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Resistance and Voltage Dividers

demofox2June 27, 20210

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 energy being used.)

[image-29]

(Circuit diagrams made at https://www.circuitlab.com/editor/)

[image-36]

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.

[image-30] [image-37]

What would happen if we put two resistors in?

[image-31] [image-38]

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 the energy 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.

[image-32]

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

[image-39] [image-40]

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.

[image-33]

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

[image-41]

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.

[image-34]

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).

555 timer block diagram

(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:

[image-42]

Two resistors in series:

[image-44]

Two resistors in parallel:

[image-43]

What Happens When Using a Voltage Divider?

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

[image-34]

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.

[image-45]

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:

[image-46]

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 Energy (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 energy and is volts multiplied by
amps. It turns out that if you want your voltage divider to have
maximum energy (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.

[image-48]

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.

[image-53]

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.

[image-55]

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.

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