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

[chirhone]

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

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

 

[Rive]

I would just use some classic USB powerbank with 2A output, and buy something like these.
Be careful with the output connector and the voltage/current selection.

Ps.: how long operation time is needed?

 

[chirhone]

I would just use some classic USB powerbank with 2A output, and buy something like these.
Be careful with the output connector and the voltage/current selection.

Ps.: how long operation time is needed?

Minimal operation time like 15 mins. For small square 9 volt battery like this, what is usually the ampere rating?

https://www.amazon.com/dp/B00MH4QM1S/?tag=pfamazon01-20

 

[Rive]

Minimal operation time like 15 mins. For small square 9 volt battery like this, what is usually the ampere rating?

Third question on the product page there:

9V alkaline batteries are meant for low current applications. If subjected to high current they will suffer voltage depression (aka droop) below the typical target range needed for a 9V battery. In other words try to keep the current under 80mA unless you just don't care about voltage droop and even then, you might be better off with a 4 x AA battery and a boost circuit to achieve the target voltage if the current is much over 150mA.

If you want it work then you need far more beef in that battery. I still suggest a powerbank. They are dirt cheap these days, and with a step-up 'cable' it'll work just fine.

 

[chirhone]

Third question on the product page there:

If you want it work then you need far more beef in that battery. I still suggest a powerbank. They are dirt cheap these days, and with a step-up 'cable' it'll work just fine.

Here the owner used 7 pcs of 1.2v battery. Do you have rough idea how many milliamps it can output?



It's too large and bulky. I just want to buy a readily available one and not build it. I'll check out your powerbank and step-up cable.

 

[Rive]

It's too large and bulky.

And all those cables...
No idea how much current can it supply (and, more importantly: how long).

Part of the reason why I suggest the powerbank+step-up cable is, that all of that is made to be portable, so you can just plug it together and it'll work without any further hassle.

 

[chirhone]

And all those cables...
No idea how much current can it supply (and, more importantly: how long).

Part of the reason why I suggest the powerbank+step-up cable is, that all of that is made to be portable, so you can just plug it together and it'll work without any further hassle.

Which is more pure and safe dc, the 7 pieces of 1.2v rechargeable battery, or 5v power bank with 5 to 9 volts step up adaptor? And why?

 

[Rive]

... pure and safe ...

Those words just does not apply.

I would consider a direct 9V-ish battery pack only for some overly sensitive, certified measurement devices.
Yours is not this class, not even close.

In quality of the supplied power the powerbank+cable thing beats the usual wall plug adapter (like the one you linked before) by far.

There is no connection to any interference from the line and the usual ripple voltage is lower and at higher frequency (easy to filter). The original adapter likely has an ugly ripple at 120Hz, since based on the provided data it's nothing more than a transformer, a graetz and a capacitor - maybe with an irrelevant fuse. The most basic thing.
If your device can work with that then it likely has some internal power supply anyway.

 

[anorlunda]

Do you know what kind of battery power pack that can output at least 9 volts 0.5 ampere?

I don't understand your difficulty. It is trivial to do an internet search. For example "AA alkaline battery specifications" I did that and within seconds I found this.

http://www.evergreencpusa.com/alkaline-battery-specifications

Just choose a candidate battery model and look up its specifications.

 

[jrmichler]

Try one of these: https://www.adafruit.com/product/248. I don't know how good it is, but should know soon because one is on its way as of two days ago. And the price is hard to beat.

 

[Borek]

9 V is rather difficult to maintain with just a battery, I would go for 3S LiPo and step down converter.

 

[chirhone]

Those words just does not apply.

I would consider a direct 9V-ish battery pack only for some overly sensitive, certified measurement devices.
Yours is not this class, not even close.

In quality of the supplied power the powerbank+cable thing beats the usual wall plug adapter (like the one you linked before) by far.

There is no connection to any interference from the line and the usual ripple voltage is lower and at higher frequency (easy to filter). The original adapter likely has an ugly ripple at 120Hz, since based on the provided data it's nothing more than a transformer, a graetz and a capacitor - maybe with an irrelevant fuse. The most basic thing.
If your device can work with that then it likely has some internal power supply anyway.

Your confidence above makes me get this 5v to 9v adaptor plugged to powerbank and it works! Thanks. Now I can bring it anywhere and split hairs or dusts.

 

[chirhone]

Those words just does not apply.

I would consider a direct 9V-ish battery pack only for some overly sensitive, certified measurement devices.
Yours is not this class, not even close.

In quality of the supplied power the powerbank+cable thing beats the usual wall plug adapter (like the one you linked before) by far.

There is no connection to any interference from the line and the usual ripple voltage is lower and at higher frequency (easy to filter). The original adapter likely has an ugly ripple at 120Hz, since based on the provided data it's nothing more than a transformer, a graetz and a capacitor - maybe with an irrelevant fuse. The most basic thing.
If your device can work with that then it likely has some internal power supply anyway.

You said the original adapter likely has an ugly ripple at 120Hz. But is it not we only use 60Hz electricity? What effect does this 120Hz has on the output? Maybe harmonics?

I was googling for the output of an ac adaptor but couldn't find one for typical ac to dc adaptor (google wildcard "ac to dc adaptor vs powerbank oscilloscope output") Can you share how the output of an ac adaptor looks like in an oscilloscope compared to the powerbank bank 5v output?

I'm deciding whether to use the powerbank as permanent power supply to the air monitor unit even in home and not use the ac adaptor anymore (to avoid surges too).

Btw. I got interested in all this when my cousin gave my parents an air purifier as Christmas present last december. Before that time. I didn't even know an air purifier exist. It has a built-in pm2.5 monitor and I bought individual monitor just to confirm it's accuracy. But even the individual monitor made in china was not accurate. So got this model as it can individually measure 0.5 and 2.5 micron particles.

 

[chirhone]

Your confidence above makes me get this 5v to 9v adaptor plugged to powerbank and it works! Thanks. Now I can bring it anywhere and split hairs or dusts.

View attachment 256435

Guess what. The 5v to 9v adaptor died after using it for 15 minutes continuous. The LED displayed 4.8v instead of 9v and multimeter tested it as 5v instead of 9v. Yesterday I only used it for 5 minutes each time and not long. This is similar to a reviewer at amazon:



"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."

This is the internal components of it:

The Dylos is still working. I tested it with a PowerMeter using the original ac adaptor and powering it. I got:

The Dylos consumes 5.1W. I'd like to know the following:

How did the adaptor above come up with 500mA in the 9V output from the input of 120v and 8.5W?

Using the formula P= VI, and I=P/V = 8.5W/120 = 0.071mA

and since 120v/9v = 13.33 x 0.071mA = 0.950A or 950mA. How is this connected to the output of 500mA and 9v?

If the Dylos consumes 5.1W at the 120v ac adaptor side. What is the equivalent ampere in the output at 9v dc?I'm looking for a 5v to 9v adapter with bigger capacity. Also I guess the ic temperature increase and burnout because of the plastic casing of the unit and lack of heat sink of exhaust?

 

[Borek]

How did the adaptor above come up with 500mA in the 9V output from the input of 120v and 8.5W?

Using the formula P= VI, and I=P/V = 8.5W/120 = 0.071mA

and since 120v/9v = 13.33 x 0.071mA = 0.950A or 950mA. How is this connected to the output of 500mA and 9v?

No idea what you are doing here. What converter does it takes some power in and puts some power - different combination of voltage/current - out. What happens inside doesn't matter - plenty of ways of converting (yours was most likely some kind of https://en.wikipedia.org/wiki/Switched-mode_power_supply - even that means dozens of solutions). One thing we can be sure is that the output power is lower than the consumed power.

 

[Rive]

If the current is too large, it will generate too much heat and thus cause burnout.

Based on the wall plug adapter the maximal current draw on 9V should be below 0.5A. The specification of the USB thing says 1.1A on 9V, which is sufficiently higher than the consumption. You have sufficient reason to return the thing or request a replacement.

The Dylos consumes 5.1W.

Sorry, no. First: those power meters has really bad measurement error on the low end of the scale: second, you measuring the loss of the plug adapter too. The real consumption is likely lower.

By the way, 5.1W at 9V means 570mA. The USB thing still should be able to supply that => you should request a replacement.

Also I guess the ic temperature increase and burnout because of the plastic casing of the unit and lack of heat sink of exhaust?

Unless you insulated the thing with something, then these details are for the manufacturer/vendor. They are selling that thing enclosed: they should present the specs accordingly.

 

[chirhone]

Based on the wall plug adapter the maximal current draw on 9V should be below 0.5A. The specification of the USB thing says 1.1A on 9V, which is sufficiently higher than the consumption. You have sufficient reason to return the thing or request a replacement.Sorry, no. First: those power meters has really bad measurement error on the low end of the scale: second, you measuring the loss of the plug adapter too. The real consumption is likely lower.

By the way, 5.1W at 9V means 570mA. The USB thing still should be able to supply that => you should request a replacement.Unless you insulated the thing with something, then these details are for the manufacturer/vendor. They are selling that thing enclosed: they should present the specs accordingly.

I ordered 2 pcs of 5v to 9v adapter so have extra. I use the 2nd one only 5 mins each time
and it doesn't get busted.

I also got an 8 compartment battery holder and will use 6 pcs of AA battery of 1.5v each. But let's assume each AA battery only output 50 mA (instead of 100mA) for total of 300mA. If the device draw 500mA. What would happen to the battery or device?

 

[Tom.G]

AA Alkaline batteries have a 1 Amp-Hour capacity, although they drop to less than a Volt if discharged that far. So go for it! In fact I was thinking of suggesting that approach until I noticed you didn't want to build anything.

Cheers,
Tom

 

[Asymptotic]

I ordered 2 pcs of 5v to 9v adapter so have extra. I use the 2nd one only 5 mins each time
and it doesn't get busted.

I also got an 8 compartment battery holder and will use 6 pcs of AA battery of 1.5v each. But let's assume each AA battery only output 50 mA (instead of 100mA) for total of 300mA. If the device draw 500mA. What would happen to the battery or device?

Guess what. The 5v to 9v adaptor died after using it for 15 minutes continuous. The LED displayed 4.8v instead of 9v and multimeter tested it as 5v instead of 9v.

In post #14 you mention the Dylos DC1100 demands 5.1 watts. Is this value supplied in the user manual?

Does your multimeter have a current range?
If yes, what you could do is set it up for current operation, and measure what current the DC 1100 actually draws.

A cheap and dirty way to make the connection for testing is to insert a piece of thin cardboard, or plastic in between one pair of batteries in the battery holder, and insert the ammeter leads (taking care to use correct polarity) at the (-) side and (+) side of the cells so isolated.

Couldn't find power requirement specs at the Dylos website, but chanced upon a review that claims a current draw of 220 milliamps. http://www.air-purifier-power.com/dylosdc1100proreview.html

The 5.1 watt value doesn't quite add up. The Jameco model 100853 supply is rated for only 4.5 watts. It is an unregulated, transformer-based supply with a full wave diode bridge and capacitor per the specification sheet, and I'd expect it to run on the hot side, and output voltage to fall below the rated 9V if demand was actually 5.1 watts. See the spec sheets at:

https://www.jameco.com/z/DCU090050Z...r-Transformer-4-5-Watt-9VDC-500mA_100853.html

The 5-to-9 V converter you have claims an output of 1 amp (9V at 1A = 9 watts), and should have no problem powering the air particle tester.

One of the selections in Amazon's "Compare with other items" list costs another $3, but has a built-in voltmeter and ammeter that'd come in handy.

 

[chirhone]

In post #14 you mention the Dylos DC1100 demands 5.1 watts. Is this value supplied in the user manual?

Does your multimeter have a current range?
If yes, what you could do is set it up for current operation, and measure what current the DC 1100 actually draws.

A cheap and dirty way to make the connection for testing is to insert a piece of thin cardboard, or plastic in between one pair of batteries in the battery holder, and insert the ammeter leads (taking care to use correct polarity) at the (-) side and (+) side of the cells so isolated.

Couldn't find power requirement specs at the Dylos website, but chanced upon a review that claims a current draw of 220 milliamps. http://www.air-purifier-power.com/dylosdc1100proreview.html

The 5.1 watt value doesn't quite add up. The Jameco model 100853 supply is rated for only 4.5 watts. It is an unregulated, transformer-based supply with a full wave diode bridge and capacitor per the specification sheet, and I'd expect it to run on the hot side, and output voltage to fall below the rated 9V if demand was actually 5.1 watts. See the spec sheets at:

https://www.jameco.com/z/DCU090050Z...r-Transformer-4-5-Watt-9VDC-500mA_100853.html

The 5-to-9 V converter you have claims an output of 1 amp (9V at 1A = 9 watts), and should have no problem powering the air particle tester.

One of the selections in Amazon's "Compare with other items" list costs another $3, but has a built-in voltmeter and ammeter that'd come in handy.
View attachment 256636

Thanks for your information. Ill ponder on them.

Im testing my Energizer Max AA now. The voltage across one battery is 1.636 volts instead of 1.5V. But does the device see 1.5v or 1.636v? 6 pcs in series is 9.8 volts instead of 9 volts. So i wonder if the dylos will see 9.8v or 9 volts. But for any device with rating of exactly 9v, doesn't the AA battery at 1.636v instead of 1.5v is overvoltage when connected in series such as 6 pieces?

 

[chirhone]

In post #14 you mention the Dylos DC1100 demands 5.1 watts. Is this value supplied in the user manual?

Does your multimeter have a current range?
If yes, what you could do is set it up for current operation, and measure what current the DC 1100 actually draws.

A cheap and dirty way to make the connection for testing is to insert a piece of thin cardboard, or plastic in between one pair of batteries in the battery holder, and insert the ammeter leads (taking care to use correct polarity) at the (-) side and (+) side of the cells so isolated.

Couldn't find power requirement specs at the Dylos website, but chanced upon a review that claims a current draw of 220 milliamps. http://www.air-purifier-power.com/dylosdc1100proreview.html

The 5.1 watt value doesn't quite add up. The Jameco model 100853 supply is rated for only 4.5 watts. It is an unregulated, transformer-based supply with a full wave diode bridge and capacitor per the specification sheet, and I'd expect it to run on the hot side, and output voltage to fall below the rated 9V if demand was actually 5.1 watts. See the spec sheets at:

https://www.jameco.com/z/DCU090050Z...r-Transformer-4-5-Watt-9VDC-500mA_100853.html

The specs above mentioned:

"

• Input: 120 VAC @ 60 Hz, 8.5W
• Output: 9 VDC @ 0.5A, 4.5W"

In message #14, the photo showed 5.1W. Was it not the power for the input which has 8.5W limit? Why did you mentioned 4.5W?

I connected the 6 pcs of AA battery in battery holder with correct polarity. But total measured 9.8 volts instead of 9 volts. I am chicken to plug this to the Dylos. Maybe i should get an AA with measured voltage of 1.3v so 7 pcs is 9.1v just like in the youtube video at msg #5? What AA battery has 1.3v measured brand new?

The 5-to-9 V converter you have claims an output of 1 amp (9V at 1A = 9 watts), and should have no problem powering the air particle tester.

One of the selections in Amazon's "Compare with other items" list costs another $3, but has a built-in voltmeter and ammeter that'd come in handy.
View attachment 256636

 

[Asymptotic]

So i wonder if the dylos will see 9.8v or 9 volts.

If the battery pack open voltage is 9.8V, it will drop by some small amount (how much depends on how much load the Dylos provides) when it is powered up. Measure it.

Study the Jameco power supply specs.

Unloaded voltage for the test subject was 13.67 volts, and dropped to 8.95V when connected to a load that draws 500 mA. Rated voltage output at 500 mA is 9V +/-5%. 5% of 9V is 0.45, so fully loaded voltage could be from 8.55V to 9.45V and still be in spec.

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.

In message #14, the photo showed 5.1W. Was it not the power for the input which has 8.5W limit? Why did you mentioned 4.5W?

Like @Rive mentions in post #14, the 5.1 watts you measured is an indication what the power supply demands from the 120VAC source. The power supply has losses, and what the Dylos uses must be less than 5.1 watts.

I mentioned 4.5 watts because that's what the supply is rated. 4.5 watts at 9 volts DC is 500 mA at 9 VDC.

In the supply specs you'll see an AC current demand of 80 mA at 120V when operated fully loaded.
120V * 0.8A = 9.6 VA, and is higher than the rated AC input wattage of 8.5W. This is due to power factor, which would be about 0.885 (8.5W/9.6VA) fully loaded.

The following isn't quite true (because power factor typically drops at lower loads, the transformer draws AC current even at no load, and unregulated DC voltage rises as load decreases), but if you've measured 5.1 watts on a supply rated 8.5 watts fully loaded, it is approximately 60% loaded, or delivering no more than 500 ma * 60%, or less than 300 mA DC.

The interplay of these factors is why it is necessary to measure load voltage and current to know what's going on.

 

[chirhone]

If the battery pack open voltage is 9.8V, it will drop by some small amount (how much depends on how much load the Dylos provides) when it is powered up. Measure it.

Study the Jameco power supply specs.

View attachment 256676

Unloaded voltage for the test subject was 13.67 volts, and dropped to 8.95V when connected to a load that draws 500 mA. Rated voltage output at 500 mA is 9V +/-5%. 5% of 9V is 0.45, so fully loaded voltage could be from 8.55V to 9.45V and still be in spec.

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.

View attachment 256681Like @Rive mentions in post #14, the 5.1 watts you measured is an indication what the power supply demands from the 120VAC source. The power supply has losses, and what the Dylos uses must be less than 5.1 watts.

I mentioned 4.5 watts because that's what the supply is rated. 4.5 watts at 9 volts DC is 500 mA at 9 VDC.
View attachment 256680

In the supply specs you'll see an AC current demand of 80 mA at 120V when operated fully loaded.
120V * 0.8A = 9.6 VA, and is higher than the rated AC input wattage of 8.5W. This is due to power factor, which would be about 0.885 (8.5W/9.6VA) fully loaded.

The following isn't quite true (because power factor typically drops at lower loads, the transformer draws AC current even at no load, and unregulated DC voltage rises as load decreases), but if you've measured 5.1 watts on a supply rated 8.5 watts fully loaded, it is approximately 60% loaded, or delivering no more than 500 ma * 60%, or less than 300 mA DC.

The interplay of these factors is why it is necessary to measure load voltage and current to know what's going on.

Thanks for the insight. I measured the voltage when it was turned on and it read 9.05 volts instead of 9.8v. Now i can finally focused on measuring.

Initially i ordered the Laser Egg which is based on the Plantower 3003 but Amazon didnt send it for days so i changed my mind and got the Dylos instead. These are reviewed in:

https://smartairfilters.com/en/blog/how-accurate-are-common-particle-counters-comparison-test/

After some reading i found out a cheaper product from china ($40) has the more latest Plantower 5003:

https://www.lazada.com.ph/products/28-car-pm25-detector-tester-meter-pm10-pm10-aqi-particle-air-quality-monitor-home-gas-thermometer-i312766766-s631198431.html?dsource=share&laz_share_info=22287687_5_100_500042076257_15363983_null&laz_token=11c153896d8eb2858ce86dff36d10e5d

Now running them together.

This is the right one in another location shown so you can see the labels:

Some puzzles. The PM 10 in the right device is higher than its PM 2.5 while in the Dylos. The bigger particles (the right number on the display) are lesser. Which is the more accurate one? Please check these sensors data.

https://aqicn.org/sensor/pms5003-7003/

http://billpentz.com/woodworking/cyclone/dylos.cfm

 

[Asymptotic]

Thanks for the insight. I measured the voltage when it was turned on and it read 9.05 volts instead of 9.8v. Now i can finally focused on measuring.

Initially i ordered the Laser Egg which is based on the Plantower 3003 but Amazon didnt send it for days so i changed my mind and got the Dylos instead. These are reviewed in:

https://smartairfilters.com/en/blog/how-accurate-are-common-particle-counters-comparison-test/

After some reading i found out a cheaper product from china ($40) has the more latest Plantower 5003:

https://www.lazada.com.ph/products/28-car-pm25-detector-tester-meter-pm10-pm10-aqi-particle-air-quality-monitor-home-gas-thermometer-i312766766-s631198431.html?dsource=share&laz_share_info=22287687_5_100_500042076257_15363983_null&laz_token=11c153896d8eb2858ce86dff36d10e5d

Now running them together.

View attachment 256684

This is the right one in another location shown so you can see the labels:

View attachment 256687

Some puzzles. The PM 10 in the right device is higher than its PM 2.5 while in the Dylos. The bigger particles (the right number on the display) are lesser. Which is the more accurate one? Please check these sensors data.

https://aqicn.org/sensor/pms5003-7003/

http://billpentz.com/woodworking/cyclone/dylos.cfm

I know nothing of the minutiae involved in air particle measurements, and cannot advise on questions of accuracy comparisons.

 

[chirhone]

I know nothing of the minutiae involved in air particle measurements, and cannot advise on questions of accuracy comparisons.

Oh, found the answer at https://www.researchgate.net/publication/320555036_Particle_Distribution_Dependent_Inaccuracy_of_the_Plantower_PMS5003_low-cost_PM-sensor
(this proves the $25000 Grim unit is more accurate than the $40 Pms 5003 in the Pm10 size):

"From these measurements, we conclude that the Plantower PMS5003 is able to measure PM2.5 more
or less correct. With a suitable calibration, it may be calibrated to a reference instrument such as the
Grimm. The PMS5003 however is not able at all to directly measure large particles > 3um. When it
outputs PM10 values, these are pure estimates based on particle sizes it can measure (d < 3um) and under the assumption of a broadly distributed particle spectrum. When the particle spectrum is not containing PM mass contributions at sizes < 3um, the PMS5003 erroneously outputs PM10 values close to zero. It therefore behaves similar to the SDS011 low-cost PM-sensor from Nova Fitness.
These findings intensify the strong suspicion that low-cost PM-sensors, mainly produced by Chinese
manufacturers and designated as PM2.5 sensors but also reporting PM10 values, aren’t able to measure particles larger than 3um and therefore solely extrapolate on PM10 values assuming a certain particle distribution"

Here is a good interactive site about pm2.5 https://www.nytimes.com/interactive/2019/12/02/climate/air-pollution-compare-ar-ul.html

Im aiming for below 5 ug/m^3 in my house. In New Delhi, it reaches 900 ug/m^3 at times. Thanks to all for the battery tips.

 

[chirhone]

If the battery pack open voltage is 9.8V, it will drop by some small amount (how much depends on how much load the Dylos provides) when it is powered up. Measure it.

Study the Jameco power supply specs.

View attachment 256676

Unloaded voltage for the test subject was 13.67 volts, and dropped to 8.95V when connected to a load that draws 500 mA. Rated voltage output at 500 mA is 9V +/-5%. 5% of 9V is 0.45, so fully loaded voltage could be from 8.55V to 9.45V and still be in spec.

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.

View attachment 256681Like @Rive mentions in post #14, the 5.1 watts you measured is an indication what the power supply demands from the 120VAC source. The power supply has losses, and what the Dylos uses must be less than 5.1 watts.

I mentioned 4.5 watts because that's what the supply is rated. 4.5 watts at 9 volts DC is 500 mA at 9 VDC.
View attachment 256680

In the supply specs you'll see an AC current demand of 80 mA at 120V when operated fully loaded.
120V * 0.8A = 9.6 VA, and is higher than the rated AC input wattage of 8.5W. This is due to power factor, which would be about 0.885 (8.5W/9.6VA) fully loaded.

The following isn't quite true (because power factor typically drops at lower loads, the transformer draws AC current even at no load, and unregulated DC voltage rises as load decreases), but if you've measured 5.1 watts on a supply rated 8.5 watts fully loaded, it is approximately 60% loaded, or delivering no more than 500 ma * 60%, or less than 300 mA DC.

The interplay of these factors is why it is necessary to measure load voltage and current to know what's going on.

For an AA battery. Its 1.5 or 1.6 volts new. When its discharged 50%. Its only 1.25 volts and getting down to 1.15 volts and below. After days of using. I measured my energizer as 1.25v already and 6pcs produced 7.5 volts. But actual draw in the Dylos is 6.84 volts as shown in the picture. Also by blocking the terminals with cardboard and testing the current in series with the 6 battery. Its about 270mA (no extra hands to take pic) .

Lets talk for example any tv remote control unit. Its run with 2 AA battery for total of 3 volts..but at 50% discharge. Its 2.5v already. For all consumer electronics, they take into account this discharging so the range of voltage is really 2 to 3 volts for the tv remote control? For the Dylos. Its maybe 6 volt to 10 volt range? What common IC in consumer circuit serve as voltage regulator or is does all IC do that able to take range of voltages? Thank you.

 

[Tom.G]

Please ignore this post, I got it confused with a different thread and posted to the wrong one! [/color] Sorry.

Its about 270mA

Are you SURE you have the polarity correct?
That number (270mA) sure looks like either polarity is reversed or the meter is defective (or on the wrong range).

Try measuring the current drawn with the 23A battery. At that current the 23A would last less than 10 minutes.

 

[chirhone]

Are you SURE you have the polarity correct?
That number (270mA) sure looks like either polarity is reversed or the meter is defective (or on the wrong range).

Try measuring the current drawn with the 23A battery. At that current the 23A would last less than 10 minutes.

See msg # 19 by Asymptotic. He said:

"
A cheap and dirty way to make the connection for testing is to insert a piece of thin cardboard, or plastic in between one pair of batteries in the battery holder, and insert the ammeter leads (taking care to use correct polarity) at the (-) side and (+) side of the cells so isolated.

Couldn't find power requirement specs at the Dylos website, but chanced upon a review that claims a current draw of 220 milliamps. http://www.air-purifier-power.com/dylosdc1100proreview.html"

The review measured 220milliamps. I measured 270milliamps. Its DC. I won't measure AC because DC is much safer. Result was close. Why is 270milliamps wrong?

 

[Tom.G]

Oops! Post #27 retracted.

 

[Asymptotic]

For an AA battery. Its 1.5 or 1.6 volts new. When its discharged 50%. Its only 1.25 volts and getting down to 1.15 volts and below. After days of using. I measured my energizer as 1.25v already and 6pcs produced 7.5 volts. But actual draw in the Dylos is 6.84 volts as shown in the picture. Also by blocking the terminals with cardboard and testing the current in series with the 6 battery. Its about 270mA (no extra hands to take pic) .

View attachment 256848

Lets talk for example any tv remote control unit. Its run with 2 AA battery for total of 3 volts..but at 50% discharge. Its 2.5v already. For all consumer electronics, they take into account this discharging so the range of voltage is really 2 to 3 volts for the tv remote control? For the Dylos. Its maybe 6 volt to 10 volt range? What common IC in consumer circuit serve as voltage regulator or is does all IC do that able to take range of voltages? Thank you.

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

But actual draw in the Dylos is 6.84 volts as shown in the picture. Also by blocking the terminals with cardboard and testing the current in series with the 6 battery. Its about 270mA (no extra hands to take pic).

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.

for example any tv remote control unit. Its run with 2 AA battery for total of 3 volts..but at 50% discharge. Its 2.5v already. For all consumer electronics, they take into account this discharging so the range of voltage is really 2 to 3 volts for the tv remote control? For the Dylos. Its maybe 6 volt to 10 volt range? What common IC in consumer circuit serve as voltage regulator or is does all IC do that able to take range of voltages?

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.

 

[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?