74LS to 74HC Interfacing with an LED Load
A 5 V system uses a 74LS00 NAND gate to drive a 74HC00 input and a red LED. The LED has V_F ≈ 2.0 V at 10 mA and is connected from the LS output to ground through a 220 Ω resistor, so it lights when the output is high. The measured LS output high is only 2.4 V, the LED is dim, and the HC gate occasionally misreads the level. The LS datasheet gives V_OH(min) = 2.7 V at I_OH = 0.4 mA, V_OL(max) = 0.5 V at I_OL = 8 mA, I_IH = 20 µA and I_IL = 0.4 mA. The HC datasheet gives V_IH(min) = 3.5 V at 5 V supply and input leakage of about 1 µA. A student proposes a 10 kΩ pull-up from the LS output to +5 V and also connecting a second identical LED in parallel to make the display brighter.
1. Calculate the LED current when the output is high, compare it with I_OH, and explain the dimness and logic failure.
Read the worked answer
With the LED from output to ground, the LS gate must source current in the high state. Even if the output stayed at 2.4 V, I_LED = (2.4 − 2.0) / 220 = 1.8 mA, more than four times the 0.4 mA I_OH rating.
The output high therefore collapses; the LED is dim and the voltage is far below the 2.7 V V_OH minimum. The 74HC input needs at least 3.5 V, so 2.4 V is not a valid high. The problem is the wrong polarity of the LED load, not just a weak pull-up.
2. Evaluate the 10 kΩ pull-up and the second parallel LED.
Read the worked answer
A 10 kΩ pull-up to +5 V can supply only (5 − 2.4) / 10,000 = 0.26 mA at 2.4 V. The LED is demanding 1.8 mA, so the pull-up cannot raise the node; solving the divider gives about 2.1 V. With the LED removed, 10 kΩ would easily overcome the 1 µA HC leakage and pull the node to nearly 5 V, but with the LED it is ineffective.
Adding a second LED in parallel doubles the load, making the high level worse and approaching the low-state sink limit. The pull-up must not be used to fight an LED connected to ground.
3. Redesign the interface for a visible LED and reliable HC drive. Give resistor values and check the currents.
Read the worked answer
Connect the LED from +5 V through a resistor to the LS output, so it is active-low. For about 6 mA with V_OL = 0.5 V: R = (5 − 2.0 − 0.5) / 0.006 = 417 Ω; use 430 Ω, giving 5.8 mA. That leaves margin under the 8 mA I_OL rating for one LS input (0.4 mA) or even two (0.8 mA).
For the HC input, add a 4.7 kΩ pull-up from the LS output to +5 V. When the LS output is high, the pull-up raises the node to nearly 5 V, satisfying V_IH = 3.5 V, and the HC leakage is only about 1 µA. Alternatively use a 74HCT input, whose V_IH is compatible with LS levels. Do not parallel LEDs without recalculating the sink current.