Buck step down convertor uses a lot of power?

[AK]

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.

 

[Tim Williams]

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

--
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Design
Website: https://www.seventransistorlabs.com/

"AK" <scientist77017@gmail.com> 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.

 

On Saturday, 27 April 2019 23:09:25 UTC+1, AK wrote:

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?

with no output current, yes.

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

Thanks.

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

 

[AK]

On Saturday, April 27, 2019 at 6:03:00 PM UTC-5, Tim Williams wrote:

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

--
Seven Transistor Labs, LLC
Electrical Engineering Consultation and Design
Website: https://www.seventransistorlabs.com/

"AK" <scientist77017@gmail.com> 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.

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



Andy

 

[Martin Brown]

On 27/04/2019 23:09, AK wrote:

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.

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.

"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?

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

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

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.

https://www.analog.com/media/en/technical-documentation/data-sheets/3632fc.pdf

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.

--
Regards,
Martin Brown

 

[Rob]

AK <scientist77017@gmail.com> wrote:

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.

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.

 

[TTman]

On 27/04/2019 23:09, AK wrote:

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.

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

---
This email has been checked for viruses by Avast antivirus software.
https://www.avast.com/antivirus

 

[AK]

On Sunday, April 28, 2019 at 3:40:40 AM UTC-5, Martin Brown wrote:

On 27/04/2019 23:09, AK wrote:
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.

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.

"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?

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

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

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.

Regards,
Martin Brown

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

 

[AK]

On Sunday, April 28, 2019 at 3:59:33 AM UTC-5, Rob wrote:

AK <scientist77017@gmail.com> wrote:
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.

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.

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

 

On Sunday, 28 April 2019 12:32:10 UTC+1, AK wrote:

On Sunday, April 28, 2019 at 3:40:40 AM UTC-5, Martin Brown wrote:
On 27/04/2019 23:09, AK wrote:
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.

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.

"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?

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

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

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.

Regards,
Martin Brown

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

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


NT

 

On Saturday, April 27, 2019 at 6:09:25 PM UTC-4, AK wrote:

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.

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 should 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.

 

[Winfield Hill]

Rob wrote...

For this purpose you may want another type of
converter for very low currents, which does
not use so much current itself.

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.


--
Thanks,
- Win

 

On Sunday, April 28, 2019 at 2:34:52 PM UTC-4, Winfield Hill wrote:

Rob wrote...

For this purpose you may want another type of
converter for very low currents, which does
not use so much current itself.

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.

Even WalMart sells this one with comparable performance:
https://www.walmart.com/ip/Pololu-3-3V-500mA-Step-Down-Voltage-Regulator/974148649

specs:
https://www.pololu.com/product/2842

--
Thanks,
- Win

 

[AK]

On Sunday, April 28, 2019 at 3:00:24 PM UTC-5, bloggs.fre...@gmail.com wrote:

On Sunday, April 28, 2019 at 2:34:52 PM UTC-4, Winfield Hill wrote:
Rob wrote...

For this purpose you may want another type of
converter for very low currents, which does
not use so much current itself.

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.

Even WalMart sells this one with comparable performance:
https://www.walmart.com/ip/Pololu-3-3V-500mA-Step-Down-Voltage-Regulator/974148649

specs:
https://www.pololu.com/product/2842




--
Thanks,
- Win

Thanks to all for the info.

Andy

 

[Martin Brown]

On 28/04/2019 12:32, AK wrote:

On Sunday, April 28, 2019 at 3:40:40 AM UTC-5, Martin Brown wrote:
On 27/04/2019 23:09, AK wrote:
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.

How long does the clock normally run on 2 AA cells? I put my dead for

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.

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.

"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?

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

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

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.

Regards,
Martin Brown

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.

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.

--
Regards,
Martin Brown

 

[Winfield Hill]

bloggs.fredbloggs.fred@gmail.com wrote...

On Sunday, April 28, 2019 at 2:34:52 PM UTC-4, Winfield Hill wrote:
Rob wrote...

For this purpose you may want another type of
converter for very low currents, which does
not use so much current itself.

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.

Even WalMart sells this one with comparable
performance: [snip]
https://www.pololu.com/product/2842

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.


--
Thanks,
- Win

 

[Martin Brown]

On 29/04/2019 10:00, Winfield Hill wrote:

bloggs.fredbloggs.fred@gmail.com wrote...

On Sunday, April 28, 2019 at 2:34:52 PM UTC-4, Winfield Hill wrote:
Rob wrote...

For this purpose you may want another type of
converter for very low currents, which does
not use so much current itself.

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.

Even WalMart sells this one with comparable
performance: [snip]
https://www.pololu.com/product/2842

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.

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.

--
Regards,
Martin Brown

 

On Monday, April 29, 2019 at 5:00:57 AM UTC-4, Winfield Hill wrote:

bloggs.fredbloggs.fred@gmail.com wrote...

On Sunday, April 28, 2019 at 2:34:52 PM UTC-4, Winfield Hill wrote:
Rob wrote...

For this purpose you may want another type of
converter for very low currents, which does
not use so much current itself.

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.

Even WalMart sells this one with comparable
performance: [snip]
https://www.pololu.com/product/2842

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.

Looks like that datasheet on the ADP5300 is somewhat deceptive. In the details they distinguish between an IQ_HYS and IQ_PWM. Those ultralow IQ are for feedthrough mode, eg. Vin =3.0V Vout=3.6V, when it's not doing any PWM. 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 limited.

--
Thanks,
- Win

 

[Winfield Hill]

bloggs.fredbloggs.fred@gmail.com wrote...

Looks like that datasheet on the ADP5300 is somewhat
deceptive. In the details they distinguish between
an IQ_HYS and IQ_PWM.

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


--
Thanks,
- Win

 

[AK]

On Monday, April 29, 2019 at 2:59:06 PM UTC-5, Winfield Hill wrote:

bloggs.fredbloggs.fred@gmail.com wrote...

Looks like that datasheet on the ADP5300 is somewhat
deceptive. In the details they distinguish between
an IQ_HYS and IQ_PWM.

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


--
Thanks,
- Win

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