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7805 Heat Problem


Active Member
Thread starter #1
I need 5V/600mA from a 12V DC power source.Main concern is the heat on linear regulators.Do not have modern Switch mode converters & needs to do with commonly available parts. I'm thinking to do with two 7805. Can it supply 5V/600mA without heatsinks?



Active Member
No, each device will dissipate 2.1 Watts ( in an ideal Balanced condition - not guaranteed in this setup) and will need a Heat Sink to dissipate the heat.
Why not use a 12 Volt bulb in series with a single device, the major voltage drop being across the bulb? Of course, the bulb will glow red and dissipate heat.


Active Member
The 7805 is dropping 7V, so 7 x 0.6 = 4.2 Watts power dissipation.

A fairly simple alternative is a "simple switcher" IC plus a few parts.
An LM2575 would be suitable.

Producing 5V from 12V at one amp, it would be 77% efficient according to the data.

That means less than one watt dissipation.

A simple solution for the 7805 would be a string of 1A rectifiers in series with the input....
Seven or eight would reduce the 7805 input to near 7V.


Well-Known Member
Most Helpful Member
Hi, how did you calculate 8 ohms?
600mA in 8 ohms= 4.8V. Then the input to a single 7805 regulator would be 12V - 4.8V= 7.2V and the regulator would heat with (7.2V - 5V) x 600mA= 1.32W. The regulator will be close to its maximum allowed temperatrure and might not last long.

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