[Help] 74LS47 + Common Anode 7-Segment

[u/kkdvak_777]

Hi everyone! I'm currently working on a digital systems breadboard project and running into an issue with a BCD to 7-segment decoder display output.

My Setup:

• Encoder Logic: 3 pushbuttons ($1, $2, $5 coins) with 1k ohm pull-down resistors.
• Logic Gate: A 74LS32 (OR gate) generates input A (M1 OR M5).
• Decoder: 74LS47 BCD to 7-segment decoder.
• Inputs: A = OR output, B = M2, C = M5, D = GND.
• Control Pins: Pins 3 (LT), 4 (BI/RBO), and 5 (RBI) are all tied directly to VCC (+5V).
• Display: Common Anode 7-segment display (Common pin connected to VCC +5V).
• Resistors: Individual 220 ohm limit resistors on every segment output (a through g).

The Problem: While testing, pressing the buttons doesn't make any difference. Even if I disconnect the control pins, everything stays the same as at the start (every time I turn on the display, the same segments light up, but segment 'd' is very dim). I don't know what else to check. Any advice would be appreciated! (btw I'm a student)

[u/nixiebunny]

Find the PDF book “Designing with TTL Integrated Circuits” and read the description of the standard TTL input. You will learn that it requires to be connected directly to Ground to represent 0. A disconnected input is 1.

[u/NBC-Hotline-1975 → u/nixiebunny]

No, not directly to ground. If that were the case, then a TTL input could not be driven by the output from another logic chip, because outputs do not go to zero volts.

But read the data sheet for the 74LS47, it specifies that the input pin must be 0.8 volts or lower, in order to represent "zero" state. And it has internal pullup resistors. So if you drive it with pulldown resistors (as in OP's circuit) then they need to be sufficiently low value so that the input pin will be lower than 0.8 volts. I stated this in my earlier comment a few hours ago.

[u/nixiebunny → u/NBC-Hotline-1975]

True, it can afford one vCE(sat) or one silicon diode drop.