Hi All,
I want to design power supply for below specification.
Could you please share your thought ?
Input : 12V-13.5V
Output : 5V or 3.3V
Current : 100mA

Case 1 : Output 5V@100mA, Power Dissipation(max) = (13.5-5)*100mA = 850mW
Case2 : Output 3.3V@100mA, Power Dissipation(max) = (13.5-3.3)*100mA = 1.02W

So both case1 and case considering the cost factor which is the best option - Linear regulator (LDO) or Buck Converter ?
Do i need heatsink in case linear regulator design ? Not sure if LDO is available to dissipate 1W power.

Thanks !

 

hi mish87,
Why do you want LDO with those relatively high input voltages, ie:12V ..13,5V down to 5V or 3.3V.?
E

 

Hi,
A 7805 in a TO220 has a thermal resistance of about 50°C/W. 1W power dissipation will heat it up to 50°C over the ambient temperature. At an ambient temperature of 40°C the IC will have 90°C. This will not destroy the IC but will reduce the lifetime of it. You can dissipate the heat better with a heat sink, but this will add to your cost.

The linear regulator is cheaper (without the heatsink) than the buck regulator.

It depends on your project which one is better.

 

hi mish87,
Why do you want LDO with those relatively high input voltages, ie:12V ..13,5V down to 5V or 3.3V.?
E
View attachment 289121

Hey ericgibbs,
I am exploring here what could be the best option and still not decided.
Only the tradeoff between LDO and Buck are cost. If choose LDO + heatsink so more or less cost would be same as LDO.
Thanks!

 

Hi,
A 7805 in a TO220 has a thermal resistance of about 50°C/W. 1W power dissipation will heat it up to 50°C over the ambient temperature. At an ambient temperature of 40°C the IC will have 90°C. This will not destroy the IC but will reduce the lifetime of it. You can dissipate the heat better with a heat sink, but this will add to your cost.

The linear regulator is cheaper (without the heatsink) than the buck regulator.

It depends on your project which one is better.

Hey,
Thanks for very good insight !
My ambient is 105degC.

Regards

 

Hi mish,
OK. As you are designing a power supply, I would also consider a variable regulator, such as the LM317.
Fit heat sinks and use some form of current limiting.

E

 

Hi mish.
My ambient is 105degC.

Do you really mean 105C ambient.??
E
https://www.techtarget.com/searchdatacenter/definition/ambient-temperature

 

Hi mish.
My ambient is 105degC.

Do you really mean 105C ambient.??
E
https://www.techtarget.com/searchdatacenter/definition/ambient-temperature

Hi,
Sorry i was talking about ambient maximum temperature so it is 105degC.
Thanks

 

Hi mish,
OK. As you are designing a power supply, I would also consider a variable regulator, such as the LM317.
Fit heat sinks and use some form of current limiting.

E

Hi E,
Do you think LM317 with Heatsink will be less costlier than a buck for 100mA current ?
Thanks

 

Not sure if LDO is available to dissipate 1W power.

For years "LDO" is used to say Low Drop Out" which is a class of linear regulator. A LM317 and 7805 are not LDO, just linear.

 

Depends on your specifications, what do you require?

Should it be:
cheapest solution?
high efficiency > low heat generation?
overall size?
Low EMC and high output purity?

Overall best performance you get with the BUCK regulator, no worries of heat and relatively simple to set up.

 

Hi mish,
The reason I am suggesting a LM317, it will give you an option to set a voltage other than 5v or 3.3v.
eg: say you needed 9Vdc voltage to check a motor or a device requiring 1.5v, it would be possible.

Use the AM1117 types fixed 5v and 3.3v , but in addition a variable supply from a LM317.
E

 

For years "LDO" is used to say Low Drop Out" which is a class of linear regulator. A LM317 and 7805 are not LDO, just linear.

TS is probably using the term "LDO" with reference at all linear regulators.

 

If the ambient temp is really 105C, a linear regulator is probably out of the question. You would likely need a cooling fan.

 

Sorry For confusion , I meant Linear Regulator instead LDO.