Hi,
You may need to explain how, and why/testing what function?, you connected it to the phone. Heating up badly is appreciated description but a very relative concept, 250mA gives approx 1.75W. Have you done the calculations provided in the datasheet pages 10 and 11 of TI version? Was that open air, or in a small enclosure or something?
What sort of heat sink is it, C/W wise? I haven't used the LM1084, but with the 7805 and no heatsink or fans I avoid using it over 100mA (I like it's very stable voltage when used lightly), as it gets hot (warmer than I like touching once safely powered down), but not that smell of "nearly burning silicon and metal hot".
Sorry to ask a stupid question: the connection to the phone wasn't accidentally reverse polarity was it? Easy to do with those barrel connector adapters.
I do not see the capacitors on your circuit described in the datasheet for the LM1084 so maybe the regulator was oscillating at a very high frequency that will make it get hot.
Hi,
Thanks, for the update, you've saved me waffling on needlessly there...
Are you trying to use it as a mobile phone (battery) charger? If so, you'd need a battery charger IC for that specific to the phone battery I imagine.
...if you think of hot water, a 50ºC shower would not be a nice experience, or holding something at that temperature, or beach sand burning feet in summer..., add 10 or 20ºC and that's very hot indeed, so it's possible the regulator is running hot but if it's working fine, then no problem. If it overheats to the point of shutting down, that is a different matter.
When you use it with your Pi, I guess you'll do the heat-sinking properly and maybe use a small fan which will help dissipate heat and help it run cooler.
The datasheet says to use an output capacitor that works properly at high frequencies like 10uF tantalum. An input capacitor also might be needed. The capacitors must have short leads and be very close to the pins on the LM1984 so do not use a solderless breadboard.
Its heating should be exactly the same as a 7805 since they are both linear and both have the same voltage across them and the same current through them, (12V - 5V) x 250mA= 1.75W.
Instead of using a 12V power supply, try 9V. Then the heating is (9V - 5V) x 250mA= 1.0W.
Did you calculate the overall thermal resistance Rjc+Rca and measure the temperature rise of case? and follow advice on caps close to regulator leads?
Did you read this ?
https://www.ti.com/general/docs/lit/getliterature.tsp?baseLiteratureNumber=spra953&fileType=pdf
unheatsunk to220 dissipating 1.75w is going to get very scalding hot...what you see is just normal, by all means put a dropping power resistor coming off the 12v to take some of the dissipation out of the to220.
The stupid circuits in Google show a series resistor between the output of the 5V regulator and the cell phone like this one:
I mean "TO220"....Common regulator package.
So you have a little heatsink?.......still 1.75w in a to220 with small heetsink is going to get a bit toasting still.....you said you could keep your fingers on it for 2 seconds so it cant have been that hot anyway
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