## Friday, June 2, 2023

I will show you how to make a simple 12V battery level indicator circuit that can be created using only LEDs and resistors. Towards the end of the video, I will briefly show the simulation of this circuit in Proteus. First, let's start with what materials are needed for this circuit.

We will make the indicator level in 4 stages. Here we can use four LEDs, 1 red, 1 blue, 1 green and 1 yellow LED. Since the highest voltage will be around 12V, we need to use four 1kÎ© resistors that we will connect for these LEDs to work without damage.

First, let's place the LEDs. The operating voltages of the red and yellow LEDs are approximately 2V. blue and green LEDs are approximately 3V. This is how the red LED is connected to the battery and 1kÎ© resistor. Similarly, we can also connect the yellow LED. Then let's connect the blue LED and finally the green LED.

Now let's assume that the battery has a voltage level of 3V. The 3V value is greater than 2V, which is the operating voltage of only the red LED close to the battery. Therefore, this voltage value will make only the red LED light up. According to conventional current direction the current will flow through the red LED and the 1kÎ© resistor connected to it, following the path I have shown here with the arrows, and will make this LED light up.

Now let's imagine that the battery has a voltage level of 5V. The 5V value is greater than the total 4V value, which is the operating voltage of the red and yellow LEDs close to the battery. Therefore, this voltage value will make the red and yellow LEDs light up. The current will pass through both the red and yellow LEDs and the 1kÎ© resistors connected to them, following the path I have shown here with arrows, and will make these two LEDs light up.

Now let's imagine that the battery has a voltage level of 9V. The 9V value is greater than the total 7V value, which is the operating voltage of the red, yellow and blue LEDs. Therefore, this voltage value will cause the red, yellow and blue LEDs to light up. The current will pass through both the red, yellow and blue LEDs and the 1kÎ© resistors connected to them, following the path I have shown here with arrows, and will make these three LEDs light up.

Finally, let's assume that the battery has a voltage level of 12V. The 12V value is greater than the total value of 10V, which is the operating voltage of the red, yellow, blue and green LEDs, i.e. all LEDs. Therefore, this voltage value will make all LEDs light up. The current will flow through the red, yellow, blue and green LEDs and the 1kÎ© resistors connected to them, following the path I have shown here with the arrows, making all four LEDs light up.

Here we can think of the red LED lighting up as 25%, the yellow LED lighting up as 50%, the blue LED lighting up as 75% and the green LED lighting up as 100% occupancy.

Finally, let's simulate this circuit in Proteus and finish. I have drawn this circuit before. Just as we have just seen, we have placed the resistors and LEDs appropriately. For example, let's set the voltage value of the battery to 12V. When we run the simulation, we can see that all LEDs light up. Let's set the voltage value of the battery to 9V. When we run the simulation, we can see that all LEDs light up except green. Finally, let's make the voltage value of the battery 6V. When we run the simulation, we can see that only the red and yellow LEDs light up.

I will show you how to make a simple 12V battery level indicator circuit that can be created using only LEDs and resistors. Towards the end of the video, I will briefly show the simulation of this circuit in Proteus. First, let's start with what materials are needed for this circuit.

We will make the indicator level in 4 stages. Here we can use four LEDs, 1 red, 1 blue, 1 green and 1 yellow LED. Since the highest voltage will be around 12V, we need to use four 1kÎ© resistors that we will connect for these LEDs to work without damage.

First, let's place the LEDs. The operating voltages of the red and yellow LEDs are approximately 2V. blue and green LEDs are approximately 3V. This is how the red LED is connected to the battery and 1kÎ© resistor. Similarly, we can also connect the yellow LED. Then let's connect the blue LED and finally the green LED.

Now let's assume that the battery has a voltage level of 3V. The 3V value is greater than 2V, which is the operating voltage of only the red LED close to the battery. Therefore, this voltage value will make only the red LED light up. According to conventional current direction the current will flow through the red LED and the 1kÎ© resistor connected to it, following the path I have shown here with the arrows, and will make this LED light up.

Now let's imagine that the battery has a voltage level of 5V. The 5V value is greater than the total 4V value, which is the operating voltage of the red and yellow LEDs close to the battery. Therefore, this voltage value will make the red and yellow LEDs light up. The current will pass through both the red and yellow LEDs and the 1kÎ© resistors connected to them, following the path I have shown here with arrows, and will make these two LEDs light up.

Now let's imagine that the battery has a voltage level of 9V. The 9V value is greater than the total 7V value, which is the operating voltage of the red, yellow and blue LEDs. Therefore, this voltage value will cause the red, yellow and blue LEDs to light up. The current will pass through both the red, yellow and blue LEDs and the 1kÎ© resistors connected to them, following the path I have shown here with arrows, and will make these three LEDs light up.

Finally, let's assume that the battery has a voltage level of 12V. The 12V value is greater than the total value of 10V, which is the operating voltage of the red, yellow, blue and green LEDs, i.e. all LEDs. Therefore, this voltage value will make all LEDs light up. The current will flow through the red, yellow, blue and green LEDs and the 1kÎ© resistors connected to them, following the path I have shown here with the arrows, making all four LEDs light up.

Here we can think of the red LED lighting up as 25%, the yellow LED lighting up as 50%, the blue LED lighting up as 75% and the green LED lighting up as 100% occupancy.

Finally, let's simulate this circuit in Proteus and finish. I have drawn this circuit before. Just as we have just seen, we have placed the resistors and LEDs appropriately. For example, let's set the voltage value of the battery to 12V. When we run the simulation, we can see that all LEDs light up. Let's set the voltage value of the battery to 9V. When we run the simulation, we can see that all LEDs light up except green. Finally, let's make the voltage value of the battery 6V. When we run the simulation, we can see that only the red and yellow LEDs light up.