Today, my son learn to use the LED for the 3V battery. As we know, a voltage of about 1.8V LEDs like the most, proper light, not heat and power consumption of approximately 10mA to 20mA only. How to use it with 3V or more voltage power supply.

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## Resistor reduce current or voltage

The Best way, so easy and cheapest is **resistor **reduce voltage. It is suitable for lower current load. And the current use stable circuit. For example, LEDs, small lamps, relays and more.

**He measures the voltage of two AA 1.5V battery in series**

He uses the breadboard and the battery holder. Then, he can read about 3V.

Then, he puts a 3mm LED across the terminal of a 3V battery, and measure a voltage it, about 2.7 volts

**The LED get too much current**

We do not should use high voltage, it can kill the LED.

In the image, the voltage across the battery is 3V. Then, the LED is hight light up and temperature too hot.

## Using limit current resistor

Currently, LED has been the voltage higher than it would be unendurable. We need to reduce the voltage down. To a level that is approximately 1.8V.

What is a popular way to reduce the current for them? The limiting current resistor is answering. We will use it in series circuit with the LED.

**How many the resistance-R1?**

In the circuit diagram, they are the series circuit.

We can find the resistance of R1 by using Ohm’s Law triangle.

R = V / I

We want the resistance (R). We need to know voltage (V) and Current (I) before.

1. Now we know the current.(IR1)

According to the principle of the circuit

The current flowing through all device is equal.

IR1 = ILED

When the LED use the current approximately 20mA.

So the current of current is 20mA, too.

2. The voltage of resistor (VR1) is what to look for!

When both resistor and LED connected in series. Then parallel or across with the 3V battery.

Thus, VR1 combined with VLED—voltage of LED— is equal to a 3V battery.

When we know the voltage of LED is 1.8V so I know the resistor’s voltage is?

= 3V – 1.8V

= 1.2V

Therefore:

Resistance of R1 = 1.2V / 2mA

= 60 ohms

But this value can be bought in all store.

So we use **56 ohms** better.

We can summarize a simple formula:

R1 = (Vin-VLED) / ILED.or

R1 = (Vin – Vload) / Iload

You look at block diagram. It is clear better.

## What Power of limiting current resistor

My child asked, how much we should use the size of the resistor?

From ohms law: P = V x I

V = voltage of resistor = 1.2V

I = ILED = 20mA = 0.02A

P = 1.2V x 0.02A

= 0.024 watts

So, we can use 0.25W resistor.

Then, he uses the ElectroDroid on mobile to find the resistor color code.

Next, draw and color it on the notebook as **Figure 5**

and later we put the 56 ohms resistor into the breadboard and measure the voltage across LED again. It makes voltage low down to 1.8V and LED is normal.

## How to convert voltage 12V to 6V relay

I would like to show you another example. Suppose you have to use 6V relay.

It is 6V 80 ohms, SPDT relay.

But you need to use it with 12V battery. It is so not good.

Because It uses the high current. Since, resistance coil is 80 ohms. When use 12V battery. The relay has too many currents flowing coil. It is about 0.15A (150mA). From

I = 12V / 80 ohms

= 0.15A

The battery is fast out of power.

And important! The relay coil is too hot.

We have many ways to reduce voltage. But using the resistor is an inexpensive way.

In the circuit is a similar to above circuit. We use the relay coil instead of LED.

Using a resistor reduce the voltage to relay.

Diode-D1 protects other parts from the high voltage pulse that is generated in the relay coil when the relay is switched OFF.

**Finding the resistor-R1**

Since Resistor-R1 = (Vin – Vload) / Iload

Vin = 12V battery

Vload = voltage of relay coil = 6V

Iload is the current flow the relay coil. But now we do not know it. Since it shows the resistance of the coil, 80 ohms.

From ohms law

I = V / R

V = 6V , R = 80 ohms

R = 6 / 80

= 0.075A or 75mA.

So, Iload is 0.075A

Put it in above formula again.

R1 = (12V – 6V) / 0.075A

= 80 ohms But this resistance cannot find in a normal store.

So we should use 82 ohms.

Next, we need to use a suitable power of resistor.

**P = V x I**

V = 6V

I = the current of relay = 0.075A

So power of resistor is.

= 6V x 0.075A

= 0.45W

We can use **82 ohms 0.5W resistor**.

**Note:** Now my son has little understanding of electronics. But he enjoyed playing with electronics.

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