12V Battery Level Indicator Circuit
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- čas přidán 17. 05. 2023
- In this video, 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 1kΩ 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 1kΩ 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 1kΩ 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 1kΩ 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 1kΩ 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 1kΩ 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.
Basically, this is the battery level indicator circuit that can be made with LEDs and resistors, guys. If you have a similar circuit example you want me to explain, you can share it in the comments section. I hope this video was useful and you liked it. Hope to see you in the next video. Goodbye.
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Fantastic video. Thanks from Bangladesh
Good basic introduction: in practice however, the steps are +3V (blue) +5V (Yellow) +7V (Green) +9V (blue) with 1K resistors. The current principal is correct, however each component must be within tolerance (LED diode leakage + matching resistors) to be a perfect voltage/current match.
I think you mean red for the +3V
How much amperage does this draw? Will it draw a 12V 5Ah sealed lead acid battery? Could an on/off switch be placed between the battery + and the first led to reduce any possible draw down?
can i connect more leds in parallel with each one but keep the circuit the same?
48v type?
Hey why are The LEDs in Series and Not in parallel
Water battery's indicator do the make
Your arrows should be showing the flow from - around to + since that is really the way electricity flows
Man, do you realize that any voltage below 9V means a that a 12V battery is damaged and dangerous to operate? You're missing the basics.
😢 This is too simple, the steps are too big. You must define a voltage step with R-series and give the single steps to a 4-comparator like uc339 and the other comp-Inputs are derived from the battery voltage. A 9V 200mAh block has 7 Ni-MH cells, the discharge voltage is 7x1,05= 7,35V and the full charge voltage is 7x1,5=10,5V and falls back to 7x1,3=9,1V.
And in four steps you have to define a cascade between 7,35 and 10,5V=abt 3,2V, this is pro step 0,8V.
Remember the uA339 has open collector outputs active high and has to be grounded to zero or minus 0V with a resistor fitting for LEDs, the cathode of the LED goes to ground.
This is a logic sequencer wiring which can be used also as timer sequencer to switch a transmitter and a receiver via relais upon the same antenna. 73 de hb9csu
Lek gawe potensi piye carane? Aku gak ngerti rek.
Best way to discharge the battery in short time. 12mA minimum discharge current, minimum. How stupid is that?