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Topic: How much current will a buck regulator draw if I'm not drawing current from it? (Read 653 times) previous topic - next topic


I know that's kind of a vague question.  What I want to know is if I have a mosfet on my board, connected to a buck regulator which I can't control, and I turn the mosfet off, how quickly is the regulator going to drain the battery?

I had a design once where I put a linear regulator on a battery and mistakenly switched the output instead of the input and as a result the battery in the thing was always dead when you went to use it.  But that was a linear regulator and a 9V battery.  Maybe a buck regulator will draw less current.  Or maybe it was the fact that the 9V was much too wimpy and a set of AA's would have lasted a week.  I don't know.

Since it's probably still a hard question to answer, I've got a datasheet here for a specific buck converter I was looking at.  This isn't the one I'm wondering about, if it was I could switch the enable pin, but it might give me a general idea of what I could expect:


So based on that datasheet, how would I determine how much current is drawn when I'm not drawing current from it?  I thought that might be the quiescent current but that doesn't seem to be the case.  Note 9 which is related to that mentions disconnecting the feedback pin, and it wouldn't be disconnected if I just stopped drawing current from the thing.


Looking deep into the datasheet there is a graph shown on page 6 called "Operating Quiescent Current" that shows a minimum current consumption from about 15 to 22ma depending on device temperature. This would be like having a LED as a only load on a battery and would of course discharge it in a calculated time frame based on the mAH capacity of the battery. If you had a 9vdc battery rated at 150mAH the standby current of the regulator would probably discharge the battery in 10 hour or less time frame.

However this chip appears to have a nice extra feature called external shutdown on pin 5.

ON /OFF -- Allows the switching regulator circuit to be shut
down using logic level signals thus dropping the total input
supply current to approximately 80 µA. Pulling this pin below
a threshold voltage of approximately 1.3V turns the regulator
on, and pulling this pin above 1.3V (up to a maximum of 25V)
shuts the regulator down. If this shutdown feature is not
needed, the ON /OFF pin can be wired to the ground pin or
it can be left open, in either case the regulator will be in the
ON condition.

This 80ua minimum current consumption when forced to off, would probably not drain a battery much more then it's shelf self-discharge rate and is the feature you should utilize if you wish to extend standby time for applications that only run for short duration and are then shutdown.



I wish I could use the enable pin, but I'm probably going to be using off the shelf switching regulators designed for remote control vehicles, because I don't have any time left to get this board ready and it would take too long to design my own.

My board is basically an arduino clone specifically designed for controlling lots of LEDs, playing sound effects, and controlling servos.  It's got an SD card on board for holding the sound effects and rather than programming it in C it'll run scripts also stored on the SD card.

I screwed up my power calculations on the linear regulator I was using several months ago and now at the last minute I need to find a solution.  I'd wanted the board to be able to be powered off anything from 6V to 12V.   But it turns out the board can only dissipate 2W, and a linear regulator powered by 12V would generate 21W of heat.

The obvious solution is to use a switching regulator, but that would mean I couldn't put the boards inside small handheld props, because I wouldn't be able to fit enough batteries to get to the 7.5-8V required.  It would also take a lot of time to design, and the potential for error is high.  In addition, for low quantities of boards, the parts would be expensive, adding around $1000 to my costs.

So, barring adding a switching regulator to my board, the next best solution is to use these ready made switching regulators. They're only around $4 each, and I can get ones which can either regulate down to a lower voltage or boost my input voltage, so I would actually have a wider range of input voltages than I originally had hoped for. 

The downside is, there's no enable pin on them.  Which means even if I switch the main board off using a mosfet, they'll still drain the battery slowly. 

I'm going to need to find a solution to that, but I'm really out of time.  Thankfully, most of my customers will be using these boards in these things:

And typically, they run on 12V so they can power an audio amplifier.  They also tend to buy crappy amps that don't have power switches.  So they will need a power switch on the battery anyway.  But it will be on the back of the pack and hard to reach, and they want to be able to turn the sound and lights off from the handheld particle thrower thing hanging on the side of it.  That's where my onboard mosfet would come in.  It would turn off my board, making the pack appear to be off, but the amp and the voltage regulator would stay on, which is ideal anyway because the amp probably would cut out part of the startup sound when they flip the pack on if the amp was not already powered up.

But I really didn't want to have to have a big power switch on the battery.  I mean in that case it's needed, but my board has a 3W audio amp onboard for smaller props, and I wanted the user to be able to use any cheap switch they like to switch power on and off on the prop.  It kinda sucks that I thought I had this perfect little solution with a voltage regulator that had an enable pin, only to be hit with a heathy dose of reality at the last minute.  Kinda screws up my board layout too.  Now I've got everything tightly packed on one side of the board and this big bare section where my voltage regulator used to be. :/

I'm ALMOST tempted to try to stick that buck regulator I posted the datasheet for on there just so I feel like I have a complete product instead of one missing a chunk, but it's so expensive and requires two large caps and a large inductor, and it would take me two weeks to design it and figure out all the math to make sure I have it all correct.  And I'd have to move a bunch of crap around on my board and make it larger so it would fit.  Ugh.

Anyway, thanks for the help, it's appreciated. 



Damn. There's a million of those modules out there and they all seem based on a single design which doesn't provide access to the enable pin!

Also, what the hell is with being able to buy 50 of these things for less than $2 each, but the chip all by itself would cost me $3.30 from Digikey at qty 250?  Even if I bought TEN THOUSAND OF THEM, it would still be more than the fully assembled board!


Damn. There's a million of those modules out there and they all seem based on a single design which doesn't provide access to the enable pin!

Also, what the hell is with being able to buy 50 of these things for less than $2 each, but the chip all by itself would cost me $3.30 from Digikey at qty 250?  Even if I bought TEN THOUSAND OF THEM, it would still be more than the fully assembled board!

I don't know how they do it. And many offer free shipment, and I get stuff here to the west coast in 10 days usually. I have been buying modules like that and other raw components from Asian sellers on E-bay for around 4 years now. Never had a no show, or not as advertised in all that time in maybe 40-50 purchases. Just twice the wrong item was sent, but when notified they sent the correct item and told me to keep the originally sent item. You will notice that most have very good if not perfect feedback rankings. So I don't know how they do it, maybe they obtain their chips from other manufactures over bought stock?


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