As often as I see the CD4050 used as a logic level converter, I'm surprised that I haven't heard much about how it seems to leak voltage from the inputs into Vdd. I hope this isn't a case where the chips I purchased are simply not working correctly.

The version I'm using was bought off eBay, and the top says "13YR26D / CD4050BE".

I was using this chip to try and work with a Nokia 3310 LCD, and I first noticed things were weird when the screen was dim. When I disconnected power to my breadboard, the LCD lit up! Where was this power coming from??

Turns out, it was coming from the 4050. The 5V coming from the Arduino outputs were actually powering the chip, and providing power to my breadboard rails.

OK, so I thought "I must be doing something wrong" (I still think that!) and I came up with a simple circuit to test the chip.

Vdd (pin 1) to 3.3V Vss (pin 8) to GND 1A (pin 2) to 5V 1Y (pin 3) to an LED/1k, then to GND

The other input/output pins I tied to GND. I don't show that in this diagram:

4050 Test circuit

I am measuring the voltage between Vdd and GND. When I plug 1A into GND, I measure 3.3V as expected. When I plug it into the 5V rail, I measure 4.13V at Vdd. The voltage at 1Y is only 3.4V though. So my question is: is it expected for the voltage at Vdd to change? To me this is not just unexpected, but a problem, because I'm powering the LCD from this output.

My workaround right now is to use one of the buffers to power my LCD. The output is closer to 3.3V, and so this seems to work. But I wish I understood this better.

  • \$\begingroup\$ Oh, an interesting follow up. I can unplug the 3.3V line from the Arduino to the 4050, and Vdd will still show voltage. I already took apart my test circuit, but my memory tells me that unplugging the arduino had no effect on the voltage, as if all the Vdd voltage was coming from the inputs instead of the 3.3V power supply. \$\endgroup\$
    – colinta
    Dec 6 '14 at 23:47
  • \$\begingroup\$ "The version I'm using was bought off eBay". Does the same thing happen on all inputs? If so then your 'CD4050' is probably a fake! \$\endgroup\$ Dec 7 '14 at 1:44
  • \$\begingroup\$ Yes, all inputs have the same behaviour. Is there any other test I could do on these to make sure it's not user error? \$\endgroup\$
    – colinta
    Dec 7 '14 at 5:42
  • \$\begingroup\$ Update: using a 74HC4050 solved EVERYthing. I tried two different orders of CD4050s, same result (leakage). 74HC's worked immediately. \$\endgroup\$
    – colinta
    Jan 30 '15 at 17:03
  • \$\begingroup\$ So what did you find in the end? What was the result? Holy hell, man. Tell us what happened. Bad chips? (I find that HIGHLY unlikely but I would like to know) Please update us plebs. Freaking hell StackExchange is useless because nobody cares, because the moderators are assholes... \$\endgroup\$
    – CR.
    Jan 15 '19 at 1:06

The CD4050 (and its cousin the CD4049) have a special input stage that allows it to accept inputs higher than Vdd without ESD diodes conducting. So, it should work just fine in this application.

enter image description here

Maybe you've got some kind of a wiring error or the chip is damaged.

  • \$\begingroup\$ I got a bunch of these when I ordered them, they all behave the same way. It is sounding like I've got a bad batch. \$\endgroup\$
    – colinta
    Dec 7 '14 at 5:43
  • \$\begingroup\$ or @Bruce is right and they're re-marked something else. \$\endgroup\$ Dec 7 '14 at 6:54
  • 1
    \$\begingroup\$ Thanks for your help, I ended up ordering replacements (from Tayda, I've had good luck with parts from them) \$\endgroup\$
    – colinta
    Dec 22 '14 at 2:03
  • 1
    \$\begingroup\$ Did the replacements perform better? \$\endgroup\$
    – CR.
    Jan 15 '19 at 1:07

There are ESD diodes on the inputs connecting the input to Vcc.

You should add a resistor in the data line to allow the voltage to drop:


simulate this circuit – Schematic created using CircuitLab

You should then get 1.7v less the forward voltage of the ESD diode dropped across the resistor instead of heading up to the 3.3V rail.

  • \$\begingroup\$ Tonight I will add a resistor to my test circuit and report back. If it doesn't fix things, I'll have to toss these. No point in holding on to bogus chips. \$\endgroup\$
    – colinta
    Dec 8 '14 at 23:58

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