How to Set and Change the Predefined Combination Code in Zen22142 Digital Lock Circuit

Hi, I need some help with a digital combination lock, the link is: http://www.zen22142.zen.co.uk/Circuits/Switching/diglock.htm

The site explains some of the concepts regard the circuit, however, what I can't understand is how to determine the predefined combination code. Could anyone assist me on this?

Any help is appreciated. Thanks

Programming with Solder. It's setup up manually by the selection of the output from one stage to input to the next stage. The output from the first stage is applied to one input on the And gate, when the second switch is triggered, it provides the clocking pulse to the second stage, and so on.

I am sorry but i have yet to learn all the fundamentals in electronics. How do I program with solder?

Since it says that it has a predefined code already, is it safe to say that the circuit has a combination code as it is? or do i have to tweak it in order to set a code for it?

Because it is based on a counter. Each pin is the designated number, i.e. pin 6 on the first chip sets the first number to pin 6. Second would be 3, third 1, last 7. The Q# is the output of the counter, and goes high when the counter is cycled to that digit. Each high affects the following stages. If you move the wire on the outputs, the selected Q# becomes the combination. Programming by solder means, no code input, but physically designating each number by selecting the lead.

Wow! you've explain it so well. I just need to confirm something. So the output on the most right side of the schematic, where the 5-15V is written, should be connected to the four outputs of each counter. I have four wires connected to it which is connected to the individual output pins of each counters. Is this right?

I mean, the output where "OUT" is written. The one below 5-15V sorry for the confusion.

On the counters, the Q# is the output of each counter. Each counter output feeds to the gate and switch that is the input for the next stage. The logic gate with 2 inputs and 1 output combine the number output from the counter and the output from the switch to make changes to the second counter. IC1 pin 6(first digit) and PBS-B inputs go to 4081, when pin 6 is high(+5v), then the push button switch causes a high pulse. That is put into the CP0 which causes IC2 to cycle. If the number 6 isn't selected on IC1, then there is no high signal, and IC2 will not cycle. Each digit has to be selected before the next one can be selected. It works like an assembly line where the first part is completed, then the next, and so on. Until you get to the last stage. The out is normally held high until the last number (7) is enabled, and the third stage has to be set back to 8, which enables the gate before the BC182B transistor. The two highs make the transistor turn on, and apply ground to light the LED, which gives you the low out for whatever you need to configure.

I understand that part, but what I don't get is what I should do with "OUT". Thank you very much for your help

The out is either High or Low, depending on Locked or unlocked. This provides the combination to trigger what ever needs a high to low transition.

I understand now, Thank you