Help me interpret this regulator datasheet please?

[cdevidal]

I'm trying to regulate 5 V down to 4 V @ 800mA. (Don't need exactly 4 V, can go under a tiny bit.) Considering using an ABLIC S-1214B00H-V5T2U7 regulator. On page 24 of its datasheet there are some charts that I don't understand.

The Vout = 5.0 V chart is the closest to 4 V and I think? it's saying that I can expect a little less than 5 V and 800mA when input voltage is 6 V, since that's 1 V over the Vout, which is what I need. But the way the curve loops back under itself like a question mark is heckin' confusing.

If I'm interpreting it correctly, I should be able to use 5 Vin to get a little less than 4 Vout @ 800mA. Just a bit confused about how to read the chart.

[Benta]

Do a web search using "foldback current limiting".

[cdevidal]

So if I understand foldback correctly, for the chart above, if Vin = 6.0 V and current is 1300mA, Vout gets reduced to about 4.5 V. Going further to 1400mA, Vout gets reduced to 2.5V. Correct?

[coppice]

So if I understand foldback correctly, for the chart above, if Vin = 6.0 V and current is 1300mA, Vout gets reduced to about 4.5 V. Going further to 1400mA, Vout gets reduced to 2.5V. Correct?

This is the overload protection characteristic of the regulator, not its normal mode of operation.

[cdevidal]

So if I understand foldback correctly, for the chart above, if Vin = 6.0 V and current is 1300mA, Vout gets reduced to about 4.5 V. Going further to 1400mA, Vout gets reduced to 2.5V. Correct?

This is the overload protection characteristic of the regulator, not its normal mode of operation.

Yes, understood. My load is 800mA which is all I need, so I don't expect it'll go into that mode. Just trying to understand the chart--did I interpret it correctly when talking about 1300mA and 1400mA?

[TimFox]

Often, when the regulator goes into foldback current limit, it is necessary to cycle power before the regulator will work normally.
Foldback is used to reduce the stress on the series regulator element with a shorted load, where the full input voltage appears across the pass element.

[floobydust]

OP, make sure start-up surge is not over 800mA i.e. have bulk capacitors they can trip CL on power up and then it just sits there stuck if your load is also there.
Also, I would suggest to be careful with heat - around 0.8W requires large copper per the IC datasheet, they give several IPC layout scenarios and it's easy to miss those.

[cdevidal]

Often, when the regulator goes into foldback current limit, it is necessary to cycle power before the regulator will work normally.
Foldback is used to reduce the stress on the series regulator element with a shorted load, where the full input voltage appears across the pass element.

Okay I think I now understand why the question mark profile. If you exceed the current on the chart, it first drops voltage, then starts dropping current along with voltage until it gets to 0.9 V and 250mA, then drops to zero. Did I get that right?

[cdevidal]

OP, make sure start-up surge is not over 800mA i.e. have bulk capacitors they can trip CL on power up and then it just sits there stuck if your load is also there.
Also, I would suggest to be careful with heat - around 0.8W requires large copper per the IC datasheet, they give several IPC layout scenarios and it's easy to miss those.

Yep I saw the copper pad but thanks for the heads up. When you say bulk capacitors, the datasheet recommends ceramic, at least 1.0uF. Would you add even more and if so what type?

[SteveThackery]

Okay I think I now understand why the question mark profile. If you exceed the current on the chart, it first drops voltage, then starts dropping current along with voltage until it gets to 0.9 V and 250mA, then drops to zero. Did I get that right?

Yes. The vertical part of the curve is where the device is current-limiting at 1400mA, and then basically if the temperature continues to be too high it reduces the current-limit value, which results in the output voltage falling as well. That's the foldback function. If the chip continues to be too hot, it will reduce the current limit right down to 300mA before throwing in the towel and switching off.

Graphs like these are really odd when you first come across them because for each value on the x axis there are two values on the y axis. Once you realise what's going on it all clicks into shape.

[floobydust]

OP, make sure start-up surge is not over 800mA i.e. have bulk capacitors they can trip CL on power up and then it just sits there stuck if your load is also there.
Also, I would suggest to be careful with heat - around 0.8W requires large copper per the IC datasheet, they give several IPC layout scenarios and it's easy to miss those.

Yep I saw the copper pad but thanks for the heads up. When you say bulk capacitors, the datasheet recommends ceramic, at least 1.0uF. Would you add even more and if so what type?

What is the 4V 800mA load? A motor/mech has inrush current that can trip the foldback and things won't start up. This is what I would worry about.
Along with the heat - are you using the HSOP-8 package? It's only capable of 1W on a 4-layer (2oz outer, 1oz inner) 4-via 2000mm² which is big.

MLCC 1uF cap is never 1uF in practice. Lucky to achieve 1/2 that when you look at the mfgr graphs and include aging. Even 10uF would not be a problem here assuming the load is not demanding 800mA+ on startup.

[cdevidal]

What is the 4V 800mA load? A motor/mech has inrush current that can trip the foldback and things won't start up. This is what I would worry about.

It's coming from 5Vmp solar to a TI BQ25176J LiFePo4 battery charger, which doesn't have sufficient thermal properties to handle any higher voltages without going over the junction temperature. Thought I would bring down the voltage first before powering that chip.

However, the BQ25176J is connected to both a 3.2V 20Ah LiFePo4 cell and the load, which can include a 3V 25mm motor with a stall current up to 1A. The BQ25176J datasheet says it is safe to connect directly to the load as long as the charge can complete within the safety timeout, and it certainly can. The motor is only powered for a few seconds per day.

Along with the heat - are you using the HSOP-8 package? It's only capable of 1W on a 4-layer (2oz outer, 1oz inner) 4-via 2000mm² which is big.

I was going to use the TO-252-5S(A) variant. There is no board sketch in the datasheet that matches but it'll be like Board E with four vias and a large copper pad but on a two layer PCB. Since no thermal properties were listed I estimated a conservative 40°C/W in thermal calculations. Ambient can go as high as 49°C (in desert conditions) plus 6°C of solar gain on the enclosure, so I estimate 55°C ambient, (800mA x 1V x 40°C/W) + 55°C = 87°C which is well below the recommended high temperature--and that's under the worst possible conditions.

MLCC 1uF cap is never 1uF in practice. Lucky to achieve 1/2 that when you look at the mfgr graphs and include aging. Even 10uF would not be a problem here assuming the load is not demanding 800mA+ on startup.

I can do some 10uF, should not be a problem.

Would you use even more capacitors for that load including the motor startup current? I'm not terribly constrained by space so I can put bigger things on the board. It'll (probably) use a TI DRV8411A motor driver, which recommends a 10uF for bulk capacitance, but given your comments about aging I may go with like 100uF or something.

[cdevidal]

Thinking about this more, the BQ25176J can only pass 800mA to both load and battery together, so I would expect that any excess current would be coming from the bypass caps and battery, not the regulator, and the regulator (which comes before the charger chip and which is not directly connected to the load) shouldn't be entering into foldback current limiting when the motor starts. I'll test this, of course. Since neither the regulator nor motor driver has a recommended upper limit on the bulk caps and since they recommend 1uF minimum, I'll use 100uF ceramics to supply lots of startup current. Cheap, and I have plenty of room for them.