I have a few 0.8" seven segment LED displays that I wanted to put to use in a project however I found that they were too dim when connected to the MAX7219 at full brightness and an I-SET resistor value of 10K.

This is because the displays I have need at least 15mA of DC current to have an acceptable brightness and really shine at 20mA. They can tolerate up to 50mA of direct current or up to 200mA at 1khz 1/10 duty cycle in a multiplexed display. The MAX7219 however can only provide a maximum of 40mA per segment current - that's peak current so the average current through each segment is around 5mA. So, 40 / 8 digits = 5mA equivalent DC current when set to maximum brightness.

The way around this problem was to use a UDN2982 as a source driver and a ULN2803A as a sink driver for the segment and digit pins respectively and suitable current limiting resistors. A hex inverter was also used as the digit pins would be at a logic low when on and the ULN2803A needs a logic 1 to 'turn on'.

With my setup I was able to get around 15mA of average current through each segment, greatly increasing the brightness. I did this by setting the peak current to around 60mA and setting the scan limit to 4 digits. LED supply voltage on the UDN2982 was 7.2V. I couldn't go over 60mA per segment peak due to the total current limit of the UDN2982 chip but that is explained in the video.

Example schematic can be found at [ Ссылка ]

The resistor values and V-LED supply will have to be calculated to your requirements. This circuit should be able to drive the big LED modules that have 2 or more LED chips per segment. As long as they are in series not parallel that is; check your display's datasheet. The ones I've looked at still need the same current but as there are more LED chips they need a higher forward voltage. I would recommend high efficiency displays for best brightness and lower power consumption.

If going down this route check that the power dissipation through either of the sink / source driver chips does not exceed the values stated in the datasheets. I've not tried this myself. The lower the current of the LED modules the better. The UDN2982 for example produces 65 degrees C of heat per watt! The power dissipation can be calculated by current x 8 digits including the decimal point x voltage dropped across the UDN2982. In my case this was 0.057x8x1.6=0.729 watts. This is worst case scenario with all digits on showing 8. which could happen during a display self test. In reality the power dissipation will be much lower as not all segments will be lit at once during typical use.

This circuit could also be used for a dot matrix display as well as for 7 segment displays. One other thing is the current limiting resistors should be at least 0.5w as they do get quite warm.

This project was really to use up some LED 7 segment modules I have spare that I wanted to use. But it demonstrates that typical LED driver chips are only intended to be used with high efficiency displays up to 0.56". Above that extra current sourcing and sinking is required.

If you require more than 15 milliamps average current per segment and / or are using more than 4 digits this circuit would not work as the power dissipation and current through the UDN2982 source driver would be too great. A bank of high power NPN and PNP transistors or even MOSFETs on heatsinks would be better in that case.

Maxim recommends using common anode displays in the case of very large displays driven by P and N channel FETs. Their example circuit can deliver 100mA peak current per segment. The link to their article is here [ Ссылка ]

UDN2982A datasheet [ Ссылка ]
ULN2803A datasheet [ Ссылка ]
74HCT04N datasheet [ Ссылка ]
74AHCT125N datasheet [ Ссылка ]

#max7219
#electronicsproject
#7segment

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