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  • alimuj Sunday, March 9, 2014 5:11 AM Reply
    I hooked up SKU: 91269 battery to B+ and B-. When I first started charging with adapter (IN+ and IN-) (5v 1A), circuit board overheated. Why? What is wrong?
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  • ebpDoug Sunday, March 9, 2014 5:53 AM Reply
    When you say "overheats", what do you mean? The board is too hot to touch or the circuit turns itself off until it cools?

    It is a simple linear regulator. At 1A charging current with 5V input, the power dissipation will be 1A * (5 - Vbat)V. If the battery voltage is 3.5V, for example, the dissipation will be 1.5 watts. That is sufficient power to make a little board like that quite warm, especially if the back of the circuit board doesn't have free air flow over it. The charge controller IC (the 8-leaded part next to the USB connector) is the only part that should be generating significant heat, and it could get too hot to touch for very long.
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  • alimuj Monday, March 10, 2014 4:07 AM Reply
    Sorry. I don't know english. Overheating means too hot.
    I measured the battery with voltmeter. The battery is 3.75 V. No problem when i soldered this battery. I switched 5v 1a universal adapter. The back side of the circuit board get warmed. I could not touch. I immediately removed the battery. I don't know why?
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  • ebpDoug Monday, March 10, 2014 5:29 AM Reply
    "Sorry. I don't know english. Overheating means too hot."
    No problem. Some people think 40°C is too hot and some know 80°C can be OK sometimes, so I needed to understand what you meant. Saying that it is too hot to touch helps.

    Can you measure the current from the adapter? It may be that the current is more than one ampere. Most adapters don't have very accurate current limiting and can supply more than their rating. The charge controller chip should limit the current quite accurately, but maybe something is wrong. If the input current is about 1A and the battery voltage is rising, everything is probably OK.

    Can you read the markings on the charge controller chip? If found some info that suggests it might be a TP4056. If it is, the IC has a metal area on the back that should be soldered to the board for heat transfer. From the pictures, it looks like the board is designed so that almost all of the heat will be transferred to the copper foil on the back of the circuit board. It could get quite hot when working normally. The TP4056 is protected against over-temperature, but at about 145°C (on the actual silicon chip).

    There is a resistor in the picture that is marked "122" which would mean 1.2k ohms. If the chip is a TP4056, that would set the current to 1A. Inspect that resistor carefully to make sure it isn't short circuited. You can probably measure it with your meter as long as the adapter and battery are disconnected. It might not read exactly 1200 ohms, but it shouldn't be very low.
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  • alimuj Monday, March 10, 2014 4:10 PM Reply
    Dear ebpDoug ;
    Thanks for your replies.
    I think, the battery can be problematic. I'll try again after the battery discharge. I'll use a different universal adapter (5v, 600 or 800 mA). I will measure the current from the adapter. I will measure the resistor. I will write the results.
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  • desolder Top 10 Forum Poster Wednesday, March 12, 2014 5:11 AM Reply
    Curiously, this one lacks a voltage dropping resistor. All the other TP4056 based charging modules include a relatively large (for a SMD) 0.4 ohm resistor which helps dissipate some of the excess energy and keeps the chip itself from having to dissipate all of it. Hopefully the engineers did their job to make sure the chip is heatsinked sufficiently. Most likely it is, as ebpDoug mentioned there's vias to the bottom copper plane.

    See 0.4 ohm dissipating resistor in this sku:

    img src: http://img.dxcdn.com/productimages/sku_205188_1.jpg

    What's up with the scrubbed LED driver IC markings?
    What are they trying to hide?

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  • JustinRomaine Wednesday, February 20, 2019 4:05 PM Reply

    Mine self destructed in a puff of smoke.

    Supply was DC 5V 1.2A. 

    ok through usb power.

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  • desolder Top 10 Forum Poster Thursday, February 21, 2019 1:13 AM Reply

    Did yours lack the 0.4 ohm voltage dropping resistor? Try to get a replacement with it.

    What's up with the scrubbed LED driver IC markings?
    What are they trying to hide?

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