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

[chirhone]

Read about battery tech at the Battery University. This article delves into battery run-time. Pay attention to the differences between low current draw and high current draw devices.
https://batteryuniversity.com/learn/article/bu_503_how_to_calculate_battery_runtime

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.

View attachment 256850

There aren't any hard and fast rules that are applied to "all consumer electronics". It comes down to each individual device - operational goals, costs, and what engineering trade-offs are required.

A TV remote design that required battery replacement every 3 months would not be looked favorably upon, but replacing battery cells after 2.5 years of average usage versus every 5 years might not even show up on a reviewer's radar.

For instance, I have an electronic home heating thermostat that must have extremely low current draw inasmuch it only began blinking a "lo battery" warning in 2019, but a pair of Energizer Titanium X91-LR6, AA cells powering it had "use by 2007" expiration dates.

When i measured the amperage. I saw -270ma. I must have used wrong polarity but this won't change the value, right? I don't want to try again to avoid any unforseened result. The Dylos has warrantee for only 3 months. Some amazon user reported it as out of order after just 10 months. Maybe surges fry the circuit. If ill buy researcheable 1.2v NiMh batteries to avoid AC surges. I may need 1.2vx7= 8.4v. I think this is better than 1.2v x 8 = 9.6v. I can't find 1.3v rechargeable anymore which is closer to 9v. 1.3vx7 = 9.1v which is in the youtube battery pack feed i shared in earlier message. Do you happen to know any 1.3v NiMh? Any ideas why they lowered the NiCd 1.3v to NiMh 1.2v?

 

[chirhone]

When i measured the amperage. I saw -270ma. I must have used wrong polarity but this won't change the value, right? I don't want to try again to avoid any unforseened result. The Dylos has warrantee for only 3 months. Some amazon user reported it as out of order after just 10 months. Maybe surges fry the circuit. If ill buy researcheable 1.2v NiMh batteries to avoid AC surges. I may need 1.2vx7= 8.4v. I think this is better than 1.2v x 8 = 9.6v. I can't find 1.3v rechargeable anymore which is closer to 9v. 1.3vx7 = 9.1v which is in the youtube battery pack feed i shared in earlier message. Do you happen to know any 1.3v NiMh? Any ideas why they lowered the NiCd 1.3v to NiMh 1.2v?

I read the reason for the 1.2v voltage is the chemistry (this is 100% accurate?):

https://core-electronics.com.au/tutorials/our-tips-for-nimh-batteries.html

"The idea behind this back of the envelope calculation is that the voltage of the battery itself comes from the chemical potential energy difference between the electrodes within. That means each NiMH battery cell will have that voltage rating of 1.2V no matter the physical size of the cell. What the physical size of the cell does indicate is the capacity of the battery. In general, the bigger the cell, the more mAh your battery will have."

Disadvantage of NiMH batteries is they can diacharge permanently if not used often. After familiaring with the pm2.5 particle counts in different locations and able to feel it by seeing the haze in the air. I will get tired of it. And i don't have other use for rechargeable AAs. I have old NiCd chargers. Perhaps i can still get brand new NICd?

"NiMH batteries do have a couple of flaws, mainly that they self-discharge. When the battery is not in use it will slowly deplete its charge and if it's left long enough your batteries can be permanently damaged. A rough estimation of the depletion in a NiMH battery is 20% of the battery level will deplete in the first 24 hours after charging, with a further 10% depleting per 30 days thereafter."

Perhaps usb powerbanks are the best after all. But the china made 5v to 9v adapters can't even handle 270mA (i assume dc amperage polarity in multimeter can be reversed without affecting the magnitude, but Tom mentioned something about reversal of polarity can affect magnitude, in what way?).

 

[Asymptotic]

When i measured the amperage. I saw -270ma. I must have used wrong polarity but this won't change the value, right? I don't want to try again to avoid any unforseened result. The Dylos has warrantee for only 3 months. Some amazon user reported it as out of order after just 10 months. Maybe surges fry the circuit. If ill buy researcheable 1.2v NiMh batteries to avoid AC surges. I may need 1.2vx7= 8.4v. I think this is better than 1.2v x 8 = 9.6v. I can't find 1.3v rechargeable anymore which is closer to 9v. 1.3vx7 = 9.1v which is in the youtube battery pack feed i shared in earlier message. Do you happen to know any 1.3v NiMh? Any ideas why they lowered the NiCd 1.3v to NiMh 1.2v?

Yes, the ammeter leads must have been reversed. No, it won't change the value from 270 mA.

Cell voltage is in large part determined by materials used for the electrodes and electrolyte, and SoC (State of Charge; how charged it is). Regarding nomenclature, although it is common practice to call an individual cell (a single AA cell, for example) a 'battery', in actuality a battery is a collection of cells. For example, a typical 12V lead-acid automotive battery is six 2.2V cells connected in series in a single case, and delivers 13.2V when fully charged.

https://en.wikipedia.org/wiki/Comparison_of_commercial_battery_types
https://batteryuniversity.com/learn/article/confusion_with_voltages

I may need 1.2vx7= 8.4v. I think this is better than 1.2v x 8 = 9.6v.

Why do you think this?

Have you measured the voltage of your Jameco power supply? Based on the specs, it ought to be just a bit more than 11 VDC with 270 mA loading. If this DC supply is the one provided by Dylos then the particle counter was designed to operate with at least this high a supply voltage.

What is the minimum supply voltage that allows satisfactory operation? You'll get a reasonable idea by simply continuing to operate it from the AA battery pack while monitoring supply voltage as the cells discharge.

 

[chirhone]

Yes, the ammeter leads must have been reversed. No, it won't change the value from 270 mA.

Cell voltage is in large part determined by materials used for the electrodes and electrolyte, and SoC (State of Charge; how charged it is). Regarding nomenclature, although it is common practice to call an individual cell (a single AA cell, for example) a 'battery', in actuality a battery is a collection of cells. For example, a typical 12V lead-acid automotive battery is six 2.2V cells connected in series in a single case, and delivers 13.2V when fully charged.

https://en.wikipedia.org/wiki/Comparison_of_commercial_battery_types
https://batteryuniversity.com/learn/article/confusion_with_voltages
Why do you think this?

Have you measured the voltage of your Jameco power supply? Based on the specs, it ought to be just a bit more than 11 VDC with 270 mA loading. If this DC supply is the one provided by Dylos then the particle counter was designed to operate with at least this high a supply voltage.

What is the minimum supply voltage that allows satisfactory operation? You'll get a reasonable idea by simply continuing to operate it from the AA battery pack while monitoring supply voltage as the cells discharge.

Thanks for all the information. I was about to edit my previous message just as you replied. My last question is the following. As the voltage of the 6 pcs AA gets lowered, does the ampere gets higher? For example. As the voltage gets down to 6.4v from 6.8v and unit still working. Does the ampere of 270ma gets higher to say 290ma to maintain certain fixed power (i didnt compute bec this page will reload if i multitask to calculator)? Does component takes certain wattage and the voltage and ampere have to adjust to maintain the wattage? Or does electronic components dependent on fixed ampere only (like 270mA for all voltage changes)?

I scratched the unit body so can't return it anymore. And don't want to constantly test for amperage lest it may malfunction.

Thanks so much for this last question.

 

[Asymptotic]

Thanks for all the information. I was about to edit my previous message just as you replied. My last question is the following. As the voltage of the 6 pcs AA gets lowered, does the ampere gets higher? For example. As the voltage gets down to 6.4v from 6.8v and unit still working. Does the ampere of 270ma gets higher to say 290ma to maintain certain fixed power (i didnt compute bec this page will reload if i multitask to calculator)? Does component takes certain wattage and the voltage and ampere have to adjust to maintain the wattage? Or does electronic components dependent on fixed ampere only (like 270mA for all voltage changes)?

I scratched the unit body so can't return it anymore. And don't want to constantly test for amperage lest it may malfunction.

Thanks so much for this last question.

The only way to know for certain is to measure the two quantities of interest - voltage and current - and observe how they behave. That said, I'd expect current to fall off as voltage decreases, and hence power as well.

Do you know anyone who has a variable voltage DC supply? If I had a gizmo like the Dylos air particle tester and wanted to know these things, I'd connect a voltmeter across the power plug, an ammeter in series, and measure and record both quantities while lowering voltage in 0.1V increments until the device stopped working.

 

[chirhone]

The only way to know for certain is to measure the two quantities of interest - voltage and current - and observe how they behave. That said, I'd expect current to fall off as voltage decreases, and hence power as well.

Do you know anyone who has a variable voltage DC supply? If I had a gizmo like the Dylos air particle tester and wanted to know these things, I'd connect a voltmeter across the power plug, an ammeter in series, and measure and record both quantities while lowering voltage in 0.1V increments until the device stopped working.

I may do it after thoroughly familar with the difference between haze, mist, fog, etc. and how to estimate pm2.5 levels by eyes only. Whats complicating it all is water vapor can increase values in those sensors, even by twice!

Concerning constant power or ampere draw. I thought the answer was clear.

Can you give examples of devices that draw constant power (wattage) by increasing amperage when voltage goes down? And devices that draw constant amperage even when voltage goes down? Thank you.

 

[Asymptotic]

Can you give examples of devices that draw constant power (wattage) by increasing amperage when voltage goes down? And devices that draw constant amperage even when voltage goes down? Thank you.

Within limits, yes. Provided that load power is constant (and speed setpoint isn't pegged at 100%), a DC motor drive in speed feedback will draw more line current as line voltage is reduced, and visa versa. No devices where line current remains constant as line voltage varies are coming to mind.

Your Dylos tester uses an unregulated DC supply. It's output voltage can range from about 13.7 VDC unloaded to 9 VDC at 500 mA load, and will be about 11 VDC when delivering 270 mA.

There must be one or more regulators within the Dylos unit, but I've been unable to find a schematic diagram for it so their nature can only be inferred. It has a fan to move air through the test chamber, a laser light source, a computer (and possibly other voltage or current sensitive components such as analog-to-digital converters).

I'd expect fan voltage is regulated (because speed variations would change the air volume, and hence the number of particles passing through the test chamber), the computer needs one or more regulated voltage supplies, and the laser diode will almost certainly have a current regulator.

None of these details are known to us, but what we can do it treat the Dylos as a "black box", measure applied voltage and current draw, and observe what the unit does. For example, it probably demands slightly more current from the supply when the number of characters increase on the LCD display. Does the fan and laser run all the time, or are they switched on only when a particle measurement is commanded? If the latter, it wouldn't surprise me if (as battery terminal voltage decreases too far) the tester would appear to be working fine, but operate strangely when the particle measurement is made.

 

[chirhone]

Within limits, yes. Provided that load power is constant (and speed setpoint isn't pegged at 100%), a DC motor drive in speed feedback will draw more line current as line voltage is reduced, and visa versa. No devices where line current remains constant as line voltage varies are coming to mind.

Your Dylos tester uses an unregulated DC supply. It's output voltage can range from about 13.7 VDC unloaded to 9 VDC at 500 mA load, and will be about 11 VDC when delivering 270 mA.

There must be one or more regulators within the Dylos unit, but I've been unable to find a schematic diagram for it so their nature can only be inferred. It has a fan to move air through the test chamber, a laser light source, a computer (and possibly other voltage or current sensitive components such as analog-to-digital converters).

I'd expect fan voltage is regulated (because speed variations would change the air volume, and hence the number of particles passing through the test chamber), the computer needs one or more regulated voltage supplies, and the laser diode will almost certainly have a current regulator.

None of these details are known to us, but what we can do it treat the Dylos as a "black box", measure applied voltage and current draw, and observe what the unit does. For example, it probably demands slightly more current from the supply when the number of characters increase on the LCD display. Does the fan and laser run all the time, or are they switched on only when a particle measurement is commanded? If the latter, it wouldn't surprise me if (as battery terminal voltage decreases too far) the tester would appear to be working fine, but operate strangely when the particle measurement is made.

Just before the battery went dead..i noticed the values of the left goes to 3 times and the right 20 times.

Switching to adaptor. The values were much less and the same as prior to the burst of numbers.

The voltage of the 6 pcs AA direct without the unit is 6.8 volts ( so each battery about 1.1 v)

Can you give examples of other devices that temporarily malfunction when voltage is very low.

Is the following another way to measure the amperage. I don't use cardboard but connect the negative lead of the multimeter to the negative of the battery pack and the positive of the multimeter to the the negative of the plug? The positive of the battery pack is connected to the positive of the plug.

But when i plugged the Dylos. It boots with dim lcd and then go off. It means the voltage is not sufficient anymore. The ampere seems to read 300ma plus quickly instead of 270ma. Next time when i get new AA battery. Ill try to get the ampere to see if it would be below 270ma. Meanwhile I am using the 5v to 9v usb adapter and powerbank for quick reading.

 

[Rive]

Meanwhile I am using the 5v to 9v usb adapter and powerbank for quick reading.

That broken adapter is bugging me for a while already. Could you please tell me the capacity of that powerbank?

If the product is just half decent, then there is no way it could break down so easily - unless it's about the dropping input voltage. Your device requires only (270mA @ 9V) 2.5W: to exceed the maximal input current of the adapter (supposed to be around 2.5A at least) requires the voltage on the powerbank to drop around 1V.
Still not a good explanation, since the protection should kick in way faster than that, but maybe an underdesigned, protectionless powerbank is a more probable...

 

[chirhone]

That broken adapter is bugging me for a while already. Could you please tell me the capacity of that powerbank?

If the product is just half decent, then there is no way it could break down so easily - unless it's about the dropping input voltage. Your device requires only (270mA @ 9V) 2.5W: to exceed the maximal input current of the adapter (supposed to be around 2.5A at least) requires the voltage on the powerbank to drop around 1V.
Still not a good explanation, since the protection should kick in way faster than that, but maybe an underdesigned, protectionless powerbank is a more probable...

Hmm. Well. To summarize. I have 2 pcs of the 5v to 9v adapter. 1st got busted after 15 mins of use. So i only use it for 5 mins now when outside. Here is my powerbank.

Here is my power meter without anything plugged in:

Here is with the jameco adaptor not plugged into the unit (1.3watts):

When i plugged it to the dylos but not turning on the unit, its consuming some power 2.3 watts):

Here is when i turned on the unit 4.9watts):

 

[Rive]

Hmm. Well. To summarize. I have 2 pcs of the 5v to 9v adapter. 1st got busted after 15 mins of use. So i only use it for 5 mins now when outside. Here is my powerbank.

9Ah should be able to supply that thing for hours. Was it plugged in the 2A output?

 

[chirhone]

9Ah should be able to supply that thing for hours. Was it plugged in the 2A output?

Yup. Maybe the 5v to 9v usb adapter is only rated for 200mA.

 

[Rive]

Not likely. The page you linked before has mentions for far higher loads, managed successfully.

 

[chirhone]

Not likely. The page you linked before has mentions for far higher loads, managed successfully.

My 1st adaptor shows only 4.8V just like in the following amazon comment:

"Woked fine at first, but after a couple of hours running about a strip of about 20 white 12V LEDs (very small load), it now outputs only 4.8V to the barrel connector, and moving the switch doesn't change anything."

The seller replied: "If the current is too large, it will generate too much heat and thus cause burnout."

Anyway i tried to connect the multimeter to measure the ampere in the powerbank and 5v to 9v adapter. Should one put it in series in the positive or negative of the 5v to 9v adaptor? I put it in the negative of it and the unit is not booting up. Only when connected without multimeter does it power up. I used the right connections or lead taps.

 

[chirhone]

My 1st adaptor shows only 4.8V just like in the following amazon comment:

"Woked fine at first, but after a couple of hours running about a strip of about 20 white 12V LEDs (very small load), it now outputs only 4.8V to the barrel connector, and moving the switch doesn't change anything."

The seller replied: "If the current is too large, it will generate too much heat and thus cause burnout."

Anyway i tried to connect the multimeter to measure the ampere in the powerbank and 5v to 9v adapter. Should one put it in series in the positive or negative of the 5v to 9v adaptor? I put it in the negative of it and the unit is not booting up. Only when connected without multimeter does it power up. I used the right connections or lead taps.

I got this new toy to test your 5v to 9v adapter:

https://www.amazon.com/gp/product/B076GPFBCQ/?tag=pfamazon01-20

The multimeter as ampere meter should work even if you use negative or positive terminal of the battery in series, is it not? Perhaps something wrong with my multimeter.

 

[Asymptotic]

Can you give examples of other devices that temporarily malfunction when voltage is very low.

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.

Is the following another way to measure the amperage. I don't use cardboard but connect the negative lead of the multimeter to the negative of the battery pack and the positive of the multimeter to the the negative of the plug? The positive of the battery pack is connected to the positive of the plug.

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.

Just before the battery went dead..i noticed the values of the left goes to 3 times and the right 20 times.

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 when i plugged the Dylos. It boots with dim lcd and then go off. It means the voltage is not sufficient anymore. The ampere seems to read 300ma plus quickly instead of 270ma. Next time when i get new AA battery. Ill try to get the ampere to see if it would be below 270ma. Meanwhile I am using the 5v to 9v usb adapter and powerbank for quick reading.

The voltage of the 6 pcs AA direct without the unit is 6.8 volts ( so each battery about 1.1 v)

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.

 

[Rive]

Perhaps something wrong with my multimeter.

Current measurement at low range is usually through a fuse. Check the value/status - it is likely you smoked it.

Based on the story so far I still insist you request a replacement for the adapter. They said it should be OK for 1.1A at 9V: and there was no time limit mentioned.

 

[Asymptotic]

Is the following another way to measure the amperage. I don't use cardboard but connect the negative lead of the multimeter to the negative of the battery pack and the positive of the multimeter to the the negative of the plug? The positive of the battery pack is connected to the positive of the plug.

View attachment 256920

The meter leads are reversed, and account for your negative current reading.

The (+) ammeter lead should be connected to the (-) lead from the battery pack, and (-) ammeter lead connected to the load.

 

[chirhone]

Current measurement at low range is usually through a fuse. Check the value/status - it is likely you smoked it.

You are right. The multimeter fuse is blown. I think something in the 5v to 9v adapter cause a surge to blow the fuse when the dylos was turned on. Anyway do you know a 5.5mm plug version of this so i can insert it between the jameco plug and dylos?

Based on the story so far I still insist you request a replacement for the adapter. They said it should be OK for 1.1A at 9V: and there was no time limit mentioned.

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

 

[chirhone]

You are right. The multimeter fuse is blown. I think something in the 5v to 9v adapter cause a surge to blow the fuse when the dylos was turned on. Anyway do you know a 5.5mm plug version of this so i can insert it between the jameco plug and dylos?

View attachment 256953
In a few sentences and from memory, can you share how exactly 5v is made to convert to 9v? Thanks.

I replaced the fuse in the multimeter and measured the amperage of the 5v to 9v adapter. It measured 269mA. The powerbank itself seems to maintain 5v pretty well even when 20% level only. Something AA batteries can't do?

 

[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.