I just received a replacement AC adapter for my laptop, since the original seems to have fried itself. However, the company seems to have semt me the wrong part. The one I received has the same plug as the broken one, and operates at the same voltage, but its rated amperage (and therefore power) is higher.
The old one was 18.5 V, 3.5 A, 65 W. The new one is 18.5 V, 4.9 A, 90 W. Their both HP/Compaq adapters, for roughly the same line of computers.
My question is this: Does that amperage rating indicate a maximum capacity, or a necessary running condition? Is the extra power rating just unnecessary capability that will remain unused? If I plug my computer into this, will it only draw what it needs (presumably up to 3.5 A)…or will it blow itself up because the drinking straw it expected is now a fire hose?
If I can’t use this safely, then I’ve got no problem sending it back. But I’m really tired of not having a laptop, and it would be nice to not have to wait a week for a round-trip UPS shipment.
Think of it as a free, but possibly un-needed upgrade. The extra current is simply capacity.
Actually, it’s possible that the original adapter was underpowered from the start, and HP’s revised the specification from needing to supply 3.5 amps to 4.9 amps.
You should be fine - voltage is what you have to worry about. Your laptop will draw as much current as it needs.
You should notice your battery charging faster when you’re using your machine while it’s charging.
The AC adaptor needs to meet or exceed the current (amperage) that the computer requires. It will only take as much as it needs.
Imagine an automotive battery that advertises ‘500 cold cranking amps’. You might need some serious oomph during a cold start but once started, the car’s power requirements aren’t all that great. Like the laptop power supply, there’s extra current vailable but won’t likely be needed and certainly is not harmful.
Here’s the deal:
The old supply was designed to produce a precise voltage of 18.5 VDC over a broad current range. The amount of current it is “asked” to deliver is entirely dependent on the load. A load might “ask” for 0.5 ADC (when the load is powered with 18.5 VDC), or a load might “ask” for 2 ADC (when the load is powered with 18.5 VDC). In either case, the supply was able to oblige.
But there is a limit… the old supply could deliver anywhere between 0 ADC and 3.5 ADC , depending on what the load “wants”.
The new supply is even better. Like the old supply, it maintains a constant 18.5 VDC on its output. But it is able to source a current anywhere between 0 and 4.9 ADC.
As long as the maximum amount of current requested by the load (we figure it must be less than 3.5 ADC, else the old supply would not have worked) is *less * than the maximum amount of current the power supply can produce (4.9 ADC for the *new * power supply), you’ll be fine.
That’s what I suspected, but I wanted the reassurances of my fellow Dopers.
Thanks to all!