It'll take a few days for a new adapter to arrive for one of the laptops which has the following input and output voltages
input: 100-240V ~ 1.6A 50-60hz
output: 18.5V 3.5A LPS
The other adapter has the following input and output voltages
input: 100-240V ~ 1.5A 50-60hz
output: 19V 3.42A
I would not use the other adapter. It provides too much voltage (these should be exact):
19V > 18.5V
And it doesn't provide enough current (it's alright if the adapter provides more, but not less):
3.42A < 3.5A
It's possible that it could work, depending upon how tolerant the Laptop is, but I wouldn't risk a laptop that I don't want to replace if the outputs on the adapters don't match up exactly
Bad voltage mismatch and bad amperage mismatch each have different kinds of consequences.
With voltage, slightly underdoing it can result in performance problems, which could lead to hard drive problems (data corruption, etc.) but slightly overdoing it (no more than 10% or 20.5v total in your case) has never caused me any problems (I repair laptops and am stuck with mismatched power supplies all the time).
With amperage, even slightly overdoing it can fry components. But if your power supply is under-amped, you won't damage anything - the laptop simply won't start.
(This is my personal experience and also the "rule of thumb" taught to me by a guy who built audio receivers - he apparently learned a lot by trial and error when it came to this.)
It may be okay to use the other adapter just by reading the specifications you've provided alone. The lower amperage should be okay because it doesn't vary by much from the amperage from the original one.
The problem lies within the voltage regulator inside your laptop. Since those 18.5 (19) volts are being reduced to microvolts for your motherboard components, you would need to see the tolerances of your voltage regulator. I would assume that a half volt difference should be tolerable, but I'm just guessing without any solid numbers. If you're the paranoid type, or don't want to risk it, I would get the specs on the voltage regulator first.
You should always match Voltage EXACTLY to avoid damage. Previous poster said it is okay to go over voltage by like 10% and I COMPLETELY disagree esp. with sensitive electronics that can overheat!
When it comes to Amperage, you need an amperage that is greater than or equal to the original label on the device which is the MAX amperage it can supply. The device will only pull the amount of current it needs but the this minimum amperage is a requirement or you may end up with insufficient amperage issues.
Actually getting a supply with double the max amperage spec is usually more efficient because the supply runs at about 50% of its rated max load and most supplies running at 50% load are way more efficient than running them at 100% load. They will pull less juice and produce less heat!
A previous poster commented that higher amperage is a problem. That is completely INCORRECT information. Only lower amperage is a problem!
Finally, no one mentioned about POLARITY and the POLARITY MARKING! Make sure the replacement adapter is the CORRECT POLARITY or it won't work or worse, damage your device!
You need to know the AC power specifications first of both items and they need to agree for best performance
Adapter-2: E2=19V I2=3.42A I2 in AC= 1.5A;
Adapter-1: E1=18.5V I1=3.5A I1 in AC= 1.6A;
E=Voltage; I=Amps; W=Watts;
W DC=E DC * I DC;
Adapter-2 capacity of electricity:
W2=E2*I2=19*3.42=64.98 Watt
Adapter-1 capacity of electricity:
W1=E1*I1=18.5*3.5=64.75 Watt
Deviation accuracy capacity of electricity:
(W2/W1)*100%-100%=+0,36%
Replacement should be of the same polarity, the difference in the supply voltage does not exceed 0.5V, and have sufficient capacity.
In general, post it became clear that the power is almost identical ... it was possible to finish the ode Electrical and arithmetic, but we have a little torment to useful conclusions.
Deviation accuracy voltage:
(E2/E1)*100%-100%=(19/18.5)*100-100=+2,7%
Deviation accuracy DC:
(I2/I1)*100%-100%=(3,42/3,5)*100-100=-2,3%
The efficiency of conversion from AC to DC:
W1~W2; E1 AC = E2 AC;
(I2 in AC/I1 in AC)*100%-100% = (1.5/1.6)*100%-100% = -6,25%
Wow! Adapter-2 - very good!!!
Use it. Good luck!
Watch for temperature. Watch for characteristics of supply power at different CPU load as possible.
