![]() But if you are using this in a darkened environment, perhaps you can get away with it. But I can see that it is designed as \$4\times 7\$ (or \$7\times 4\$, depending on your perspective.) So the first step for me is to examine it sufficiently well to decide how the manufacturer probably sees its operation.įrom this, found below the Absolute Maximum Rating section, I see that the peak forward current is given as typically \$I_\text\$ running through them, continuously. With the LED’s anode connected to a digital pin, the cathode is connected to ground: Note: All LEDs need a current limiting resistor placed on either the anode side or cathode side to prevent the LED from burning out. ![]() I've never used one of these types of bargraph displays before. The LED can be turned on or off by switching power at the anode or the cathode. Set up a 4 byte area of memory that contains a bitmap of the desired LED states and use that to communicate with the ISR (your program writes to it and the ISR only reads it). ![]() It is best to do this in a timer interrupt routine, though it's fine to test it with simple-minded delays- but that will use up 100% of the MCU cycles. By bringing the Q1-Q4 high while the C1.C7 are changing you avoid "ghosting" (dimly lit unwanted segments). Also invert the state to high=ON, of course.įirmware wise, the algorithm is to start with all 4 drives to Q1-Q4 high (off), set the GPIOs corresponding to C1.C7 LOW for each LED out of 7 you want to have "on", bring the drive for Q1 LOW, hold for perhaps 250-1000usec, then bring all Q1-Q4 high again, set up for Q2 and repeat. If that does not yield sufficient brightness you could add a ULN2003A or 7 MOSFETs to the 7 resistors, however you would have to recalculate the resistor values to yield the higher current and take into account the relatively high drop of the Darlington transistors into the ULN2003A. If your LED has a Vf of 2.0V (green or yellow) that's about 200 ohms. You pick RXi to yield the peak current of 14.3mA, so about (4.8V-Vf)/0.0143. You would pick RSi to yield about 1/20 of the 100mA total LED current through the transistor base, so about 5mA. With this circuit you are driving the entire display current through the MCU ground pin so you are limited to an absolute maximum of 200mA, and you should stay well away from that, so let's assume 100mA total current (meaning an average LED current per LED of only 3.6mA and a peak LED current of 14.3mA) Simulate this circuit – Schematic created using CircuitLab The color may be a massive close to this primary color (red, green or blue), and once 2 parts have an equivalent strongest intensity, then the color may be a hue of a secondary color (a shade of cyan, magenta or yellow).The easiest way to drive this multiplexed display is like this (although the designation is K, what you have is driven the same as a 4-digit common Anode numerical LED display): with the anode and cathode part during testing, it may damage Arduino Board. ![]() Respectively we can get the two others basic colours, green and blue When one in all the parts has the strongest intensity. It will look like this We can also use different displays like LED Display. In order to get red light on the LED we will call the setColor() function and set value of 255 for the redValue argument and 0 for the two others. ![]() So now in the loop function we will make our program which will change the color of the LED each second. ![]()
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