Linear Power Supply Design Thermal Control

Question

I am a electronics hobbyist trying to make a linear power supply/phone charger. This is a schematic that I have created. Optimally, my load would be a phone charging at 1A and 5V from a usb jack with the D+/D- lines shorted. From some experimenting, a multimeter measurment of the voltage on the wire that gets filtered by the caps and goes into the 5V regulator is ~18VDC. I know that 1A current draw from a 7805 will need serious cooling so I figured a fan would be able to both provide a voltage drop that will reduce the wattage of the 7805 as well as provide some cooling for all the components. My fan is a 12vdc 70mA computer fan. I am stuck finding a way to safely run the fan from an 18V line. I figured a voltage divider would be a good way to create 12v and also take more wattage away from the regulator. I am simply stuck finding values for those resistors. I did some quick math and found that R1 should be ~3.6K for 5mA thru the LED and R2 should be ~1.3k. Does those numbers seem reasonable? Would running the fan straight off of the 18V line be a feasible alternative?

Update: Would something like this form a resistor divider, dropping the voltage going into the regulator?[![enter image description here][2]][2]

[![enter image description here][2]]

Answer 1

Well, 12V fans usually run without issue at 18V. A drop resistor is feasible too, a 100 ohm 0.5W (drop 6 volt at 70mA, rounded up) would do the trick. You are dropping from 18 to 5 with a linear regulator so I guess efficiency is not you main target!

Since you have a center tapped transformer, have you considered running an LDO instead of an high dropout like the 7805 from the rectified 6V winding? In fact from a 9V supply the 7805 would be quite happy to work (with much less heating)

And, of course, they sell pin compatible switching modules for replacing the 7805, too

Answer 2

Put a linear 7812 regulator to make 12V for the fan. Running it directly from 18V can exceed fan ratings.

However, otherwise the design makes thermally no sense. If the 5V regulator provides 1A charging current (5 watts), it needs to dissipate 12-13 watts as heat when doing so.

If the junction temperature is allowed to go up to 125 °C, and the thermal resistance of the regulator between junction and case is 3°C/W, the case temperature must be held below 86 °C at all times. It really does not sound practical.