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I'm looking to use a simple USB power bank as a portable charger for my laptop (just to get a little extra boost if the battery starts running low at an inopportune time).

My plan is to use something like this (http://amzn.com/B00LRK8JDC), which has two USB type A ports, and I can draw up to 2.4 amps from each simultaneously. If I splice together two USB cables to make one that is a Y with a USB-C PD plug on one end (my laptop can charge through its USB-C/Thunderbolt 3 port) and two USB type A ports on the other, would it work? The output would be 4.8A at 5V, right?

Rough MS Paint Diagram: enter image description here

Nelson
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3 Answers3

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Actually: No

The listing, as should be expected with a cheap power bank, clearly states a Total Maximum of 3.4 A. That's 2.4 A and 1 A.

Moreover the device does not specify whether both outputs are generated from the same boost converter internally, or if it's two different voltage boosters. Connecting two voltage sources in parallel with no regards to proper engineering means absolute unmitigated disaster, as nothing is 0% wrong. One source will be 5.05V the other 4.95V, it'll be more likely they compete or destroy each other than offer more current.

Apart from that the USB-C specification does NOT allow for more than 3A on a single 5V supply, it offers quick charging and advanced charging through devices that offer a special higher voltage on the same connector. With these higher voltages it's possible to get 30W through a cable without it being problematic.

30W is enough for the lower power types to charge, intel i3-5005, for example, should happily do so in most cases.

The reason they do not use 5V for that, is because trying to get 6A at 5V (required for 30W) through a thin USB cable is absolute nonsense, as at the other end it will be 4V or less, due to resistive losses.

So, yes, if you want to do that, you do need a boost converter. One of which will not cost a lot of hardware or space at all, this is unnecessary hyperbole. After all the power pack is already floating itself, and even if it weren't; 30W isn't a lot of power anyway (a hand full of cubic cm is enough to convert 30W isolated or not at 500kHz ~ 3MHz switched frequency operation). The main reason you don't want that is the power pack won't be able to supply anywhere near enough power to make that even remotely interesting, due to afore mentioned facts.

Asmyldof
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Yes, you are correct. Connecting power sources in parallell as per your drawing will increase the current capacity while voltage remains the same.

However, your make and model of the laptop will by far be the biggest factor in this equation. 4.8A at 5V is only 24Watts, which is far less than noticeable for a conventional laptop. Additionally, most laptops need a lot more than 5V supply, so unless you have a laptop smaller than most, you're going to need to increase the voltage somehow, and this cannot be done without significant loss when it comes to DC.

Lastly, if I was to get the voltage high enough, I wouldn't feed it into my laptop without a DC-DC converter to ensure a steady voltage with proper galvanic insulation. At the end of the day, the rig I would feel comfortable using would necessitate a backpack, and at that point I might as well use an AGM battery for increased voltage and capacity.

+1 for the MSpaint Schematic

Jarmund
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As I see, you are trying to build a legacy TypeA -> Type-C cable by yourself. Which means that the Type-C end will not have any CC controller compliant to Power Delivery specification. Even if your unspecified laptop says Type-C PD (power delivery), your cable will not respond to laptop's PD inquiries.

Since your cable will not respond to PD, the laptop must turn to the standard Type-C advertising mechanism and measure the signal level on CC pin. The only thing in this situation can occur if you have 10k pull-up (to VBUS) inside your cable. Then the laptop should take maximum allowable charging current, which is 3A per Type-C specifications, not more. All this assumes that the laptop fully follows Type-C/PD specifications, which could be a very strong assumption.

So, you will have 3A at 5V. That's it.

Maximum a PD-enabled Type-C device can get from 5V is 5A, but you would need to use a PD "electronic marker" instead of simple pull-up. These chips are already available from Texas Instruments, NXP, Cypress, and few other manufacturers.

Ale..chenski
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