shaji@sbgrad5.cs.sunysb.edu (Shaji) (03/22/90)
In article <1990Mar19.154615.22362@tree.uucp> mirandor@.PacBell.COM (W. Le Roy Davis (1-23-90)) writes: >I am looking for the circut for a very inexpensive three event detector. >I work with the Boy Scouts, and they have this "pine box derby" I couldn't send mail, so here goes: I seem to have a no-frills circuit that uses 6 2-input nand gates, 6 resistors, and 6 LEDs. This is assuming that your trigger switches stay on once they are triggered (ie, they don't just provide a pulse). If not, you will need three latches to capture the trigger inputs. The basic idea is to compare pairwise, who beat who. To compare A and B, --------------------------------------------------------- | | | |-------------| |-----------| | |-----| | LED to VCC | | | | NAND |--------------------| NAND |--- LED to VCC GND-^^^^--|-| | GND---^^^^--|--| | 100ohm | |-------------| 100ohm | |-----------| | | | | / / / / | | | | VCC VCC Normally, both NAND gates are high, since the trigger switches are open and the corresponding inputs are low. (the resistors have to be low enough to ensure this condition). Suppose A beats B. Then A's Nand gate output goes low. This disables B's NAND gate. The gate of the winner is low. All you have to do now is compare the three pairwise. The winner is the one with two wins, the second has one win, and the third none. For convenience, you could hook up all three LEDs of a contestant near each other. Or you can get to be as fancy as you want - put in digital displays, colored lights etc depending on whether a contestant had 0, 1 or 2 wins. Not too complicated. This is cheap - at the expense of assuming ideal components. The disadvantage is that if the cars are within a fraction of a microsecond of each other, you could get contradictory results. A small price to pay for a simple, ckt, I think. Shaji Bhaskar shaji@sbcs.sunysb.edu