# L293D Question

L293/D Datasheet: http://users.ece.utexas.edu/~valvano/Datasheets/L293d.pdf

Page 4:

Continuous output current, IO: L293D . . . . . . . . . ±600 mA

Is the current mentioned in the datasheet 600mA per pin (pin 3,6,11,14), per 2 pins (motor) 3,6 or
in total the chip can handle?

That's what I don't understand, can someone explain to me please.

D.60

I believe it is 600ma per pin, and you use a pin on each side of a motor. Therefore, the current flows though both. So 600ma per motor.

I believe it is 600ma per pin, and you use a pin on each side of a motor. Therefore, the current flows though both. So 600ma per motor.

So it's 600mA per pin or per motor? Do you mean like pin 3 600mA, 6 another 600mA?
We can use from 2 up to 4 motors with the chip.

600ma per pin. I choose not to eschew obfuscation.

600ma per pin. I choose not to eschew obfuscation.

So that means if i use 4 motors the chip will be able to handle in total 2.4A? isn’t that kinda a lot?

2.4A * 36V = 86.4W,

Wrong calculation. If running, the motor drops most of the 36V.

But at 2.4 A total, it is very unlikely that the L293D won't overheat and shut down.

Therefore, try piggybacking the chips. It doubles the current that you can handle.

Or go to more efficient MOSFET drivers, browse at www.pololu.com

Or the TLE5206 from DigiKey. 5A continuous, but only 2 half-H drivers.

try piggybacking the chips

I don't understand what you mean.

Or go to more efficient MOSFET drivers

MOSFET are good but expensive sometimes depends on what you need, there is cheap ones as well.
It's nice to build a MOSFET H-Bridge but it's better to have a compact H-Bridge microcontroller, less space
and easier to use in diagrams and circuits.

EDIT:

TLE5206

7\$ for a single one? dam that's expensive, yes 5A but does it have the posibility of controlling
the PWM?

D.60

Piggybacking is soldering an identical chip on top of the driver. It will double the current capacity while using the same connections.

Piggybacking is soldering an identical chip on top of the driver. It will double the current capacity while using the same connections.

Well never had that kind of idea thanks for telling me, sounds like a horrible idea and kinda strange
but if its working is not so bad

btw just for the record, i got 10x L293D for 2.40\$ it's big difference between 7\$ TLE5206

L293 from Ebay? I was just browsing for motor drivers for my specific needs

What if you Piggybacking like 10x chips that will like 6A per pin, will they be able to share the
current equally?

edit:

L293 from Ebay?

L293D from aliexpress.com cheapest than ebay

L293 is old, inefficient technology that runs hot because the output drops something like 2V, and with 2A of current flow that's 4W. You want nice hefty MOSFET parts with low Rds that won't heat up with high current flow. Go browse Pololu and you can see what I'm talking about. Rds of 0.1 ohm will only drop 400mW, and 0.1 ohm is high.

L293 is old, inefficient technology that runs hot because the output drops something like 2V, and with 2A of current flow that's 4W. You want nice hefty MOSFET parts with low Rds that won't heat up with high current flow. Go browse Pololu and you can see what I'm talking about. Rds of 0.1 ohm will only drop 400mW, and 0.1 ohm is high.

L293D Inefficient, why?

I'm planing to use them to control 2 motors per chip, if the motors will not be under load the motors
will only need up to 30~50mA each, maximum voltage and speed the motors will run at 100mA and
if they are under and only under load or stall they will need up to 350mA current each, that will be
maximum 2W per motor so yea 4W maximum.

Now I'm not familiar with the power consumption and how hot it will be.

Inefficient because they were designed in 1986 using the technology of the day.
Look at sheet 5:
VOH High-level output voltage
VOL Low-level output voltage
At 600mA, the output voltage drop is 1.2 to 1.8V.
That’s 1.2v to 1.8V that your motor will not see, so it will be less powerful.
The part dissipates Power = IV = IIR or VV/R (V=IR, Ohms law, sub in IR for V, or V/R for I)
IV: 1A1.8V = 1.8W, your motor gets Vsource -1.8V, may be enough loss to slow your motor down.
IIR: 1A1A low Rds = low power, and V=IR = low voltage drop. 1A*.1ohm =0.1V, so your motor can run faster.

New technology MOSFETs with low Rds are the way to go.

L293d.pdf (372 KB)

If you can't get parts in Brazil, that's understandable too, do the best you can with what you have. Some heatsinking may be required if the motors at higher currents for a longer time.

Looking forward to the Olympics? Are you far from there? (I think you have mentioned Brazil in the past, please forgive me if I have that wrong).

Well 1st thanks for the info I will look up for MOSFETs next time, would be nice if you can suggest
me a MOSFET of the same abilities as the L293D, controlling the direction and PWM/speed of the
motor.

If you can't get parts in Brazil, that's understandable too, do the best you can with what you have. Some heatsinking may be required if the motors at higher currents for a longer time.

Looking forward to the Olympics? Are you far from there? (I think you have mentioned Brazil in the past, please forgive me if I have that wrong).

Heheh

I'm not from Brazil, I'm from Europe (country? I would like to keep it private if that's ok )

From my point of view L293D are good even if they are old, it's a good start to gain experience
and learn.
And really cheap

I believe TI made a succesor to the L293- something like SN775410. But I have searched for different combinations of that and found nothing.