Buck IC Thermal management calculation

Thread Starter

anishkgt

Joined Mar 21, 2017
458
Hey all,
I am desigining a buck converter to step-down a 12 100A powersupply to 8.1v 10A AOZ2264QI-11. I am some trouble with determining the Tjunction result
as per datasheet in pg:12
Ptotal_loss = Vin x Iin – VO x IO
Io is 10A but Iin ? how do i determine Io ? 100A is the power supply rating, to use that, there would be some loss to it.
 

jpanhalt

Joined Jan 18, 2008
9,435
That is an efficient buck converter. Page 11 gives some approximations:

1590516749821.png

You will likely have to actually measure the Iin , that those equations should get you in the ball park. Also, when the unit shuts down, you will know you exceeded the temperature/
 

ronsimpson

Joined Oct 7, 2019
684
Ptotal_loss = Vin x Iin – VO x IO
They did not give you much to go on. There is too much not known.
Power out is 8.1V x 10A. That we know.
If the power loss was zero: power in = 12V x 6.75A
There is a good formula for power loss in the inductor. "Iout^2 X R X 1.1" Out current squared x Inductor DC resistance x 1.1.
There is not a good formula for power loss in the IC.
The resistance of the inside the IC is 0.009 ohms when the power transistor is on and 0.004 when it is off. Because the transistor is on most of the time I will use 0.009 ohms all the time to simplify the math. Power loss in the transistors is about Iout^2 x 0.009 x 1.1. (math in my head; about 1 watt)
The power loss should be slightly lower because of the lower 0.004 ohms during the off time but there is power loss in the brains of the IC that is not counted.
Back to the data sheet. The IC will increase in temperature by about 40C/watt. Room temperature = 25C. So the IC will be about 65C on the silicon. (with 1 watt loss in the IC)
----edited---
P=V^2xI that we know. So why the x1.1? The first formula is for DC but we have a current that is not DC but AC. I don't know exactly what the current looks like but it might be a ramping from 9A to 11A and back to 9A. (average=10A) This will cause a little more loss. About 10% more.
I have simplified the math a little to save a headache and we lost a little in the accuracy. But close enough.
 

Thread Starter

anishkgt

Joined Mar 21, 2017
458
and these and maybe during the initial charge the IC may go into short-circuit mode and maybe later it may start recharging.

I think i may go with something off the shelf like the BQ2640RVAT. Claims to handle 10A and several other bells and whistles comes with a trade-off of at price of 4$ and extra passives.
 

jpanhalt

Joined Jan 18, 2008
9,435
This is the part in the datasheet to which I refer here and in your other thread:

Source: Datasheet
The actual peak current is greater than the current-limit threshold by an amount equal to the inductor ripple current. Therefore, the exact current-limit characteristic and maximum load capability are a function of the inductor value as well as input and output voltages. The current limit will keep the low-side MOSFET ON and will not allow another highside on-time, until the current in the low-side MOSFET reduces below the current limit. After 64 switching cycles, the AOZ2264QI-11 considers this is a true failed condition and therefore, turns-off both high-side and low-side MOSFETs and latches off. Only when triggered, the enable can restart the AOZ2264QI11 again.
 
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