My plan for an S-100 linear power supply PCB may be a bit more complicated than it first appears.  I am following the method described in the classic book by Steve Ciarcia "Build Your Own Z80 Computer" on pages 1-29

http://www.scribd.com/doc/13388965/Build-Your-Own-Z80-Computer

I've run though the calculations a few times varying assumptions. Getting the selection of transformers right is critical since everything that follows depends on their parameters.

As I suspected, an "off the shelf" transformer with the proper voltages does not appear to be available.  Using a custom unit defeats my intent of making something that is relatively low cost, uses standard commercially available components, and other hobbyists can build with a reasonable amount of effort.  Also, I don't want to supply parts rather specify items that the builders can buy themselves for their home brew systems.

It appears the two transformer method the MITS Altair is probably best.  I've identified some candidate transformers at Digikey.  There are two possible transformers that would work, I think using the method Dwight suggests of peak voltage minus diode voltage.  I believe 80VA is probably sufficient for the current requirements of a small S-100 system.

For the +8V rail, this model which gives 10A in the parallel output configuration

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=237-1262-ND

Maximum voltage from the transformer would be Vrms * 1.4 = 8V * 1.4 = 11.2V

Subtract two diode voltages from the bridge leaves Vpeak = 11.2V - (0.6 * 2) = 10V

If 8V is the minimum, then Vripple is 10V - 8V = 2V

For the capacitor on the +8V rail, using Steve's formula, I get

C1 = dt * I / dv = .0083s * 10A / 2V = ~42 mF

42mF is quite large for PCB mount but a couple of these in parallel should do it (22mF 16V electrolytic capacitor)

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=493-1054-ND

For the +14V/-14V rails, this model gives 2.8A each

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=237-1280-ND

Maximum voltage from the transformer would be Vrms * 1.4 = 14V * 1.4 = 19.6V

Subtract two diode voltages from the bridge leaves Vpeak = 19.6V - (0.6 * 2) = 18.4V

If 15V is the minimum, then Vripple is 18.4V - 15V = 3.4V

For the capacitor on the +15V and -15V rails, using Steve's formula, I get

C2 = C3 = dt * I / dv = .0083s * 2.8A / 3.4V = ~7 mF

7 mF is OK for PCB mount so one of these for each the +15V and -15V rail (10 mF 25V electrolytic capacitor)

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=493-1071-ND

The transformers are chassis mount units.  I was hoping to get PCB mount transformers but the largest I can find is 56VA which is probably not enough current capacity.  The capacitors are all PCB mount although the voltage ratings are a bit less than I'd like. 

Digikey sells 10A capable diodes that I can use in two groups of four to form the bridge rectifiers.

http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=10A01CT-ND

Put a 10A fuse on the +8V rail, and 2.5A fuses on the +15V and -15V rails and that should do it.  The PCB doesn't have many components but will still be sizeable due to the scale of the components and the thick and wide traces they'll require.

The configuration will be two chassis mount transformers connected to a PCB.  All three pieces should be mounted in a kit box with a fan, a fuse, power cord, and a switch for safety.

Thanks and have a nice day!

Andrew Lynch