simulate this circuit – Schematic created using CircuitLab

Hey guys

I've been trying to use a 4bit adder IC, but there are 2 main problems I've found which prevent me from using it:

  • If power is connected to the VCC pin of the IC, and GND is not connected, then every single output pin gives an output, without any connected inputs
  • Also, no outputs work when I plug GND into GND and so on, even when I connect inputs in with pull-down resistors.

This schematic is only using the first bit of the adder as a test, yet literally nothing turns on. I know that 10k resistors may be too much, however, they have worked perfectly with other ICs i've used

Heres the datasheet: https://pdf1.alldatasheet.com/datasheet-pdf/view/5700/MOTOROLA/SN74LS283N.html

Thanks for the help guys

PS I'm not good at creating circuit diagrams so try to forgive anything that doesn't make sense

  • 1
    \$\begingroup\$ Well... very little makes sense. Is the negative side of the 5 volt supply connected to the 74LS283 GND pin? Do the apparently unconnected ends of R1 and R2 connect anywhere? Inputs on 74LS parts source current, and need to be pulled down by a 510 Ohm (or so) resistor - better to connect a switch between input and Ground to make a definite Low. An unconnected input will be considered a High. If there is no Ground connected to the chip, the outputs will be High, because there is nothing to pull them Low. \$\endgroup\$ Commented Jun 11, 2020 at 1:19
  • \$\begingroup\$ Yes, sorry every GND symbol on the schematic is a connection to negative of the power supply, my bad. \$\endgroup\$ Commented Jun 11, 2020 at 1:29
  • \$\begingroup\$ Also, could you explain what you mean by adding a switch in between input and ground? I don't understand, thanks man \$\endgroup\$ Commented Jun 11, 2020 at 1:39
  • \$\begingroup\$ I only see one Ground symbol - that on the 'ls283. There are other triangle symbols on the ends of resistors, but the way they are drawn they don't look like Ground symbols to me. You haven't said if the negative end of the 5 volt supply is intended to be grounded. If you show a schematic, you MUST draw it carefully, showing all connections, and using conventional symbols correctly, otherwise few of us will try to understand it. If the top of R4 is supposed to be grounded, the LED is backwards - but the LED cathode isn't connected to anything. \$\endgroup\$ Commented Jun 11, 2020 at 1:39
  • \$\begingroup\$ With 74LS inputs, you should use a switch from input to Ground to ensure that the input is really Low when it should be. Your 10K resistors to Ground will only pull the input down to a "maybe" state. You should use a 10K to +5 V to ensure that the input is High when the switch is open. \$\endgroup\$ Commented Jun 11, 2020 at 1:43

1 Answer 1


I know that 10k resistors may be too much

Yes, they are too much. An LSTTL input may go to several internal diodes and/or transistors which are operated by drawing current from them. The input voltage should be below 0.8 V for a 'low' (logic '0'), so the pull down resistance must be low enough to drop less than 0.8 V at the specified input current. 74LS283 A and B inputs may source up to 0.8 mA, so the maximum resistance that guarantees logic '0' is 0.8 V / 0.8 mA = 1 kΩ.

Other logic types have different current requirements depending on the semiconductors used and internal operating currents. Most modern logic IC's use MOSFETs which are driven by voltage and require negligible DC input current, so very high resistor values can be used. This also applies to the older CD4000 and 74HC series, and NMOS LSI chips.

  • \$\begingroup\$ Thanks man! This worked, thanks for explaining as well \$\endgroup\$ Commented Jun 11, 2020 at 16:01

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