Check what is wrong with the handheld vacuum cleaner

[TheRosssiiii]

This handheld vacuum cleaner broke, I would be curious to fix it but I don't know how to do it...looks very simple inside and for it i would be curious to try:

https://imgur.com/QPWJrXe

https://imgur.com/SSKCZ7r

https://imgur.com/rCTC4rm

it was nearly fully charged (2/3 on it) but during the work time, it went off suddenly...I thought I heard a little "poof" before the silence.

What i could check ?

-battery ?
-boards

i have a multimeter but i don't know how to conduce the tests.

 

[Harry Bloomfield]

Set the meter on 20v dc and report the measurement between the two terminals of the battery. Switch it on, measure and report the voltage across the motor terminals.

 

[TheRosssiiii]

Set the meter on 20v dc and report the measurement between the two terminals of the battery. Switch it on, measure and report the voltage across the motor terminals.

i have took the measure on the +/- of the battery:

https://imgur.com/b92Zi2E

and uscually is 0v indicated on the multimeter.
While if i plug the battery, it reach 5v that is very low value i think considering that the light indicator show 3/3 lights and mean that the bettery should be fully charged.

i did resistence test on +/- pads of the motors and it start to sound, so there is continuity on the motor

https://imgur.com/e3ROreT

 

[FrodoOne]

On the board where it is marked "CHARGER" you have two terminals where a pair of White wires are incoming from the charger (I presume).
Next to these are the two terminals for the Battery and then there are two terminals for the White wires to the Motor.
All these terminals are covered with a soft plastic substance to "protect" them (for some reason) but this "stuff" can be removed by pulling it off,
so you did not need to cut open the plastic on the battery to access the terminals.

Set the meter on 20v dc and report the measurement between the two terminals of the battery. Switch it on, measure and report the voltage across the motor terminals.

If you can expose those terminals you could measure the voltage coming from the charger - during charging,
the voltage across the battery - during charging,
the voltage across the battery - after charging,
the voltage across the battery when the device is switched ON and
the voltage applied to the motor.

(Clip leads attached to your multi-meter would assist with such measurements - and you should equip yourself with these.
The more expensive "Hook" test leads shown in the following video are often better for this purpose but "Alligator" Clip Leads can have their uses.
However, be warned that cheap Alligator Clip Leads are often badly made and should always be tested before use.
[They should have zero resistance from clip to clip.]
It is often necessary to undo the "crimp" connection of these leads and solder the connection before use, to get them working properly.)

You wrote "While if i plug the battery, it reach 5v that is very low value i think considering that the light indicator show 3/3 lights and mean that the bettery should be fully charged." but we don't know if this is during charging, after charging or while attempting to supply the "Load", by switching ON.
(I gather that the ON-OFF switch is part of the "Board" shown in your photos.)
Are there any markings on the Battery to indicate what type of battery it is ?
(It appears to contain only 2 Cells. Is that correct?)
5 V could be low if each cell is a Lithium Polymer (“LiPo”) type.

[In future, you should use the "Insert Image" tab (The "rectangle" five places from the Right at the top) or "CTRL + P" to post photos.}

 

[TheRosssiiii]

On the board where it is marked "CHARGER" you have two terminals where a pair of White wires are incoming from the charger (I presume).
Next to these are the two terminals for the Battery and then there are two terminals for the White wires to the Motor.
All these terminals are covered with a soft plastic substance to "protect" them (for some reason) but this "stuff" can be removed by pulling it off,
so you did not need to cut open the plastic on the battery to access the terminals.

If you can expose those terminals you could measure the voltage coming from the charger - during charging,
the voltage across the battery - during charging,
the voltage across the battery - after charging,
the voltage across the battery when the device is switched ON and
the voltage applied to the motor.

(Clip leads attached to your multi-meter would assist with such measurements - and you should equip yourself with these.
The more expensive "Hook" test leads shown in the following video are often better for this purpose but "Alligator" Clip Leads can have their uses.
However, be warned that cheap Alligator Clip Leads are often badly made and should always be tested before use.
[They should have zero resistance from clip to clip.]
It is often necessary to undo the "crimp" connection of these leads and solder the connection before use, to get them working properly.)

You wrote "While if i plug the battery, it reach 5v that is very low value i think considering that the light indicator show 3/3 lights and mean that the bettery should be fully charged." but we don't know if this is during charging, after charging or while attempting to supply the "Load", by switching ON.
(I gather that the ON-OFF switch is part of the "Board" shown in your photos.)
Are there any markings on the Battery to indicate what type of battery it is ?
(It appears to contain only 2 Cells. Is that correct?)
5 V could be low if each cell is a Lithium Polymer (“LiPo”) type.

[In future, you should use the "Insert Image" tab (The "rectangle" five places from the Right at the top) or "CTRL + P" to post photos.}

Sorry i wrote wrong, "While if i plug the battery, it reach 5v that is very low value i think co" here i should have said that it reach 5v when i connect the device to the usb charger, multimeter indicate 5v.

Battery are 2x 3.7v 18650 litium

I have tried to remove that white protection, but is very difficult.

Here the movie of the tests i did:

https://www.youtube.com/watch?v=OBShnfsvz1M&list=PLbknhnjWQGJV7kyVfMID_6oihf4y6MniQ&index=1

(if i past the move as media the website indicate that the movie is not available)

1) i have tested the positive and negative pads of any single cell and multimeter indicate 4v for one but for the other start from 4 and drop drop...

2) after i have connected the power with usb cable (i can't start the engine by default while charging) and one cell indicate 4v while the other 0.65v.

3) the measure while charging on the red and black cable of the battery indicate always 5v.

i think that one cell of that 2s pack is broke.

@ How i can separete the battery pack from the board without doing damages and take elettrick shock ? i have to cut wires one to one and not cut both simultaneously otherwise it go in short and i take current even if is not hight

@ does i have to buy 2 new 18650 lion battery and solder them at the same way ? A 2s lipo pack could be fine or have a bit different voltage ? i don't remeber too much, i use these lipo for fpv drones.

Could i try to use that set of 18650 lion battery that i have at home ?

are the same type but have different head and short circuit protection.

 

[FrodoOne]

Each of those "Cells" in the "Battery" is rated at 3.7 V.
(In English, a "Battery" means a "Group of Things/Devices" - although [unfortunately] many current-day "English Speakers" may not realise that !)
A "Battery" of two "Cells" (each of 3.7 V) would have a total voltage of 7.4 V

The "Vacuum Cleaner" should have been supplied with a "Device" to provide a "Charge" from a Voltage Supply higher than 7.4 V

One cannot charge a "Battery" of two 3.7 V "Cells" in series by using a USB supply - of 5.1 V, or thereabouts.
(The charger must supply a higher Voltage.)

You wrote
"1) i have tested the positive and negative pads of any single cell and multimeter indicate 4v for one but for the other start from 4 and drop drop...

2) after i have connected the power with usb cable (i can't start the engine by default while charging) and one cell indicate 4v while the other 0.65v."

which seems to indicate that ONE of the TWO cells of the existing battery is faulty.

 

[Harry Bloomfield]

One cannot charge a "Battery" of two 3.7 V "Cells" in series by using a USB supply - of 5.1 V, or thereabouts.
(The charger must supply a higher Voltage.)

The small pcb looks as if it may include a little inverter to boost the voltage. The pair of 18650 cells will each need a voltage of around >4.2v across them, to put a charge into them.

 

[FrodoOne]

@ How i can separete the battery pack from the board without doing damages and take elettrick shock ? i have to cut wires one to one and not cut both simultaneously otherwise it go in short and i take current even if is not hight

@ does i have to buy 2 new 18650 lion battery and solder them at the same way ? A 2s lipo pack could be fine or have a bit different voltage ? i don't remeber too much, i use these lipo for fpv drones.

1. If you really cannot get to the soldered connections to the "board", cut the individual conductors concerned (one at a time) and connect them appropriately to the conductors to the new Battery of Cells, with soldered connections and "heat-shrink.
You won't get a "shock" from any such "Extra Low Voltage" and cutting the conductors together could damage only the existing (faulty ?) cells and (possibly) the "cutters concerned".

2. The Cells in the exiting Battery pair appear to be "C" size cells.
While smaller Cells ("A" or "AA") of the same type should work, they will not have the same milliampere hour capacity as larger Cells.
Use the largest and highest milliampere hour Cells that will fit into the space available.

 

[FrodoOne]

The small pcb looks as if it may include a little inverter to boost the voltage. The pair of 18650 cells will each need a voltage of around >4.2v across them, to put a charge into them.

OK.
But, what was the voltage supplied from the original "Charger" (if any) or was this device designed to be charged from a USB supply ?

 

[Harry Bloomfield]

OK.
But, what was the voltage supplied from the original "Charger" (if any) or was this device designed to be charged from a USB supply ?

Obviously intended to be charged from a USB supply. You do know what an 'inverter' is don't you?

 

[FrodoOne]

Obviously intended to be charged from a USB supply. You do know what an 'inverter' is don't you?

Why is that "Obvious" ?
Yes" I do know what is an "inverter".

While you stated "The small pcb looks as if it may include a little inverter to boost the voltage.", I am not sure that I can discern sufficient associated components shown in the photos provided to achieve that outcome.

Please advise.
(You do know the associated components which would be required, do you not ?)

 

[Harry Bloomfield]

Why is that "Obvious" ?

The cells are too tall and slim to be C cells, the op is measuring approx 4v which suggests they are a pair of Lithium 19650's. I would have expected them to be in parallel, except the op measures the voltage of one reducing, but not the voltage of the second one, therefore they must be in series. Series would require a voltage of around >8.4v to recharge them, USB only supports 5v.

The op mentions being able to plug a USB adaptor into the vac to charge it, I personally have never come across any USB type lead and socket being utilised for anything but something needing a USB supply.

Yes" I do know what is an "inverter".

While you stated "The small pcb looks as if it may include a little inverter to boost the voltage.", I am not sure that I can discern sufficient associated components shown in the photos provided to achieve that outcome.

You appeared confused that a supply of greater than 5v, could be derived from 5v USB. I spy a ferrite component on the on the pcb, which is likely the transformer for an inverter, plus a couple of IC's - more than adequate to suggest there is an inverter on the pcb.

 

[FrodoOne]

The cells are too tall and slim to be C cells, the op is measuring approx 4v which suggests they are a pair of Lithium 19650's. I would have expected them to be in parallel, except the op measures the voltage of one reducing, but not the voltage of the second one, therefore they must be in series. Series would require a voltage of around >8.4v to recharge them, USB only supports 5v.

The op mentions being able to plug a USB adaptor into the vac to charge it, I personally have never come across any USB type lead and socket being utilised for anything but something needing a USB supply.

The ratio of height to width as "pictured" does not (to me) seem to support your case.
Only the OP can tell us.
The first Photo at Post #3 seems to indicate that the Cells concerned are in Series and NOT in Parallel.

You appeared confused that a supply of greater than 5v, could be derived from 5v USB. I spy a ferrite component on the on the pcb, which is likely the transformer for an inverter, plus a couple of IC's - more than adequate to suggest there is an inverter on the pcb.

If you can "spy a ferrite component on the on the pcb, which is likely the transformer for an inverter, plus a couple of IC's"
good luck, because I cannot see any such item in the photos posted.

However, this seems to have become an esoteric argument in which you have engaged when all that I was trying to do was to assist a person with (possibly) limited skills in the English language asking for assistance in that language.

I tried my best at "explanations" in English - without being either "didactic" or "condescending".
If I have failed in these attempts, I apologise to TheRosssiiii

​

 

[Harry Bloomfield]

If you can "spy a ferrite component on the on the pcb, which is likely the transformer for an inverter, plus a couple of IC's"
good luck, because I cannot see any such item in the photos posted.

The second photo posted in this thread, close to the top edge of the pcb, is what appears to be a ferrite, likely a transformer, covered with heat shrink. I also spy diodes, to rectify the output of an inverter.

I tried my best at "explanations" in English - without being either "didactic" or "condescending".

I did neither, I simply tried to point out where you seemed to be misinterpreting, or were unsure about what you were seeing. I dropped out of the discussion as it seemed remote diagnosis was doomed to failure, due to the op's limited skill set.

 

[FrodoOne]

The second photo posted in this thread, close to the top edge of the pcb, is what appears to be a ferrite, likely a transformer, covered with heat shrink. I also spy diodes, to rectify the output of an inverter.

Maybe !

it seemed remote diagnosis was doomed to failure, due to the op's limited skill set.

Agreed.