Hi ,

I am new here so please let me know if I am not in the right forum , thanks
I have a surveillance movable camera (it has two motors) and it stop working and I have opened and checked its circuit board for repair I have noticed that the AZ1117D 3.3v become very hot???? . thanks

 

Hi ,

I am new here so please let me know if I am not in the right forum , thanks
I have a surveillance movable camera (it has two motors) and it stop working and I have opened and checked its circuit board for repair I have noticed that the AZ1117D 3.3v become very hot???? . thanks

Overheating of linear VRs is most generally upshot of overload (excessive current drain) you are advised to check any electrolytics and go from there... Please keep us posted

Best regards
HP

 

Hello,

AS HP mentioned over current will heat-up the regulator.
Are the motors also powered from the 3.3 Volts?
If so, check if they are not stalled.

Bertus

 

Overheating of linear VRs is most generally upshot of overload (excessive current drain) you are advised to check any electrolytics and go from there... Please keep us posted

Best regards
HP

Thanks.

 

Hello,

AS HP mentioned over current will heat-up the regulator.
Are the motors also powered from the 3.3 Volts?
If so, check if they are not stalled.

Bertus

YES the motors also powered from the 3.3 Volts I have disconnected the motors cable and WIFI module the IR diodes the regulator still overheated.. the power supply supposed to give 5v output (as stated in the manual) but it give about 8v output instead??

 

Hello,

When the powersupply is giving 8 Volts in stead of the mentioned 5 Volts, the regulator will have to drop 4.7 Volts in stead of 1.7 Volts.
This voltage drop is multiplied by the drawn current.
This power must be dissipated by the voltage regulator.
The regulator will have to dissipate 2.76 times more as when the powersupply was 5 Volts.

Bertus

 

Hello,

When the powersupply is giving 8 Volts in stead of the mentioned 5 Volts, the regulator will have to drop 4.7 Volts in stead of 1.7 Volts.
This voltage drop is multiplied by the drawn current.
This power must be dissipated by the voltage regulator.
The regulator will have to dissipate 2.76 times more as when the powersupply was 5 Volts.

Bertus

Hello Bertus,

I will try another 5v powersupply and see , thanks