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I need some information on this inverter

I just picked up this power inverter for $11 to replace a 400W 3.5A one that isn't good enough to run my primary laptop.

While the 3.5A unit would comfortably run a standard laptop, the one I previously had won't run my primary laptop (130W/6.7A). Because of my primary laptop I decided to get a better inverter.

The unit is made by Tripp-Lite and the only specs I have on it is it runs on 12V and has a 30A fuse installed with 2 outlets. The model number is PV0536. However, when I look that up it gives me a 375W model and I have a 300W model.

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An easy fix for a big power boost.

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@nick there is definitely a PV0536 300W power inverter. What is your question about it? The current I in amps (A) is equal to the power P in watts (W), divided by the voltage V in volts (V):

I(A) = P(W) / V(V)

I(A) = P(300) / V(110)

meaning that this would most like deliver no more than 2.5A (2.72 by calculation).

To run your 6.7A laptop you need an 800W inverter.

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My question was related to how many amps it puts out. While it runs a normal laptop like a T420 normally, the Dell that uses 6.7A of power is notoriously hard to find an inverter for.

At this point, it seems like running the 6.7A Dell off of an inverter (easily) is out of the question. You can't run that much current off of a 12V cigarette lighter AFAIK.


@ick "To run your 6.7A laptop you need an 800W inverter."


Hi @nick ,

This link may be of some help.

What if any are the AC Voltage and current input values that are hopefully printed on the laptop's adapter? It may be easier to work it out that way.

If you know what the AC adapter needs for an input you may be able to work out what the inverter has to supply for an output and then use the link to work it out.

Another consideration is if the inverter output is a pure sine wave or a modified sine wave.

Regardless of the fact that the inverter may be able to handle the output load, if the adapter does not "accept" a modified sine wave input it won't work anyway.


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