Problem with Voltage rogulator

[amith.srivatsa]

convert 12vdc to 5vdc

Hello all,

I need to convert 12V DC to 5V DC. Current requirement can go upto 1.5A. I am currently using the L4940 which should support my requiement. The data sheet says 5V,1.5 A. I have mounted it on a breadboard.

Unfortunately, I can see that its getting heated a lot and then goes to thermal shutdown mode. I see that there is a current of about 0.8 - 1A max.

Why is this happening ?
How can I reduce the heat dissipated ?
Will having 2 or 3 L4940 in parallel help me solve the problem ? or is there any other solution to solve this problem ?

Any help would be greatly appreciated. Many thanks in advance...

 

[nandhu015]

Mount a Heat Sink to the regulator.

Regulator heating is normal if you draw more than .3A

Nandhu

I need to convert 12V DC to 5V DC.

If so why are you using low drop out regulator? you can use ordinary 7805 which is low cost.

Another reason for heating is the voltage difference is high (12-5= 7volts to be dropped)

 

[pauloynski]

As any other linear regulator, it dissipates a lot of heat. Think of it as being a controllable resistor added in series with your load. The package has a junction to ambient thermal resistance of 50(TO220) or 62°C (DPACK) per dissipated Watt. At 12V input and 5V output @1.5A it will dissipate more than 10W (P=(12-5) x 1.5). Please take a look at the TOTAL POWER DISSIPATION curve shown in the datasheet of the L4940. So, a huge heatsink is required (less than 10°C/W). Also, take a look at this heatsink calculator
**broken link removed** You´ll find others if you google for a while.
Regards

 

[Ave_Rapina]

Hello,

L4940 is an low dropout series regulator, but you are pushing lots of current from an series regulator, and that result in heat ( bad efficiency).

You can use an Dc/Dc regulator from texas, its has samples two.

Go to this link and choose one: **broken link removed**


Regards

 

[cameo_2007]

parallel two regulators and provide a heat sink for both,your problem is solved.

 

[doraemon]

Hello!

As others already replied, there is no point using this LDO and on top
of that going from 12V to 5V with a linear regulator is just wasting more than
half of the power.

Now if you have this regulator in your drawer and want to use it, then
put a heat sink on it, but check the characteristics of the heat sink. You want
to draw 1.5 amp, and the voltage difference is 7V, so you have to dissipate
more than 10W. I would go for a 15~20W heat sink. Remember that a heat
sink dissipation ability depends on how air circulates around it.

Dora.

Hello all,

I need to convert 12V DC to 5V DC. Current requirement can go upto 1.5A. I am currently using the L4940 which should support my requiement. The data sheet says 5V,1.5 A. I have mounted it on a breadboard.

Unfortunately, I can see that its getting heated a lot and then goes to thermal shutdown mode. I see that there is a current of about 0.8 - 1A max.

Why is this happening ?
How can I reduce the heat dissipated ?
Will having 2 or 3 L4940 in parallel help me solve the problem ? or is there any other solution to solve this problem ?

Any help would be greatly appreciated. Many thanks in advance...