Hi all, I need a small help regarding L293D. I am making a small project using L293D. Bo motor which hardly consumes 60ma at 5v. i am powering my L293D using a step up circuit 3.7V to 5V from 3.7v li-ion battery. output current from step up circuit is 200ma which is not a issue. my problem is the motor voltage pin no 8 shows me 5.12v. when i drive my motor the voltage accross shows me only 3.00v... due to which the motor drives very slow. can anybody please help why this happen....
i have even checked that when motor is in motion... the voltage across motor is 3v but pin 8 of L293D shows 5.12v...
The datasheet for the driver says that at -0.6A it will output min VCC2 – 1.8 typical VCC2 – 1.4
5-1.8V = 3.2V 5-1.4V = 3.6V
And the datasheet says that at 0.6A it will output typical 1.2V max: 1.8V
So - how do you feel your design deviates from the datasheet "contract"?
yes.... you are right... thanks for the help...
what if i use L298N... can it solve my problem.. i have gone through the datasheet of L298N... but sorry did not understood anything
You can normally manage better with a FET-based solution. With two FET with 1 ohm Ron each, the total voltage drop would be 2*1*0.06 V = 120 mV. Or 12 mV if each FET has a Ron of 0.1 ohm.
Bipolar bridges normally has larger voltage drops.
But much FET alternatives expects a larger voltage than 5V to make sure the FET opens/closes fully and doesn't operate as a programmable resistor.