TTL NAND driving an LED indicator LOW (active-LOW sink configuration) A standard TTL NAND output sinks current through a 470 Ω series resistor and LED to VCC = 5 V. What current flows when the LED is ON (output LOW)?

Introduction / Context:When a TTL gate output pulls LOW, it can sink current from a load tied to VCC. For an LED indicator with a series resistor connected to VCC, the LED current is set by the voltage across the resistor, which depends on VCC, the LED forward drop, and the LOW-level output voltage of the gate (VOL). The question tests understanding of active-LOW LED driving and basic current calculation.

Given Data / Assumptions:

• Logic family: standard TTL NAND, VCC ≈ 5 V.
• Series resistor: R = 470 Ω in series with the LED, tied to VCC.
• Gate output is LOW when LED is ON, so current flows from VCC → resistor → LED → gate output to ground.
• Use typical textbook values for LED forward drop and VOL to obtain the closest numerical option.

Concept / Approach:Ohm's law sets current I = V_R / R, where V_R is the voltage across the resistor. V_R = VCC − V_LED − VOL. Typical small red LED forward drop is about 2.0 V to 2.2 V under a few milliamps. A TTL VOL under modest sink currents is typically a few hundred millivolts. The closest standard value among the options generally corresponds to around 5 mA to 6 mA.

Step-by-Step Solution:

Verification / Alternative check:Trying nearby typical values (for example, V_LED between 2.0 and 2.3 V and VOL between 0.2 and 0.4 V) yields currents in the 5 mA to 6.5 mA range. The given options include 5.32 mA, which matches the calculation using mainstream assumptions.

Why Other Options Are Wrong:

• 7.02 mA: implies a much lower LED drop or VOL, not typical at this operating point.
• 8.51 mA: requires an unrealistically low combined drop across LED and VOL for 470 Ω.
• 10.63 mA: far too high for the stated component values and typical TTL levels.

Common Pitfalls:Forgetting to include VOL. Assuming the LED is tied to ground rather than VCC in an active-LOW sink configuration leads to the wrong reference polarities.

Final Answer:5.32 mA