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Suppose we have connected an X volt (X=10) as the supply of a 4051 MUX (which is active LOW), can we disable the device by applying a B volte (assume B is 5 volt). In other words, how high is HIGH relative to the supply voltage?

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  • \$\begingroup\$ Read the data sheet - it will tell you the digital logic levels for 1 and 0. \$\endgroup\$ – Andy aka May 6 '17 at 15:07
  • \$\begingroup\$ thanks Andy, I've did already, but I couldn't find anything on that. I'll take a look again now. \$\endgroup\$ – MimSaad May 6 '17 at 15:08
  • \$\begingroup\$ Just look at the datasheet. Any good one will have logic levels on it. \$\endgroup\$ – Hearth May 6 '17 at 15:08
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    \$\begingroup\$ For 4000-series CMOS logic, the standard is: logic low is 0..30% Vdd; logic high is 70..100% Vdd. Stay out of 30..70% Vdd, the chip's entitled to logically misbehave. (There are applications that do things in this range to get 'linear mode' from the IC but put that aside for pure logic circuits.) Take nothing for granted, always check the data sheet. \$\endgroup\$ – TonyM May 6 '17 at 15:22
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    \$\begingroup\$ The 4051 has internal level shifting- the guaranteed logic level 1 is 0.7*(Vdd - Vss) relative to Vss, and Vee does not affect it, at least for Vee <= Vss. \$\endgroup\$ – Spehro Pefhany May 7 '17 at 13:36
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Well here it is (at least for the CD4051/52): -

enter image description here

This is from the first 4051 data sheet found. On a 10 volt supply the high level must be at least 7 volts. It's a 70%:30% thing.

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  • \$\begingroup\$ exactly, you got it! \$\endgroup\$ – MimSaad May 6 '17 at 15:15

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