Doubts about synchronous buck output current - Page 1
Forum section: electronics / beginners

author: Benjam
author_rank: Regular Contributor
timestamp: April 23, 2022, 11:30:40 am
content: When I was learning the synchronous buck circuit, I read some schematics, some are 24V to 12V1A, some are 48V to 5V3A, what I am curious about is why the output current is 1A or 3A, what is the basis for these ? Why not 12V2A, or 5V1A?My intuitive guess is that how many amps need to be output is based on the total amount of current that all other devices need to consume after that. For example, take 48V to 5V3A as an example, assuming that on the board, there is IC1 (consumes 0.1A), IC2 (consumes 0.05A), IC3 (consumes 0.04A), Q1 (consumes 0.06A), Q2 (consumes 0.2A), Q3 (consumes 0.05A), the sum of these devices is:A (total consumption) = 0.1A + 0.05A + 0.04A + 0.06A + 0.2A + 0.05A = 0.5ASo, if you take the above 24V to 12V1A, is it enough?If you take 5V3A (of course we ignore the voltage value here), it is not so much use, even if it is 5V1A, it can be used, am I right?If my guess is correct, then a new doubt arises. If there are a lot of devices on a circuit board, do they all need to be added up one by one like this to know how much ampere current is required? ?Maybe for professionals, my doubts are a bit stupid, even ridiculous, but I can understand and I am also trying to learn more.

author: golden_labels
author_rank: Super Contributor
timestamp: April 23, 2022, 12:36:06 pm
content: NOTE: Re-reading the question and seeing Doctorandus_P’s answer, I realized I might’ve misunderstood it. I asumed it’s about why the power supplies themselves have limited current output. I will leave the post in case it was the right interpretation.Components in the power supply, just like anything, can handle only limited current. The power supply itself is also experiencing its own power losses and must be able to handle associated heat. Finally, there are physical limitations, like the size of the inductor and output capacitors: they can only store so much energy and going for higher power supply rating would require larger components.That last part complicates design decisions. You can’t blindly slap a larger inductor and let things run. That’s because the PSU current rating is the maximum current it can deliver. But it must also work at lower currents, including changing currents. If the inductor stores more energy, it also can release more energy. If the load suddenly draws less current, the inductor will by itself not care: it will happily dump all it has into your poor device like a water hammer:The only thing that stands between that mass of energy and whatever you power, is a quick reaction from the PSU controller. It’s also not better, though not catastrphic, in the other direction. Inductors oppose change in current. If your device suddenly becomes hungrier, the inductor impedes delivery of energy, causing voltage drop.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: April 23, 2022, 12:55:05 pm
content: Indeed, you add all the currents of the consumers to get to a current that the power supply must be able to deliver.And your power supply can deliver a smaller current if not all the current is needed of course.With buck modules there is an extra thing to take into consideration though.With some of them, the current capability of the swithch in the SMPS module is mentioned in the datasheet, and not the averaged output current, and this averaged output current is lower then the inductor peak current (and thus also the current through the switch).

author: Benjam
author_rank: Regular Contributor
timestamp: April 23, 2022, 07:53:07 pm
content: I still do not fully understand, looking forward to a clearer answer, where is the master? Please solve the confusion.An additional point of conjecture is that, for example, all the components on the PCB board, only need to consume 1.5A current, through the DCDC (for example, external 48V PSU to provide 48V DC, and continuous current 50A) to turn out can be 5V10A, or 5V20A to provide all the components to use, are stable to use, right? Because 10A or 20A is more than 1.5A, enough. But it can not be lower than 1.5A, otherwise it will lead to voltage drop, am I understanding it right?But I additionally heard a statement "the current is too big, resulting in device burnout" Why is this? There is even a more technical term called "overcurrent protection". For example, a MOSFET its Id (Max) maximum is 3A, but I passed to the gate voltage is too high, resulting in the flow of current through the MOSFET exceeds the Id (Max) 3A, so the MOSFET may be damaged because of overcurrent. Is that right? Did I understand correctly? (Need clear guidance from a guru).Another case, for example an IC1, which has the following data in its datasheet.Then if it is connected to the power supply from the above mentioned 5V10A (through DCDC), that 10A obviously exceeds the Vdd1 Current (Idd1) Max 10mA mentioned in the datasheet, then what will be the problem of such?May be a wrong understanding, forgive me for being a novice (although I am trying very hard to learn), another understanding (just my own speculation)The current is like a reservoir (e.g. 5V10A from 48V PSU via DCDC mentioned above), it has a total of 10 tons of water (similar to 10A current), IC1 is like a family x living under this reservoir, when he wants to wash the dishes, he will turn on the switch, then the water flows out of the pipe with the size of the switch. Add he uses up 1 ton of water, then at this time the reservoir changes from the original 10 tons of water to only 9 tons of water (ignore for a moment this premise of continuous current supply), if the remaining 9 families, consume all the remaining 9 tons of water, meaning that the 11th family wants to use water, there is no water (mapping to my previous doubts, if the first 10 components on the PCB, all consumed 10A, then the remaining other components will have no current to use (insufficient current), so it may lead to voltage drop problems).So, even if the 5V10A current source is connected to IC1, it will not damage it, although the datasheet mentions Vdd1 Current(Idd1) Max 10mA, meaning it will consume that much current at the maximum, right?So my whole guess is.1: Connecting 5V10A current to IC1 will not damage IC1, IC1 will run fine.The reason: Because IC1 only consumes the current it needs from 10A, for example, Max 10mA, it consumes 10mA from 10A, so whether it is 5V10A, or 5V20A, 5V30A, 5V40A, there is no problem.Share your understanding, or tell me if my understanding is correct? I must say sorry for the long-windedness of my description, and perhaps some of it is not clear, but I hope you can understand and help me as a newbie, I need to clarify this confusion thoroughly, because it is the cornerstone!Thank you all!

author: golden_labels
author_rank: Super Contributor
timestamp: April 23, 2022, 10:50:57 pm
content: Benjam,please clarify: what are you asking about?Your first post looked as if you ask about design of synchronous buck converters. Then, after Doctorandus_P’s post, it seemed as if it could be either that or choosing the right PSU. Though limiting the question to synchronous buck converters makes that interpretation quite weird. Now you ask about… dunno what, and you refer to designations of some components on a board we do not even see.At this point I will answer as if you asked about choosing a PSU, completely ignoring its topology.The current rating of a power supply is the maximum current it is designed to deliver. If a power supply is rated for the maximum of 10A, it can as well deliver 1.5A. Most power supplies require some minimum load, but that is in range of milliamperes to hundreds of milliamperes,(1)and not really related to your question. Though note that a 10A power supply may be quite inefficient if delivering only 1.5A. They are usually designed to have the top efficiency closer to the maximum current.The comparison with a water reservoir is right.The power requirements section of ADuM7702 tell what voltage and current that particular device requires. The left part of ADuM7702 requires any voltage between 4.5V and 5.5V on its power supply pin, and consumes current of up to 10mA. The right part, which is galvanically separated and has its own power supply, requires any voltage between 3V and 5.5V and consumes up to 3mA.Connecting a 5V power supply to either of its VDD1pin will not damage ADuM7702. 5V is between 4.5V and 5.5V. The currents are not directly relevant for possibility of damage here.Yes, ADuM7702 will consume 10mA through its VDD1. Up to 10mA will be consumed and the power supply will produce 10mA. It doesn’t matter if the power source can deliver 100mA, 1A, 10A or 100A. However, on a practical note, power supplies have minimal load (as mentioned above). Under a dozen mA is a quite low current, so you should check what is the minimal load. A 10A converter providing only 10mA will also be very inefficient.(1)With the upper endpoint certainly not being a power supply rated for a few amps.

author: Benjam
author_rank: Regular Contributor
timestamp: April 23, 2022, 11:37:57 pm
content: I have understood, thank you very much for your patient answer. Thanks!