LED parallel current sharing problem

小太阳yy

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For a structure like this where two LEDs are connected in parallel, if the voltages of the LEDs are different, it is easy to cause different currents flowing through the left and right sides. Some people have given a solution by connecting a resistor in series under each of the two LEDs, saying that this can help equalize the current. The resistance values of the two resistors are equal. In my opinion, only when the resistance values of the two resistors are unequal can this help equalize the current. Why can equal resistances also help equalize the current? Teacher, can you explain this to me?

maychang

If the two LEDs are of different colors, for example, one is red and the other is green, then in order to make the luminous intensity meet the actual needs, the series resistors should of course be different. This should actually adjust the resistance until it feels appropriate to the human eye. If the two LEDs are products of the same manufacturer and model, then in order to avoid the trouble of measurement, the series resistors of the two light-emitting tubes are usually the same.

小太阳yy

maychang posted on 2023-11-25 17:59 If the two LEDs are of different colors, for example, one is red and the other is green, then in order to make the luminous intensity meet the actual needs, the series resistors should of course be different. This...

Even if the LEDs are from the same manufacturer and the same model, there are still differences in voltage. It seems that the series resistor is meaningless. For example, the left one is 3.2v 20mA. The right one is 3.1V 20mA. How can I eliminate this 0.1V voltage difference?

maychang

小太阳yy posted on 2023-11-25 18:07 Even if the LED voltage is the same model from the same manufacturer, it is still different. The series resistor seems meaningless. For example, the 3.2v 20mA on the left. On the right...

[Even if the LEDs are from the same manufacturer and model, the voltages are still different, so it seems meaningless to add a series resistor.]

Yes, there are still some small differences in the voltage of LEDs of the same model from the same manufacturer.

However, series resistors still make sense. When the same resistor is connected in series, the difference in current between two LEDs with different forward voltages is smaller than when there is no series resistor. The higher the supply voltage, the smaller the current difference.

小太阳yy

maychang posted on 2023-11-25 18:15 [Even if the LED voltage of the same model is from the same manufacturer, it is still different. The series resistor seems meaningless.] Yes, the same manufacturer and the same...

Why is the current difference smaller when the same resistor is connected in series than when the resistor is not connected in series? I don't understand why the higher the voltage, the smaller the current difference is. Can you explain this to me?

maychang

Xiaoshangyangyy posted on 2023-11-25 18:23 Why is the current difference smaller when the same resistor is connected in series than when the resistor is not connected in series? I don't understand this. The higher the voltage, the smaller the current? Please explain...

The voltage-current characteristic of the light-emitting tube is nonlinear, and its shape is roughly similar to the forward characteristic of an ordinary diode. When the forward voltage of the light-emitting tube is very small, there is almost no current. When the voltage at both ends increases to a certain value, the current begins to increase rapidly with the voltage.

The voltages at both ends of the same type of LEDs also vary slightly under the same current, but not significantly. For example, for the same red LED, with a current of 5mA, one voltage may be 1.65V and the other 1.66V. If the two tubes are connected in parallel, with the same voltage at both ends, then one may be 4.5mA and the other 5.5mA. If the two tubes are connected in series with a resistor of 680 ohms, the currents of the two LEDs can be calculated to be nearly the same by using the graphical method.

Let's take an extreme example: if one circuit has a red LED in series and the other circuit has no LED, then no matter how much voltage is applied, the current in the circuit without the LED is theoretically infinite. If a 680 ohm resistor is connected in series, then the current in the circuit with the LED is approximately 5mA, and the current in the circuit without the LED is 7.35mA, and the difference between the two currents is less than 1.5 times.

It is inconvenient to draw pictures, so I hope you can understand through the text description.

小太阳yy

maychang posted on 2023-11-25 19:11 The voltage-current characteristic of the light-emitting tube is nonlinear, and its shape is roughly similar to the forward characteristic of an ordinary diode. The forward voltage of the light-emitting tube is also very small...

It's a bit hard to understand. If the current is close to the same after the resistors are connected in series, then the voltage on the two resistors should be approximately equal, and then the total voltage in parallel is equal. In this way, the voltage across the two LEDs should be approximately equal. In this way, if they are all equal, the current must be different. It's a bit hard to understand.

maychang

小太阳yy posted on 2023-11-25 19:27 It is a bit difficult to understand. If the currents are close to the same after the resistors are connected in series, then the voltages on the two resistors should be approximately equal, and then in the end...

I modified the post on the 6th floor and added an example where one circuit has a LED and the other circuit does not. You can have a look at it. You can increase the voltage to 10V or even 100V, and increase the resistance to twice or even 20 times, and do the calculation again.

maychang

小太阳yy posted on 2023-11-25 19:27 It is a bit difficult to understand. If the currents are close to the same after the resistors are connected in series, then the voltages on the two resistors should be approximately equal, and then in the end...

It is easier to understand by actually drawing the forward voltage-current curve of the light-emitting tube and then drawing the resistance for calculation.

小太阳yy

小太阳yy posted on 2023-11-25 18:23 Why is the current difference smaller when the same resistor is connected in series than when it is not connected in series? I don't understand why the higher the voltage, the smaller the current difference. Please explain...

I have the picture here, but I still don't understand. If I want the current flowing through two parallel LEDs to be basically the same, it seems contradictory no matter how I think about it.

maychang

小太阳yy posted on 2023-11-25 19:50 I have the picture here, but I still can't figure it out. If I want the current flowing through two parallel LEDs to be basically the same, it's contradictory no matter how I think about it. Alas

[I have the picture here, but I still can’t figure it out]

As I said, it is easier to understand if we draw the voltage-current curve of the light-emitting tube and then draw the resistance.

Wait a moment while I draw the picture.

小太阳yy

maychang posted on 2023-11-25 19:33 It is easier to understand if you actually draw the forward voltage-current curve of the light-emitting tube and then draw the resistor to calculate.

Teacher, can you please help explain it? It is really hard to understand.

小太阳yy

maychang posted on 2023-11-25 20:11 [I have the picture here, but I still don’t understand it] I said that it is easier to understand if you draw the voltage-current curve of the light-emitting tube and then draw the resistance. And...

Thanks you!!!!

maychang

Xiaoyangyy posted on 2023-11-25 19:50 I have the picture here, but I still can't figure it out. If I want the current flowing through two parallel LEDs to be basically the same, it's contradictory no matter how I think about it.

According to the figure on the 10th floor, when a 3.0V voltage is applied to LED1, the current reaches 20mA, and when a 3.5V current is applied to LED2, the current reaches 20mA.

According to the figure on the 10th floor, if two LEDs are connected in parallel and a 3V voltage is applied, the current in LED1 is 20mA, while the current in LED2 is less than 2mA.

maychang

小太阳yy posted on 2023-11-25 19:50 I have the picture here, but I still can't figure it out. If I want the current flowing through two parallel LEDs to be basically the same, it's contradictory no matter how I think about it. Alas

The curve in Figure 10 is still used.

We connect LED1 and LED2 to a 5V power supply with a resistor in series. The resistor value is assumed to be 330 ohms. Since 5V/330 ohms is 15mA, the light blue line in the figure above represents the 330 ohm resistor.

So, this 330 ohm resistor is connected in series with LED1. How much current will flow through LED1? We can see that the voltage at the intersection of the light blue straight line and the orange curve of LED1 is 2.8V and the current is 5.7mA (black vertical line in the figure).

Similarly, another 330 ohm resistor is connected in series with LED2. The voltage at the intersection of the light blue straight line and the green curve of LED2 is 3.3V and the current is 5mA (black vertical line).

Let's take a look at the current difference when two LEDs are connected in parallel. The red line on the left side of the figure is the current difference when a 3V voltage is applied to the two LEDs (in parallel).

After the two LEDs are connected in series with a 330 ohm resistor, the difference in current between the two LEDs is the red line on the right. Obviously, the red line on the right is much smaller than the red line on the left.

maychang

小太阳yy posted on 2023-11-25 19:50 I have the picture here, but I still can't figure it out. If I want the current flowing through two parallel LEDs to be basically the same, it's contradictory no matter how I think about it. Alas

Higher voltages cannot be drawn because the horizontal axis of the graph only goes up to 5V. But we can imagine that the blue straight line intersects the horizontal axis further to the right. In order to keep the current at 15mA when there is no LED, the straight line should be closer to horizontal, that is, the slope is smaller. Then the straight line closer to horizontal intersects with the two LED curves, and it is obvious that the red broken line on the right side of the 15th floor figure will be shorter in height. This is the source of "the higher the voltage, the smaller the current difference", because the higher the voltage, the greater the resistance. If the voltage is infinite and the resistance is infinite, then the light blue straight line will become horizontal, and the current flowing through the two LEDs will be absolutely equal.

小太阳yy

maychang published on 2023-11-25 21:02 Still using the curve in the 10th floor figure. We connect a resistor in series with LED1 and LED2 and connect them to a 5V power supply. The resistor value is assumed to be 330 ohms. Because 5 ...

Teacher, should this diagram be drawn like this? You just said that 5V is 15mA. Also, what is the voltage on the resistor and the voltage on the LED at this time?

小太阳yy

maychang posted on 2023-11-25 21:09 Higher voltages cannot be drawn because the horizontal axis of the graph only goes up to 5V. But we can imagine that the blue straight line intersects the horizontal axis further to the right. To keep the LEDs ...

This is too difficult. I don’t know why we have to draw the picture like this to find the focus when connecting in series. I will think about it again. Thank you for the teacher’s answer! ! ! !

maychang

Xiaoyangyy posted on 2023-11-25 19:50 I have the picture here, but I still can't figure it out. If I want the current flowing through two parallel LEDs to be basically the same, it's contradictory no matter how I think about it.

Originally, to represent the resistor in the voltage-current coordinate, its image should be a straight line passing through the origin. The larger the resistor, the smaller the slope, and the smaller the resistor, the larger the slope. The drawing of the blue straight line in the figure on the 15th floor is actually a left-right flip of the resistor image . The reason for this flip is that in the series circuit, the voltage across the LED plus the voltage across the resistor is the power supply voltage. In the 15th floor, the power supply voltage is 5V, so the intersection of the light blue straight line and the horizontal axis is drawn at 5V. In this way, the intersection of the 330 ohm resistor (blue) and LED1 (orange) represents the voltage across LED1, and the horizontal distance to the right of the intersection represents the voltage across the resistor. Similarly, the intersection of the 330 ohm resistor and LED2 represents the voltage across LED2, and the horizontal distance to the right of the intersection represents the voltage across the resistor.

小太阳yy

maychang posted on 2023-11-25 21:36 Originally, to represent resistance in voltage-current coordinates, its image should be a straight line passing through the origin. The larger the resistance, the smaller the slope, and the smaller the resistance, the larger the slope. 15 ...

I will think about it carefully. Thank you for your patient guidance. Thank you!!