Arduino + motor driver (L293D) power consumption

Original question : My chip consists of 1 arduino nano and 3 L293D motor driver. This chip is for controlling motors and other equipments. My question is about how to calculate power consumption of my chip. I searched for about voltage settings of arduiono nano and L293D motor driver, but there are many phusical thinks that i cannot understand. How many hours 2000 mah battery enough for.

Note : Motors and other equipments will have own power supply. That i say it is just for Arduino Nano, Bluetoot adapter for arduino (HC-05) and 3 motor driver (L293D).

• If you are powering all modules from the bench power supply, note down the voltage and the current. Product of the two will give you power consumption.
– Umar
Commented Mar 7, 2017 at 6:09
• It also depends on the voltage provides by power bank.
– Umar
Commented Mar 7, 2017 at 6:09
• I cannot got power settings of L293D. I read technical specification but cannot understand Commented Mar 7, 2017 at 6:15
• Sorry, I misread your question as you already have it connected and working, correct me if i am wrong. you are only looking for the battery life time estimation.
– Umar
Commented Mar 7, 2017 at 6:23
• are you driving three motor with one L293D or you are using 3 L293D motor driver ICs?
– Umar
Commented Mar 7, 2017 at 6:28

To calculate L293D power usage, you need to lookup the datasheet and check the Voh and Vol. These are the High level voltage and Low level voltage.

On Texas Instruments datasheet, we have:

So you will lose about 1.4V when your output is -0.6A, and 1.2V when your output is 0.6A. If your motor uses 2 outputs (bi-directional) then you have to add the losses.

Then, to get the power dissipated on the power side of the L293D, you take this voltage drop and multiply by the motor´s current.

Example:
Lets say your motor is using 0.6Amps and your power source is 5V + 24V @ 25°C (just like in the datasheet).

The power loss in the L293D is aprox.: (1.4V + 1.2V) x 0.6A = 1,56W.
Since you´ve lost 2.6V in the L293D, your motor will use: (24V - 2.6V) x 0.6A = 12,84W.

To measure this directly on your real circuit, use a multimeter in Ampere mode to measure current, and Volt mode to measure the voltage on the motor terminals when it is on.

As for the arduino, your battery has 2000mAh and I assume it is a 7,4V battery. So it has 2000mAh x 7,4V = 14800mWh of energy stored in it.

If you are running your arduino at 5V, then: 14800mWh / 5V = 2960mAh at 5V.
Now let´s say your arduino and it´s modules, the L293D´s and the motors are using 500mA average, then your battery will last: 2960mAh / 500mA = 5.92 hours in theory.
Why "in theory"? because in real life it is impossible to extract all the energy from inside a battery. A minimum ammount of energy has to remain inside it and some power is lost when you extract it´s energy.

Of course, this is a very raw calculation. In reality, you have to calculate power used in every component. But by calculating the "big guns" first you can get an idea of where you stand.

Basic power loss calculation is done by multiplying voltage drop by the current in each circuit. Voltage drop is the difference between your volts before and after the component.