Electronic – Reducing voltage drop over 7805 regulator

7805power supply

I have several projects (clocks) that require 12V and 5V supply. The 5V is for TTL ICs and the 12V powers a SMPS(Switching Mode Power Supply) to give HV(High Voltage) for a tube. The 12V pulls less than 100 mA, the 5V about 400 mA. The power source is a 15V 1A wall wart.

For my first project, I wired both the 7805 and 7812 with heatsinks to the 15V supply. The 7805 heatsink got to around 43.3C (110F). After the first build, I discovered the the heat produced by the 7805 depends on the voltage drop, and the current drawn, and I want to reduce the 10V being dropped over the 7805.

For my 2nd build, I was going to wire the 15V to the 7812, and run the 7805 off of 12V instead of 15V (giving 7V drop instead of 10V). I was wondering if there was a more efficient way to do this? Maybe using a Zener diode, or a voltage divider? I already have the 7805 and 7812 soldered, as well as the output caps, but I haven't soldered the inputs yet.

For the next clock, I am open to suggestions for using other regulator ICs.

Best Answer

Simply use an SMPS replacement for the 7805. These are readily available ...here's a commercial variant that plugs into the TO 220 pinout you have with no heatsink required:

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There are many variants available on Ebay or from Digikey/Mouser.