I have a few questions about 7-segment LED displays.

When we build a circuit using 7 segment (common-anode) and decoder (7447), we also use 4 tactile switches to make combinations. Should I directly connect switches to Vcc or ground (I have no knowledge about that) or what should I do with switches, after completing rest of circuit i.e. resistors to IC, IC to Vcc and display to Vcc?

I still need the answer but I think I did it!

Picture of my 7-segment display circuit


1 Answer 1


It is not recommended lo leave logic inputs floating. You should use a pull-down / pull-up resistor (see Spehro Pefhany's comment).


simulate this circuit – Schematic created using CircuitLab

This circuit inputs low level by default (switch open). If you press the switch then it inputs high level.


simulate this circuit

This circuit inputs high level by default (switch open). If you press the switch then it inputs low level.

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    \$\begingroup\$ In the first image it is actually called a pull-down resistor. \$\endgroup\$
    – Gerben
    Commented Jun 7, 2014 at 19:26
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    \$\begingroup\$ This is wrong. A 1K pull-up resistor should be used. 4.7K will not bring a TTL input into the region for a valid '0', since the input current is -1.6mA. A pull-down would have to be 250 ohms or less, and would waste a lot of power. \$\endgroup\$ Commented Jun 7, 2014 at 19:32
  • \$\begingroup\$ So you say ı need to use 150 ohm resistor or something less than 470 ohm ? \$\endgroup\$
    – Vitasset
    Commented Jun 7, 2014 at 20:31
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    \$\begingroup\$ @Vitasset Is "250 ohms or less" not clear? 240 ohms will work and is a standard E24 value. Two 470 in parallel will work. 150 ohms will dissipate 0.17W when the switch is closed, so don't use a 1/8" resistor, but it will work. \$\endgroup\$ Commented Jun 7, 2014 at 20:46

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