Hello Electrical engineering community.

I am using motor drivers to get 5v to 12v and I am controlling each motor with each IC L293D.

Each motor driver is controlling each motor and the outputs is taken from only one side of the IC.

12v battery -> 7805 voltage regulators -> two L293D IC's with 12v given and the output is taken via diodes to resist back emf-> 100 rpm 12v DC motors.(the other two output pins of the IC are left floating).

Everything is working fine but from 1st pin I am getting an output of 10.23 and 0.72.. and from the second IC I am getting an output of 9.1 and 0.61.

I don't understand why is there a difference of 1v.

Sorry I can't upload a picture of the ckt or make the question in more detailed manner as I am using my phone.

Any help would be appreciated.

I don't understand what tags to be given

  • \$\begingroup\$ L293 H bridges are pretty poor if you expect full rail-to-rail performance. See this: electronics.stackexchange.com/questions/108686/… - they use Darlington transistors that won't efficiently switch on to a low volt drop. \$\endgroup\$
    – Andy aka
    Commented Sep 30, 2014 at 10:44
  • 1
    \$\begingroup\$ yes, as Andy said the darlington transistor drivers will drop anything between 1.2-3 volts, as you've noticed. Sad right? That loss of voltage is burned as heat, basically. You should find MOSFET based motor drivers whenever possible. \$\endgroup\$
    – KyranF
    Commented Sep 30, 2014 at 10:54
  • \$\begingroup\$ Yes I also find one of the IC with less output voltage, getting heated up more than the other. \$\endgroup\$
    – Jasser
    Commented Sep 30, 2014 at 11:00
  • \$\begingroup\$ What should be done for the output to be the same. @KyranF \$\endgroup\$
    – Jasser
    Commented Sep 30, 2014 at 11:03
  • 1
    \$\begingroup\$ the difference in voltage is due to heat, current, or some other poor tolerance issue. Don't stress over just 1 volt. I don't think you understand enough of this to see what the issue is here. It is the fact the L293 series (no matter what nice little letters are after it) use bipolar junction transistors in their output drive stage, so they will not put out a nice ~12V - 0V and they will also have various output voltages from eachother depending on each individual IC and also the motor it's driving. \$\endgroup\$
    – KyranF
    Commented Sep 30, 2014 at 12:56

1 Answer 1


You seem to expect that each exemplar of a particular chip type behaves the same, but that is a misconception. What you can expect is that each exemplar behaves within the specifications guaranteed by its datasheet (if you use it within the limits that are also stated in said datasheet).

(But take care: different manufacturers might produce chips with the same type number, yet the each have their own datasheet, which might not fully agree!)

For the TI L293D the relevant part of the datasheet is shown below.

enter image description here

It states that the maximum voltage drop (for a current of 0.6A) over both the high side and the low side of the output drivers is 1.8 Volt. Note that it does not state a minimum, so the drop voltages you observe are all in accordance with the datasheet.

Side note: strictly speaking these values are valid for Vcc2 = 24V only, but it is likely that they apply for other reasonable Vcc2 values. The fact that the datasheet mentions these figures for 24V only is an indication that this chip was never meant for low voltages, and indeed it doesn't perform too well at lower voltages. But it is cheap and widely available...

PS the D after the L293 only means that the chip has built-in freewheel diodes. It has nothing to do with the voltage drop of the drivers.


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