# Resistance and Voltage Dividers

When i first started working with electronics, i tended to think of my circuits, or even parts of my circuit, in isolation. The horror of it though is that your circuit is plugged into other things – at minimum a battery, but commonly other devices, or your house and the power grid – and those things can affect how your circuit works.

Beyond being physically connected with wires to other things, your circuits also have a connection to the rest of the world through electromagnetic fields.

In this post we are going to talk about voltage divers, which on one hand can be useful if made on purpose, but can also be made on accident and cause you strange behaviors.

## Voltage Dividers

Voltage dividers are a way of giving you a lower voltage. If you have a 9 volt battery and only want 6 volts, a voltage divider can do that for you. There is a downside to voltage dividers that we’ll explore in this post, but they are incredibly simple to make: you only need two resistors.

First let’s look at a single resistor in a circuit. Lets put a 1000 ohm resistor in a circuit with a 9 volt battery. If we connect our multimeter probes to the wire on the same side of the resistor and measure volts we’ll get zero volts (see diagram below). This is because volts is a measurement of electric potential between two points. Our multimeter is measuring the difference in electric potential between two points right next to each other on a wire, and the difference is essentially zero. The red and black arrows on the circuit diagram are where we connect the red (+) and black (-) probes of our multimeter.(Tangent: 9 milliamps is going through this circuit since there is 1000 ohms of resistance and 9 volts. The power supply says 8 but it has limited accuracy, resistors are not exactly their labeled value, wires have resistance, etc. It also reads that there are 9 volts * 8 milliamps = 72 milliwatts of power being used.)

What if we put our multimeter on different sides of the resistor? In that case, we read 9 volts. The resistor makes it more difficult for electricity to cross, and thus there is a difference in electric potential of 9 volts, on each side.

What would happen if we put two resistors in?

If we measure at the red and black arrows again, we’ll still have 9 volts. If we measure at the red and orange arrows though, we’ll see 4.5 volts. If we read at the orange and black arrows, we’ll also see 4.5 volts. We know that the whole circuit needs to go from 9 volts to 0 volts since that is what is provided by our battery, but it dropped by half on the first resistor, and then dropped the rest of the way on the second resistor. (Tangent: the total resistance here is 2000 ohms, so 4.5 milliamps would flow through the circuit)

Let’s change the value of the resistors and see what happens.

I didn’t have a 2000 ohm resistor so i just put two 1000 ohm resistors in series (more on that further down).

If we measure between the red and black, we still have 9 volts. If we measure between the red and orange, we get 3 volts though, and if we measure between the orange and black, we get 6 volts. Weird! (Tangent: The total resistance here is 3000 ohms, so there should be 9 volts / 3000 ohms = 3 milliamps flowing through the circuit but my power supply isn’t showing that correctly.)

Similarly, you can change the second resistor to be half instead of double and get the opposite result.

I didn’t have a 500 ohm resistor so i put two 1000 resistor ohms in parallel (more on that further down).

What is going on here is that the 9 volts are dropping off across the resistors based on their relative values. When the resistors are equal in value, they each get half of the voltage. When they are unequal, the voltage across the R2 resistor is calculated like this:

$V_{R2} = V * \text{R2} / (\text{R1}+\text{R2})$

To actually use this as a power source, you would connect new wires as the positive and negative power for a sub circuit.

Note in the above, I’m not saying that is -6V and +6V, which would be 12 volts total, I’m just labeling the positive and negative sides of the 6 volts of power available.

You could use the top part as a 3 volt source if you wanted instead, or in addition to the 6 volts you are using from the bottom part. You could even split the voltage into more than just two levels, but instead could put in N resistors to have N voltage levels.

The famous 555 timer for instance internally uses a voltage divider with three 5K resistors to make three different power levels, and that is why it’s called a 555 interestingly. You can see it at the top of this diagram of a 555 timer, between the ground (pin 1) and the +Vcc supply (pin 8).

(This image is from this 555 timer tutorial: https://www.electronics-tutorials.ws/waveforms/555_timer.html)

## Resistors in Series vs in Parallel

When I needed a 2k Ohm resistor in the last section I put two 1k Ohm resistors in series. When you put resistors in series, their values add together, allowing you to additively create whatever resistance you need.

When I needed a 500 Ohm resistor and didn’t have one, I put two 1k Ohm resistors in parallel. This is because putting resistors in parallel gives electricity more than one path to get through, and thus has lower resistance than if there was only one of the resistors. The exact equation for the resistance of resistors in parallel is:

$1/R = 1/R_1 + 1/R_2 + ... + 1/R_N$

Where $R_i$ is the value of a specific resistor.

This means that if you put two of the same valued resistors in parallel, the resistance will be cut in half. If you put three of them in parallel, the resistance will be cut in three.

This formula comes up again in electronics. For capacitors, when you put them in parallel, their capacitance adds. When you put them in series, their capacitance follows the parallel resistor equation. It’s the same formulas, but parallel / series reversed. Strange huh?

$1/C = 1/C_1 + 1/C_2 + ... + 1/C_N$

Where $C_i$ is the value of a specific capacitor (in Farads).

Something else strange is that this is why thicker wire has less resistance too. There are more paths for electricity to travel through the thicker wire, compared to thinner wire, so resistance goes down.

Below are some images of two 1k Ohm resistors in series and in parallel, with the multimeter showing the total resistance value.

One resistor:

Two resistors in series:

Two resistors in parallel:

## What Happens When Using a Voltage Divider?

Ok so let’s start with the voltage divider we set up before.

Now let’s say we actually use that 6 volts to power something. That something will have a resistance of 2k Ohms. Maybe it’s some kind of light bulb.

We can simplify this circuit though. The 2k Ohms of our load, and the 2k Ohms of the voltage divider are in parallel so we can use our formula for parallel resistance, or remember that two capacitors of equal value in parallel get half the resistance. So that means we could describe our circuit this way, as far as resistance is concerned:

The problem with that is that our voltage divider has changed. The resistors are equal now, which means that our 6 volts has dropped down to 4.5 volts!

If we decreased the resistance of what we were powering, the voltage would drop too. Intuitively, imagine if you had a short circuit so had zero resistance across the load, the electricity would completely bypass the 2k Ohm resistor in the voltage divider as if it weren’t there, so there would be zero volts difference between the top and bottom of the 2k Ohm resistor.

If we increased the resistance of what we were powering, we would raise the combined parallel resistance there on the 2nd part of the voltage divider, but luckily would at most have 2k ohm resistance. For instance, using a 1 mega ohm resistive load, the parallel resistance formula gives us a resistance of 1.996 k Ohms. So, if we had a high resistance load, we’d get nearly our full 6 volts, but would never quite have the full 6 volts. At the limit, if our load was disconnected, and thus had infinite resistance, we would get the full 6 volts.

If you know the resistance of the load you are plugging into the voltage divider, you can take it into account and choose a resistor for the voltage divider that gives you the desired parallel resistance amount and thus the right voltage. Some loads have variable resistances though, and then you have a problem and should look at other methods of changing the DC voltage level, such as a buck converter.

Some loads have no resistance though, and a voltage divider can come in really handy. Supplying power to a transistor’s base, or to an op amp’s input, or to an optocoupler’s input for instance can make great use of these because they just “read” the voltage signal there without putting any extra load on it.

The lesson here is that whenever you plug things together, you might get strange drops in voltage because you’ve accidentally created a voltage divider. If your resistance is sufficiently higher than whatever internal resistance what you’ve plugged into has, you can ignore the voltage drop, but that also decreases the amperage so may not be desirable.

This effect even comes up in batteries (and other power sources) which essentially can be modeled as an ideal voltage source, with a small resistance (like 10 ohms). If you use a low valued resistor on a battery, the voltage will drop because you are secretly part of a voltage divider involving the internal resistance of the battery (and in fact, that “internal resistor” can’t take that much power and will start heating up, which can be dangerous! So don’t short circuit batteries!). Since a battery’s resistance is so small, your resistance level is likely to be much higher when using the battery to power something, and this isn’t something you really have to worry about in normal situations.

Of course, all this talk only deals with DC and resistors. Things get more complex when you have capacitors, inductors or AC power.

## Maximum Power (Watts)

So we saw that as R2’s resistance gets larger, the voltage across R2 becomes larger, and at infinite resistance, it gets all the voltage available.

We also know that the larger the resistance, the lower the amps in the circuit, so getting that voltage comes at a cost.

Watts is a unit of measurement of power and is volts multiplied by amps. It turns out that if you want your voltage divider to have maximum power (watts), that R1 should equal R2. Wikipedia has more about that here: https://en.wikipedia.org/wiki/Impedance_matching

Here are some graphs showing this, where if resistor R1 is 1k Ohms, that you get the highest amount of watts when R2 is also 1k Ohms, despite the behavior of the volts and amps.

## Calculating Resistance (and Voltage) of an Unknown Circuit

Since plugging your circuit into other things can make an implicit / unintentional voltage divider, you probably want to know how much resistance some other black box circuitry might have. Luckily you can figure this out using Ohms law (see last post: Voltage, Amps, Resistance and LEDs (Ohm’s Law)) and some simple algebra.

First, connect a resistor to the + and – and measure the amps in the circuit. If you use a resistor that is too low value, or has too low of a wattage rating, the resistor will get hot, possibly start glowing or burst into flames (resistors have a rating in watts and the common ones for small electronics like those seen in this post can handle 1/4 of a watt). So basically be careful if doing this with high voltages – and in fact, if my blog is your primary source of knowledge, please don’t mess with high voltage 🙂

So let’s say we connect a 1k Ohm resistor and read a value of 0.01 amps or 10 milliamps.

Ohms law says:

$I = V/R$

where I is current, V is volts and R is resistance.

So we now have this formula:

$0.01 = V / (1000 + R_1)$

We have one equation with two unknowns, so we need another equation to make it solvable by having two equations and two uknowns. Let’s say we take an amperage measurement using a 500 Ohm resistor and get 0.017 amps or 17 milliamps.

That gives us a second equation:

$0.017 = V / (500 + R_1)$

We now have two equations with two unknowns!

We can solve the first equation for V and get:

$V = 0.01 * (1000 + R_1)$

From there we can plug V into the second equation to get:

$0.017 = 0.01 * (1000 + R_1) / (500 + R_1)$

Solving for R1, we get:

$R_1 = (0.01 * 1000 - 0.017 * 500) / (0.017 - 0.01) = 214.28 \Omega$

If you do the calculations, you get 214.28 ohms, which means the unknown circuit has that much resistance.

What’s nice is that you can also use this to get the total amount of voltage available to this circuit by plugging this resistance into the first equation that we solved for V:

$V = 0.01 * (1000 + 214.28) = 12.14 \text{volts}$

This was a toy example i made up, using 12 volts and 200 ohms of resistance, so our answer is pretty close. The inaccuracies came from rounding off the numbers, but you’ll get the same problems in real life from not completely accurate measurements and imperfect electronic components.

For convenience, here are the equations to calculate the resistance of an unknown circuit, without having to do the algebra each time.

$R_1 = ( I_A * R_{2A} - I_B * R_{2B}) / (I_B-I_A)$

Where $R_1$ is the resistance of the unknown circuit. $R_{2A}$ is the first resistor value you connected and measured to get $I_A$ amps. $R_{2B}$ is the second resistor value you connected and measured to get $I_B$ amps.

Once you have the $R_1$ value, you can plug it into this to get the voltage available to the circuit:

$V = I_A * (R_{2A} + R_1)$

Let’s take these equations for a spin with a battery. I accidentally popped the fuse on my digital multimeter and can’t use it to measure amps so i’ll use my analog multimeter.

First i’ll measure the amps with a 1k Ohm resistor. The knob is set to 10 milliamp measurements so the bottom row of readings (that are labeled 0 to 10) are where you read from. I drew some yellow to show you where to read from. I read 8.6 milliamps.

Next i’ll put two 1k Ohm resistors in series to make 2k Ohms of resistance and measure amps to get what looks like 4.6 milliamps.

Ok so let’s plug our values into the equations!

$R_1 = ( I_A * R_{2A} - I_B * R_{2B}) / (I_B-I_A)$

$R_1 = ( 0.0086 * 1000 - 0.0046 * 2000) / (0.0046-0.0086) = 150 \Omega$

So it looks like this 9 V battery has 150 ohms of resistance. I’ve heard that as a battery is used, it’s resistance goes up, so maybe this battery is nearing needing to be replaced having such large resistance.

Let’s calculate how many volts it has.

$V = I_A * (R_{2A} + R_1)$

$V = 0.0086 * (1000 + 150) = 9.89 volts$

So, the battery has 9.89 volts inside of it. Either they made the battery have higher than 9 volts inside of it, to account for internal resistance dropping the output voltage, or my 5$analog multimeter is not very accurate and these are just ball park figures. ## Closing Thanks for reading and hopefully you found this interesting or useful. Have any requests or ideas for other topics to write about? Drop me a message on twitter at @Atrix256. # Voltage, Amps, Resistance and LEDs (Ohm’s Law) I’ve taken up learning electronics during the pandemic, and have enjoyed it quite a bit. I’ve been programming for 25+ years, so it’s nice to have something different to learn and work on that is still both technical and creative. It’s cool getting a deeper understanding of how the fundamental forces of nature work, as well as being able to MacGyver a hand crank powered flashlight from an old printer if needed (Check out a 40 second video of that here!). It’s also nice having something physical to show at the end of the day, although it does require consumable parts, so there are pros and cons vs making software. Friends (Hi Wayne!) and YouTube have helped me learn a lot, but I found the subject pretty alien at first and wanted to try my hand at some explanations from a different POV. This post starts that journey by taking the first steps into DC electronics. ## Ultra Basics Electricity flows if there is a path for it to flow in and the flow is made up of electrons. Electrons are negatively charged so travel from the negative side of a circuit to the positive side. Conventional current flow is backwards from this though, and says that electricity flows from the positive side to the negative side. In this case, it’s not electrons flowing but “holes” flowing. Holes are a weird concept, but they are just a place that will accept an electron. Here is an open circuit, which means that there is a gap. Since the circuit is not closed, electricity cannot flow. (made in https://www.circuitlab.com/editor/#) If you close the circuit, like the below, electricity is able to flow. The circle on the left is a power source with a + and – terminal. It’s labeled as a 1.5 volt double A battery. Here is a diagram of a circuit with a switch that can be used to open or close the circuit. Being able to read and make circuit diagrams is real helpful when building things or trying to understand how circuits work. Of note: the higher the voltage, the farther that electricity can jump across gaps. So, while at low voltage, a circuit may be open, turning up the voltage may make it closed when the electricity arcs across! ## Ohm’s Law IMAGE CREDIT: Eberhard Sengpiel The most useful thing you can learn about DC electricity is Ohm’s law which mathematically explains the relationship between voltage, amperage and resistance. Ohm’s law is: $I = V/R$ In the equation I stands for Intensity and means current aka amps, V stands for voltage and R stands for resistance. If electricity was water, voltage would be the water pressure, amperage would be how much water was running through the pipe, and resistance would be a squeezing of the pipe, like in the image above. Current is measured in amperes (amps) or the letter A. 500mA is 500 milliamps, or half of an amp, and 1.2A is 1.2 amps. Note: electricity is dangerous! It can take only a few hundred milliamps to be fatal, but voltage is needed to be able to let those amps penetrate your skin. Voltage is measured in volts or the letter V. If you see 9V on a battery, that means it’s a 9 volt battery, and is capable of providing 9 volts. Resistance is measured in Ohms or the omega symbol $\Omega$. So if you see $5\Omega$ that means 5 Ohms of resistance. If you see $5k\Omega$ that means 5 kiloohms which is 1000 times as much resistance. If you see $5M\Omega$ with a capitol M, that means 5 megaohms, which is 1000 times as much resistance again. Where Ohm’s law comes in handy is when you know two of these three values and you are trying to calculate the third one. As written, the formula showed how to calculate amps when you know voltage and resistance, but you can use algebra to re-arrange it to a formula for any of the three: $I = V/R$ $R = V/I$ $V = I*R$ This comes up quite often – if you know how much voltage a battery has, and you know how many amps you need, you can use this to calculate the value of the resistor to get the desired amps. ## Diodes, LEDs and Resistors LED stands for Light Emitting Diode. A diode is something which lets electricity only flow in one direction and it has a couple of common uses: • Protecting circuits from electricity flowing in the wrong direction. • Turning Alternating Current (AC) into Direct Current (AC) by rectifying it (preventing the negative part of AC from getting through. Same as last bullet point) • Lots of cool tricks, like stabilizing uneven power levels by letting voltage over a specific value “spill over” out of the circuit. Here is a pack of various diodes i bought from amazon for 10$. There are a quite a few different types of diodes, which are useful for different situations.

Here are some diodes close up. The black one is a rectifier diode IN4001, and the more colorful one is a switching diode 1N4148. Those part numbers are actually written on the diodes themselves but are a bit hard to see. You can use these numbers to look up the data sheet for the parts to understand how they work, what their properties are, how much voltage and amps they can handle, and often even see simple circuit diagrams on using them for common tasks. Data sheets are super useful and if doing electronics work, you will be googling quite a few of them! Here is the data sheet for 1N4148 which i found by googling for “1N4148 data sheet” and clicking the first link. 1N4148 Data sheet.

Here are two circuit diagrams with diodes in them. The black triangle with the line on it is a diode. The arrow shows the direction that it allows conventional flow to travel. The line on the arrow corresponds to the bands on the right of the diodes in the image above, which is the negative side of the diode (cathode). The left circuit is a closed circuit and allows electricity to flow. That diode is forward biased. The circuit on the right has the diode reverse biased which does not allow electricity to flow.

LEDs can do many things regular diodes can do, since they are diodes, but they have the property that when electricity flows through them, they light up. Since they are diodes, and only let electricity flow in one direction, LEDs have a + side and a – side and you have to hook them up correctly in a circuit for them to light up. If you hook them up the wrong way, it doesn’t damage them, but they don’t light up and they don’t close the circuit for electricity to flow. The symbol for an LED is the diode symbol, but with arrows coming out of it.

Here are a pack of LEDs i have that came as part of a larger electronics kit. You can get a couple hundred LEDs in a variety of colors from amazon for about 10$. Some LEDs are in colored plastic cases, some are in clear cases. There are even LEDs that shine in infrared and ultraviolet. LEDs also come in different sizes. This pack has 3mm and 5mm LEDS. Here is an up close look at a white LED. The longer leg is the positive side, which means you need to plug the positive side of the circuit into it if you want it to light up. the negative side has a shorter leg, but the negative side also has a flat side on the circular ring at the bottom, which can’t really be seen in this picture. All diodes have a voltage drop, which is a voltage amount consumed by the diode. If you are providing less than that amount of voltage, the diode will act as an open switch, and electricity won’t flow through it. The specific voltage drop for diodes can be found in data sheets, but i’ve found it difficult to find data sheets for LEDs. Luckily I picked up a “Mega328” component tester from amazon for 15$. It lets you plug in a component, press the blue button, and then tells you information about the component. It’s super handy! Here you can see the voltage drop of 2 different LEDs. The smaller red LED has a voltage drop of 1.88V while the larger green LED has a voltage drop of 2.5V. If you supply them with less than that amount of voltage, they will not light up!

So what would happen if we tried to connect the LEDs to the batteries below?

The large green LED has a 2.5V voltage drop, while the AAA battery only has 1.5V as you can see on the label. That means the LED doesn’t light up.

The smaller red LED has a 1.88V voltage drop and is connecting to a 9V battery so it has enough voltage and should light up. Let’s use Ohm’s law to calculate how much current – in amps – are going through the LED.

I = V/R and in our case V is 9 and R is 0 because we have no resistance.

$I = \frac{9}{0} = \infty$

Oops we have infinite current! The LED is destroyed pretty quickly after you plug it in.

There isn’t actually infinite current, because the metal wires connected to the LED have a very tiny amount of resistance to them, just like all wire, and the battery has a limit of how many amps it can give. So in any case, it isn’t infinite amps, but it is a very large number, limited by how many amps the 9V battery can actually deliver. The LED would actually be destroyed. You should basically always use a resistor with an LED to limit the current and keep it from being destroyed. Here is an interesting read about how to calculate the internal resistance of a battery which will then tell you how many amps it can give you: Measuring Internal Resistance of Batteries.

When you have a circuit with this low of resistance, it’s considered a short circuit, and if the LED didn’t get destroyed, the battery would start getting hot and it could become a dangerous situation. This is also why short circuits themselves are bad news. They have a LOT of current running through them which can cause things to heat up, melt and catch fire.

3mm and 5mm LEDs typically want 20 milliamps maximum (20mA or 0.02A) to be at full brightness. If you give them less, they will be less bright but still function.

We can calculate then how much resistance they want to be maximally bright if we know the voltage of the power source we are using and the voltage drop off of the LED we are trying to power.

Let’s take the larger green LED with a 2.5V voltage drop, and power it with a 9V batter, aiming to get 20mA.

First we subtract the voltage drop from the supply to see how much voltage we have to work with: 9V – 2.5V = 6.5V.

Next, we know we want 20mA and we have 6.5V, and we are just trying to solve for resistance so we use Ohm’s law: R = V/I.

$R = 6.5V / 0.02A = 325\Omega$

So, we need 325 ohms of resistance to get 20mA in our LED from a 9V battery. Here is a pack of resistors i got from amazon for 12$. Resistors have funny colored bands on them which tell you their rating. You can find charts for decoding them all over the place, but again, the “Mega328” will tell you this too. In fact, a multi meter will tell you as well. Multi meters aren’t very expensive. Here’s one i got from amazon for 35$ which has tons of features and works really nicely.

I don’t have any 325 ohm resistors, but i do have 470 ohm resistors, so i’ll just use one of those. That’s 14mA if you do the math, which is a bit lower than 20mA, but it still works just fine despite not being as bright as it could be. You can get different resistances by connecting resistors in parallel or series and doing some math, but this works for now. I used a mini breadboard (the green thing) to hook this circuit up. Every horizontal line of 5 holes is connected to each other electrically. It’s a nice way to play with circuits without having to solder things together. By convention, red is used for the positive terminal and black or blue is used for the negative.

By the way, quick fun fact. A 1.5V AA battery is considered dead when it has dropped down to 1.35V. At this point, it still has energy in it though! If you are clever with electronics, you could make circuitry to use this power from dead batteries to give you 1.5V or higher, and you could drain so called dead batteries even further.

## LEDs Turning Light Into Power

Many things in electronics turn out to be reversible. Speakers work as poor microphones, and microphones work as poor speakers. Similarly, LEDs can work as poor solar cells and turn light into energy. Want to see? Here i hook my multimeter up to an LED, and have it set to read volts. It reads 48.7mV. Energy is flowing all around us from radio waves, etc, so it’s picking up some of that.

When i put the LED in the beam of the flashlight, it jumps up to 1.644V. Pretty cool huh?

## Did You Like This Post?

It’s a little different than what I usually write about, but hopefully you liked it. Careful though, this stuff escalates quickly. Before you know it you’ll be harvesting optocouplers and coils from old printers to make a rail gun.