Member #502939

Member Since: December 27, 2013

Country: United States

  • Product COM-08530 | about 3 months ago

    You need greater than 12 volts to get it to light. I’m using nominal 14.8V (max 16.8V) Lithium Ion battery. Connect supply(+) to 5 (middle) trace of display. Connect any one of the other display segment traces to a current limit resistor. Connect the other side of the current limit resistor to supply(-). Segment will light. Resistor should be picked to set up around 20mA current. Forward voltage for display is around 11.6V. For my battery this gives (16.8 - 11.6) = 5.2V / 0.020A = 260ohms. When you are ready to hook up the TPIC6B595, break the resistor’s connection to ground and connect it to one of the drain pins on the TPIC6B595 (this will become your segment ground when active). You of course need to have the other pins of the shift register connected correctly and you have to understand how to use it. The project I’m building has four digits and I plan to use a second TPIC6B595 connected to the first (and P channel MOSFETs and pullup resistors ) to do digit selection for multiplexing as well.

    EDIT: Actually now that I think of it… it seems to me that since a red LED is typically driven at 20mA and on the 7 segments there are two parallel paths, you would want pick resistors to drive each segment at 40mA so that each parallel branch of LEDs gets 20mA. The decimal point however is a single branch of LEDs so you would want to pick a resistor that would drive that branch at 20mA. Perhaps others can comment on whether they believe that thought is correct or not.

  • Product COM-08530 | about 4 months ago

    I’m an Arduino beginner and more of a software type so please forgive me but… I don’t understand why you would need a current limit resistor when powering one of these with 12VDC. The datasheet shows that the 12volt drop will effectively occur over 6 LEDs. That means 2volts per LED. What am I missing here?

    EDIT: Nevermind, I understand now.

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