Safe to use a higher Voltage/Amp AC adapter on notebook?

PremiumG

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Jun 4, 2001
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I recently lost my charger for my laptop. The charger was rated 18.5V and 3.5 AMP, 65W.

I came across a replacement adapter that is 19V and 4.74A, 90W. It fits and it works, but Im afraid I might be slowly degrading my laptop battery or laptop using it.

I googled it and some web pages/ebay auctions list my laptop as compatible and some do not list it on their compatibility list.


 

Jiggz

Diamond Member
Mar 10, 2001
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It should be OK. The amperage and wattage rating is actually better than your original. The only thing that is out of spec is the voltage rating which is 0.5V. This is nothing considering the voltage droops when the adapter is under load. So you are OK and actually got a good replacement.
 

PremiumG

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Jun 4, 2001
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if the amp and watt is too high, won't it hurt the battery?

or are those specs max amp and max watt?
 

JackMDS

Elite Member
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Oct 25, 1999
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These are Max capacity.

0.5Volt should be troublesome.

Make sure that the replacement DC plug and polarity are indeed the same as the original.
 

MoMeanMugs

Golden Member
Apr 29, 2001
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The amperage difference is HUGE. There's no way I'd use that adapter if you value your laptop. I foresee your voltage regulator frying.
 

JackMDS

Elite Member
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Oct 25, 1999
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Current (amperage) is state as max possible.

Under normal electronics Current is Not forced, it is the product of Voltage and Resistance.

I= V / R (R is the battery resistance state and depend of its level of its charge).

If the input to the battery is controlled inside the laptop, then there would not be any difference.

If it does not controlled, the current would be the regular current + few % due to the extra 0.5Volt.

 

Aluvus

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Apr 27, 2006
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Originally posted by: PremiumG
if the amp and watt is too high, won't it hurt the battery?

or are those specs max amp and max watt?

They are maximums.

Originally posted by: MoMeanMugs
The amperage difference is HUGE. There's no way I'd use that adapter if you value your laptop. I foresee your voltage regulator frying.

The difference in current ratings is absolutely no problem at all.

Originally posted by: JackMDS
Current (amperage) is state as max possible.

Under normal electronics Current is Not forced, it is the product of Voltage and Resistance.

I= V / R (R is the battery resistance state and depend of its level of its charge).

If the input to the battery is controlled inside the laptop, then there would not be any difference.

If it does not controlled, the current would be the regular current + few % due to the extra 0.5Volt.

Ohm's Law is not a useful model for a device as complex as a laptop, and especially so for a Lithium-Ion battery charging system. A laptop is not a resistor.

LiI chargers go to great lengths to control the voltage seen by the battery, so no a higher input voltage will not translate into more current into the battery (up to some limit). Similarly, I would not be surprised if the laptop immediately regulates down whatever voltage comes in from the supply.

The voltage range that all of this hardware can operate at is the limiting factor here. A 2.7% difference is probably within spec.

Assuming the OP's connector makes good contact and is the correct polarity, using this power supply with this laptop is probably OK.
 

Jiggz

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Mar 10, 2001
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Originally posted by: MoMeanMugs
The amperage difference is HUGE. There's no way I'd use that adapter if you value your laptop. I foresee your voltage regulator frying.

Sir, current in a circuit is "drawn" by the load and not "pushed" by the source. Same is true with the wattage which is nothing but a measure of power. Rating of a current or power source is always the maximum that can be provided by the source without overloading which causes overheating and fire.

As for the voltage and polarity concern, read and heed the OP. The OP is already using the adapter and is working perfectly. So this should not be a concern anymore. The OP's concern now is if this over voltage of 0.5V is harmful or of concern. AND I SAY NO. And I made this call as an electrician and as an electronic tech. From experience, it's never unusual for adapters to have voltage droop under load, which is also typical with Alkaline 1.5V cell batteries which can droop as low as 1.2V under load. This is the reason the Ni-Cd and NiMH cell batteries are rated at 1.2V, because they have little voltage droop under load.
 

MoMeanMugs

Golden Member
Apr 29, 2001
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I'm an electrical engineer, but thanks for trying to explain things to me. ;) By my experience, his VR will not last as long. Look at the nature of the transformer in the power pack.
 

MoMeanMugs

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Apr 29, 2001
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I guess I should explain a little further. That VR is going to be working harder dissipating heat. Depending on what kind of safety factor was designed by the laptop manufacturer (he doesn't state which), that VR could be pumping out quite a bit of heat already. I'm not saying he's going to burn it up right away, but I've seen VR's speced for 15 V burn up at 13 V (on more than one occasion - supposed to be quality ones). It should be noted that a voltage regulator is actually regulating current since it's a transistor. Now the stepdown transformer is going to be providing more current to the VR, and thus more heat. I was never arguing that his battery or the motherboard itself would be damaged since it sits on the other side of the VR. However, I just don't like to risk things. We all know manufacturers cut corners to cut costs = cheap crap inside. I guess it all depends on how long he plans on using it. I'm not screaming impending doom, but rather issuing a word of caution.
 

mindless1

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Aug 11, 2001
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Originally posted by: MoMeanMugs
I guess I should explain a little further. That VR is going to be working harder dissipating heat.

No, certain types of electronics (linear regulators not present in this power supply stage(s)) would have higher heat but not in this situation.

Depending on what kind of safety factor was designed by the laptop manufacturer (he doesn't state which), that VR could be pumping out quite a bit of heat already. I'm not saying he's going to burn it up right away, but I've seen VR's speced for 15 V burn up at 13 V (on more than one occasion - supposed to be quality ones).

Certainly in different types of equipment with a different regulation type than seen in a notebook, a larger input voltage can exceed the thermal margins, as related to the amount of current the regulator is handling. It just isn't applicable in this case.

It should be noted that a voltage regulator is actually regulating current since it's a transistor. Now the stepdown transformer is going to be providing more current to the VR, and thus more heat.

With all due respect, you are thinking of a different circuit topology not applicable here. The laptop brick is a basic switching type that will actually be providing slightly less current at this barely higher voltage. At the laptop, on either a separate power regulation board or that circuit integral to the mainboard, there is switchign step down circuitry that has no significant difference in heat generation from this small difference in voltage, and will actually have slightly lower heat generation if anything due to shorter on-state. If there were a larger difference in voltage we might have concern about coming too near the ratings of some particularly susceptible (tantalum) capacitors, but this time the difference is so slight it's negligible. Plus, in the past few years manufacturers have started to shun tantalums in favor of other organic, MLC and other solid types - there may not be any tants in it at all.

I was never arguing that his battery or the motherboard itself would be damaged since it sits on the other side of the VR. However, I just don't like to risk things. We all know manufacturers cut corners to cut costs = cheap crap inside. I guess it all depends on how long he plans on using it. I'm not screaming impending doom, but rather issuing a word of caution.

From the details mentioned thus far there is no cause for concern. Edit: (Irrelevant content removed).

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In reply to the OP, so long as the replacement supply is of reasonable quality, meaning it is not some piece of junk more likely to fail (which we have no reason to believe, but there's a wide world full of all different quality goods out there) then it should work fine. The battery is not direct charged but rather through a regulation and control circuit. Someday your battery will die from aging as all do but it won't be from using an AC-DC supply that was 0.5V higher than the original, and as others mentioned the higher current rating is not only OK, it is often good to have a little more margin between the actual load and the peak rating of the supply, but for size and cost (profit) reasons the manufacturers only give you what you started with instead.
 

ding5550123

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Jan 3, 2006
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With That voltage it would seem fine. Be care ful that you son't severly overcharge your battery, because they can catch fire and explode (and I mean explode) but the half volt higher, and the extra amp won't hurt.
 

Jiggz

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Mar 10, 2001
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Originally posted by: MoMeanMugs
I guess I should explain a little further. That VR is going to be working harder dissipating heat. Depending on what kind of safety factor was designed by the laptop manufacturer (he doesn't state which), that VR could be pumping out quite a bit of heat already. I'm not saying he's going to burn it up right away, but I've seen VR's speced for 15 V burn up at 13 V (on more than one occasion - supposed to be quality ones). It should be noted that a voltage regulator is actually regulating current since it's a transistor. Now the stepdown transformer is going to be providing more current to the VR, and thus more heat. I was never arguing that his battery or the motherboard itself would be damaged since it sits on the other side of the VR. However, I just don't like to risk things. We all know manufacturers cut corners to cut costs = cheap crap inside. I guess it all depends on how long he plans on using it. I'm not screaming impending doom, but rather issuing a word of caution.

With all due respect with your EE Degree, a voltage regulator does not regulate current, it regulate voltage as implied in the name. What regulate current are inductors, chokes and current limiting device. True there is a direct relationship with V and I, but in this particular case, an increase of 0.5V in V will have very little increase in I from the load side. As for the adapter, it will actually run cooler because of its higher designed carrying load (90W) and a lower actual load of approx 60W.

The amperage difference is HUGE. There's no way I'd use that adapter if you value your laptop. I foresee your voltage regulator frying.

A good example that contradicts your first post above is a nuclear electrical power plant. This plant is rated at Megawatts of power with Megaamps of current. Yet, when you plug in a 60 watt house bulb it does not blow up! Because the bulb which is the load only draws current it was designed for, in this case 60 W. This is no different from the 90W power adapter and supposedly 65W laptop.
 

imported_Baloo

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Feb 2, 2006
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Originally posted by: MoMeanMugs
The amperage difference is HUGE. There's no way I'd use that adapter if you value your laptop. I foresee your voltage regulator frying.

The amp is a rating for the power adaptor, means the power adaptor can handle up to that amount, it does not mean the laptop will consume that amount.

What matter's here is the voltage, and nothing else. 19 is close enough to 18.5 that it won't be a problem. So long as the DC plug fits (Polarity is the same for all, it's an industry standard).

EDIT: other's beat me to it with better explanations than mine.
 

MoMeanMugs

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Apr 29, 2001
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I fully understand what everyone is saying. I was never arguing that you people were wrong. My only point was that cheap parts are often used in laptops (look at Dell for example), and it depends where the voltage regulator is placed in the circuit. I've seen many placed in the laptop itself, and I guess I jumped to conclusions. I apologize. Jiggz, if you look at the VLSI scale of a voltage regulator, it's regulating the current. I guess it's all how you look at it. Anyway, sorry if I came off the wrong way.
 

Jiggz

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Mar 10, 2001
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Originally posted by: MoMeanMugs
I fully understand what everyone is saying. I was never arguing that you people were wrong. My only point was that cheap parts are often used in laptops (look at Dell for example), and it depends where the voltage regulator is placed in the circuit. I've seen many placed in the laptop itself, and I guess I jumped to conclusions. I apologize. Jiggz, if you look at the VLSI scale of a voltage regulator, it's regulating the current. I guess it's all how you look at it. Anyway, sorry if I came off the wrong way.

Mugs, if you are referring to Large Scale Integrated circuits, it probably does. Since IC chips are also considered LSI or maybe nowadays even VLSI, it's probable it also includes a current regulator. A good example of these IC chips is the VR and CR on auto alternator's or charging system. I still remember when this regulator device were electro-mechanical but nowadays is nothing but a single IC chip.
 

PremiumG

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Jun 4, 2001
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so...... general consensus is its safe to use???

Looks like 5 vs 2 (safe vs unsafe)

 

RebateMonger

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Dec 24, 2005
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Originally posted by: PremiumG
Looks like 5 vs 2 (safe vs unsafe)
Make that 6 vs 2 (safe vs unsafe).

1/2 volt extra in the charger is unlikely to cause any charging circuit problems. And some laptops (some of the Dells, for instance) offer two different chargers: a low current one (65W, for instance) and a higher curent one (90W, for instance).