Calculate expected voltage around a resistor [duplicate]

Question

It's been a few years since I first studied the Ohm's law at school. Now I'm really getting into electronics, and I must admit a part of it still puzzles me.

I know this is kind of a strange question, but I have no clue how to express it otherwise, so I'll just lay out my reasoning. The following train of thoughts is wrong and/or incomplete. Please stop me whenever I'm wrong (even for the sligthest thing), and complete this reasoning.

Here it comes:

Let's make a circuit with:

• 6V battery (typically 4 AA batteries)
• 2k ohms resistor
• LED, for the sake of doing something

Circuit map:

^{simulate this circuit – Schematic created using CircuitLab}

From Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)

\$ I = \dfrac{V}{R} = 3mA \$

I do understand that the current is limited due to the resistor... crystal clear.

The book tells me that, if I measure the potential difference between both sides of the resistor, I will have something between 0 and 6 volts.

Sounds legit:

• can't be 6 volts because the two ends of the battery are connected, therefore lowering the potential
• can't be 0 volts, since the current does not freely move around, due to the resistor; resulting in one side being more negatively charged, and the other more positively charged, since the electrons can move faster to an end of the battery than through the resistor.

What I don't understand (and if my whole reasoning is right until now), is: How can I calculate the expected voltage around the resistor ?

Answer 1

From the Ohm's law, we know that the current in the circuit will be of (considering only resistor has any resistance)

Na-ah, the LED has a resistance too. Common LEDs use 2.5V at 20mA. We can calculate the LED's resistance (on 2.5V) with \$R=\frac{V}{I}=\frac{2.5}{0.02}=125\Omega\$.

Now, when it comes to your circuit, you shouldn't just use a 2K resistor, you should think about that and calculate the needed value. Now, the voltage over the LED probably won't be 2.5V, which means the LED has a different resistance, different current is flowing, etc. You cannot calculate the current flowing in this circuit and therefore you cannot calculate the voltage over the resistor (we'll come to that later). Here's how you should design your circuit:

You want a 0.02A current through the circuit (the current is the same everywhere as it is a series circuit). That means the total resistance has to be \$R_{tot}=\frac{V}{I}=\frac{6}{0.02}=300\Omega\$. We already saw the resistance of the LED was \$125\Omega\$, that means your resistor has to be \$300-125=175\Omega\$.

Now, how do you calculate the voltage over the resistor? Take Ohm's law, the current through the resistor (0.02A) and the resistance (\$175\Omega\$):

$$V=I\cdot{}R = 0.02\cdot175 = 3.5V$$

Note that this also equals \$6V-2.5V\$.

Answer 2

The sum of all voltage drops must equal the source voltage.

\$V_{source} = V_{resistor} + V_{led}\$

Depending on what wavelength the LED the operates at, it will essentially drop the same the voltage (when you are within the current restrictions of the LED).

If we assume that the LED drops 1.2V, then that means that the resistor must drop 4.8V.

The LED forward voltage will always be a constant (might deviate slightly), when you are within the current limits.

If the LED can take a max of 20mA, then at 15mA and 10mA, the voltage drop across the LED will always be the forward voltage drop (in this example, the 1.2V).

Make sense ?

Answer 3

Your LED will have about 2 volts across it for a standard LED which means you'll have about 2mA flowing in the circuit. It won't be very bright.

When you said: -

I = V/R -> 3mA

That's where you made an error. Even though the LED may be talked-about as having little resistance it still drops a voltage across it and for your average LED it will be about 2V whether you're feeding it with 2mA or 20mA.