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I am currently struggling a lot with a tedious issue - I have to control my 7-segment display with a CD4511 decoder. However, my display is Common Anode, and it is not changable in my case. The best I got out of the situation is that the numbers were inverted(lights that should be on are off and vise versa). Later on, I have connected 7 transistors in order to reverse all the outputs of the chip, which resolved the issue, but only partially. Yes, the lights were on when they needed to be, however, my project REQUIRES to use only one transistor.

So, I used the PNP transistor, connecting the collector to the common of 7-segment display, base to arduino digital pin(and setting this pin to low) and the emitter to 5V. The result I am getting is that the needed lights DO light up, but only for a mere second, instantly turning off afterwards. Is there anything I can do to properly invert the display?Here is a circuit I have built. Also, I am working on TinkerCad

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    \$\begingroup\$ The "I have to control my 7-segment display with a CD4511 decoder" and the "my project REQUIRES" both suggest to me this could be homework. Could you clarify on this point? Is this a puzzle you've been instructed to solve and therefore is limited (we'd want to know your explicit project limitations more fully) or are you open to a variety of options (we'd want to know your personal limitations more fully.) I'm asking because I don't want to poorly invest time. (Schematic could be laid out better, too. Any chance?) \$\endgroup\$
    – jonk
    Jan 16, 2022 at 21:20
  • \$\begingroup\$ Not a homework, rather a part of a bigger project. The options are limited, yes, I am forced to use a 4511 decoder, 1 display, and 1 transistor. Unfortunately, I am not able to edit the schematic itself, as it is automatically done by the Tinkercad(simulator I am working on) \$\endgroup\$
    – Shodan-AlphaVersion
    Jan 16, 2022 at 21:27
  • \$\begingroup\$ this smells like schoolwork, even if you say that it is not ........ replace the display with one that works .... or try hacking the one you have \$\endgroup\$
    – jsotola
    Jan 16, 2022 at 21:58
  • \$\begingroup\$ I can think of at least one answer, but it's really only suitable for a homework question and unfortunately yours is not such a question. \$\endgroup\$
    – Spehro Pefhany
    Jan 16, 2022 at 22:22
  • \$\begingroup\$ (CD4511 decoder…Common Anode…only one transistorscrewed for good.) A 1k resistor in the power supply line of a display driver won't work with displays needing (many) mA. One second of operation would seem plausible if the display driver had a buffer capacitor in the mF range. \$\endgroup\$
    – greybeard
    May 30 at 13:07

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Why do you need a transistor? thats the nice thing of that IC, its all inside.

If you are constrained on component count for whatever reason, then just invert the 7-segement mapping in software, its really that simple.

Also, why is there only one resistor? 1K is enough to not damage the leds or the driver, but the segments will get dimmer if you turn on multiple of them. That's really bad practice. Power the 7-seg to the positive supply as you would normally, then the cathode of each segment gets a resistor to the input of the driver IC. Pick the resistors to stay under the max sinking current of that driver.

Also, this circuit diagram hurts my eyes. Dont make big inter-connecting lines on power supply nets. Thats why we have net labels, they are free to use ;) same with the grounds

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  • \$\begingroup\$ The point of the question is that you can't invert the mapping in software, because the mapping is in the IC. \$\endgroup\$
    – pipe
    May 30 at 12:25
  • \$\begingroup\$ He stated that he got it working however the segments where inverted. Perhaps the original application intended the leds to be driven by a high-side driver instead of a low-side one, explaining the inversion. Simply flip the on and off states in firmware, if you are able to modify it. No need for 7 transistors. The CD4511 is able to sink current on its own up to 50mA for all outputs combined. Enough for a 7 segment. There are some other issues with this schematic though. \$\endgroup\$
    – Thijs
    May 30 at 14:20

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