Maybe you can in series with a 10Ω/15W resistor to the input of LM7805 to reduce the heat.

 

Murata I have some of these. These are a Switching Power Supply that is mostly LM7805 compatible. They do not get hot! In your case they will pull 1/2 as much power from the 12V supply. The price is high. There are other companies making similar.
R-78HB5.0-0.5L
These two are at digikey.com

 

Murata I have some of these. These are a Switching Power Supply that is mostly LM7805 compatible. They do not get hot! In your case they will pull 1/2 as much power from the 12V supply. The price is high. There are other companies making similar.
R-78HB5.0-0.5L
These two are at digikey.com
View attachment 332282

I like these and the can fit right into the existing pcb holes.

 

Maybe you can in series with a 10Ω/15W resistor to the input of LM7805 to reduce the heat.

Shouldn't be needed by rights. IMO.
Why try to correct a situation the should not be there in the first place if the application rules are followed? !!

 

Would suggest a different approach; a Buck Converter.
The 7805 has to waste 8.8 volts whereas a buck converter will switch on and off rapidly. Depending on the desired output you adjust it down. In the end the buck will be OFF more than it is ON. This results in an average temperature. A rough estimate of percentage of ON time is approximately 40%. It will run a whole lot cooler too.

I definitely agree that a buck converter will be a much more efficient choice. You will need to add a filter capacitor, though.Many years ago I did try adding a series resistor with a 7812 regulator dropping from 24 volts. The result was a 15 volts peak to peak oscillation on the 12 volts line. I had to add both a 20 FD capacitor to the input PLUS the 0.1 MFD on both input and output, to halt the oscillation.

 

I have been using the TO92 and the TO220 versions for some years now and never ran into a problem, observing the manuf. design criteria. !

 

??With a series dropping resistor??

 

??With a series dropping resistor??

If the .34 amp current is constant a 10 ohm, 5 watt series resistor at the input side of the regulator will reduce the wattage across the 7805 from 2.38 to 1.22 watts with 1.16 watts transferred across the resistor.

 

https://docs.rs-online.com/4488/0900766b814b241f.pdf
Figure 12 shows the series resistor, and that the input capacitor goes after the series resistor.

 

??With a series dropping resistor??

NO dropping resistor!

 

Listen to @MaxHeadRoom. Just put a heatsink and appropriate capacitors on it and stop over-thinking the problem.

 

I like the series resistor - it has other benefits apart from reducing the power dissipation in the regulator. With a good big capacitor on the input it reduces the input ripple, removes high frequency noise (a 7805's PSRR isn't great) and removes short-duration spikes on the power supply. I put a TVS diode across the capacitor for good measure.

 

https://docs.rs-online.com/4488/0900766b814b241f.pdf
Figure 12 shows the series resistor, and that the input capacitor goes after the series resistor.

It's in Figure 8 of the page 5.

 

It's in Figure 8 of the page 5.

Actually it’s figure 23, page 27.
figure 8 is the same circuit, but with extra ripple injected through A capacitor as a PSRR test.