If you’ve ever tried to understand amps, voltage, and resistance you’ve no doubt heard the water analogy. It goes something like this…

Consider a tank of water. The amount of water in the tank is comparable to the amount of electrical charge in a power source. Now consider the tank has a hole in the bottom of it. The more water in the tank, the higher the pressure is of the water coming out of the hole. This pressure is comparable to electrical voltage. The flow of water coming out of the hole is comparable to electrical current. Finally – the size of the hole is comparable to electrical resistance. The bigger the hole, the less resistance there would be to water leaving the tank.

I had read that analogy many times and still just didn’t get it. I think I thought it was an over generalization that just couldn’t be that simple. As it turns out, it really is that simple (in most cases). As Im a big believer in learning by doing, let’s do a quick example to show how this analogy actually works out. Consider you have a simple series circuit that looks like this…

While it might be hard to see here, these lights are trying to turn on, but just can’t quite do it. If we measure the current going through the circuit we can see that it’s only .05 milliamps. So why is that? Let’s check the voltage drop of each LED…

Like we learned in the last post, voltage in a series circuit is always split between all of the components on the circuit. In this case, our 5 volt power supply is split between these 3 LEDs. If we add up the voltage drops we get 4.936 volts which is pretty darn close to our input voltage. So if we’re supply voltage to each LED why aren’t they lighting? The problem is we aren’t reaching the LED’s forward voltage. If we look at the spec sheet for these LEDs we can see that forward voltage is 2 volts…

If we don’t reach the forward voltage, the LEDs can’t conduct current which means there is hardly any current at all flowing through this circuit. So back to our water analogy…

In the top picture, we are showing what’s currently happening. Since the voltage is split between all 3 LEDs none of them are reaching their forward voltage so none of them are conducting current. If you think of the 5v power supply as a water tank, the analogy works out pretty well here. There just isn’t enough water (voltage) in the tank (power source) to fill the LEDs. Now if we took an LED out of the path, each LED would get 2.5 volts which would be the LEDs max voltage. Then we’d have this…

Each LED would be able to reach above its forward voltage of 2 volts and actually reach it’s max voltage of 2.5 volts. Let’s try that out..

It works! We can see the circuit is now drawing almost 13 milliamps of current. If we check the voltage of each LED we’ll see that both are over 2 volts…

Note that in this case, one of the LEDs is taking more of the voltage but that the voltage drop across both still equals the total input voltage of the power source.

Now to take the water analogy a step further, let’s look at a parallel circuit. Parallel circuits are interesting because the voltage across all the components in parallel will always be the same. That’s a major change from series where all LEDs shared the total voltage. Wikipedia summarizes this by saying the following…

Simply put, in a parallel circuit current increases but the voltage stays the same, and in a series circuit current stays the same but the voltage decreases

As we saw in our series circuits, the current is the same in all locations in the circuit but the voltage decreases as we add components that cause a voltage drop. Let’s look at a basic parallel circuit now…

You might need to enlarge the picture for detail so here’s a schematic of what I have…

So it’s a pretty simple parallel circuit. I have 3 LEDs connected in parallel and a total of 30 ohms of resistance in the positive lead from the power supply. Where did I get 30 ohms from? Why from Ohms law of course! These are green LEDs that have a voltage drop of right around 3.3 volts. They can support around 20 milliamps a piece. If you do the math there, you should find that you need a resistor of ~28 ohms when you solve (R = 1.7/.06). In this case, since the resistor is in the main line before any of the parallel branches, we need to account for the current draw of all of the LEDs (20 milliamps a piece). So let’s turn it on and see what we get…

So it’s hard to see, but I have my multimeter inline on the positive side of the leftmost LED. Here you can see that the current is almost 20 milliamps. So our equation worked! If we checked the current draw of the other LEDs, we’d find similar numbers. Actually, from left to right I found 19, 15, and 14 milliamp readings. The difference here is due to the fact that Im using cheap LEDs that have low tolerances. If we then check the current reading from the power supply we would see a total of 49 milliamps of current being consumed by the circuit…

If we checked the voltage of any of these LEDs, we should see them staying relatively the same…

Above we checked the voltage drop of the first two LEDs and found them to be relatively the same. That’s expected. Now let’s start plugging in more LEDs into the circuit and see what happens…

While all the LEDs lit up, they certainly aren’t as bright and the voltage of the first LED has dropped considerably. So how can this happen in a parallel circuit? Admittedly I was confused the first time I did this. I next checked the current draw…

Huh? 56 milliamps? Remember up above when we only had 3 LEDs plugged in and we were drawing 49 milliamps? So this doesn’t seem to be adding up. In a parallel circuit voltage should stay the same across all components and current should increase. To make things even stranger, I happened to notice this when I accidentally picked up the wrong LED and placed it in the circuit…

What was going on? Now putting a different color LED in the circuit made all the others go almost completely out? Well – as it turns out – I’ve already talked about why this is. I even did the math to prove it. You’ll recall above that I did the math to determine the resistance of putting 3 green LEDs in parallel…

It worked great. But I was also putting my resistors right off of the power supply. So going back to my water analogy, I was making the hose small for all of the LEDs at the same time. Putting the resistor here means that Im limiting the current and voltage for all of them making it harder for the circuit to draw more current for each new LED I added. So imagine the power supply is the water tank again. Imagine the tank has one main line coming off of it into a manifold that splits into 3 shower heads. The manifold has a valve on it so I can adjust the water flow through all the heads at the same time. This works great initially, all of the shower heads have perfect flow. But if I add more heads, I would need to adjust the manifold again to allow more water to flow. If I didn’t, the existing heads would get less and less water. That’s exactly what we see here.

Could this have worked correctly with 8 LEDs? Sure – but I would have needed to adjust the resistance (the manifold if we’re talking water) to account for 160 milliamps of draw instead of 60. Not exactly an ideal parallel circuit.

In addition, if the components arent 100% identical as far as voltage drop and current draw go Im looking for trouble. Let’s consider the image above where I inserted a yellow LED by mistake and all the green LEDs got dim. The yellow LED has a much lower forward voltage. So when the electricity comes into the circuit the path through the yellow LED is much easier to take than the other paths available via the green LEDs. If I had sized the resistor to allow for much higher current draw I’d likely burn out the yellow LED, which would then remove if from the circuit allowing the other LEDs to draw more than they should causing a cascading failure of LEDs.

So is this how parallel circuits should work? No. A better design for a parallel circuit looks like this…

Here, each parallel branch has it’s own resistor. This allows for the circuit to act much differently. First off, the voltage at each branch will be identical since we aren’t sharing the same resistor (yeah a real parallel circuit!). Secondly – the current draw can increase as needed as I add LEDs since the resistor is sized for each individual components. Let’s take a look at a setup like this…

Here we can see the circuit with 3 blue LEDs plugged in. Note the voltage drop of a single LED is 3.16 volts. Now let’s check the overall current going through the circuit…

With 3 LEDs in path its drawing about 50 milliamps. Ok – looks good so far, let’s add some more…

With 8 LEDs our current draw went up wot almost 120 milliamps. This is what we expect with a parallel circuit! The current increases to meet the need, but the voltage stays the same…

If we check the voltage drop again we’ll see that it is unchanged. The other benefit of this setup is that we can put items with different electrical characteristics in the circuit. For instance, while hard to tell, I have a red, yellow, and green LED in the the circuit below…And they all work! So I think if you’re going to build a parallel circuit, it’s best to do it in this manner. The first iteration ‘should’ work in some cases, but I think theres likely too much variation in components and their voltage drop and current requirements to make it feasible.