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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

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

schematic

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.

schematic

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 Jun 7 '14 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\$ – Spehro Pefhany Jun 7 '14 at 19:32
  • \$\begingroup\$ So you say ı need to use 150 ohm resistor or something less than 470 ohm ? \$\endgroup\$ – Vitasset Jun 7 '14 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\$ – Spehro Pefhany Jun 7 '14 at 20:46

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