how can solve 'LM7805 heat problem with Relay'

Thread Starter

Kim-JiHoon

Joined May 3, 2020
25
Hello, first of all I have to sorry about my english.

I wan study electronics with AVR(now atmega328) but there is no community in my country...

I have heat issue with LM7805. Here is my circuit

제목 없음.png

If relay on, LM7805's temperature rises over 50'C (24v 3A)

but relay off, LM7805's temperature under 40'C (about 37'c)

I tried

1. SW1 - Diode - Relay
(3pin) - (1N4148) - (5pin)

2. Relay - Diode - Relay
(2pin) - (1N4148) - (5pin)

3. Relay - Capaciter - Relay
(2pin) - (47uF/16v) - (5pin)

I googled and figure out 'Relay protection circuit' or 'Coil reverse voltage' is the reason

but i don't know what i have to do.

so i want down LM7805's temperature. (I want fixed 5V out for Atmega328p)

1. If there is asolution without heatsink what is that?

2. sould I use LM2576?
 

bertus

Joined Apr 5, 2008
20,537
Hello,

The in and output capacitors do have a wrong value:

L78XX_decoupling.png

The dissipated power is (24 Volts - 5 Volts) times the used current by the relay.
A switching regulator will likely be more efficient.

Bertus
 

AlbertHall

Joined Jun 4, 2014
9,873
The reason for the heat is that there is 19V across the LM7805 and that multiplied by the relay coil current (the power dissipated in the '7805) is enough to heat up the regulator.
What is the relay coil current or resistance?
Using a switch mode regulator, like LM2576, will significantly reduce the heat and power wastage.
 

ronsimpson

Joined Oct 7, 2019
680
The in and output capacitors do have a wrong value:
I use large capacitors and small capacitors in parallel.

Can you change the relay? Can you get 24 volt relay or 12 volt relay? This change will have the relay coil current coming from 24V and not going through the regulator. And a 12 or 24V relay will use less current.
 

BobTPH

Joined Jun 5, 2013
2,391
Diodes have nothing to do with it. It is all a matter of current. How much current are you drawing with the relay on? Without a heat sink, the max current you can safely draw is about 50ma, which results in about 1W.

You would be much better off with a buck converter when dropping a voltage by so much.

Bob
 

Thread Starter

Kim-JiHoon

Joined May 3, 2020
25
Diodes have nothing to do with it. It is all a matter of current. How much current are you drawing with the relay on? Without a heat sink, the max current you can safely draw is about 50ma, which results in about 1W.

You would be much better off with a buck converter when dropping a voltage by so much.

Bob
Relay off = 23.56 V on main , 0v on Relay (coil)
Relay on = 23.56 V on main , 4.8v on Relay (coil)

I want 24V control with using 5V
 

ronsimpson

Joined Oct 7, 2019
680
I can not find the data sheet for your relay. Run the relay from the 24V supply. I added a high power resistor that is 3X what I think the coil in the relay measures. Relay might be 20 ohms so 20x3=60.
1590500119181.png
 

Thread Starter

Kim-JiHoon

Joined May 3, 2020
25
this is my final result but when Relay on LM7805's temperature is too high
than if I change Relay, it will solve problem?
is the reason temperature too high, Relay's coil currented too high?

IMG_2.jpgIMG_1.jpg
 

ElectricSpidey

Joined Dec 2, 2017
1,156
Relay coil is 72mA so that's 1.34 watts on the regulator.

A small heat sink will do the trick.

Or you could drop some of the voltage with a series resistor placed in the regulators positive input.
 
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