karikamiya,

It is rated for 180 A continuous... but that depends on how it is integrated into your project. Also, I don't recommend any "burst" over 180A no matter how good the cooling is... the fact is that this board has LOW thermal capacity due to very small size and if the waste heat generation were to jump from a max of 8 Watts to 10 or 12 for even a fraction of a second, the temperature could exceed the limits of the shunts or INA169. The shunts have masses of mere milligrams - so even a very brief excursion of over-power (2 Watts each shunt) could conceivably raise their temperature by several 10's of Celsius. The shunt manufacturer is wise in stating 2 Watts as limit... they know the minutia of detail regarding how quickly heat is conducted out and away from the metal film resistive element. Also, think in terms of an exploding bridgewire detonator.... a little wire heats from ambient to > boiling point of the metal and into a plasma on the order of a millisecond or less; it emulates the shockwave of a chemical explosive well enough to initiate a chemical explosive. Thin metal films of 100 micrograms can heat and melt VERY rapidly if heat generation outstrips heat loss. In other words, I know you were asking about "burst" in terms of using the full stated 180 amp range, but here I want to 1) assure you it's OK to use this sensor at 180 amps if component temperatures are kept in spec, but 2) never try a burst > 180 amps.

If you wish to use this unit up to 180 Amps, you certainly can, however this requires proper engineering work in the integrated design of your project + this PCB. There some guiding principles and helpful equations detailed below, and employing good engineering methods will guide integration of this PCB. The process might require extra cooling be added above and beyond your initial guess of what an adequate cooling amount is.

Based on spec sheets of sub-components, this PCB is rated to a maximum 180 Amps provided that heat is removed so as to not exceed upper temperature limits of the components. The shunts are rated to 2 Watts each and there are 4 of them. It doesn't matter that they are stacked in this PCB, the fact is their electrical conductivity is high and it follows that the thermal conductivity within the shunt itself will be high also (heat conduction in conductor is dominated by the "gas" of conduction electrons). I realize "high conductivity" is a subjective term here. The shunt metal film resistance element will generate heat across the voltage drop area and those are on an insulating substrate, so in fact the hottest part of the shunts will be the metal film mid-point between the leads, and for the PCB will be the area between the stack of shunts. However this area is only about 2mm wide and spread out over a nice longer distance of about 6mm (I'm glad the chip was designed this way). As a general principle of PCBs, the unit will heat above surrounding ambient until heat loss = heat generation. The only requirement is to have this temperature not exceed upper temperature limit of the components. Spec sheets of the parts show: INA169 +85C, LVK25 0.001 Ohm shunt +125C. The 0603 passives (0.1uF, 10k, 1k, 4.7k, 73.2k) are rated to +85C. Thus, the limitations are +125 for shunts and +85 for everything else. Because the shunts are the heating source, you might find in your board they equilibrate hotter than the other parts, and most probably will be.

Waste heat power of a resistor scales as the 2nd power of current (power at 180 amps = 4x that at 90 Amps, or in this case 2 Watts at 90 amps, and 8 watts at 180 amps). In practice, if your installation has this sensor confined in shrink wrap and/or the heavy gauge power leads are short, then you might not be able to use it at 180 amps. Loss of heat to the environment is only by 1st power of temperature difference between hot object and the environment. This applies to heat conduction out through the heavy leads, and from board to the air. The air and heavy leads might settle at different temperatures, but the loss equation is still by 1st order temperature delta through each route respectively. So the fact that heat loss is 1st order, and heat generation is 2nd order by current level, I can deriving a simple back-of-napkin equation to estimate the temperature of the board at various current loads... you could use this to take a measured temperature at a known current load in order to estimate the temperature at some other current... a guesstimate is possible to determine max operating current load without exceeding upper temperature limits of the components based on measured current and temp at some lower value. I stress "guesstimate" because complicating factors arise and in reality systems like this are not completely linear.... greater heat delta = greater convection of air and this affects heat loss to environment... other affects change etc ad inifinum.

Heat Loss (Watts) = k1(Delta Temp) where k1 is a constant that defines relationship between the units. Heat Creation (Watts) = k2(I^2) k2 is a constant defining relationship between units... in this case would be resistance in Ohms.

The temperature equilibrates to stable value when the loss = creation, thus at that equil temp: k1(Delta Temp) = k2(I^2)
Rearrange to: DeltaTemp = (k2I^2)/k1, and we now see the temperature will vary as the square of current to a decent approximate (excluding 2nd order affects). Going a little farther, and because k1 and k2 are constants (more or less), rearrange to (I^2)/(DeltaTemp) = k1/k2, thus plugging in DeltaT1 at Current1 and wanting estimate for DeltaT2 at hypothetical Current2, we'd us this relationship: DeltaT2 = ((DeltaT1)(Current2^2))/(Current1^2). If instead you want to estimate current required to cause another DeltaT, then use: Current2 = ((DeltaT2*Current1^2)/DeltaT1)^0.5

If ambient is +25C and the PCB stabilizes at +35 under a load of 100 Amperes, we have a DeltaT of +10. Considering an upper limit of +85C (DeltaT of +60), then the estimated current flow to reach this DeltaT would be around: ((65C*(100Amps^2)/10C)^0.5 = 254 amps... which being above 180 amps means that we can safely pass 180 Amps because 180 amps would result in an estimated DeltaT < +60 and thus Temp2 less than 85C. In fact, the estimate of Temp2 at 180 amps in this case is 57C if ambient is 25 C. Yes 57C would cause a little pain to the touch (again DON'T touch a live circuit!), BUT is almost 30 deg C below upper limit of the INA169.

If you desire to push towards upper limit of 180 Amps, then I suggest a thorough evaluation of the integrated design under max load. For this you'd need a temperature sensor, such as one of those IR guns, or you could adhere a thermistor to both the INA169 and the shunt stack. Measure the temperature under full load and typical cooling airflow conditions, and simply ensure the INA169 doesn't exceed +85 C and the shunt stack +125C.

Note: +50C is roughly the threshold of heat pain to your flesh... so testing if the part is too hot by merely touching it (I don't recommend touching any powered circuit) is not a quantitative analysis to be relied on if you need to use the upper current range. Consider that +125C is well above boiling point of water and +85 would rapidly cause a nasty 1st/2nd degree burn... it would be very easy to wrongly judge the part is 'burning' up or too hot but it might still be 20 C under max limit. Of course magic smoke would be a clear sign of overheating, but that's more of a concern for semiconductor materials because of a semiconductor's inverse temperature coefficient of electrical conductivity and their ability for thermal runaway. But in this circuit the shunts are the heat source, and because they're normal conductors, they have positive resistance temperature coefficient. Rather than thermal runaway, they would tend to pinch off current as they heat up... they are in this sense self-limiting to some degree versus how semiconductors work. I'm NOT saying they do in fact self limit too much current, I'm simply stating that I'd be surprised if they didn't have lower conductivity at higher temperature like normal conductors. On the other hand, by virtue of their function as a tight-spec shunt and wide operating temperature range, the manufacturer is forced to use an alloy for the metal film that is relatively flat for resistivity as function of temperature. At high enough current the metal film would eventually generate heat faster than it is cooled out, and the film would melt. I don't know if that would be gentle or an explosive failure... wait a sec I'm going to short one of these across a large LiPo 3s pack and let you know....