heck yea. that is simpler. if i can find the hef4514 in the stores...
Just to verify:
1) now, it's more important i stay with a regulated power supply, right? or should i put the regulator back in (yea, i agree with you, it's low cost) or did you take it out because i said i would use a regulated supply?
If you use the HEF4514 it works from +12V, so you dont need the +6Vreg
2) And the "1KO" is 1K ohm resistors, right? all 4 goes to 12v right?
Using the 4514 the 1KOs goto +12V
3) what wattage for the 1K's?
Quarter Watt [0.25W] ,, eg: 12V/1000R= 12mA.... 12V * 12mA = 0.144W
4) on the '514, pin 1 and 24 go to 12v and pin 12 and 23 goes to ground?
Thats correct
And lastly, since I think all my current lamps are 6v (240ma was what one was drawing at 6v), and i would have to buy 8 new ones, but before i do, i want to see the brightness with one as a test. **OR** - is there now an easy way to step the voltage back down to 6v - maybe simply using a 6v power supply? but i know the 2019 need 6v...... so..... any other options easily? I forget what's the point of using the 12v lamps instead of 6v besides lower power?
Why dont you calculate a resistor value that will drop the +12V to the lamp voltage?
eg: R=V/I .... 6V/0.24A = 25R.... W= V * I = 6V * 0.24A =1.44W say 2Watt.
Connect the single, 2W, 25R resistor between the +12V and the lamp supply.
Remember there is only one lamp lit at any one time.