Discussion in 'General Electronics Chat' started by SPQR, Jan 9, 2013.

1. ### SPQR Thread Starter Member

Nov 4, 2011
379
48
Load considerations in linear power supply design

Me again.

Recently while trying to help another forum member with a power supply, I learned quite a bit about calculations on the various components of linear power supplies. I’ve build four or five over the last many years, but didn’t really “get” the details.
Now I feel a little better about them.

A – Always using RMS values for the secondary coil for subsequent calculations
B – There is a formal way to calculate voltage drop across a diode/diode bridge
C – There is a formal calculation for voltage loss across the filter section
D – The need for at least 2V above the 78XX voltage level
But I’d like to get your thoughts on load considerations (the “E” in the diagram)
In other words, what type of loads will change what I do in designing a linear power supply (assuming adequate voltage regulation).
I bought a 9V wall wart on Saturday, and for fun, I measured the voltage across the DC plug, and it was 13.4 volts (unloaded). What were they thinking?

The answer to my question might be something like:
“If you had XXX type of load, you need to think about YYY in the power supply”.

Again, I thank you in advance.

Mod Note - this is the thread that I couldn't upload because of the long title, so I've placed the full title at the top of this post.

File size:
30.6 KB
Views:
58
2. ### MrChips Moderator

Oct 2, 2009
12,440
3,361

If you have a 5V output regulator, you need a minimum of about 7V for a standard LM7805 regulator. What you have to look at is the ripple voltage after the filter C. The voltage must not drop below 7V. Hence you are looking for an rms voltage of about 9V.

You do not need a filter with an inductor and two storage capacitors. One capacitor alone will do. You just need to know how to determine the value of the capacitor. Conversely, what is usually done, you select a value such as 1000μF and then calculate the ripple voltage.

Check this out:

http://www.zen22142.zen.co.uk/Design/dcpsu.htm

As I posted on the other post, a 9VDC wall wart ought to be the rms rating when loaded at the rated current. Not all ratings are the same. You really need to test each wall wart separately. An output of 12 to 15VDC on an unloaded 9VDC wall wart is not unusual.

So the question is which wall wart should you use?

You want a voltage and current rating that is high enough to keep the ripple voltage above the minimum input voltage of the regulator when fully loaded.

You want a voltage that is not so high that the regulator gets too hot to touch, otherwise you have to add a heat sink to the regulator.

3. ### MrChips Moderator

Oct 2, 2009
12,440
3,361
What is also missing in your design is a 0.33μF cap to GND on the input pin and a 0.1μF cap to GND on the output pin of the LM7805 regulator. Without these the regulator could oscillate.

4. ### JMac3108 Active Member

Aug 16, 2010
349
66
You should include a thermal calculation in your list of linear regulator calculations. In my experience, this is the most often missed step in linear regulator design.

(1) Calculate power dissipation in the regulator. P = (Vin-Vout) x Load current

(2) Look up the thermal resistance of the regulator in its datasheet. Its specified in degreesC/Watt

(3) Calculate the temperature rise of the regulator. Trise = P x Rthermal

(4) If the temperature rise added to the maximum ambient temperature your circuit will ever see exceeds the maximum rated junction temperature of the part (as specifed in datasheet, usually 125C or 150C), then you need a heatsink. A good rule of thumb is never to exceed 100C, and I like to keep things well below that.

(4) NEVER assume the regulator can put out its rated current in your design. The rated current is simply the max that it can put out before it shuts down. In other words, a linear regulator with a 1A datasheet rating may not be able to put out 1A in your circuit depending on the input and output voltages. The limiting factor is the temperature of the device.

5. ### MrChips Moderator

Oct 2, 2009
12,440
3,361
Good point about the thermal analysis. I use my finger. If its too hot to touch, something needs to be changed, such as adding a heat sink.

Apr 5, 2008
15,647
2,346
Last edited: Jan 9, 2013
Metalmann likes this.
7. ### SPQR Thread Starter Member

Nov 4, 2011
379
48
As usual, thank you all for the superb comments.
I learn tons from you all.

@MrChips - Yes I learned about those capacitors around the regulator from you...I just didn't add them for simplicity. I add capacitors to ALL my circuits now, whether they need them or not!

I guess my question is fairly unsophisticated - perhaps I should state it more directly.

Why is it that the engineers on that "9V" wall wart engineered it to give 13V on no load?
I understand that the load on a power supply should not exceed the current supplied by the supply, and the supply should supply "enough" voltage.

But I'm wondering if all power supplies are "over engineered" to supply a higher voltage that stated on the "wart" itself.

If you are driving DC motors are you thinking differently about the supply than if you are driving TTL circuits?

@Jmac3108 - ohhh...excellent point. I'll study a bit about "thermal regulation" in linear supplies.

@Bertus - thanks for the info. I just kludged that diagram together quickly. I use Powerpoint a lot, and got a bunch of free symbols from the eehomepage.

Thanks again.
I'll think a little more about how I might better pose the question.

8. ### MrChips Moderator

Oct 2, 2009
12,440
3,361
Just the opposite. Cheap transformers cannot supply the rated power. They reach saturation and the voltage drops.

A well known transformer manufacturer is more conservative with their ratings and their transformers are able to hold the voltage a bit better when under full load.

9. ### SPQR Thread Starter Member

Nov 4, 2011
379
48
Ok, this is moving in the right direction.

So I need to consider the "quality" of the transformer if I really want to assure adequate voltage output - "over-engineer" the transformer to a higher voltage than necessary to cover any potential voltage losses on loading. (I assumed that all transformers would do what they say they would do - - stupid me)

So some directed questions:

Would you change something in the power supply for a high impedence load vs a low impedance load?

Would you change something in the power supply for a constant voltage/current load vs a load that has constantly varying current needs?

I think you see what I wrestling with.

10. ### JMac3108 Active Member

Aug 16, 2010
349
66
Hahahah, yes, I use the finger test too

But in industry sometimes you have constraints and tradeoffs that don't allow you to run something that cool, and you're forced to do the anaylsis and run a part hotter than you would like.

11. ### WBahn Moderator

Mar 31, 2012
17,743
4,789
It's not that they engineered it to have a higher voltage when unloaded, it's that they designed it to have close to the rated voltage when the rated current is being pulled from it and to be as cheap as possible to build. The result is that they have pretty poor regulation and as the load goes down the voltage climbs. This is one of the reasons why you should, whenever possible, use the wallwart that came with the device and, if you can't, use one that is not rated to deliver a whole lot more current than the device it is powering needs.

SPQR likes this.
12. ### JMac3108 Active Member

Aug 16, 2010
349
66
Thermal analysis, not thermal regulation. And it applies to all semiconductors, not just linear regulators. The current handling capability of all semiconductors is limited by the temperature of the die.

In particular, transistors, mosfets, diodes, and linear regulators, used in any sort of power application, should have a thermal calculation performed on them as part of the design process.

13. ### ScottWang Moderator

Aug 23, 2012
4,853
767
The datasheet shows that the typical Dropout Voltage is 2V, so it could be more, you can measure yours.
Dropout Voltage = 2V, but at least used it at 2.5V or more, my friend have been wroked in the instrument company, and they used it up to 3V, so I also used it up to 3V if I can.

The rating current is 1A, but you better used it under 750mA or less, and add a heat sink, if you want to get more current then you can use other IC as below.

L78S05CT 2A
78H05 5A
μA78P05 10A

In the full load that you better used it working under 140 ℉ continuing, and no more heat increasing, that is the industrial standard of our products, and it won't damage in the normally situations, maybe you don't need to reach the standard.

I only used it had a little warm with heat sink when I used finger to touch and it feel a little warm not hot, if you can feel the heat very strong, it isn't good, but probably it won't damaged the IC.

7805 datasheet, fairchild.
http://www.datasheetcatalog.org/datasheet/fairchild/LM7805.pdf

No regulated power supply that they came from the devices no need the power regulation, as radio or some other devices, because they can working a big range of voltage or they had a regular IC inside, so the No regulated power supply still have their markets.

14. ### WBahn Moderator

Mar 31, 2012
17,743
4,789
I recommend using worst case limits on the specs. Assume your regulator will have the maximum dropout listed in the data sheet.

Whether you use min or max values depends on what results in the worst condition for your design. If your design holds up under those conditions, you've got a real good shot at it working properly and reliably. You might even consider tacking on an additional 10% of headroom. So, for instance, if they say the typical dropout is 2V but that the max is 2.2V, then tack on an additional 10% of the difference and make it 2.22V for your analysis.

Now, there will be times when this will result in a more conservative design than you can tolerate, usually due to size or cost constraints. So adjust accordingly. But in many low volume designs, the cost of doing a sufficiently detailed analysis to justify trimming the performance margins is more than the incremental cost associated with the more conservative design.

15. ### SPQR Thread Starter Member

Nov 4, 2011
379
48
This has been an excellent discussion.
I've learned:
1 - Don't necessarily trust wall warts. They are often underdesigned, and/or designed for a specific device that may not be relevant to a circuit in question.
2 - I've added "thermal analysis" to my list of things to do.
3 - If I want to build a good power supply, I should buy good components, slightly over the cacluated values needed, and put them together right.

Thanks again!