This is a simple but effective code lock circuit that has an automatic reset facility. The circuit is made around the dual flip-flop IC CD4013.Two CD 4013 ICs are used here. Push button switches are used for entering the code number. One side of all the push button switches are connected to +12V DC. The remaining end of push buttons 2,3,6,8 is connected to clock input pins of the filp-flops. The remaining end of other push button switches are shorted and connected to the set pin of the filp-flops.
The relay coil will be activated only if the code is entered in correct sequence and if there is any variation, the lock will be resetted. Here is correct code is 2368.When you press 2 the first flip flop(IC1a) will be triggered and the value at the data in (pin9) will be transferred to the Q output (pin13).Since pin 9 is grounded the value is â€œ0â€ and so the pin 13 becomes low. For the subsequent pressing of the remaining code digits in the correct sequence the â€œ0â€ will reach the Q output (pin1) of the last flip flop (IC2b).This makes the transistor ON and the relay is energised.The automatic reset facility is achieved by the resistor R11 and capacitor C2.The positive end of capacitor C2 is connected to the set pin of the filp-flops.When the transistor is switched ON, the capacitor C2 begins to charge and when the voltage across it becomes sufficient the flip-flops are resetted. This makes the lock open for a fixed amount of time and then it locks automatically. The time delay can be adjusted by varying the values of R11 and C2.
Circuit diagram with Parts list.
- Assemble the circuit on a good quality PCB.
- The circuit can be powered from 12V DC.
- Mount the ICs on holders.
- The L1 can be a 12V, 200 Ohm SPDT relay.
- Capacitor C1 should be tantalum type.
- The C1 and C2 must be rated at least 25V.