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This question already has an answer here:

To preface; I am aware that there are a lot of questions that are similar to mine. But I am finding mixed signals as well as jargon I don't understand; and I was really just hoping somebody could give me a yes/no because I'm really paranoid about plugging it in based off of my reading

I have an LED sign that says that it takes "Class 2, 18VDC, 2A" from a wall adapter with "100-240VAC, 50-60HZ, 1.2A"

I lost the adapter, and it's surprisingly difficult to buy a new one that matches those specs. I did find an old laptop power supply that has these ratings:

Input: 100-240VAC, 50-60HZ, 1.2A Output:19V, 3.42A

There is a weird symbol between the 19V and the 3.42A, but I have no idea what it means.

Here is a picture of the sign:

LED Sign

And here is one of the laptop power supply:

enter image description here

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marked as duplicate by Passerby, PeterJ, Daniel Grillo, JRE, Dave Tweed Jan 30 '16 at 2:52

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    \$\begingroup\$ You should try to determine what is inside the device. An ugly solution for dropping 1V would be to add two series power diodes \$\endgroup\$ – TEMLIB Jan 28 '16 at 1:36
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    \$\begingroup\$ Just FYI, the symbol in 19V⎓3.42A (two horizontal lines, top line solid, bottom line broken) is Unicode U+2393 ⎓ "direct current symbol form two". It means 19 volts DC. If the supply was labeled 19V~3.42A that would mean 19Vrms AC instead of 19V DC. \$\endgroup\$ – MarkU Jan 28 '16 at 3:21
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The weird symbol indicates its a DC output. This laptop adapter supplies an output voltage of 19 V and a maximum output current of 3.42 A, well above your original adapter's 2A maximum. This doesn't mean your device will consume 3.42 A when its powered with this adapter; it is only an indication of the maximum current that can be drawn through the adapter by any device that it is connected to, above which the adapter gets damaged. Any device will only draw as much current as it needs, so long as its power source can supply it.

However, the laptop adapter's voltage is a full volt above the specified 18 V; this will cause more current to flow into your device, since the voltage has been increased. Whether this difference is significant enough to destroy your LED sign is a matter of how much tolerance was built into it; the 1-volt increase may merely increase the brightness of the LEDs or burn them, if 19 V is outside the device's range, or have no noticeable effect at all (if the 19 V is being stepped down further within the device).

TLDR, don't try using this adapter unless you are willing to risk damaging your device from the extra current.

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    \$\begingroup\$ That's what I figured. Would the damage be immediate? Like could I plug it in, see what it looks like? or would it potentially just fry immediately. Also, could you recommend a place where I could buy a 18v adapter? \$\endgroup\$ – A O Jan 28 '16 at 1:34
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    \$\begingroup\$ @AO Any adapter you find on ebay with output voltage 18 V and maximum current >= 2 A will be satisfactory. The input current of the adapter isn't relevant, only an input AC voltage of 100-240 V \$\endgroup\$ – TisteAndii Jan 28 '16 at 1:44
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    \$\begingroup\$ Awesome thanks for the answer! Could you explain why the input of the adapter isn't relevant? Why would the sign include that part? \$\endgroup\$ – A O Jan 28 '16 at 1:53
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    \$\begingroup\$ @AO practically every adapter is designed to run off mains (house power), and the nature of "switching power supplies" makes it easy for them to support a wide range of input voltages with little impact on performance. The damage is likely not immediate, the 18V is fed to the charge controller for the laptop motherboard, typically these will have capacitors and switching components with a voltage rating (say 25V caps), I have used higher voltage supplies without issue after verifying that the front facing components are tolerant to the higher voltage. \$\endgroup\$ – crasic Jan 28 '16 at 2:10
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    \$\begingroup\$ The max input current is a sort of safety rating. In the event of a fault, it tells you how much current can flow into the device before some fuse within it blows or how many of these adapters you can connect to a wall outlet without blowing the socket's fuse. \$\endgroup\$ – TisteAndii Jan 28 '16 at 2:10
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First off, that symbol on the laptop pwr adaptor simply means that its output is DC current (as opposed to AC mains power). Now, for usage, the LED sign very likely (but not guaranteed) has an internal voltage regulator, in which case, the small (about 6%) overvoltage would be handled there. However, if you want to be more cautious with it, then you could add your own highly stable voltage regulator (a great reference for this can be found at: http://www.rason.org/Projects/discreg/discreg.htm ) or, for a "quick and dirty" approach, you could try simply adding a 5w or 10w 0.5ohm resistor inline between the adaptor and the sign. To come up with this, dividing the sign's required current (2A) by its required voltage (18V) yields an expected load resistance of 9ohms; to get the same 2A current from a 19V source requires a resistance of 9.5ohms; then, accounting for a 1V drop across you 0.5ohm resistor at 2A, we see that the resistor should regularly be dissipating 2W of power, so must make sure its rating is sufficiently above 2W to insure against damage; yielding the 0.5ohm 5-10 watt resistor (cheaply available in a ceramic potted wire-wound type). NOTE: The resistor will not stop the light from "seeing" move than the expected 18v supply, but it will limit the voltage/current to the sign down to 18V/2A at the maximum-power point. Since 19V is so close (+6%) to 18v, this should prevent damage to the sign as it is unlikely that any components used would fail at 19v with minimal current, but I would NOT recommend a similar approach id you were using, for example, a 24v source, since the significantly higher voltage could blow capacitors, transistors, diodes, etc in the sign that may not be spec'd to withstand that voltage.

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    \$\begingroup\$ Thanks for the informational answer! I think I may just play it safe and try to buy a new adapter. Does it matter what the input to the power supply is? The sign literally asks for a 1.2A input, 2A output from a power supply. But this adapter on eBay says 0.2A input, 2A output ebay.com/itm/… \$\endgroup\$ – A O Jan 28 '16 at 1:38
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    \$\begingroup\$ @AO 1.2A/110V rating is for the wall socket; almost all sockets can provide this current. 0.2A is the actual current consumed by the supply. \$\endgroup\$ – ilkhd Jan 28 '16 at 2:11
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    \$\begingroup\$ That adaptor should work fine for your sign. since the voltage is being stepped down by a factor of 6:1, an ideal converter could give you the sign's spec'd 18V/2A from only 0.33A input...the difference in input amperages is just really a matter of efficiency. Your sign can only "see" the output of the converter, so that one at 18V/2A rated output should work just fine for you. :) \$\endgroup\$ – Robherc KV5ROB Jan 28 '16 at 2:23
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    \$\begingroup\$ Also, for future reference, the symbol after the 2A rating on both adaptors (with the - sign, then the C and dot, then the + sign) tells you that the output plug has the DC - on the outside of the plug & the DC + on the inside. NOT all adaptors have that same "polarity" on their plug, so it's important to notice where the DC + and - flows are on both your supply & your device, so you can make sure you're not hooking something up backwards, which can cause immediate & major damage to devices without reverse polarity protection. \$\endgroup\$ – Robherc KV5ROB Jan 28 '16 at 2:36

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