NOR-Only NOT, OR, and AND
Six NOR gates implement NOT A, A OR B, and A AND B as three visible functions. The circuit demonstrates functional completeness through explicit gate-level transformations.
What You'll Learn
- Construct NOT A with one tied-input NOR gate.
- Construct A OR B by inverting a NOR result.
- Construct A AND B from inverted inputs using De Morgan's law.
- Verify the three simultaneously visible outputs for every A/B vector.
- Explain functional completeness without treating NOR-only form as universally optimal.
How It Works
This production circuit uses six NOR gates to construct three functions from switches A and B. Three output lights show NOT A, A OR B, and A AND B. There is no separate light for a raw two-input NOR result.
The NOT branch ties both inputs of one NOR to A, producing ¬(A + A) = ¬A. The OR branch sends A and B into one NOR, then ties both inputs of a second NOR to that intermediate result, canceling the inversion and producing A + B. The AND branch first makes ¬A and ¬B with two tied-input NORs, then computes ¬(¬A + ¬B) = A·B.
These constructions demonstrate NOR's functional completeness. The circuit exposes the three completed logic functions at lamps and leaves the intermediate NOR nodes inside their respective branches.
Truth Table
All three lamps are circuit outputs. For each A/B vector, read them in the order NOT A, A OR B, and A AND B.
| Inputs | Output | ||||
|---|---|---|---|---|---|
| A | B | NOT A | A OR B | A AND B | |
| 0 | 0 | 1 | 0 | 0 | A is low, so NOT A is high; neither OR nor AND is asserted |
| 0 | 1 | 1 | 1 | 0 | Only B is high; OR is asserted while AND remains low |
| 1 | 0 | 0 | 1 | 0 | Only A is high; NOT A is low and OR is asserted |
| 1 | 1 | 0 | 1 | 1 | Both inputs are high; OR and AND are asserted while NOT A is low |
Boolean Expression
Tying both inputs of the first NOR to A makes its visible output ¬A.
A second tied-input NOR inverts NOR(A, B), recovering A + B.
Two tied-input NORs form ¬A and ¬B; NORing those values gives A·B by De Morgan's law.
Try It Step-by-Step
Set the listed controls in the embedded circuit, then compare the observed modeled state with the expected result.
- 1A=0B=0Expected:
NOT A = 1, A OR B = 0, A AND B = 0What you'll see:At 00, the tied-input inverter branch is the only high output. - 2A=0B=1Expected:
NOT A = 1, A OR B = 1, A AND B = 0What you'll see:At 01, the two-NOR OR branch goes high while the three-NOR AND branch remains low. - 3A=1B=0Expected:
NOT A = 0, A OR B = 1, A AND B = 0What you'll see:At 10, OR remains high and NOT A changes low, showing that the first output depends only on A. - 4A=1B=1Expected:
NOT A = 0, A OR B = 1, A AND B = 1What you'll see:At 11, the reconstructed AND output goes high; all outputs match their named functions.
Components Used
Applications and Design Context
Standard-cell synthesis. NOR cells are available in common logic libraries and are selected when their polarity, fan-in, and timing fit the optimized network.
Active-high latch structures. Cross-coupled NOR gates form the canonical active-high SR latch, subject to its forbidden simultaneous-assertion condition.
Logic normalization. Boolean equations can be transformed into NOR-oriented forms for systematic implementation and verification.
Discrete logic design. A limited inventory of NOR packages can implement several functions without separate NOT, OR, and AND devices.
Digital-logic education. The three visible branches provide direct evidence of De Morgan transformations and universal-gate synthesis.
Frequently Asked Questions
Why is NOR called a universal gate?
A gate set is functionally complete when every Boolean function can be composed from it. NOR can realize NOT and then use that inversion to realize OR and AND, so arbitrary combinational logic can be expressed with NOR alone.
How does a tied-input NOR become NOT?
When both inputs receive A, the NOR computes ¬(A + A). Since A + A = A, the result is ¬A. One NOR is therefore sufficient for the visible NOT A branch.
Why does the OR branch require two NOR gates?
The first NOR produces ¬(A + B). The second has both inputs tied to that intermediate signal and inverts it, yielding A + B. The intermediate NOR value is not presented as a separate circuit output.
How does the AND branch work?
The circuit first generates ¬A and ¬B with two tied-input NORs. A third NOR produces ¬(¬A + ¬B), which equals A·B. This is a direct application of De Morgan's law.
Should every circuit be implemented with NOR gates only?
Not necessarily. NOR-only form proves that an implementation is possible, but a physical design is selected from the available library according to delay, area, power, fan-in, and loading. A mixed-cell network may be superior.