74LS Output Driving CMOS Inputs and an LED
In a breadboard project, the output of a 74LS00 NAND gate is required to drive three 74HC00 inputs and to light an LED as a 'logic 1' indicator. The supply is 5.0 V. The LED has a forward voltage of 2.0 V and is connected from the gate output to ground through a 330 Ω resistor. On the bench the LED is dim, and the gate output high is measured at 2.1 V. The three HC inputs sometimes behave as if they are receiving a low signal, especially when the LED is connected. The student adds a 10 kΩ pull-up resistor from the gate output to +5 V; the LED brightness changes only slightly and the high level rises to 2.2 V. The student then proposes removing the 330 Ω resistor and connecting the LED directly between the output and ground, arguing that the gate will then 'definitely' pull high. The 74LS00 data sheet gives IOH(max) = −0.4 mA and IOL(max) = 8 mA.
1. Calculate the LED current if the output were an ideal 5.0 V source. Compare it with the 74LS00 output ratings and explain the measured 2.1 V and dim LED.
Read the worked answer
With an ideal 5.0 V output and a 2.0 V LED, the resistor current would be:
ILED = (5.0 − 2.0) / 330 Ω = 9.1 mA.
That is far beyond the 74LS00 high-level source rating of 0.4 mA. A 74LS output can sink 8 mA when low, but it cannot source 9 mA when high. The output transistor and internal pull-up cannot hold 5 V into that load, so the output high collapses. At a measured 2.1 V the LED current is only about (2.1 − 2.0)/330 = 0.3 mA, which explains the dim glow. The 10 kΩ pull-up can add only about (5.0 − 2.1)/10 kΩ = 0.29 mA, which is negligible compared with the 9 mA the LED is trying to draw. The problem is a load placed on the wrong side of the logic output, not a weak logic gate.
2. Explain why the 74HC00 inputs may misread the 2.1 V level and calculate the required VIH at a 5 V supply. Will the 10 kΩ pull-up fix the problem? Propose two better fixes.
Read the worked answer
For a 74HC CMOS gate at VDD = 5 V, the guaranteed high-level input threshold is typically:
VIH = 0.7 × VDD = 3.5 V.
A 2.1 V high is below that threshold, so the HC inputs are not guaranteed to see a logic 1. Their actual switching point may be around half-supply, so they may appear to work on one board and fail on another, especially with the LED loading the node.
The 10 kΩ pull-up does not fix this. When the LS output is high, it is actively driving the node; the pull-up only adds about 0.3 mA and cannot supply the LED current or raise the node to 3.5 V. Better fixes are:
- Move the LED to the low side: connect +5 V → 330 Ω → LED → gate output, so the LED is on when the output is low. The gate then sinks current, which it is rated to do; use 390 Ω if the current must stay below 8 mA.
- Keep the LED off the logic node entirely and drive it through a small transistor, or buffer the logic output. A 74HCT00 can also be used for the HC inputs because HCT has a TTL-compatible VIH of about 2.0 V, but the LED must still be moved or buffered so the logic level is not degraded.
3. Evaluate the proposal to remove the 330 Ω resistor and connect the LED directly between the output and ground. What would likely happen to the gate and the LED?
Read the worked answer
The proposal is unsafe. With the LED directly from output to ground, there is no series element to set the current. When the gate tries to drive high, the LED clamps the output at about its forward voltage, around 2 V, and the current is limited only by the gate output impedance and the LED's internal resistance. That current can exceed the 74LS00 ratings and damage the output stage, and it can also overheat the LED. The logic high would still be far below the 3.5 V needed by the HC inputs, so the original fault would remain. When the output is low, the LED would be off, but that does not make the high state safe. A current-limiting resistor is required on either the high side or the low side, and the LED should not share the logic node with CMOS inputs unless the currents have been checked against the data sheet.