# 7805 Heat Problem

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#### Suraj143

##### Active Member
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?

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#### Ramussons

##### 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.

#### rjenkinsgb

##### Well-Known 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.

#### JonSea

##### Well-Known Member
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

##### Well-Known Member
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.

#### Suraj143

##### Active Member
I like simple solutions. Would an 8 $$\Omega$$ 5W resistor do the same job?

Mike.
Hi, how did you calculate 8 ohms?