Buck step down convertor uses a lot of power?

I have a DC-DC Buck Step Down Converter Module LM2596 Voltage Regulator set to supply 3.0 volts to an atomic clock using a 8 Ah 12 V battery. The clock normally uses 2 AA batteries.

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The battery started out at 12.07 V. After 10 days, it's at 8.45 Volts.

It does seem like it should have dropped that much.

It claims Static power: 20 ma. Does that mean power used?

The converter has a digital voltage readout. Does it use a lot of current?

Thanks.

Reply to
AK
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with no output current, yes.

You didn't mention your battery capacity or how charged it was beforehand. If we guess the regulator eats 25mA, in 10 days that's 10x24x0.025 = 6Ah.

NT

Reply to
tabbypurr

Yes, and 8Ah / 10 d / 24h/d = 0.033A or 33mA, not far from 20mA.

Those modules aren't terribly efficient even at full power, and are way off what you'd need to power a battery-sipping module (if that's what your clock is).

Consider a higher efficiency regulator, usually with some kind of burst mode feature. (This is as much a bug as a feature -- burst mode means the regulator is capable of so-and-so current, but it's only efficient when running at modest throttle. So it gets good efficiency that way, but the downside is the output voltage varies up and down -- relatively high output ripple. Most loads don't care about this, so it's okay. The other downside is, because it's turning on and off quickly, it makes the switching inductor sing at the burst frequency, which can be annoying. A lot of commercial devices that make a high-pitched whine, are doing this.)

Tim

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"AK"  wrote in message  
news:75f92201-5109-4cd2-bded-781e56a19edc@googlegroups.com... 
>I have a DC-DC Buck Step Down Converter Module LM2596 Voltage Regulator set  
>to supply 3.0 volts to an atomic clock using a 8 Ah 12 V battery. The clock  
>normally uses 2 AA batteries. 
> 
> "https://www.ebay.com/itm/DC-DC-Buck-Step-Down-Converter-Module-LM2596-Voltage-Regulator-Led-Voltmeter-M3/301724177038?epid=10003955778&hash=item464029928e:g:bAwAAOSwdsFUNdsC" 
> 
> The battery started out at 12.07 V. After 10 days, it's at 8.45 Volts. 
> 
> It does seem like it should have dropped that much. 
> 
> It claims Static power: 20 ma. Does that mean power used? 
> 
> The converter has a digital voltage readout. Does it use a lot of current? 
> 
> Thanks.
Reply to
Tim Williams

I found a switch that turns the digital voltmeter off. That should extend the life.

Andy

Reply to
AK

How long does the clock normally run on 2 AA cells? I put my dead for high current purposes AA cells into my clocks and mice to use them up! They often still last a year in the clocks - much less in my mice.

It draws that much current to do nothing at all and more if loaded.

Perhaps around 500uA per segment lit.

You could always do something radical like *measure* the current the thing draws and the current that it supplies to your atomic clock.

My guess is that the clock draws so little current that if you must power it with a big lead acid accumulator you would be much better off with a crude series resistor dropper, 3v zener diode and a modest capacitor in parallel with it.

An IC based solution probably won't be any more efficient since AD's lowest current step down chip 3632 has a quiescent current of 12uA. It is likely that an LCD atomic clock draws much less than that.

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You could do a lot better with a custom built joule thief like circuit that is designed to step down and still be efficient at tiny currents. Or a modification of the 3632 that shut it down after charging up a capacitor and sits in shutdown until the terminal voltage falls.

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Regards, 
Martin Brown
Reply to
Martin Brown

Use your meter to measure the current at the different points in the circuit. (between converter and clock moduke, with and without the options that you can turn on and off, and between battery and converter, same)

Then you know what is going on and what you can and cannot do.

For this purpose you may want another type of converter for very low currents, which does not use so much current itself. Or you may want to go back to using AA batteries or using 2 batteries of a larger type (C, D) without using the converter.

Reply to
Rob

How about powering your clock from a single cell LIPO? A full charge might last 10 years....

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Reply to
TTman

So, what would I need specifically in terms of series resistor dropper and capacitor.? I know where I can get the zener diode.

I have resistors and some 1n 4001.

Andy

Reply to
AK

I wanted to use the large battery since I am not using it. It was formerly used as a UPC when I upgraded it.

I thought with it's much larger capacity than aa or 18650, it would last for a long time.

Andy

Reply to
AK

You haven't told us what current the clock draws yet.

NT

Reply to
tabbypurr

et to supply 3.0 volts to an atomic clock using a 8 Ah 12 V battery. The cl ock normally uses 2 AA batteries.

?

The battery self-discharge current is probably greater than the loading of the atomic clock, which means it makes no sense to use it. The 2x AA shoul d power the clock for a few years, and will therefore outlast the lead acid . Move on to another project, you're just throwing money away on this one.

Reply to
bloggs.fredbloggs.fred

I have gotten some pretty-low-current buck converter PCBs from China, but there are lots of better IC choices available, if one is ready to do a little simple engineering. For example, an ADP5300 only consumes 0.2uA to regulate with loads up to 50mA. It works in hysteresis mode, switching on only long enough to re-energize the output capacitor.

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 Thanks, 
    - Win
Reply to
Winfield Hill

Even WalMart sells this one with comparable performance:

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specs:

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Reply to
bloggs.fredbloggs.fred

Thanks to all for the info.

Andy

Reply to
AK

I think you really need to answer this question and explain why you need your atomic clock to run for more than two years continuously. The whole purpose of an "atomic" clock is that it self sets from time service radio transmissions. Changing its battery should not be traumatic.

I'm assuming here that you don't mean a caesium or rubidium fountain clock or an H-maser - they require more power so I can't see anyone running them off 2 AA cells for more than a few minutes.

You should measure the current your load consumes and set the resistor to drop about 9v at the required current.

Try 1M, 1uF and a 3v zener (or 5 1N4001 in series to get about 3v).

It is possible that the clock CPU bootup will not take kindly to the slow rise time of the voltage so you may need to connect to clock to the zener after waiting a few seconds for the capacitor to charge up.

We remain mystified why you would want to do this.

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Regards, 
Martin Brown
Reply to
Martin Brown

A nice little buck converter, but its quiescent current is 200uA, 1000 times the AD part, could be up to 1000 times harder on the OP's battery.

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 Thanks, 
    - Win
Reply to
Winfield Hill

That chip has amazingly low standby current - but at what point does it all become academic because of the battery internal self discharge.

One minor snag for the OP is that ADP5300 Vmax is 6.5v.

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Regards, 
Martin Brown
Reply to
Martin Brown

Looks like that datasheet on the ADP5300 is somewhat deceptive. In the deta ils they distinguish between an IQ_HYS and IQ_PWM. Those ultralow IQ are fo r feedthrough mode, eg. Vin =3.0V Vout=3.6V, when it's not doing any PW M. If the chip has to PWM then IQ jumps up to 435uA. I was wondering how it could do so well switching at 2.0MHz. Then the input voltage range is limi ted.

Reply to
bloggs.fredbloggs.fred

The low-power hsyteretic control works to 50mA, the high-power low-ripple PWM control is for up to 500mA.

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 Thanks, 
    - Win
Reply to
Winfield Hill

I fully charged battery up.

In a day, it dropped from 12.6 -> 12.4 volts.

I am happy with that. If it lasts a year, I will be happy.

Andy

Reply to
AK

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