High side smart FET features?

Hello
We are wanting to use the BTS6163 high side smart FET.
Please could you help interpret the datasheet..?

BTS6163 smart FET datasheet
https://www.infineon.com/dgdl/Infin...n.pdf?fileId=db3a3043183a95550118606cdf2336a2

Page 3 says that turn-ON slew rate of output voltage is 0.65V/us, but it also says that it gets to 90% of Vout in 150us……Do you know which is it to be?

The first page says it has a “nominal current” of 5.5A, but then page 3 gives “RON” spec at 7.5A.
Also page 3 says that RthJC is only 1.1K/W, so why on earth is the “nominal current” stated as being just 5.5A? [Rds(on) is only 38 milliohms]

It is a tiny little Mosfet so of course its max current and max power dissipation are also low.

Hi,

Page 3 says that turn-ON slew rate of output voltage is 0.65V/us, but it also says that it gets to 90% of Vout in 150us……Do you know which is it to be?

Click to expand...

Did you do some calculations by yourself? Where are your results?

****
Did you see Figure1a?
There you see that Vout at first has a high slew rate, then a slow slew rate.
The first part is marked "dV/dton" whis is given with 0.65V/us. (test conditions are between 25% and 50% of Vout)
(50% - 25%) of 24V are 6V. With a slew rate of 0.65V/us this takes: 6V / 0.65V/us = 9.2us typically.

But datasheet does neither mention how long it takes from 0% to 25%, nor how long it takes from 50% to 90%.
In the Fig. you see a knee. Maybe at 75% (not to scale, just an asumption)

Summary:
So what do we know and what do we not know: (beginning at rising edge of IN signal)
* delay until OUT starts to rise. (not mentioned in the datasheet)
* slew rate from 0% to 25% (not mentioned in the datsheet)
* slew rate from 25% to 50%: given with 0.65V/us typ
* slew rate from 50% to 90% (not mentioned in the datsheet)
...Total time from IN to 90% of VOut: given with 150us typ

So the datasheet is not wrong and not inconsitent, but it is incomplete.

My question: do you need the information for your application?

Klaus

When one finds such incomplete info, the best one can do is to take the worst case scenario. In this case, the maximum Ton time, which is 300 us.
With a maximum toff time of 550 us, this means that the highest frequency that you will be able to switch reliably is about 1 Khz.

It is a tiny little Mosfet so of course its max current and max power dissipation are also low.

Click to expand...

Thanks, but the Rth(jc) is just 1.1 K/W, so we can thermal via it down to a heatsink and get good cooling....so there is surely no reason that "nominal current" needs to be limited to just 5.5A?

Mosfet's current ratings are thermally limited.

Whether the required load current can be sustained, a complete thermal analysis and I would add validation testing, must be performed.

The good news about SmartFets is that they have internal die sensing which would prevent damage, but you would have to deal with nuisance turn-offs.

Thanks,
Actually my apologies, the 5.5A “Nominal” is for when its just on a single sided PCB, and with no thermal via’ing down to a heatsink.
The other point was the short circuit protection, this doesn’t actually seem to act until there’s 3.5V across the internal FET’s drain-source voltage?.....that’s above 70 Amps
So Presumably the thermal switch off would do the job for a prolonged overload current of say 50A, and presumably the Rdson for that kind of current would rise very high so as to end up triggering the short circuit protection anyway

the 5.5A “Nominal” is for when its just on a single sided PCB, and with no thermal via’ing down to a heatsink.

Click to expand...

Yes, the point is well explained in data sheet footnote 6.

Is it the case that the BST6163 doesn’t actually shutdown due to short circuit unless the drain_souce voltage of the internal FET goes above 3.5V for 2us?
So in other words, since the rds(on) of the BST6163’s internal FET is 30 milliohms, then for example 70 Amps could flow and the BST6163 wouldn’t even recognise this as an overcurrent because the Vds of the internal fet remained below 3.5V?
(I appreciate that the rds(on) would probably increase well above 30 milliohms if it had 70 Amps flowing in it, but could anybody hazard a guess at the graph of rds(on) vs drain_source current for currents up to 70Amps?)

Hello,
We are using the BTS6163 smart FET because of its short circuit protection.
On page 11 of its datasheet it appears to suggest that we will only reap the benefit
of its short circuit protection if we have 300 milliohms of 'stray' output resistance and
9uH of 'stray' output inductance.
Is this correct?
We do not have that much stray L and R in our output wiring...so does this mean we won't
benefit from the short circuit protection as described in the datasheet?

BTS6163 datasheet:
https://www.infineon.com/dgdl/Infin...n.pdf?fileId=db3a3043183a95550118606cdf2336a2

On page 11 of its datasheet it appears to suggest that we will only reap the benefit of its short circuit protection if we have 300 milliohms of 'stray' output resistance and 9uH of 'stray' output inductance.

Click to expand...

You apparently misunderstand the meaning of L/R curve. It tells maximum inductance versus minimum load resistance or vice versa. You can have lower load resistance if the inductance is also lower.

Thanks, I must admit we are using the BTS6163 downstream of a PSU which has a built in current limit to 19A in the event of output short circuit...so therefore a short circuit downstream of the BTS6163 would never actually activate the BTS6163 to shut down, since the current through the BTS6163's internal FET would never get above 19 Amps.
(and thence the drain-source voltage across the internal fet of the BTS6163 would never get above 3.5V....and so "short circuit" would never be recognised.)

So it is a shame for us that these smart FETs rated to 60V dont have a definite short circuit protection current shutdown level..

You apparently misunderstand the meaning of L/R curve. It tells maximum inductance versus minimum load resistance or vice versa. You can have lower load resistance if the inductance is also lower.

Click to expand...

Thanks, and presumably this is because they want to limit the amount of energy stored in the inductor by limiting the peak current of the shorted current…and thereby allow the BTS6163 to survive the short circuit event…..the situation is though that we may well have some 3uH of stray inductance in our wiring , though we definitely don’t have anywhere near 230milliOhms of stray resistance…we have much less than that…so I really am thinking that the BTS6163 is making too many stipulations for our system, and it actually isn’t even going to work for us at all.
Its just a shame there are no other smart fets rated to 60V with such low rds(on)…anywhere in the world.

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On page 5 of the BTS6163 datasheet, it says that the BTS6163 limits its current to 22A if it has 45V across its internal FET. Does this current limiting happen straight away?.....it seems very good for the BTS6163 to be able to limit the current down to 22A very quickly….would there be a much higher initial current than 22A whilst the BTS6163 attempts to limit the current down to 22A?

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Fig 3c, page 11 of the BTS6163 datasheet appears to show that the BTS6163 shuts off 2us [“td(sc2)”] after the V(ON) voltage goes above 1V.
This does not correspond to what it says on page 5 of the datasheet, where it says that the BTS6163 shuts down 2us after the V(ON) goes above VON(SC).
Where….
V(ON) = voltage across drain-source of the internal fet of the BTS6163
VON(SC) = 3.5V

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Fig 6b of the BTS6163 datasheet shows that the BTS6163 can be used to monitor the current through it. (Because the current flowing out of the I(IS) pin is a known fraction of the actual current in the internal power fet of the BTS6163….) The problem is that the ratio of “actual FET current” to I(is) current is very widely toleranced, at anywhere between 11000:1 to 8000:1
So the BTS6163 appears not to be an accurate current monitor.