Seanix monitor

[Undead]

Re: Seanix monitor

It's unlikely any of these regulators are 5A - with the input of 5v, to get 1.8v at 5A it would dissipate 3.2 watts * 5A = 15 watts.

Pay attention to this particular one because the pin order is different than the one on most common regulators.
Yours appears to be 1. GND 2. OUT 3. IN with the TAB output

Most regulators are 1. IN 2. GND 3. OUT with the TAB ground.

PS. See if ca.newark.com can be of help. I don't know if the shipping charges are convenient for you, I think someone on this forum complained about that but I'm not sure.

I don't really follow why you say it's impossible when the data sheet has it. I'm still very new to all of this.

I did notice the ground seemed to be on the left instead of the pin 2 and the tab, but didn't know that it was that unusual.

Thanks for the newark suggestion, I did look there for other electronic items in the past and the shipping did seem to be a bit steep.

????

AP1084K18L-13

into mouser search field brings up

http://ca.mouser.com/ProductDetail/D...Y7YI7dCw%3d%3d

$1.78 (754 in stock).

Sorry retiredcaps I don't mean to frustrate you with my lack of knowledge, the number I put into the search was the AP1084D18L-13 number that you first posted, I don't know enough to know that the AP1084K18L-13 can be used instead. The first one gives a result of:
http://ca.mouser.com/ProductDetail/D...F4BLLY8A%3d%3d

which shows not in stock and a minimum order of 1620. I did notice the tab is slightly different in this last link you posted but it doesn't look like a significant difference.

Thank you for the new link by the way.

[mariushm]

Re: Seanix monitor

Linear voltage regulators work by dissipating the difference of voltage as heat through a heatsink.

So if the voltage regulator receives 5v and outputs 1.8v like it happens in your case, this means the difference of 3.2 volts needs to be dissipated as heat.
This means for each 1A of power, the regulator will dissipate 3.2v x 1 A = 3.2 watts of power.

The regulator IS rated for maximum of 5A, but if you look at page 5 at the bottom of the chart, you can see it has a "Thermal Resistance Junction-to- Case" of 3.5 C per watt

The chip itself will stop working at 150 C, so let's do some math for 5A, following these formulas, and let's say we want to chip to not exceed 50c (we don't want to cook the capacitors around) :

Let's use the math from the app note here:

VOUT = 1.8V
TJ MAX = 150°C
θJC of the TO-263 = 3.5°C/W
θCS ≈ 0°C/W (soldered directly to board)

Vout = 1.8V

Pd = (Vin – V out) x I out = [5.1V – 1.8V] x 5A = 16.5 watts

We're already OVER the maximum power dissipation the chip can do: check the datasheet, in particular the footnotes, the regulation is guaranteed up to 15w


Maximum temperature rise, ∆T = TJ(MAX) – TA = 150°C – 50°C = 100°C
Thermal resistance requirement, θJA (worst case):
∆T / Pd = 100°C/16.5 = ~ 6.1°C/W

Heat sink thermal resistance, θSA = θJA – (θJC + θCS)
θSA = 6.1 – (3.5 + 0) = 3.4°C/W (max)

So if you look now at the application note at left bottom corner at the graph, you'll see that in order to dissipate those 5A of power, the pcb area would be unrealistically large... the chip simply can't be cooled without better cooling (metal heatsink or fan above it)

Now look at the picture... you barely have half an inch by half an inch for that 1.8v regulator... it's simply impossible for that chip to output 5A without going to 150c.
It can probably output 1-1.5A safely with that little heatsink under itself, under the tab

The chip CAN do 5A, but for that it needs to dissipate less power, to receive LESS input voltage. It needs just 1.4v or more to regulate, so had the chip actually use 1.8v + 1.4v + 0.1v for margin error = 3.3v , it may actually do 5A. But on your board the input voltage is 5V.

If the input was 3.3v, it would output 1.8v and dissipate only 1.5 watts per A or 7.5 watts at 5A. with some serious pad on both sides of the pcb it would work.
Which is still more than what your pcb has alloted to this regulator, so even in ideal conditions the regulator wouldn't output 5A like this. It just shows they probably bought these regulators cheaply and didn't really care about the maximum current output.

If the input was 3.3v and you relax the max temp to about 65c then the whole thing changes:

Pd = (3.3v - 1.8 ) x 5 = 7.5w
delta t = 100c
delta t / pd = 115 / 7.5 = 15.3 c / w
θSA = 15.3 - 3.5 = 11.8 c / w

and you'd need about 5000-6000 square mm of pcb to cool it, if it was to output 5A all the time.

So keeping all this in mind, if you can't find a cheap regulator rated for 5A, pick one that can do 1.5-2A - the regulator physically can't do 5A on that pcb.

[Undead]

Re: Seanix monitor

Wow that's quite the explanation, I have had to read it several times to even get the basic idea of it. It does seem to make sense if I am following correctly. Thank you very much for taking the time to go into so much detail.