0

I was using a motor driver breakout board with a L293D IC on it, along with a 7805 voltage regulator and a couple of capacitors.

enter image description here

enter image description here

I have been using it for 4-5 months and the wire connections stayed the same. Everything was well connected. Last night it suddenly stopped working and there was huge amount of smoke coming out of the board. Upon inspection I figured out that the IC was burnt.

enter image description here

I am unable to figure out why. Maybe I'm missing some huge part in my knowledge. Any help in figuring out what went wrong would be really helpful.

Would it be safe to use the same board after replacing the wasted IC?

ocrdu
  • 8,705
  • 21
  • 30
  • 42
jash101
  • 133
  • 1
  • 6
  • [Try this](https://electronics.stackexchange.com/questions/108686/l293-l298-and-sn754410-h-bridge-drivers-on-low-voltage-power-supply) - they just can't be expected to perform at low voltages with circa 1 amp loads. – Andy aka Jul 19 '20 at 18:37
  • Hey Andy, thanks for the link! I was using them with 12V voltage supply, and the motors draw somewhere around 200mA. The max current the IC can handle as far as I know is 600mA, so I think what you are referring to doesn't really apply in my case I've added a few pictures for clarity :) – jash101 Jul 19 '20 at 18:48
  • Why did you mention the 7805 regulator? They are still $hit devices BTW. – Andy aka Jul 19 '20 at 18:53
  • Is it? Why is that? – jash101 Jul 19 '20 at 19:03
  • Why did you mention the 7805 regulator? Ditto the previous. – Andy aka Jul 19 '20 at 20:02

2 Answers2

0

Least thermal resistance with a L293 is to the centre/ground pins.
These should connect to a heat sink - think 6 cm² (1 square inch) of 35 micron copper - this should reduce resistance from about 80°C/W to about 15.
You report using 1/3 rated current. Try to establish case temperature while the going is good.
There used to be solder-on heatsinks.

Mounting a Power-DIP in a plastic socket looks contraindicated without advance warning.

greybeard
  • 1,469
  • 1
  • 4
  • 17
0

No doubt about it the IC could not dissipate enough heat so it decided to slowly fry itself. It must have a heat sink, follow the information on the data sheet. Remember that chip is dropping about 3V 1.4V per darlington pair, and there are two in each motor circuit.

The L293D is designed to provide bidirectional drive currents of up to 600-mA at voltages from 4.5 V to 36V (TI datasheet https://www.ti.com/lit/ds/symlink/l293d.pdf?ts=1672323097370&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FL293D. Both devices are designed to drive inductive loads. If you replace it get rid of the socket and add a heat sink. The PCB copper will get rid of some heat. Glueing one to the plastic is not very effective but will help.

Gil
  • 4,951
  • 2
  • 12
  • 14