broken 4.8vdc power adaptor. is it better to replace with a 5v 1 a or a 5v 2a?

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For cordless hair clipper the power adaptor stopped working and I can’t find a replacement one locally and don’t wanna spend extra $ if I don’t have to. It has a wierd type of connector on the end so it’s not really common.

The original power adaptor is a DC 4.8v at 1.5a.

I have some spare 5v 1a and or 5v 2a that I can splice and use. Just wondering what the better bet would be?

I tested the 5v 2a adaptor and it’s pulling 5.5-5.6v with my multimeter. I spliced into the special hair clipper cord and it seems to be working alright.

Sorry I am not a electrician and don’t know all the specifics but with my math that means it would be giving my hair clipper battery 11w of power compared to 7.2w of power that the normal power adapter was giving. I could also splice into a different 5v 1a adaptor which would provide my clipper with 5-5.5w of power.


So if you were in my shoes would you use the 5v 2a (which seems to actually be 5.5v) or would you use the 5v 1a and splice Thst which would be around 5.5w. The original charger was 7.2 w (4.8x1.5)

Or am I completely overthinking?


the battery with some internet research seems to be a 18650 lithium Bat

Thanks!
 
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Lithium batteries have a charger IC and as such, I would expect them to be somewhat tolerant of over-voltage (by a little bit, like in this case). The IC has to protect the battery from over-charge and whatnot.

I'm guessing neither charger is (or was) well regulated. Just a hunch.

If it was me, if it wasn't much money, I'd be tempted to buy a proper 4.8V charger, or somehow regulate down, just to sleep better at night. [This coming from someone who lost a 12V wall cube in a storm months ago and has been running the oh-so-ever critical WiFi router off a 13.8V supply instead.]
 
There are a few variables to consider. It's not as simple as wattage.

1) What battery type and count is in the clipper?

2) Does the clipper have a charge detection and cutoff circuit to stop charging, or just a faux circuit that turns a light green and keeps trickle charging? How long does it take to charge, that should provide a little more info about what you're dealing with.

3) Is the 4.8V charger, one of those old school (heavy due to large E-core transformer inside) types, where at no load the voltage goes up to about 1.4V the nominal rated voltage? If you have never measured it, kind of hard to figure out now besides the hint I just provided, but if it's really dead you could crack it open (even if you destroy the casing in a vice or with a saw, what does it matter if it's dead?) and it is very easy to distinguish between the two types with a picture of the internals.

4) It may not be hard to repair it, if the casing is not so utterly destroyed that it can't be glued back together. One way to less-destructively get one apart (if no screws and tabs alone holding it together), is put in a bench vice and apply pressures to opposing corners, then when it starts to bulge the plastic a "Little", if it doesn't crack the seam, gently use a hammer and chisel to pop the seam, then rotate 180' in the vice and do it again till there's enough of a cracked seam to move the chisel around to pop it the rest of the way open.

There are a few things that could go wrong but if it is the unregulated type PSU, one of the most common is a worn out thermal fuse on the primary side transformer windings. It is possible that fuse blew due to a bad diode, or short in the battery (not unlikely at all if old NiCd or NIMH with 4 in a series), but thermal cycling will eventually make them fail with no other fault because it is sealed in the PSU and runs hot (on "most" types of clippers, we don't know exactly what you have).

5) The way to determine current used running is either measure it or do a rough approximation based on how long it runs on a full charge vs the capacity of the battery. That will tell you if 1A current is enough, or more is beneficial, except that odds are that the charger is a higher voltage than the battery and pulled down by the battery, and charging the battery faster as a result, so it could overcharge the battery if there is no overcharge protection circuit, or if you were to remove the battery, running at a higher voltage could burn it out if ran for very long at a time.

6) If these replacement PSU are both regulated output then the easy answer is that the 2A gives more margin and overall a better choice, except it can't be all that well regulated if producing 5.5V.

7) Since there are so many variables, without further information my best guess at what you should do is use the 5V, 2A PSU but put a silicone diode in series with it, which will drop about 0.6V from the 5.5V you meausred, or if that was an unloaded voltage and under power it settles down to 5.0V, then I'd want to determine battery and charging method to determine if that's a problem, or do you only want to forever forward, run it from the PSU because the battery capacity is too low to be useful at this point? If that is the case then I would put a silicone diode in series anyway because at higher voltage, and higher current, the motor could end up a fair % beyond it's design limit.

8) It's also possible that the best thing to do is just hook up the lesser 1A PSU and see how it works, planning that when your clipper blade gets dull that it may be time to get a new clipper unless you have the means to sharpen it yourself. :)
 
i opened it up and found a
ICR 18650 2000 MAH 3.7V 18KPH20 battery inside. i assume this is a lithium ion battery
with my brief research on lithium ion batteries it seems to be that it needs 4.2v to charge however i am not sure why the power adaptor that came with it was a 4.8v 1.5a. it looks to be a regular power adaptor. or is it possible that the small circuit board inside the hair clipper deals with the charging and voltage?

it got me thinking, does that mean the original power adaptor is actually one thats made to charge lithium ion batteries? if so, cant i just find a old device that had a 3.7v lion battery and use that charger?

thx!
 
There are a few variables to consider.

1) What battery type and count is in the clipper?

2) Does the clipper have a charge detection and cutoff circuit to stop charging, or just a faux circuit that turns a light green and keeps trickle charging? How long does it take to charge, that should provide a little more info about what you're dealing with.

3) Is the 4.8V charger, one of those old school (heavy due to large E-core transformer inside) types, where at no load the voltage goes up to about 1.4V the nominal rated voltage? If you have never measured it, kind of hard to figure out now besides the hint I just provided, but if it's really dead you could crack it open (even if you destroy the casing in a vice or with a saw, what does it matter if it's dead?) and it is very easy to distinguish between the two types with a picture of the internals.

4) It may not be hard to repair it, if the casing is not so utterly destroyed that it can't be glued back together. One way to less-destructively get one apart (if no screws and tabs alone holding it together), is put in a bench vice and apply pressures to opposing corners, then when it starts to bulge the plastic a "Little", if it doesn't crack the seam, gently use a hammer and chisel to pop the seam, then rotate 180' in the vice and do it again till there's enough of a cracked seam to move the chisel around to pop it the rest of the way open.

5) The way to determine current used running is either measure it or do a rough approximation based on how long it runs on a full charge vs the capacity of the battery. That will tell you if 1A current is enough, or more is beneficial, except that odds are that the charger is a higher voltage than the battery and pulled down by the battery, and charging the battery faster as a result, so it could overcharge the battery if there is no overcharge protection circuit, or if you were to remove the battery, running at a higher voltage could burn it out if ran for very long at a time.

6) If these replacement PSU are both regulated output then the easy answer is that the 2A gives more margin and overall a better choice, except it can't be all that well regulated if producing 5.5V.

7) Since there are so many variables, without further information my best guess at what you should do is use the 5V, 2A PSU but put a silicone diode in series with it, which will drop about 0.6V from the 5.5V you meausred, or if that was an unloaded voltage and under power it settles down to 5.0V, then I'd want to determine battery and charging method to determine if that's a problem, or do you only want to forever forward, run it from the PSU because the battery capacity is too low to be useful at this point? If that is the case then I would put a silicone diode in series anyway because at higher voltage, and higher current, the motor could end up a fair % beyond it's design limit.

8) It's also possible that the best thing to do is just hook up the lesser 1A PSU and see how it works, planning that when your clipper blade gets dull that it may be time to get a new clipper unless you have the means to sharpen it yourself. :)
thanks for the reply!

1. i found 1 ICR 18650 2000 MAH 3.7V 18KPH20 battery inside seems to be a lithium
2. i honestly have no clue, it does have a light that turns off when full
3. it seems to be a normal power supply adaptor, idk if lithium ion has a larger power adaptor but its totally not one of those very old heavy ones
 
If a Li-Ion cell like that, it must have a charge termination circuit.

If new enough to use an 18650 cell, it "probably" (but it's not a rule or anything) is new enough that EU mandates cause it to also have a regulated output PSU.

I doubt it would run for a full two hours off a 2000mAh battery, so it probably draws more than one amp, unless designed to go a week between charges or something. Anyway, the odds are that the 1A would work fine for purposes where you just need to recharge it, or have a little charge in it while trying to use it, but if you want to use it while the internal battery is completely out of the picture (from a low voltage cutoff circuit, also needed (included) with Li-Ion cells, then you should use the 2A PSU.

I'd measure voltage while the PSU is charging the battery. If it's still at 5.5V to 5.6V, I'd put a silicone diode in series. If it has dropped to 5V, I'd run it without a diode, or put a schottky diode in series to drop that 0.2V or so, keeping in mind that the original 4.8V PSU might not have been tightly regulated either.

It's a bit odd that they didn't just use an off the shelf 5V PSU in the first place, though in quantity, it is a trivial change to make a 5V design into a 4.8V, probably just a single resistor value. If you can non-destructively get one of your two psu open and reverse engineer it, you might be able to do same to it to have a target 4.8V output.

If all else were equal, if both PSU had output at 5.0V instead of one at 5.5V (what is the other's output under load?), it would be easiest to just suggest to use the higher current rated PSU of the two, but if the PSU is directly powering the motor, a jump from the 3.7V nominal of the battery to 5V or more, is a big increase, particularly on a standard brushed motor where input voltage x current = wattage but it's not the current of the PSU that matters, rather the resistance of the motor windings determines the current. For example if such a motor drew 1A at 3.7V, so about 3.7ohms avg resistance and 3.7W, then at 5.5V with 3.7 ohms resistance, that would be 8.2W, over twice the power through the motor.

The easier answer is whip out a multimeter and measure everything, PSU voltage while charging, PSU voltage while running the clippers, current from battery alone, current with PSU hooked up.

I'd still first try putting a silicon diode in series with the 2A PSU, see if it charges and runs like that. It is less chance of problems then, more conservative to start with lower power than over power.
 
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i opened it up and found a
ICR 18650 2000 MAH 3.7V 18KPH20 battery inside. i assume this is a lithium ion battery
with my brief research on lithium ion batteries it seems to be that it needs 4.2v to charge however i am not sure why the power adaptor that came with it was a 4.8v 1.5a. it looks to be a regular power adaptor. or is it possible that the small circuit board inside the hair clipper deals with the charging and voltage?

it got me thinking, does that mean the original power adaptor is actually one thats made to charge lithium ion batteries? if so, cant i just find a old device that had a 3.7v lion battery and use that charger?

thx!

I doubt it's anything other than a generic power adapter that meets a spec. The device should be able to tolerate a small amount of voltage difference. A lot of battery powered devices have to deal with large changes in battery output as the battery draws down. There's almost nothing out there that's meant to charge a battery directly from the power adapter. Especially not for anything charging a lithium-ion battery. That's going to require a reasonably sophisticated charging circuit to prevent it from catching on fire.

Quite a few devices that charge a single 3.7V lithium-on battery operate off a 5V nominal USB-A supply. And by nominal, that means the standard allows for anything between 4.75 to 5.25V. I wouldn't really worry about using a 5V nominal power supply. That's just way too little to make it unsafe. It's got to have a regulated charging circuit that can at least handle 5V plus or minus 0.25V.
 
If a Li-Ion cell like that, it must have a charge termination circuit.

If new enough to use an 18650 cell, it "probably" (but it's not a rule or anything) is new enough that EU mandates cause it to also have a regulated output PSU.

I doubt it would run for a full two hours off a 2000mAh battery, so it probably draws more than one amp, unless designed to go a week between charges or something. Anyway, the odds are that the 1A would work fine for purposes where you just need to recharge it, or have a little charge in it while trying to use it, but if you want to use it while the internal battery is completely out of the picture (from a low voltage cutoff circuit, also needed (included) with Li-Ion cells, then you should use the 2A PSU.

I'd measure voltage while the PSU is charging the battery. If it's still at 5.5V to 5.6V, I'd put a silicone diode in series. If it has dropped to 5V, I'd run it without a diode, or put a schottky diode in series to drop that 0.2V or so, keeping in mind that the original 4.8V PSU might not have been tightly regulated either.

It's a bit odd that they didn't just use an off the shelf 5V PSU in the first place, though in quantity, it is a trivial change to make a 5V design into a 4.8V, probably just a single resistor value. If you can non-destructively get one of your two psu open and reverse engineer it, you might be able to do same to it to have a target 4.8V output.

If all else were equal, if both PSU had output at 5.0V instead of one at 5.5V (what is the other's output under load?), it would be easiest to just suggest to use the higher current rated PSU of the two, but if the PSU is directly powering the motor, a jump from the 3.7V nominal of the battery to 5V or more, is a big increase, particularly on a standard brushed motor where input voltage x current = wattage but it's not the current of the PSU that matters, rather the resistance of the motor windings determines the current. For example if such a motor drew 1A at 3.7V, so about 3.7ohms avg resistance and 3.7W, then at 5.5V with 3.7 ohms resistance, that would be 8.2W, over twice the power through the motor.

The easier answer is whip out a multimeter and measure everything, PSU voltage while charging, PSU voltage while running the clippers, current from battery alone, current with PSU hooked up.

I'd still first try putting a silicon diode in series with the 2A PSU, see if it charges and runs like that. It is less chance of problems then, more conservative to start with lower power than over power.
thanks for the reply, well i took it apart, so this is how i currently have it all wired up. i attached a pic of the board
5v 2 a power supply is spliced to the old power cord for the hair clipper.
so i plugged it all in and got it charging and checked the voltage at the splice site. it was around 5.1 to 5.15v which seems to be lower now compared to the 5.6 when it wasnt charging.

first the charing port prongs connect to the circuit board, i tested the voltage during charging and it was at 4.7v at those points

i then moved to where the battery was connected to the circuit board, the voltage there was 3.9v ish.

so does this sound like its charging properlly how a lithium battery is supposed to? seems like the board is lowering the voltage eh?

thx!
 

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The charge IC has to lower the voltage because it switches it on and off through a transistor (internal to the IC in this case). It may eve be a charge IC designed for 5V use because of the USB era, making that a common application.

Not much can be determined about the 3.9V at the battery, while it is still charging. Once it indicates fully charged, it should approach 4.2V at the battery, never above that (a hundredth of a volt might be excusable) or perhaps a little under that for higher battery lifespan.

That is the key factor, that fully charged battery does not get much above 4.20V, and that the charger IC does not get excessively hot, but since as mentioned above, it's probably targeted at a 5V application, at 5.15V you should be okay in that regard.

Did they scrub the markings off those two ICs in the picture or could the camera just not focus on them? Q1 in the pic is probably something like a TP4056 with a max voltage rating of 8V, but looking at the pic I can't tell if there's a parallel PSU input to the motor (then isolated by diodes) or all motor power comes from the battery and the power is isolated.

My point is that with it connected, I'd measure what the voltage is getting to the motor, although this thought was when it was possibly 5.6V, while down at 5.1V, you'd not so much higher than the 4.8V was.

Odds are you'll be fine with it as set up, but odds are also good that the 1A PSU would have worked fine too, as long as it doesn't claim that it tries to rapid charge the battery in 2 hours or less. If it does make that claim then stick with the 2A PSU. If it's a TP4056 or similar, it won't be charging at over a 1A rate, without an external transistor involved which isn't present (at least on the side of the PCB pictured), but it is always nice to have a PSU spec'd for a little more than the max rated current to promote long lifespan.
 
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4.8V is within 5% of 5V. Typically design tolerance is 5% so using 5V for 4.8V should be reasonable.

You need 1.5A and therefore the 2A power supply. Bigger is fine but you shouldn't go lower. I once got a FireTV stick from a coworker who incorrectly gave me a 1A power supply instead of the 2A one and it went into reboot loop, switching from 1A to 2A power supply fixed it.
 
The charge IC has to lower the voltage because it switches it on and off through a transistor (internal to the IC in this case). It may eve be a charge IC designed for 5V use because of the USB era, making that a common application.

Not much can be determined about the 3.9V at the battery, while it is still charging. Once it indicates fully charged, it should approach 4.2V at the battery, never above that (a hundredth of a volt might be excusable) or perhaps a little under that for higher battery lifespan.

That is the key factor, that fully charged battery does not get much above 4.20V, and that the charger IC does not get excessively hot, but since as mentioned above, it's probably targeted at a 5V application, at 5.15V you should be okay in that regard.

Did they scrub the markings off those two ICs in the picture or could the camera just not focus on them? Q1 in the pic is probably something like a TP4056 with a max voltage rating of 8V, but looking at the pic I can't tell if there's a parallel PSU input to the motor (then isolated by diodes) or all motor power comes from the battery and the power is isolated.

My point is that with it connected, I'd measure what the voltage is getting to the motor, although this thought was when it was possibly 5.6V, while down at 5.1V, you'd not so much higher than the 4.8V was.

Odds are you'll be fine with it as set up, but odds are also good that the 1A PSU would have worked fine too, as long as it doesn't claim that it tries to rapid charge the battery in 2 hours or less. If it does make that claim then stick with the 2A PSU. If it's a TP4056 or similar, it won't be charging at over a 1A rate, without an external transistor involved which isn't present (at least on the side of the PCB pictured), but it is always nice to have a PSU spec'd for a little more than the max rated current to promote long lifespan.

thanks so much for the reply, i just wanted to follow up
Q1 seems to be a DTM4606 DJ43M, i dont really know what that means but maybe it confirms your thinking that its a volt regulator?

i will for sure charge it later and make sure it doesnt overvolt
 
Okay, DTM4606 is an external mosfet so the microcontroller (other IC higher up in pic) is handling the multiple functions... not that it really matters, same principles apply.
 
Just use the 5v 2a power supply and call it a day. It is only 0.2v higher and the charging IC will compensate, it will be fine.
 
IMO, lower is better when it comes to battery charging. Charging his battery is not like running a CPU-powered component and I’m assuming the OP won’t be using this thing while it’s charging, so why tax and possibly damage the charging circuit with too of a voltage/current?

The 1A charger either will or will not work.

@Dave9 No offense, but you’re making all of this way too complicated. If you couldn’t answer all of your questions based on the information in the OP (and asking the OP what kind of gadget he has - not sure why this hasn’t been asked yet) then what makes you think he can figure it out?

Put a diode in series to drop the voltage? For a trimmer? Are you serious?
 
The charge IC has to lower the voltage because it switches it on and off through a transistor (internal to the IC in this case). It may eve be a charge IC designed for 5V use because of the USB era, making that a common application.

Anything designed for USB would probably have to deal with less than maximum input current. For example, my USB power packs can just charge slower if I plug them into a standard 0.5A port but of course charge faster if I use a 2.1A plug-in supply.

4.8V sounds really odd though.
 
IMO, lower is better when it comes to battery charging. Charging his battery is not like running a CPU-powered component and I’m assuming the OP won’t be using this thing while it’s charging, so why tax and possibly damage the charging circuit with too of a voltage/current?

The 1A charger either will or will not work.

@Dave9 No offense, but you’re making all of this way too complicated. If you couldn’t answer all of your questions based on the information in the OP (and asking the OP what kind of gadget he has - not sure why this hasn’t been asked yet) then what makes you think he can figure it out?

Put a diode in series to drop the voltage? For a trimmer? Are you serious?

Usually the charge circuits on these are reasonably complex when there's a li-po battery involved (which there appears to be) so the risk of overdriving the battery with a higher current power source shouldn't exist. The charge system will only take the amount of current it's setup to use to charge the battery, so the extra headroom afforded by the higher amperage charger is a good thing. On the other hand, if it tries to draw more from the charger than the charger can give (the lower amperage charger) it may nuke the wall wart.
 
Anything designed for USB would probably have to deal with less than maximum input current. For example, my USB power packs can just charge slower if I plug them into a standard 0.5A port but of course charge faster if I use a 2.1A plug-in supply.

4.8V sounds really odd though.
If it is designed to be plugged into a PC. If it comes with a power supply then it can be all over the place.

Like the FireTV stick that will boot loop with a 1A supply so they slap a 2A one instead.
 
Usually the charge circuits on these are reasonably complex when there's a li-po battery involved (which there appears to be) so the risk of overdriving the battery with a higher current power source shouldn't exist. The charge system will only take the amount of current it's setup to use to charge the battery, so the extra headroom afforded by the higher amperage charger is a good thing. On the other hand, if it tries to draw more from the charger than the charger can give (the lower amperage charger) it may nuke the wall wart.

Sure. A well regulated voltage source doesn't depend on the load. You've got a lot of tiny components on any board or chip being driven by the same regulated voltage source. A solid 5V power source that can reliably drive up to 100A isn't going to be any more dangerous than one that drive 2A. Ohm's Law doesn't change because of how powerful the power supply is.

If it's not well regulated and relies on an exact combination of power adapter and device, then all bets are off. That was the problem with a lot of vaping devices that caught on fire. They used a USB micro-B connector out of convenience, but they were only designed to use the included power adapter. When users grabbed other types of USB power adapters it didn't work as intended. However, those were really a bad idea.
 
If it is designed to be plugged into a PC. If it comes with a power supply then it can be all over the place.

Like the FireTV stick that will boot loop with a 1A supply so they slap a 2A one instead.

I've got a 2nd generation Echo Dot that comes with a 1.8A power adapter and a USB cable. I've played around with different power adapters and it doesn't seem to be an issue using something else. I don't know if maybe it's possible to overload an underpowered adapter, but I haven't seen it happen.
 
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