7805 Heat Problem

[Suraj143]

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?

 

[Ramussons]

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.

 

[rjenkinsgb]

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.
https://www.ti.com/lit/ds/symlink/lm2575-n.pdf

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.

 

[JonSea]

The CUI VXO78-500 series is a switching regulator designed to replace a standard 78xx linear regulator which almost meets your needs. These modules supply up to 500mA and are less than two bucks at Digikey. These solved a tough problem for me, so others may be interested in them too.

 

[Pommie]

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.

I like simple solutions. Would an 8 \( \Omega \) 5W resistor do the same job?

Mike.

 

[Cicero]

Like Jonsea said, quite a few of these 78xx replacements around, I often use this one https://uk.rs-online.com/web/p/switching-regulators/1247657/

 

[Suraj143]

I like simple solutions. Would an 8 \( \Omega \) 5W resistor do the same job?

Mike.

Hi, how did you calculate 8 ohms?

 

[audioguru]

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.