Beginner safety questions regarding voltage and current

Question

I'm a complete beginner to electronics so please excuse the simplicity here :)

I have a phone charger that outputs at:

5V - 3A
or 9V - 1.6A
or 12V - 1.2A

I would like to use another charger while at work, which outputs at:

5V - 1.8A

I don't know what the input for the phone itself is.

I assume it's safe to use the second charger because the phone will only take the current that it needs?
If the second charger had a voltage higher than the phone's charger, should it be avoided?
In general, how would I determine if something is safe to use or not?
When do things get to the point of damaging the device, or even worse, starting a fire?

Thanks!

Related: Choosing power supply, how to get the voltage and current ratings? — The Photon, Aug 21 '17 at 18:27
"I assume it's safe to use the second charger" ... And what if the phone needs full 2.5A of current if it sees the DCP (Chineese) signature? (which is quite unlikely, but as an alternative assumption)? — Ale..chenski, Aug 22 '17 at 18:29

score 0 · Accepted Answer · answered Aug 21 '17 at 18:36

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I assume it's safe to use the second charger because the phone will only take the current that it needs?

The second charger's current specification is lower anyway, so this isn't an issue. But your assumption is right -- devices draw the current they require.

If the second charger had a voltage higher than the phone's charger, should it be avoided?

No, because this is USB where smart devices negotiate the voltage to be supplied.

In general, how would I determine if something is safe to use or not? When do things get to the point of damaging the device, or even worse, starting a fire?

In general, you need to match the voltage and use a supply that can supply at least as much current as the device requires. But this is a special case because USB has a specific charging protocol where the devices negotiate the voltage to be supplied and the current to be drawn. Unless you have a device, charger, or cable that doesn't follow the USB specification, you don't have to worry.

answered Aug 21 '17 at 18:36

David Schwartz

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"If the second charger had a voltage higher than the phone's charger, should it be avoided? No, because this is USB" The OP never said it was USB, although that is quite likely. Anyway, if the voltage is higher it absolutely should not be used. If the phone expects USB then it still should not be used since the phone would expect 5 V prior to negotiation. – Matt Aug 22 '17 at 05:27
@Matt It's not just quite likely, it's all but certain. Did you read the specs of his charger? That's clearly a QC charger. QC chargers do output 5V prior to negotiation. – David Schwartz Aug 22 '17 at 16:52
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Yes, but the OP asked if a higher voltage charger would be okay. The correct answer without extra information is "No" because you dont know that whatever other random charger the OP comes across will be a USB QC. It very well might just be a DC power supply. – Matt Aug 22 '17 at 16:56
@Matt A DC power supply with a voltage other than 5V and a connector that mates with a phone that works with a QC charger? Do you have any reason the OP is likely enough to encounter such a thing that it's worth mentioning? – David Schwartz Aug 22 '17 at 17:01
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Whats the chance that someone comes across such a charger? Probably quite small. Whats the chance that someone reading EE.SE reads this and thinks they can make one without fully understanding how it works? Probably a good bit higher. – Matt Aug 22 '17 at 17:13
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Fortunately, I already explained the conditions under which the answer applies, "No, because this is USB where smart devices negotiate the voltage to be supplied.". – David Schwartz Aug 22 '17 at 17:18
-1. This is "USB", and your answer is largely a wishful thinking. Most current USB devices, even smart one, don't negotiate higher voltage. They mostly stick at 5V, but are only smart enough to recognize "charger signature", and take corresponding current out of USB port or cable. – Ale..chenski Aug 22 '17 at 17:46
@AliChen They don't negotiate a higher voltage because they don't support one and as a result, they get the highest voltage they can support. That's absolutely perfect, the best case outcome. – David Schwartz Aug 22 '17 at 17:51

Ale..chenski · Answer 2 · 2017-08-22T18:23:01.490

From the description of voltages/currents, your phone is designed for Qualcomm QuickCharge specifications. As such, the phone expects certain responses on D+/D- wires upon connection.

If your other charger isn't designed for QC specs, your phone likely won't take more than 500 mA and will charge very slow.

However, if your other charger conforms to Chinese specifications (D+ shorted with D-), which is very likely, and your phone also has a built-in intelligence to recognize this "charger signature" as an alternative, it will try to take 2.5A from the other charger, which will be too much for it, and the voltage will sag (if the other charger is designed correctly). Then the phone might scale down its consumption, and everything will be fine and safe.

As one can see, there are so many "ifs", and each if carries probability of not being true. If you like to gamble, go for it. In your case I would recommend getting a replacement charger for your phone, and keep it at work.

To start, you may want to find actual operational range of your phone, using a in-line tester like this one, and then determine if the other charger works or not.

This is precisely how these devices are designed to be used! — David Schwartz, Aug 22 '17 at 16:54
@DavidSchwartz, no, this is how FUTURE devices are supposed to be designed, to specifications that are either constantly and drastically evolving (as USB PD), or not disclosed to public (as QuickCharge). Very few devices in today's market have implemented PD, and QC chargers are 3X in price, and hard to come by. Please don't push your wishful thinking on reality. — Ale..chenski, Aug 22 '17 at 17:40

Beginner safety questions regarding voltage and current

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