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I'm looking into making a 7 segment LCD display (runs at 7.4 V) powered by a car battery. In order to do it I'll use a CD4511, since it can handle up to 18V. But upon reading the datasheet, I did not quite understand which is the appropriate input logic voltage to control the IC (if it's 5v or the same as the VCC applied to the IC). Thus, I made a diagram of what I think that would work, but I wanted some feedback before actually building it.

enter image description here

Rafael Mattei
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    It won't work. Current from Vbat would be pushed via transistor into MCU data pins, limited by Vbe drop of about 0.7V and the base resistance. There is also no way to pull data wire low to 4511. Most likely MCU goes up in a puff of smoke. The 4511 does need data input range based on supply voltage. – Justme Apr 26 '21 at 16:50
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    Logic levels are typically <1\3 and >2\3 Vdd but not >Vdd – Tony Stewart EE75 Apr 26 '21 at 16:51
  • Is the uC 5V? Or 3.3V? Is the LED 5, 10 or 15mA? Or 20mA?? Does it really need 7,4V? Don’t design anything until all interface specs with tolerances for temperature are defined – Tony Stewart EE75 Apr 26 '21 at 16:58
  • it's a 5v uC and the LED datasheet only says it's a red LED and has a voltage drop of 7.4V. – Rafael Mattei Apr 26 '21 at 17:03
  • What current @ 7.4V. Show datasheet in Question – Tony Stewart EE75 Apr 26 '21 at 17:08
  • Two additional things. 1) Study the CD4511 datasheet. It will describe what input conditions are considered low and high. 2) "18v" is not good enough for automotive applications. There routinely are large spikes on the 12V power, sometimes as much as 60VDC and even 80V reported. You will want to study ways to ruggedize this for automotive applications. – rdtsc Apr 26 '21 at 18:07

2 Answers2

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I assume you mean LED rather than LCD, as the 4511 is inappropriate for LCD and your schematic shows an LED display. If the actual circuit is a single LED per segment you would be better off adding a regulator to the 4511, but I'll assume it's a "big digit" type with several series LEDs per segment.

The input voltage should be a minimum of about 70% of Vdd, so for a 7.4V Vdd you would need a bit more than 5V in.

Your level shifter will not really work- try this:

schematic

simulate this circuit – Schematic created using CircuitLab

enter image description here

Spehro Pefhany
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Assuming that you use a CD4543B - which has the same input range specs as the CD4511 - the input voltage requires:

  • Vil < 1.5 (Vcc=5), < 3 (Vcc=10)
  • Vih > 3.5 (Vcc=5), > 7 (Vcc=10)

Linear interpolation to Vcc=7.4V means Vil < 2.22, Vih > 5.18. It is possible to make a passive level-shifter so long as:

  • your microcontroller ports support open-drain mode
  • when the drain is open, the passive ladder presents less than 5V to the MCU
  • the linear interpolation above is sound (this is kind of sketchy)

schematic

simulate this circuit – Schematic created using CircuitLab

This does what's expected:

simulation

and is marginally simpler than an active level shifter, though I figured this out mostly out of interest and if you have the real estate and budget I would still recommend an integrated, active level-shifting IC.

Reinderien
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