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....
Prio one would be to verify the voltage input to the driver - does your step-up manage to supply 5V when the motor is powered? Don't just assume 200mA is enough - measure the voltage when loaded.
the voltage shows across pin 8 is 5.12v
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.