Driving LEDs from logic gates: Why is it usually recommended to sink LED current (pull LOW) from a logic output rather than source current (pull HIGH)?

Difficulty: Medium

Correct Answer: LOW-level current is larger.

Explanation:


Introduction / Context:
Many logic families (especially TTL and some 5 V CMOS variants driving TTL-compatible loads) can sink more current at a valid LOW than they can source at a valid HIGH. Correctly orienting the LED and resistor takes advantage of this capability for brighter and more reliable operation.


Given Data / Assumptions:

  • Logic output provides defined VOL and VOH limits.
  • Sinking current generally has a higher capability (mA) than sourcing.
  • LED with series resistor can be placed to VCC (for sinking) or to ground (for sourcing).


Concept / Approach:
Configure the LED to VCC with its resistor, and let the gate pull the cathode LOW to turn it on. This uses the stronger sink path, ensuring the LED current meets brightness targets while keeping output within valid logic levels.


Step-by-Step Solution:

1) Review datasheet: IOH (source) is often lower than IOL (sink).2) Choose sink configuration: VCC → resistor → LED → output pin.3) When output = LOW, current flows, LED turns ON; when output = HIGH, LED is OFF.4) Verify VOL stays under maximum at desired current.


Verification / Alternative check:
Compare brightness/current achievable when sourcing versus sinking using the same output device limits.


Why Other Options Are Wrong:

  • LOW-level current is smaller.: Opposite of typical capability.
  • HIGH-level current is larger.: Not generally true for TTL-style outputs.
  • LOW smaller & HIGH larger: Self-contradictory to common datasheets.


Common Pitfalls:
Ignoring VOL rise with excessive sink current, not recalculating resistor value, and forgetting per-pin and per-package current limits.


Final Answer:
LOW-level current is larger.

Discussion & Comments

No comments yet. Be the first to comment!
Join Discussion