Hiya, ran across this (and a previous post asking if someone could do you a build- which you clearly don't need anymore hehe). Have you also solved this problem
If not, I have "a" solution, although I'm not really clever enough for a full how-to and implementation. But the idea is straightforward enough.
some sort of "adder" like a data AND gate (probably using some sort of 4xxx style chip. You could set that so that it only throws an output once it has an "input of 2". Not sure if I've made that make sense, but in principal it requires TWO input pulses before it ouputs ONE, with a bypass switch, you could have this do a 10p to 20p conversion
This will run into an issue if you have any ODD size prizes (10p, 50p £1.50 etc) which I know those old machines do. You'd end up with that HALF state being "stored". You could either just live with that and the effect it would have - sometimes an odd prize would have 10p added, sometimes it would lose 10p. Or you could add a "clear" function to this 'adder/buffer' circuit/chip that would reset the count
Sorry I can't be of much help in the full technicals of how to actually build this, but hopefully it's planted a seed for you
Personally, I'm hoping to build an alternative solution to the mixed-age machines problem. I plan on having the usual £1 and 20p hoppers, (for modern-ish machines) and to go old-school with 10p and 2p solenoid tubes for my lovely 80s and early 90s games (probably will not bother with 5p but maybe...) I know the usual problem with payout tubes is the 50V issue, but I might try and find a way of "constructing" my own payout system for this. I've not investigated yet how I'd get the 2 differing sytems to interface through the ipac, but probably just with switching. I'd also probably end up with a second coin mech that would feed/stock those tubes (as well as adding the 2p accept).
