Choice of voltage regulators?

Thread Starter

metermannd

Joined Oct 25, 2020
472
What is the advantage of using an LM317 to derive a +5V supply versus using a plain old 7805?

Just something I noticed while looking over a schematic and was curious.
I should add, the input was +28V from a switching supply (which in turn is fed from 240VAC mains).
 

bertus

Joined Apr 5, 2008
22,882
Hello,

How much current is drawn from the 5 volts?
Keep in mind that the linear regulator must dissipate (28 - 5) volts X used current.

Bertus
 

Tonyr1084

Joined Sep 24, 2015
9,744
Both regulators have to dissipate over 20 volts which translates to wasted heat energy. If you're looking for a better solution then a buck converter would be far more efficient.
 

dl324

Joined Mar 30, 2015
18,216
What is the advantage of using an LM317 to derive a +5V supply versus using a plain old 7805?
It depends on the situation.

An LM317 takes two additional resistors, but it can get closer to 5V than a plain old 7805.

LM317 has better protection, and its maximum current is typically greater than 1.5A because the 1.5A is at worst case conditions.
 

BobTPH

Joined Jun 5, 2013
11,463
At 1.5 A, the LM317 would have to dissipate

23 x 1.5 = 34.5W, requiring a very large heat sink. Hopefully, you need way less than that, in which case the 7805 would be my choice.
 

Thread Starter

metermannd

Joined Oct 25, 2020
472
I'm not sure current draw really comes into play, as the 5V supply only powers the PIC and a few op-amps.
It seems the version used in this application is the LM317AEMP, which is apparently a SOT-223...
I think dl324 may be on to something in speaking of the protection factor, as this was actually for a register out of a smart meter.
 

MrChips

Joined Oct 2, 2009
34,626
There are many alternatives to LM7805.
Three things you need to consider:
1) Current draw. 1 A output is common. You can get 1.5 A and even 3 A regulators.
2) Voltage overhead. You can get low drop out regulators or standard 3 V or higher overhead.
3) Power dissipation. You want the input voltage to be about 3-5 V higher. 28 V input is excessive. For that, use a buck DC-DC converter.
 

Tonyr1084

Joined Sep 24, 2015
9,744
3) Power dissipation. You want the input voltage to be about 3-5 V higher. 28 V input is excessive. For that, use a buck DC-DC converter.
That's why I suggested a buck converter. Far more efficient. But that doesn't really answer your question. And your question has certainly been answered. I could repeat it but then I'd just be a parrot.

But if you're actually considering a project - the buck is the best.

Is your 28V AC or DC derived? If it's 28VAC through a full wave rectifier (bridge) with filter caps then your final DC voltage will be 28x1.414=39.592. 39.592-1.4Vf=38.192. You have to count for the drop in voltage across two rectifiers, which are commonly 0.6 to 0.7 volts forward drop (or Vf). 38VDC may be too much for the LM and the 7805 to handle.

So is your 28V AC or DC?

You can ignore this comment.
 
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Thread Starter

metermannd

Joined Oct 25, 2020
472
So it looks to me like it was mainly designer's preference that the LM317 was chosen... probably had more confidence in its ability to drop that much voltage versus the 7805.
I could see that a buck converter would be a better choice than a regulator (one of the boards that comes across my bench regularly at work actually uses miniature SMPS modules to derive 5V and 3.3V as needed).
It then seems to me that the current draw of the 5V portion of this board has to be somewhere down in the milliamp range if it is only needed for some op-amps and one PIC. If I were to say 250mA, that works out to 5.75W...
 
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