What type of battery pack can power a device with 9V 0.5A output?

[Asymptotic]

The powerbank itself seems to maintain 5v pretty well even when 20% level only. Something AA batteries can't do?

The powerbank probably hides direct battery voltage information in favor of a discharge status percentage.
It is a DC-DC converter built around a battery, for instance, a 3.7V Li-Ion, and a specialized chip designed to correctly charge the battery and provide a regulated output voltage when in use.

In this case, your powerbank will deliver 5 VDC output from 100% to 0% charge "level" as the terminal voltage of the internal battery pack drops from 3.7V to whatever reduced battery voltage the powerbank designer decided to consider 0% charge.

 

[Rive]

The powerbank itself seems to maintain 5v pretty well even when 20% level only.

As @Asymptotic said, the powerbank has some internal electronics to maintain stable output voltage (and provide protection): so the remaining only suspect of the trouble is the converter. It is perfectly justified to request a replacement (if you decide to do so).

In a few sentences and from memory, can you share how exactly 5v is made to convert to 9v? Thanks.

That's not just a few sentences. For this application what (not how, but: what) it does is to conserve power (with some losses): I_out * U_out /0.8= I_in * U_in

 

[chirhone]

As @Asymptotic said, the powerbank has some internal electronics to maintain stable output voltage (and provide protection): so the remaining only suspect of the trouble is the converter. It is perfectly justified to request a replacement (if you decide to do so).That's not just a few sentences. For this application what (not how, but: what) it does is to conserve power (with some losses): I_out * U_out /0.8= I_in * U_in

The one on top is the busted one showing only 4.9v at either 9v or 12v. It should display 9 or 12. Not 4.9. I have to prove to manufacturer i didnt overload it. But bought each direct from china market for only $3 each. Shipping it back may cost $20.

 

[Rive]

The one on top is the busted one showing only 4.9v at either 9v or 12v.

That's a common failure mode for boost converters: what's coming in will go out without any 'boost' (with some slight voltage drop). If you can identify the controller chip then you might attempt a repair (needs sufficient skills).

Shipping it back may cost $20.

That's why I said replacement and not return
It's entirely up to you.

 

[chirhone]

My Maglight (LED version) has a sharp cutoff between low (yet acceptable) light output to completely off when the batteries are low. Computers do bad things when supply voltages are low; exactly what these bad things are depends upon the computer, but include memory corruption, locking up, continuously rebooting, starting to boot then shutting down, and so on. An industrial PLC (Programmable Logic Controller) is likely to lose it's program if it is turned off and the memory back-up battery is low. Many such examples of low voltage issues exist.
Yes. In fact, it is the preferred method. The "cell isolation" technique I outlined has one pro (it doesn't involve cutting into the wiring), but among it's cons is awkwardness.Readout numbers on the left are small particle, and those on the right are the large particle count. The following is pure guesswork, but (provided the fan voltage supply, thus fan speed, is regulated)

when supply voltage drops below the point where that regulator no longer functions, the previously fixed fan speed will slow down to track with the reduction in supply voltage. Slowing the fan reduces both air volume and velocity through the test chamber.From what I've recently read regarding laser-based particle count schemes, they determine particle count and size based on how many times the laser beam to photoreceiver path is blocked (count), and how much time it takes until the beam is sensed again (an indication of particle size). The unusual increase in large particle count due to low supply voltage would be consistent with a reduction in fan speed leading to reduced air velocity, and increasing the amount of time particles block the laser beam.

But even if it increases the amount of time particles block the laser beam, it should count lesser particles not 20 times more.

I want to focus now how laser based pm2.5 works in details. I am looking for paper or website that shows the full details which can include 16 channel GRIMM monitors that costs $25000.

I saw this in this page.

https://translate.googleusercontent.com/translate_c?depth=1&hl=nl&nv=1&rurl=translate.google.com&sl=nl&sp=nmt4&tl=en&u=https://www.fijnstofmeter.com/&xid=17259,15700021,15700043,15700186,15700191,15700259,15700271&usg=ALkJrhhlB_KukvpqefBEht4SEM4T9t9xvw

"How does a particulate matter meter work?

A particulate matter meter / particle counter works on the basis of the 'light scattering' or 'light blocking' principle. The air is passed through a measuring chamber through a fan. A laser beam illuminates the particles. The light that comes in contact with a particle is absorbed or scattered in specific directions. The direction of scattering is characteristic of the size of the particle, whereby the particle counter is able to determine the number of particles of a specific size."

This observation suggests when battery open circuit voltage has dropped to 6.8V it is enough voltage to get the computer running, and (just barely) power the LCD display, but when the computer commands the laser and/or fan to turn on (increasing current demand decrease battery voltage) the voltage drops below what the computer needs to operate, and it shuts off.You might observe a brief time of 300 mA+ current every time the unit is powered up. A fan requires more current to bring it up from zero speed to normal running speed (typically, one second or less) than it does to maintain normal running speed.

 

[Asymptotic]

But even if it increases the amount of time particles block the laser beam, it should count lesser particles not 20 times more.

I want to focus now how laser based pm2.5 works in details. I am looking for paper or website that shows the full details which can include 16 channel GRIMM monitors that costs $25000.

I saw this in this page.

View attachment 256993

https://translate.googleusercontent.com/translate_c?depth=1&hl=nl&nv=1&rurl=translate.google.com&sl=nl&sp=nmt4&tl=en&u=https://www.fijnstofmeter.com/&xid=17259,15700021,15700043,15700186,15700191,15700259,15700271&usg=ALkJrhhlB_KukvpqefBEht4SEM4T9t9xvw

"How does a particulate matter meter work?

A particulate matter meter / particle counter works on the basis of the 'light scattering' or 'light blocking' principle. The air is passed through a measuring chamber through a fan. A laser beam illuminates the particles. The light that comes in contact with a particle is absorbed or scattered in specific directions. The direction of scattering is characteristic of the size of the particle, whereby the particle counter is able to determine the number of particles of a specific size."

You probably want to start another thread to discuss the inner workings of air particle testers. It strays far afield of the original subject of this one ("9V 0.5A battery pack").

I'd be in the same boat as you regarding answers on this question. The sum total of what I know about them was picked up along the way while researching how the Dylos is powered.

 

[chirhone]

That's a common failure mode for boost converters: what's coming in will go out without any 'boost' (with some slight voltage drop). If you can identify the controller chip then you might attempt a repair (needs sufficient skills).That's why I said replacement and not return
It's entirely up to you.

How accurate do you think is this?

I bought one and tried it on the 5v-9v adapter and dylos. But it showed 490ma instead of 269ma I measured using a multimeter.

Showing 490ma instead of 269ma using multimeter

Without any load, the usb ammeter is showing 0. Is it simply not accurate to show 490ma instead of 267ma?

 

[Asymptotic]

Is it simply not accurate to show 490ma instead of 267ma?

You are measuring 0.49 amps at 4.96V from the powerbank, and 9.1V at 267 mA output from the 5-9V boost converter to the Dylos load.

4.96V*0.49A = 2.43 watts
9.1V*0.267A = 2.42 watts

These power values agree to approximately 0.5%.

 

[chirhone]

You are measuring 0.49 amps at 4.96V from the powerbank, and 9.1V at 267 mA output from the 5-9V boost converter to the Dylos load.

4.96V*0.49A = 2.43 watts
9.1V*0.267A = 2.42 watts

These power values agree to approximately 0.5%.

Thanks. I didnt think of it. I thought it was simply unaccurate for small amperage. Now I am waiting for this connector from china.

Ill plug the jameco adaptor to its 5.5mm female. Then connect the usb ammeter to the above. And then use this to connect to the dylos.

It will work isn't it? This is to get the voltage and ampere from the jameco adaptor and unit itself.

 

[Asymptotic]

Thanks. I didnt think of it. I thought it was simply unaccurate for small amperage. Now I am waiting for this connector from china.

View attachment 257113

Ill plug the jameco adaptor to its 5.5mm female. Then connect the usb ammeter to the above. And then use this to connect to the dylos.

It will work isn't it? This is to get the voltage and ampere from the jameco adaptor and unit itself.

View attachment 257114

Make a sketch of what you intend to do. If I'm following what you are saying I think the answer is no.

 

[chirhone]

Make a sketch of what you intend to do. If I'm following what you are saying I think the answer is no.

Why won't it work? I already have the male plug/USB in green and USB detector. I tested using continuity tester the outside plug of the green is negative with respect to the USB pins standard. I'm just awaiting the female USB connector (unless this polarity is reverse but conventional is negative is always the outside of the plug).

 

[Asymptotic]

View attachment 257115

Why won't it work? I already have the male plug/USB in green and USB detector. I tested using continuity tester the outside plug of the green is negative with respect to the USB pins standard. I'm just awaiting the female USB connector (unless this polarity is reverse but conventional is negative is always the outside of the plug).

This monitor isn't the one picture in earlier posts #45 and #49. If this Keweisi is the one you have

https://www.amazon.com/dp/B06X6K7NK9/?tag=pfamazon01-20

it lists specifications of:
. Input voltage : 3.0 - 9.0V
. Measuring current: 0.00 -3.00a

Unloaded, the Jameco supply voltage will be about 13.7V, and approximately 11V at 270 mA load. I don't know what the USB Detector will do if supplied substantially more than 9V. Provided it isn't damaged, voltage display is in x.xx format with what appears to be a fixed decimal point, and is at best is capable of displaying up to 9.99V

 

[chirhone]

This monitor isn't the one picture in earlier posts #45 and #49. If this Keweisi is the one you have

https://www.amazon.com/dp/B06X6K7NK9/?tag=pfamazon01-20

it lists specifications of:
. Input voltage : 3.0 - 9.0V
. Measuring current: 0.00 -3.00a

Unloaded, the Jameco supply voltage will be about 13.7V, and approximately 11V at 270 mA load. I don't know what the USB Detector will do if supplied substantially more than 9V. Provided it isn't damaged, voltage display is in x.xx format with what appears to be a fixed decimal point, and is at best is capable of displaying up to 9.99V

Thanks. I didnt think of the voltage limit. Checking this exact product i got

https://www.lazada.com.ph/products/digital-led-usb-charger-tester-amp-ma-charge-current-volt-meter-monitor-gauge-intl-i128640716-s138744448.html?dsource=share&laz_share_info=22615686_5_100_500042076257_15656003_null&laz_token=2f0e1830ecb3fb1494a00aa992bcdad7

I saw the voltage is only up to 8v! Thanks for the warning. I won't try it anymore. Do you know a similar configuration with larger voltage range? At least i could make use of the usb male and female connector by connecting my multimeter to the male below in the picture by cutting one wire of the pair. It would work either one of the pair when measuring ampere with the multimeter? Thank you.

 

[chirhone]

Thanks. I didnt think of the voltage limit. Checking this exact product i got

https://www.lazada.com.ph/products/digital-led-usb-charger-tester-amp-ma-charge-current-volt-meter-monitor-gauge-intl-i128640716-s138744448.html?dsource=share&laz_share_info=22615686_5_100_500042076257_15656003_null&laz_token=2f0e1830ecb3fb1494a00aa992bcdad7

I saw the voltage is only up to 8v! Thanks for the warning. I won't try it anymore. Do you know a similar configuration with larger voltage range? At least i could make use of the usb male and female connector by connecting my multimeter to the male below in the picture by cutting one wire of the pair. It would work either one of the pair when measuring ampere with the multimeter? Thank you.

About the link above. The actual unit shipped to me is the same one in your amazon link. Tnx for the real spec of up to 9v. I guess i can use it with the 6 pcs of AA for 9v and see if i can add one more 1.5v aa (without connecting it to the unit).

 

[chirhone]

You are measuring 0.49 amps at 4.96V from the powerbank, and 9.1V at 267 mA output from the 5-9V boost converter to the Dylos load.

4.96V*0.49A = 2.43 watts
9.1V*0.267A = 2.42 watts

These power values agree to approximately 0.5%.

The multimeter reads 0.269A, not 0.267A. But after 10 seconds using the USB ammeter, it gets steady at 0.50A so the wattage is still close between the 2 testers.

Now let's combine the concepts in powerbanks, AA battery and AC charger. You wrote in message #30

"270 mA at 6.84V is 1.85 watts.

270 mA is a fairly high current draw for AA alkaline cells, and will provide about 7.4 hours service life at 270 mA continuous demand at 21°C per the Eveready E91 spec sheet."

The AA battery reads 270mA at 6.84.
The powerbank reads 269mA at 4.96v.

This is why i thought the unit draws 270mA constant. That's why I thought the 0,49A in the USB ammeter was wrong at 4.96V because i thought it should be a constant 270mA.

So is the 270mA at 6.84 using AA above and 269mA using USB meter at 4.96v above just a coincidence??

About the ac adaptor. You wrote in one site, it was reported at 220mA and this would give voltage of about 11v + (I read the entire thread many times to comprehend everything you said). I understand this current value is for the particular adaptor and not related to the current values in the powerbank and AA battery, right?

About the AC adaptor. I won't find a separate USB-like ammeter because I just need to know the voltage and current values once and the male and female usb connectors can give me means to measure it once.

Btw... i found an old NiMH energizer charger in my old things and I think it still works. It has this specs:

http://www.emtcompany.com/energizer-chp41us-battery-charger/101322337.html

Do you believe that the USB ammeter can find great use in the AA battery pack to monitor when voltage gets down from 9v to 6v (discharge voltage)?

Let's wrap up all this battery thing for the Dylos because amazon has just refunded my cheap inaccurate china multi pollution meter today (see 1st msg in https://www.physicsforums.com/threa...ion-particulates-2-5-microns-or-below.983287/ ), giving me a go ahead to buy separate accurate monitors for Co2, and TVOC.

Appreciated so much all your battery tips. I can apply them to my Co2 and Tvoc testers if they don't use batteries. Thanks.

 

[chirhone]

You are measuring 0.49 amps at 4.96V from the powerbank, and 9.1V at 267 mA output from the 5-9V boost converter to the Dylos load.

4.96V*0.49A = 2.43 watts
9.1V*0.267A = 2.42 watts

These power values agree to approximately 0.5%.

To illustrate my questions in my previous message. Consider your message number 22:

"It is unregulated, and when the Jameco supply is operated with less than 500 mA load, output voltage will be higher than 9V. If the Dylos draws 220 mA (as indicated in the review article) power supply output voltage will be on the order of 11.6 volts. Measure it.

"

Notice your 220ma times 11.6v = 2.554v. It's not the same as in your 2.43 watts result from the powerbank. This was what I meant whether the wattage in the adaptor is related to the wattage in the powerbank.

Or is it the adaptor and powerbank should have separate total wattage as seen by the product?

As for the AA battery. I assume it can't maintain constant power because it discharges or has no regulator like the powerbank.

Also where did you get the figure above? I couldn't find it in the jameco pdf file for the product.

Thank you,

 

[chirhone]

You are measuring 0.49 amps at 4.96V from the powerbank, and 9.1V at 267 mA output from the 5-9V boost converter to the Dylos load.

4.96V*0.49A = 2.43 watts
9.1V*0.267A = 2.42 watts

These power values agree to approximately 0.5%.

I finally measured the voltage and ampere of the jameco adaptor as you suggested. It's 10.36v and -269.9mA (see pictures above). Power is about 2.8watt. It's different from the powerbank 2.43watt. Conclusion? power is dependent on the source? Note the 6 pcs of AA battery measured also 270mA but 6.84v.

I connected the positive of the lead to the negative of the jameco adaptor terminal, why the display showed -269.9mA with negative? My last measurement will be the 6 pcs AA battery pack but for now they are all discharged already. Thanks for all tips.

 

[Rive]

I finally measured the voltage and ampere of the jameco adaptor as you suggested. It's 10.36v and -269.9mA
...
Note the 6 pcs of AA battery measured also 270mA but 6.84v.

Well, that's a bit disappointing.

There are two basic types of voltage regulators: DCDC converters: this type is used in that USB-to-9V converter. This type keeps power as constant: you could observe that. The second type is 'linear regulator' , which keeps only current as constant: the one in your device seems to be this type. Although this type sometimes cheaper, especially in this low power range, it works with relatively high losses (the loss depends on the input voltage).

There is no consequence, just a remark.

By the way, based on the behavior of that regulator there is a high chance that your device would be able to work directly from USB - in case you would dare to disassemble and modify it

 

[chirhone]

Well, that's a bit disappointing.

There are two basic types of voltage regulators: DCDC converters: this type is used in that USB-to-9V converter. This type keeps power as constant: you could observe that. The second type is 'linear regulator' , which keeps only current as constant: the one in your device seems to be this type. Although this type sometimes cheaper, especially in this low power range, it works with relatively high losses (the loss depends on the input voltage).

There is no consequence, just a remark.

By the way, based on the behavior of that regulator there is a high chance that your device would be able to work directly from USB - in case you would dare to disassemble and modify it

So that's what dc to dc converter does.. maintain power. I just realize it now and not able to absorb it before. Thanks for the tips. Btw.. I think I know why the first 5v to 9v USB adapter got busted. The other day. I left my second adaptor plugged to the unit and left in the room for 30 minutes. I almost forgot about it. when I touched the USB adaptor. It's so hot. So I guess it depends on the temperature of the location. In china or some parts of US, it's so cold so the USB doesn't overheat. In hot places, it can overheat and get busted. Remember it's not military spec. This makes sense now.

The reason I was so interested in batteries at start was because I thought different streets or buildings have different levels of pm2.5 particulate. But I went to different places and measured almost uniform levels as most were near busy street. I also invested in one of the most powerful air purifier that was why the counts above is only 200 to 300 plus. Average is 2000 to 4000 in the city.

So getting the baseline. I am getting tired using the dylos and plan to return it to amazon, but it has some scratches when I tried to plug the usb adaptor without the plastic casing. I wonder if amazon will still accept it with the scratches.

 

[chirhone]

Well, that's a bit disappointing.

There are two basic types of voltage regulators: DCDC converters: this type is used in that USB-to-9V converter. This type keeps power as constant: you could observe that. The second type is 'linear regulator' , which keeps only current as constant: the one in your device seems to be this type. Although this type sometimes cheaper, especially in this low power range, it works with relatively high losses (the loss depends on the input voltage).

Here is something puzzling. The wattage in the jameco adaptor output is 2.8w (10.36vx270mA). Powerbank is 2.43w, battery is 1.84w. If it can run it at 1.84w, the almost 1 watt is being dissipated inside the dylos or inside the jameco adaptor as heat? If inside the jameco adaptor. But i measured the wattage from the output of the adaptor.

Also for more info. Input no-load power of the adaptor is 1.3watts as shown in pictures a few days ago. Input power with load is 4.9watts.

There is no consequence, just a remark.

By the way, based on the behavior of that regulator there is a high chance that your device would be able to work directly from USB - in case you would dare to disassemble and modify it

 

[Rive]

The wattage in the adaptor output is 2.8w (10.36vx270mA). Powerbank is 2.43w, battery is 1.84w. If it can run it at 184w, the almost 1 watt is being dissipated inside the dylos or inside the jameco adaptor?

Actually, since these numbers represents the consumption of the Dylos, almost all the power dissipates within the Dylos (some power escapes as backlight of the display, and as air movenemt due the fan) in every case: it is just in case there is lower input voltage, less dissipation happens.
The linear regulators are just like this. The current kept ~ constant, so the closer the input voltage to the output voltage, the lower the loss.
I suspect the actual 'wattage' of the Dylos to be around 1.35W. Above that, it's the regulation loss.

 

[chirhone]

Actually, since these numbers represents the consumption of the Dylos, almost all the power dissipates within the Dylos (some power escapes as backlight of the display, and as air movenemt due the fan) in every case: it is just in case there is lower input voltage, less dissipation happens.
The linear regulators are just like this. The current kept ~ constant, so the closer the input voltage to the output voltage, the lower the loss.
I suspect the actual 'wattage' of the Dylos to be around 1.35W. Above that, it's the regulation loss.

1.35w/270mA= 5 volts. The unit can't boot up in the busted 5v to 9v adaptor with 5v reading in the led.

Whatever, so it means if I don't use the jameco adaptor. It can cause less stress on the dylos? I think I may still need the unit when opening windows. The rate of the air purifier strengh has to counter the diffusion of the 2.5 micron particulate from outside the windows. So if I use batteries. The life of the unit can be longer? It has warrantee of 3 months only and some reported it goes broken in 10 months. Do you know other examples of devices or gadgets where the life is longer using battery than using adaptor?

Also have you seen this vintage 10 year old charger?

It is supposed to have timer based on strength of battery. But when it reached zero. The batteries were so hot. No problem by mixing NiMh and alkaine batteries on gadgets ? I want to try AA batteries on the USB ammeter when my female USB connectdor arrives from china. All flights were banned so more delays in getting it.

 

[Rive]

1.35w/270mA= 5 volts. The unit can't boot up in the busted 5v to 9v adaptor with 5v reading in the led.

Every linear regulator requires a minimal voltage drop present, usually around 1.5-2V. The reason I suspect the internal voltage of the Dylos being 5V is that you could run it at 6.8V, but no lower => that's exactly an 5V internal regulator and a minimal voltage drop together.

Whatever, so it means if I don't use the jameco adaptor. It can cause less stress on the dylos?

In theory, yes. But: it should not matter. I would just keep the voltage at 8-9V DC. That's already lower dissipation than with the original adapter.

No problem by mixing NiMh and alkaine batteries ?

I would not try, nor in that battery pack, nor in that charger.

 

[chirhone]

Every linear regulator requires a minimal voltage drop present, usually around 1.5-2V. The reason I suspect the internal voltage of the Dylos being 5V is that you could run it at 6.8V, but no lower => that's exactly an 5V internal regulator and a minimal voltage drop together.In theory, yes. But: it should not matter. I would just keep the voltage at 8-9V DC. That's already lower dissipation than with the original adapter.I would not try, nor in that battery pack, nor in that charger.

I have 6 discharged AA alkaline battery. I bought 2 NiMH rechargeable so I can remove the 2 alkaline and replaced them with the NiMH so I can see what happens in the USB ammeter/voltage when voltages goes below 6.8v. I just googled now it is hazardous. Thanks for the warning, I'm so ignorant.

Btw.. I have more than a dozen 12v dc switching adaptor like the following which can take in 100-240v ac and output of 12v. It powers my cctv cameras, routers, tv box, cable unit and almost all gadgets. It's so rare to have a big adaptor like the Jameco.

It is only about $2 to $3 maximum. Do you have ideas why dylos didn't just use one like it with 9v? why does it have to use unregulated adaptor? Do you know the advantage of unregulated adaptor? Is it just lower price and nothing more? If i can get 9v version of the switching adaptor, its safer?

Thanks very much.

 

[chirhone]

Well, that's a bit disappointing.

There are two basic types of voltage regulators: DCDC converters: this type is used in that USB-to-9V converter. This type keeps power as constant: you could observe that. The second type is 'linear regulator' , which keeps only current as constant: the one in your device seems to be this type. Although this type sometimes cheaper, especially in this low power range, it works with relatively high losses (the loss depends on the input voltage).

I've been trying to analyze this for a couple of days. This was what happened. Weeks ago. I glued the switch lever of the 5v-to-9v adaptor so the 9v can't be moved to 12v as I didn't want it to be exposed to 12v.

2 days ago. I tried to remove the glue because i want to set it to 12v to try a wifi router, but I ended up cutting a cooper in the PCB that connects the first terminal of the switch to another part of circuit.

Imagine a switch has 3 terminals. Terminal 1 & 2 makes it 9v. Terminal 2 and 3 makes it 12v. While the switch is in Terminal 1 & 2 making it 9v. I accidentally cut the cooper path to terminal 1. This means the switch momentary is neither 9v nor 12v. When I realized it and reconnected the cooper path. The device no longer works.

Do you have any idea what kind of design is the dc-dc converter (how many designs are there, I'm trying to google it but can't find the behavior) based on the youtube video above? When you turned it on (just turned on in the video), the LCD in the 5v-9v adaptor gets from 18v then down to 4.9v. The multimeter shows 15v to 5v when the output of the 5v-9v adaptor was connected. Note the dylos isn't connected to the above, of course because the output is connected to the multimeter. Thank you.

There is no consequence, just a remark.

By the way, based on the behavior of that regulator there is a high chance that your device would be able to work directly from USB - in case you would dare to disassemble and modify it

Here is the Dylos internal. I saw it shared by one user at amazon. The laser at the middle is aligned to the compartment below such that air is suck from the top to the bottom via the fan.

 

[Tom.G]

There are two basic types of voltage regulators...
... The second type is 'linear regulator' , which keeps only current as constant

Hmm... I always thought voltage regulators regulated voltage. If they keep current constant, say at 100mA for instance, wouldn't a 10,000Ω load resistor have 1,000 Volts across it? Rather impressive for a 9V supply.

 

[chirhone]

I've been trying to analyze this for a couple of days. This was what happened. Weeks ago. I glued the switch lever of the 5v-to-9v adaptor so the 9v can't be moved to 12v as I didn't want it to be exposed to 12v.

2 days ago. I tried to remove the glue because i want to set it to 12v to try a wifi router, but I ended up cutting a cooper in the PCB that connects the first terminal of the switch to another part of circuit.

Imagine a switch has 3 terminals. Terminal 1 & 2 makes it 9v. Terminal 2 and 3 makes it 12v. While the switch is in Terminal 1 & 2 making it 9v. I accidentally cut the cooper path to terminal 1. This means the switch momentary is neither 9v nor 12v. When I realized it and reconnected the cooper path. The device no longer works.

Do you have any idea what kind of design is the dc-dc converter (how many designs are there, I'm trying to google it but can't find the behavior) based on the youtube video above? When you turned it on (just turned on in the video), the LCD in the 5v-9v adaptor gets from 18v then down to 4.9v. The multimeter shows 15v to 5v when the output of the 5v-9v adaptor was connected. Note the dylos isn't connected to the above, of course because the output is connected to the multimeter. Thank you.

Here is the Dylos internal. I saw it shared by one user at amazon. The laser at the middle is aligned to the compartment below such that air is suck from the top to the bottom via the fan.

View attachment 257411

This is home wifi router. Many use the 5v to 9v adaptors to use the router it at car or coffee.

This was why I tried to use my 2nd working 5v to 9v adaptor and adjusted it to 12v when the cooper cut was made.

Here is what's so puzzling. The 5v to 12v adaptor is very useful in a lot of items. But how come they are all made in china. Doesn't the USA ever produce one? How come? I can't understand how the product is not useful in the United States.

Pending a USA made 5v to 12v adaptor. I connected the following battery pack with voltage and current meter.

I used the 6 pcs of partially discharged Energizer AA batteries and one ordinary battery. Just to test the usb volmeter & ammeter. I still need to buy 7 pcs of rechargeable NiMH and a fully automatic NiMH with negative delta capabilities. However. If I can buy a made in USA 5v to 9v adaptor. I may opt for the latter. So please advise on any product like it. Thank you.

 

[FRANCOIS SIMON]

This is not a battery charger. It is use to power various electronic pack such as kidtron (??) experimental board. As for battery meeting the requirement, they usually not rechargeable: Alkaline; Zinc-carbon, Lithium - 550mAh @ 9V.

 

[chirhone]

`I bought the latest NiMH charger that can individually measure the charge and stop charging when it is full (with negative delta something capability).

With my 7 AA rechargeable NiMH batteries I connected my USB ammeter/voltmeter.

I realized the USB ammeter/voltmeter is only calibrated for 5v. This is because when the voltage is 8.9v. The current is wrong. Using a multimeter. The current is 270mA instead of 200mA.

Do you know another device like the USB ammeter/volmeter that can accurately display the current?

What kind of circuit is it that can only show accurate current if it is 5v only?

At least the voltage is more or less accurate and I can predict if the rechargeble AA NiMH batteries are about to get discharged.

I ordered another 5v-to-9v adaptor to power my router and the unit if i can't get a mini ammeter/voltmeter with display. It doesn't have to be USB.

(for some background of what is pm2.5 and stuff, see this i just read today https://edition.cnn.com/2020/02/25/health/most-polluted-cities-india-pakistan-intl-hnk/index.html )

 

[Knowledge Seeker]

I have this adaptor. Do you know what kind of battery power pack that can output at least 9 volts 0.5 ampere?

View attachment 256320

This is the spec sheet:

https://www.jameco.com/Jameco/Products/ProdDS/163628.pdf

It's the adaptor included in the world's cheapest (and only) 2 channel pm2.5 monitor. A 16 channel unit costs about $25000. This only costs $260 but it should be plugged into wall. If I can get a battery pack, I can bring it anywhere like outside. The manufacturer doesn't have any battery pack for it. They recommended using 20 meter extension to bring it outside, but it's too long, and didn't want to comment what kind of battery pack available for it because they have none.

https://www.amazon.com/gp/product/B004AWEG0Y/?tag=pfamazon01-20

The simplest solution would be to put 2 li-ion 1 A/h in series and try your device at that lower voltage (8 volts fully charged and 7.2 volts nominal). As far as I can tell, this will work with almost all devices that need 9 V.