theorem not only finds use in electronics, but also in physics, economics, and more.
This principle helps us analyze linear circuits with more than one independent source by calculating the contribution of each independent source separately. It also provides some insight into in determining the contribution of each source to the parameter under investigation.
What’s an independent source?
Glad you asked.
Unlike a dependent source, an independent voltage or current source’s output does not depend on some other parameter or source in the circuit.
For example, a battery is an independent voltage source, while something like an amplifier (among other things) is usually considered a dependent source. Right now, we’ll stick with independent sources in our study of circuit analysis.
The procedure for using superposition in circuit analysis is as follows:
- Remove all independent voltage sources and replace with a short.
- Remove all independent current sources and replace with an open. In other words, just remove the current sources.
- You should now have only one remaining source. Calculate the contribution that this source makes to the parameter of interest.
- When all independent voltage and current sources have been considered, add the results obtained for each.
- If you do run into a dependent source in your travels, these stay active. Do not short or remove them. Grab a bottle of aspirin, you may need it.
Before we do a few examples, realize that superposition may result in more work if the circuit you’re analyzing has more than two or three sources. However, it is useful for reducing the complexity of certain circuits.
Another thing to bear in mind is that to use superposition, the circuit or parameter you’re looking for must be linear. So, you couldn’t use superposition to find the individual powers components consume and add them up. Power (P = I2R) is not linear due to the I2 term. You could, however, use superposition to calculate the voltage or current and then use Ohm’s Law to get the power.
Here’s a few examples to clear things up.
Let’s revisit that fictitious circuit from the last circuit analysis post. This time, we’ll find the current I using superposition.
We pick one of the two voltage sources, short it out, then use KVL.
Let’s short the 20 V source. The circuit now looks like:
50 = -2I – I —-> 50 = -3I
I = -16.67 A
Now, we’ll put the 20 V source back and short the 50 V source.
Using KVL again:
-20 = -2I – I —-> -20 = -3I
I = 6.67 A
Now we’ll add the two results together:
6.67 A – 16.37 A = -10 A which is the same answer we got last time.
Example 7 is another interesting circuit analysis problem using superposition.
Find the current I in the figure below.
Here we have a voltage source and a current source. Recall that to use superposition, we short voltage sources and remove current sources. We’ll start by considering the voltage source by itself. Removing the current source, we have the circuit below.
The first thing to note about the figure above is that R3 is just hanging out there in space. Since no current flows through R3, no voltage drops across it and we can safely ignore it. This gives us a series circuit composed of the voltage source, R1, and R2.
Using Ohm’s Law, we can easily find the current the voltage source contributes to the circuit:
I = V/R = 10/3k = 3.33 mA
Now, we’ll reinsert the current source and short the voltage source which gives us the circuit below.
This nut’s a bit tougher to crack.
Let’s use KCL and add another current (call it I2) flowing into node 0 (I would’ve used node 1 but I’s and 1’s look a lot alike, ya know). I didn’t add I2 in the picture, so you’ll just have to imagine it flowing into node 0.
Using KCL, we get:
1 mA = (V0 / 1k) + (V0 / 3k) where V0 is the voltage at node 0.
If you’re good at algebra, you can solve this by hand. I’m lazy, so I used an online equation solver to get V0:
V0 = 0.67 V
Now that we know V0, let’s find I2:
I2 = 0.67/1k = 0.67 mA
Since the total current in this circuit is 1 mA, we just need to subtract 0.67 mA from 1 mA to get 0.33 mA.
However, notice that I is flowing the opposite direction it should be as the 1 mA from the current source should split at the bottom then flow up towards R1 and R2.
No big deal — we just need to slap a negative sign in front of our answer: I = -0.33 mA.
This is important, because now we need to add up the individual contributions to I from each source and if the signs are incorrect, we’ll get a wrong answer.
Itotal = 3.33 mA – 0.33 mA = 3mA.
The current from one source slightly opposes the current from the other for a total net of 3 mA.
Ninja Circuit Analysis Concluded
This is probably one of the longest posts I’ve written so far, but I wanted to not only be sure my circuit analysis skills were up to snuff, I wanted to select the right examples.
The next post on circuit analysis will either cover Thevenin and Norton or it’ll go over nodal and mesh analysis. I’m not sure yet but maybe that’ll keep you on your toes until next time ?
While you’re waiting leave a comment and tell us:
Which circuit analysis technique is your favorite?
Which don’t you like? Why?
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