# Plug power adapter into 19V laptop [duplicate]

I know questions similar to these have been asked many times before. However, I could not deduce the answer to this question, because they only did tell me what I should do, but not why I should do that.

I have a laptop which needs 19V DC power, but I do not know whether I have the original adapter. I have some old adapters which seem to fit, but, you could've already guessed it, they give different output voltages.

Now I have multiple options:

• I plug an adapter with 17 V / 1200 mA into the device.
• I plug an adapter with 18 V / 1250 mA into the device.
• I buy a new adapter which outputs 19 V, and plug that into the device.

Now I have the following questions:

1. What will (most likely) happen with the laptop if I plug in abovementioned adapters? Is it okay, is it bad, does it harm the device?
2. Does Amp. matter?
3. This might be basic physics; can anyone post a link where this is perfectly explained?
• You can try the 18V one, it will most likely work but I still recommend buying the correct voltage & current PSU as soon as possible. – user256743 May 27 '14 at 0:02

As a general rule of thumb, the supply voltage should be just right and the amperage just needs to be high enough.

1. What will (most likely) happen with the laptop if I plug in abovementioned adapters? Is it okay, is it bad, does it harm the device?

If the supply voltage is too low, the laptop may not work at all or could fail unpredictably. If the supply voltage is too high, you would risk damaging the laptop.

In practical terms, I would recommend you find or buy a power adapter with the same voltage, but if you must use one of the adapters you already have, then try the one with the closest voltage, provided that it provides enough current. Also, its probably more risky to supply too high a voltage than to supply too low of a voltage.

1. Does Amp. matter?

Short version:

Yes, the amperage does matter. Amperage, measured in amperes, is the measurement of the electric current. The amperage indicated on the power adapter is the maximum current rating of the power supply. You need a power adapter with a current rating greater than or equal to the amperage of the laptop. If you have too low a supply current, your laptop may not work or it could fail unpredictably at high load. This is because under normal operation, your laptop will drawing less power than the amount given in its specifications and it may draw more or less power depending on what its doing.

Long version:

The voltage of the power adapter should match (power adapters supply a constant voltage over a range of currents), whereas the current rating (in amps) of the power supply needs to be equal to or greater than the load of the device it powers.

The power-adapters you are speaking of are called regulated AC-to-DC power supplies, and the type you have is almost certainly a Switched-mode power supply (SMPS). Generally speaking, DC power supplies such as the SMPS are variable current, fixed voltage devices. The current rating (in amps) of the power supply is the maximum amount of electric current that it can deliver. This means that the power supply may power any electrical device that consumes no more than the designated current rating.

If the power supply can provide more current than the device requires, then the device simply won't draw as much power as the supply can provide. The supply might be a little less efficient, but no harm is done.

However, if a device (such as your laptop) requires more current than the supply can provide, then the device (your laptop) won't be able to draw as much current as it requires to operate normally.

To make these judgements, you will need to know the amount of current you laptop requires. If you laptop doesn't give you its amperage directly, you can calculate it from its power rating.

electric power is given by: `P = I*V`

therefore:

`I = P/V`

where: I is the amperage / current in amperes, P is the electric power in watts, and V is the voltage in volts

On a related note, Ohm's law states:

`V = I*R`

where: V is the voltage (potential difference) in volts, I is the current in amperes, and R is the resistance of the conductor in units of ohms