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### Topic: Why is there such a large difference in the current capacity of these vregs? (Read 4504 times)previous topic - next topic

#### scswift

These are both 3.3v regs, they have the same package, they appear to have nearly identical specs, yet the first one is rated for 250mA and \$1.50, and the second is rated for 150mA, and \$0.50:

The first one seems to have a slightly lower dropout.  350mV @ 125 degrees, versus 400mV @ 125 degrees for the second.

But the second one has a 150 degrees max junction temperature, while the first has only a 125 degrees max junction temperature, and I know this is used in the power dissipation calculations.

They both can handle up to 16v, and both list their PEAK output current as 350mA.

So what's the difference?  Is it the dropout voltage?  If so, where does that factor into any power dissipation calculations?  I looked up how to calculate the power dissipation, and found this equation:

PD = IOUT  * (VIN - VOUT) + VIN*IQ
where VIN = input voltage; VOUT = output voltage; IOUT = output current, A; and IQ = quiescent current, A

I assume quiescent current is also known as ground current as that seems to match up with the minumum quiescent current listed in the absolute maximum ratings and there's no other quiescent current listed.  In which case for both vregs, it's 850uA typical at 150mA output.  Which means as far as the power dissipation equation goes, the two regulators are identical.

So why is the current rating so different for these parts?

#### johnwasser

Maybe they just justify the price difference by degrading the specs on the lower-cost option.
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#### James C4S

It's a result of TI acquiring National.  Two different companies, two different products, two different manufacturing processes, two different designs, for the essentially the same end product.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

#### scswift

So you're saying they're actually identical?

I took another look at the spec sheet and I noticed they had some graphs labeled "short circuit current".  If I'm not mistaken those show a worst case scenario of the reg trying to put out as much current as it can at 6v and 16v.  I think the graph also shows the thermal regulation kicking in shortly after the start of the test.

If that's the case, then for the first reg @ 6v, it peaks at 450mA, and drops to 300mA over the course of 2 seconds, then takes a nose dive, leveling out again just under 250mA.
For the second reg, it peaks at 350mA, drops to 250mA after half a second, then over the next 1.5 seconds slowly drops to 225mA.

That would seem to indicate the first part is slightly better.  It can provide an average of 325mA for 2 seconds before thermal regulation kicks in, whereas the second part really can't provide more than an average of 250mA over that same period.

But it also seems to indicate the parts are largely the same, if all you're interested in is their long term current handling ability, and that the second part can handle quite a bit more than 150mA if you've got the input voltage low enough.  (Seems at 16V they both crap out around 100mA.)

On the other hand, I just noticed the graphs for the two regulators aren't measuring exactly the same stuff.  The first is measuring 2.8v out, and the second 3.3v out.  I'm not sure how the difference would affect the graphs.  It could either make the two regs more equal or more different.  But still, with the second reg putting out 3.3v it's serving up quite a bit more current than I expected.

#### johnwasser

So you're saying they're actually identical?

No, he said they are different because they were designed and made by different companies before the companies merged.
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#### scswift

Yes, but he also said they were essentially the same.  When I said "identical" I meant in specs, not internals.

#### jwatte

They are not identical in specs, because they have difference current capacity, and different cost. One was designed for lower cost, lower current capacity, and the other for higher cost, higher current capacity. They may look the same on the outside, but they are not on the inside!
Also, the fact that they both can go to 16V is probably more because of use cases (car batteries go to 13.8V, say.)

Btw: 400 mA drop-out compared to 250 mA of drop-out means 60% higher power dissipation at a minimum. Coupled that with a potentially different way to couple the hot parts to the outside, and you get parts that perform differently.

#### zoomkat

Yes, but he also said they were essentially the same.  When I said "identical" I meant in specs, not internals.

Perhaps you can get somebody to help you interpet the data sheets.
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#### scswift

Quote

Btw: 400 mA drop-out compared to 250 mA of drop-out means 60% higher power dissipation at a minimum.

That's what I would have thought myself, but none of the power dissipation equations I have seen in the datasheets or online appear to include drop out in the power dissipation calculations.  And since the Iq for both parts is the same, the amount of power that needs to be dissipated for both is equal.

I think I found the difference between the two regulators though.

It's not that the two packages are mounted differently.  The landing pad sizes are virtually identical.

The difference can be seen on page 5 note 3 in the first datasheet, and in the Absolute Maximum ratings on page 4 of the second datasheet.  The junction-to-ambient thermal resistance: ?j-a

For the first part, ?j-a = 220°C/W
For the second, ?j-a = 206°C/W

And the equation for calculating the amount of power that can be dissipated at a specific ambient temperature is:

p(max) = ( Tj(max) - Ta ) / ?j-a

So, the power which can be dissipated is the maximum junction temperature, minus the ambient temperature, divided by the junction-to-ambient thermal resistance.

This power calculation isn't new to me, but yesterday when I was tired I could not for the life of me find those thermal impedance ratings and I saw something in one of the datasheets indicating what a typical value would be on FR4 pcb, and I thought that it must be the same for both parts because the two packages are the same size and neither has any kind of heat sinking.

This leaves me with a little problem though.  A lower thermal impedance is better.  And doing the calculations for am ambient temperature of 25 degrees, I get:

(125-25) / 220 = .455
(150 - 25) / 206 = .607

Which means the first part, the one which is 3x as expensive and has a higher current rating, can actually dissipate less power than the second part.

But that's for the same package.  The first part comes in a second type of package with a heat sink on the bottom.  And that package has a thermal resistance of 282 C/W on a 2 layer board with no heat sink (worse than the first two) and 62 C/W on a 4-layer board with 6 thermal vias.

Plugging that into my equation I get:
(125-25) / 62 = 1.61

Which is 2.65x better than the 50 cent regulator.  So by all rights it ought to be rated for 400mA if the cheaper one is 150mA, but I don't know what criteria they used to decide to rate it... ie what amient temperature, what current, what input and output voltages.  So that could explain the difference.

Anyway after all this, assuming I don't go with the third option, the part with the heat sink on the bottom:
http://search.digikey.com/us/en/products/LP2992ILD-3.3%2FNOPB/LP2992ILD-3.3CT-ND/567557

It would seem that the 50 cent part would be the better of the two choices.

Now I just have to figure out how much current it can really carry by doing the rest of those power calculatons.

#### scswift

Just did some calculations for the 50 cent 3.3v regulator:

Assuming a max PD of .607:

A = .607 / (Vin - Vout)

Vout = 3.3v, so:

.607 / (5 - 3.3) = .357 (Piggybacking off my 5A 5v regulator)
.607 / (8.4 - 3.3) = .119  (7.2v NiMh -> 8.4v at full charge)
.606 / (12 - 3.3) = .070 (12v battery - probably more like 13v+ but whatever)
.606 / (16 - 3.3) =  .048 (16v worst case)

So based on this information, I think I should be okay using the same 50 cent 3.3v regulator I used for my SD card.  Using a 7.2v rechargable or 6v alkaline will be what I reccomend, and if someone absolutely must run at 12v, they can plug the expansion board with the regulator into one of my servo ports which supply a regulated 5v instead of Vcc, which will put some more stress on the 5v 5A regulator but take a lot off the 3v.

#### scswift

Decided to do the calcs for my 5v reg as well:
http://www.micrel.com/_PDF/mic29150.pdf

Junction temp max - Tj(max) = 125°C
Ambient temp - Ta = 25°C
Thermal resistance - ?j-a = 2°C/W

Pd(max) = (Tj(max) - Ta) / ?j-a
Pd(max) = (125°C - Ta) / 2°C/W
Pd(max) = 25W @ 75°C
Pd(max) = 50W @ 25°C

I'm not sure which ambient temp to use, the regulator docs mention 75°C but 167°F seems excessive to me.  Still, I'll go with 75 for my calculations.

This regulator gives a slightly different calculation for power dissipation which takes into account the ground current:
Pd = Iout (1.01Vin - Vout)

So if Vout = 5v, and Pd = 25W per my above calculation, and I rearrange the equation like so:

Iout = Pd / (1.01Vin - Vout)

Then, @ 75°C ambient and:
6v in, Iout(max) = 23.5A
8.4v in, Iout = 7.2A
12V in, Iout = 3.5A
16V in, Iout = 2.24A

#### oric_dan

Quote

A = .607 / (Vin - Vout)

Vout = 3.3v, so:

.607 / (5 - 3.3) = .357 (Piggybacking off my 5A 5v regulator)
.607 / (8.4 - 3.3) = .119  (7.2v NiMh -> 8.4v at full charge)
.606 / (12 - 3.3) = .070 (12v battery - probably more like 13v+ but whatever)
.606 / (16 - 3.3) =  .048 (16v worst case)

Let me toss in a few ideas here. These last calculations are the most pertinent. They show
you are quite limited in how much current the v.regs can "realistically" supply, without melting
your pcb. Assuming the NiMH cells are the most likely to be used, you've got ~120 mA max.

HOWEVER, and it's a BIG however, you're allowing 150degC temp on the v.reg. How hot is that?
100degC is the temp of boiling water, and how long could you keep your hand in boiling water
for? 0.5 sec, that's it. IOW, you're really pushing it here.

Quote

PD = IOUT  * (VIN - VOUT) + VIN*IQ
where VIN = input voltage; VOUT = output voltage; IOUT = output current, A; and IQ = quiescent current, A

I assume quiescent current is also known as ground current as that seems to match up with the minumum quiescent current listed in the absolute maximum ratings and there's no other quiescent current listed.  In which case for both vregs, it's 850uA typical at 150mA output.  Which means as far as the power dissipation equation goes, the two regulators are identical.

Secondly, you're initial ideas here were offbase. The PD eqn is simply how you calculate PD from
Ohm's Law and circuit theory, and has nothing to do with the characteristics of the v.regs. The 1st
term is the important one, and has nothing to do with what's "inside" the box, only with the
input-output characteristics. The 2nd term is basically insignificant, since IQ is > 150X smaller than
IOUT.

Thirdly, for my part [and I've designed a lot of pcbs],  I wouldn't even think about using such a tiny
smt part at all, unless I was sure PD would always be maybe < 50% of your 0.6W value. Then the v.reg
would not overheat so much, and it would still be a 75degC = 167degF on the part. Still pretty
darn hot.

Fourthly, I don't even like the SOT223 parts [160degC/W], let alone the tiny ones you're using. I
go for the NCP1117 DPAKs [67 degC/W].

Fifthly, I think the answer to your initial question ["difference in current capacity"] mainly has to
do with the physical characteristics of the PNP output transistor in the parts, and not with the
size of the package, etc. Eg, the smaller the collector area of the transistor, the lower the
max current.

#### scswift

Thanks for the info Dan. :-)

Quote

Let me toss in a few ideas here. These last calculations are the most pertinent. They show
you are quite limited in how much current the v.regs can "realistically" supply, without melting
your pcb. Assuming the NiMH cells are the most likely to be used, you've got ~120 mA max.

HOWEVER, and it's a BIG however, you're allowing 150degC temp on the v.reg. How hot is that?
100degC is the temp of boiling water, and how long could you keep your hand in boiling water
for? 0.5 sec, that's it. IOW, you're really pushing it here.

Well, I was calculating absolute maximums, not necessarily the average current the regs would be putting out.

In my particular setup for example, I would like to power a 3v vibration motor or a few lasers.  A massive 10G vibration motor or a green laser might need 200mA.  A smaller vibration motor or three red lasers might need 150mA.  And a small vibration motor or a single laser might need onl 60mA.  But in any case, they would only be on for brief periods of time.  A few seconds at most, usually.

Quote

Thirdly, for my part [and I've designed a lot of pcbs],  I wouldn't even think about using such a tiny
smt part at all, unless I was sure PD would always be maybe < 50% of your 0.6W value. Then the v.reg
would not overheat so much, and it would still be a 75degC = 167degF on the part. Still pretty
darn hot.

Well, my plan as of this moment is to forget trying to run the 3.3v off the battery because even a NiMh doesn't leave me with much amperage to work with before the 3.3v reg would overheat.  So the plan now is to run both 3.3v regs on my board off the 5v regulator.  That will allow them to put put almost 350mA, which is 150mA more than what I ever expect the reg will need to put out, and even if does get to 200mA, it will only be for brief periods as I mentioned earlier.

Oh, and in case you're wondering why I have two 3.3v regs instead of just one bigger one, it has to do with the fact that the board is modular.  The 3v3 reg that would drive the motor/lasers would be on its own board, and the other one drives an SD card.  It would be nice if I could just use one big 3v3 reg on the main board, but I don't really have the space for it, and it would make the way everything connects to the main board a lot more complicated for people.  Trying to keep things simple.  The board is already studded with pinouts:

http://shawnswift.com/arduino/layout2.png

Quote

Fourthly, I don't even like the SOT223 parts [160degC/W], let alone the tiny ones you're using. I
go for the NCP1117 DPAKs [67 degC/W].

That looks like a nice part, but the problem is, that regulator doesn't have an enable pin, and the plan is to switch the motor and laser using the enable pin on the regulator.  Keeps things simple.  I've calculated the regulator should be able to switch at the PWM frequency of the Arduino (around 500hz) but if not, it's no big deal, I can do software PWM more slowly or not at all.  It's just a vibration motor.  Can't do PWM on the laser anyway; the driver boards usually don't like being switched faster than 10-100hz from what I've read.

Now usually, someone would use a mosfet for switching, but like I said, I don't really need to switch fast, I think the reg can handle PWM speeds, it saves me a component, and I don't have to worry about finding a mosfet which can handle the same current the regulator can.  I had that problem with the first 5A reg I chose, because it didn't have an enable pin, and I need to be able to turn the power off, and I didn't want to have to use a 5A switch, but a mosfet that could handle 5A was the same size as the regulator.  In the end I decided to just choose a larger regulator with an enable pin, and I probably got more current capacity as a result to boot.

I'd be interested in your thoughts on this plan though.

#### scswift

On another note, I'm looking for a suitable flyback diode for the aforementioned vibration motors.

This motor here is probably the worst case scenario:
https://catalog.precisionmicrodrives.com/order-parts/product/320-100-20mm-vibration-motor-25mm-type

So that would be run at 3.3v around 170mA, but I'd like to assume 200mA just to be on the safe side.

I found this diode on Digikey:

And I found this thread on flyback diodes:
http://www.electro-tech-online.com/electronic-projects-design-ideas-reviews/85310-when-flyback-diode-necessary.html

Retrolefty there states the diode should be 4x the voltage rating of the coil, and 40v is well over 12v.  And it's got a 5.5A surge current rating which is over 25x as much as I think I need.

#### takao21106

the cheaper one's maybe just will make "pop" one day if you work them too much towards the margin.

It happened to me, using very tiny SMD diodes for a dc/dc converter.
Fine at 1.2 volts, at 5 volts it only survived for some minutes.

Then I changed to a big, way overdimensionated diode (2A rating, max. current maybe 100 mA).

I would not bother too much about it.
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