Electronics

• LEDs
  • Resistance
  • Anatomy of a green 5mm LED
  • Wiring schematics
  • Pinouts
• Stepping down voltage with resistence
• Potentiometers
• References

LEDs

The forward current of a LED depends on its color. The regular cheap 3-10 mm LED diodes range from $1,8-2,2V$ (red, orange, yellow, etc) to $3,2-3,8V$ (white, green, warm white, etc). They have a typical forward current of $20mA$, and a maximum forward current of $30mA$¹ (for reference, a GPIO pin of an ESP32 provides up to ca max $40mA$).

Resistance

Since these LEDs all generally have consistent forward currents and forward voltages, we can just use Ohms law and work out a reference table for common supply voltages.

The resistor values are assume the typical forward current $I_f$ of $20mA$.

The voltage across the circuit is equal to the sum of the voltage across the components in the circuit, the resistor and the LED.

So for an LED with a forward d voltage $V_f=1.8V$ that is connected to a supply voltage $V_s=3.3V$, the voltage across the resistor is $V=V_s-V_f=1.5V$.

>>> I = 0.02     # 20 mA == 0.02 A
>>> Vs = 3.3
>>> Vf = 1.8
>>> V = Vs - Vf
>>> V/I          # R = V/I
75               # 75 Ohm

So to connect LED with a forward voltage of 1.8 V to 3.3 V, it should be protected by a 75 Ω resistor.

Anatomy of a green 5mm LED2

Wiring schematics

Multiple LEDs connected in series.

Proper way to connect multiple LEDs in parallel, each LED has its own current limiting resistor.

Pinouts

Regular and RGB 3mm and 5mm LED diodes³:

LED pinout

RGB LED pinout

• The cathode leg ($-$ or $K$) is the shorter leg (usually $17mm$).
• The anode leg ($+$ or $A$) is the longer leg (usually $19mm$).

Stepping down voltage with resistence

There are multiple different ways to step DC Voltage down. Using a buck converter is simple, but they are bulky. But resistence reduces voltage (voltage falls as resistance increases).

To do that, you use two resistors $R_1$ and $R_2$ and "divide" the circuit between them⁴, connecting $R1$ to $V_{IN}$ and $R_2$ to $GND$. If $R_1$ and $R_2$ are of equal value, $V_{IN}$ is reduced by half ($V_{OUT}=V_{IN}/2$).

This circuit is called a voltage divider, $R_1$ is a pull-up resistor and $R_2$ is a pull-down resistor.

The general equation for reducting $V_{IN}$ to $V_{OUT}$:

$$R_2=\frac{V_{OUT}\times R_1}{V_{IN}-V_{OUT}}$$

Since both $R_1$ and $R_2$ are unknown (also we aren't in school, and experimentation is more interesting than algebra anyway) this involves some experimentation and picking some value resistor for $R_1$ and then calculating ca. what value resistor we need to get for $R_2$.

For example $10k\Omega$ is a reasonable starting point for $R_1$. Stepping $7V$ down to $5V$:

>>> v_out = 5
>>> v_in = 7
>>> r1 = 10
>>> (v_out*r1)/(v_in-v_out)    # 5*10 / (7-5)
25.0                           #   50 / 2

So to step $7V$ down to $5V$ using a $10k\Omega$ pull-up resistor $R_1$ to $V_{IN}$, we need to use a pull-down resistor $R_2$ that is $25k\Omega$.

Potentiometers

Potentiometer
Pinout of a regular potentiometer.

References