RE: [N8VEM-S100:1139] S-100 SMB and S-100 bus extender PCBs status

["John Monahan" <mon...@vitasoft.org>]

That's what I thought too Neil, then I was thinking perhaps there is more heat generated by the chip in the actual electronics dropping the voltage. However it's hard to believe it could be that significant. In any event I order a few L4940-5's from Mouser and will check them out.

BTW I just added a short note  about increasing the heat sink area on the web page.  See bottom of page here:-
http://s100computers.com/My%20System%20Pages/SMB%20Board/S100%20Bus%20SMB.htm

John


John Monahan Ph.D
e-mail: mon...@vitasoft.org
Text:    mon...@txt.att.net



-----Original Message-----
From: n8vem...@googlegroups.com [mailto:n8vem...@googlegroups.com] On Behalf Of nbreeden
Sent: Thursday, October 11, 2012 12:25 PM
To: n8vem...@googlegroups.com
Subject: Re: [N8VEM-S100:1139] S-100 SMB and S-100 bus extender PCBs status

This makes no sense to me. We need to drop 3VDC to get from 8 volts to 5 volts. 3 volts drop at 1.5. amps is 4.5 watts, period. This is completely independent of the minimum voltage delta between input and output for the regulator to work. Ultimately the regulator presents a series resistance with the load to drop the excess voltage across (the load being the other resistor in the voltage divider). Both regulators will disapate the same energy as heat. P=E*I

For my SMB I added a second 7805 for the displays. Ultimately both regulators on my SMB disapate the same heat as a single regulator would, however with two heat sinks I have twice the surface area to disapate the heat. 

You can't get around the physics, these are both linear regulators, they both need to drop 3 volts, at 1.5 amps that drop generates 4.5 watts, the load at 5 volts at 1.5 amps generates 7.5 watts, total heat is then 12 watts.

-Neil