Electronic – L293D overheating

What power does the parts list specify for R4? The output current is up to 2 A, then a 2.2 Ω resistor will dissipate 8.8 W. A 3 Ω resistor will dissipate even a lot more: 12 W. Even at 1 W a small resistor will get hot.

Now the thyristor only conducts for short periods, in order not to have a higher voltage on the smoothing capacitors than necessary. And here Wouter brings up an important point. If the current is pulsed it will cause a lot more dissipation. At a constant 2 A current the 3 Ω resistor dissipates 12 W. If we pulse 20 A at a 10 % duty cycle we get the same average current, but when we calculate the dissipation:

\$\mathrm{P = 10\% \times I^2 R = 0.1 \times 20^2 \times 3 \text{ W} = 120 \text{ W } !! }\$

Note that the circuit was designed to reduce power dissipation in the voltage regulator's power transistor.

Exceeding the ratings continuously is not a good idea. The part will fail. And possibly catch fire (I am not kidding). A heat sink will not improve things by much on a plastic DIL. The junction isn't thermally connected well enough to the case to bring the heat that far, even if you do manage to firmly attach a heatsink to the surface.

However, you seem to be fairly close to the limit, and you claim to have a bagful of L293s lying around. You can simply use multiple channels in parallel (one chip per motor instead of half chip per motor). This should be enough to keep things within the acceptable limits.

The L293D can have multiple channels connected in parallel. This is not a generally applicable approach, though. Don't try this with, say 7805s or LDOs without first considering the consequences. Power regulation ICs which allow this will have mention of it in the datasheet as load balancing. However, the L293D is not strictly a power regulating IC, and your load is fairly specific (a motor). There will be minor separation in (a) turn on time and (b) on resistance. (a) will mean that there will be a brief transient on one channel before the other turns on and (b) means that the current drawn through the two channels will be unequal.

The differences will usually be small enough to not be a cause for too much worry except if you've also got highly sensitive systems on the same supply lines or physically very close to it. For most practical purposes it's fine, and personally I've generally preferred to use the L293 in parallel instead of the L298 if the current is under an ampere per motor because of the headache L298's package brings to the table.