Need help in arranging the voltage regulator for my servo

[bajji93]

I'm giving my servo's specs here
current drain - 8 mA (idle)
160 mA (no load work)
stall current - 600 mA
operating voltage - 4.8 to 6 V

I'm using 10 servo motor for my robotic leg, which will be controlled by arduino. I tried using a common 7805 for all the 10 servos. But, later i realized that if all the motor are working, it needs 10*160 mA= 1.6 A of current. But, 7805 can out only 1 A of current. So, i decided to use a dedicated 7805 for each of the servos. So, i'll need 10 7805 for that. I'll be using a 11.1 V 3000mAh Li-Ion battery for this.

My question is will each of my 7805 be able to deliver 500mA at 5V if all the servos are working with some reasonable loads.

:shock::shock:

 

[chuckey]

According to :- https://www.sparkfun.com/datasheets/Components/LM7805.pdf , the 7805 has an internal current limit at 750 mA, though they quote I max at 1 A! It looks a reasonable idea.
Frank

 

[bajji93]

According to :- https://www.sparkfun.com/datasheets/Components/LM7805.pdf , the 7805 has an internal current limit at 750 mA, though they quote I max at 1 A! It looks a reasonable idea.
Frank

so, you say that using dedicated voltage regulator will work. Am I correct ?
:?:

 

[chuckey]

Yes one per motor,.
Frank

 

[bajji93]

Yes one per motor,.
Frank

thanks from my heart !

 

[crutschow]

Is the battery capable of readily supplying 1.6A of continuous current?

Note that 500mA @ 5V output will dissipate about (11-5) * 0.5 = 3W in each regulator since you are wasting more than half the battery's energy so they will need to be on a heatsink. It would be more efficient if you used a switching regulator or a lower battery voltage.

 

[bajji93]

Is the battery capable of readily supplying 1.6A of continuous current?

Note that 500mA @ 5V output will dissipate about (11-5) * 0.5 = 3W in each regulator since you are wasting more than half the battery's energy so they will need to be on a heatsink. It would be more efficient if you used a switching regulator or a lower battery voltage.

Yea. My battery can discharge 3A readily!!!
I can use a common voltage regulator if I'm using a switching regulator huh??

 

[bajji93]

Is the battery capable of readily supplying 1.6A of continuous current?

Note that 500mA @ 5V output will dissipate about (11-5) * 0.5 = 3W in each regulator since you are wasting more than half the battery's energy so they will need to be on a heatsink. It would be more efficient if you used a switching regulator or a lower battery voltage.

okay, we'll take the worst case. consider all my motors are in need of 500 mA, then the total current needed would be 500mA*10=5000 mA=5A . LM2596 can out only 3A of current at max.
I think using dedicated 7805 for each motor will not take out more power loss if use 7.4 V as the input voltage to the 7805. The loss in this case will be around (7.4-5)*0.5A= 1.2 W per regulator.
I'm totally confused here !!!! :-?

 

[crutschow]

okay, we'll take the worst case. consider all my motors are in need of 500 mA, then the total current needed would be 500mA*10=5000 mA=5A . LM2596 can out only 3A of current at max.
I think using dedicated 7805 for each motor will not take out more power loss if use 7.4 V as the input voltage to the 7805. The loss in this case will be around (7.4-5)*0.5A= 1.2 W per regulator.
I'm totally confused here !!!! :-?

You originally stated the battery was 11.1V. Now you are using 7.4V. I'm confused also. :-?

 

[bajji93]

You originally stated the battery was 11.1V. Now you are using 7.4V. I'm confused also. :-?

hehe.. is it okay if use a 7.4 input ??
cuz its a 3 cell battery.. 3.7 V * 3=11.1 V @ 1500mAh !!

 

[crutschow]

hehe.. is it okay if use a 7.4 input ??
cuz its a 3 cell battery.. 3.7 V * 3=11.1 V @ 1500mAh !!

It's okay for the 5V regulators but where are you getting the 7.4V? Are you thinking of using just two of the three cells? Is that possible with your battery?

 

[Audioguru]

A Lithium rechargeable battery cell voltage AVERAGES at 3.7V. It is 4.2V when fully charged (a 3-cell battery will be 12.6V) and a discharged cell needing a charge is 3.2V (a 2-cell battery is only 6.4V which is too low to feed an ordinary 5V regulator).

 

[bajji93]

It's okay for the 5V regulators but where are you getting the 7.4V? Are you thinking of using just two of the three cells? Is that possible with your battery?

ya. i thought of using only 2 of available 3 cells. My battery totally has 4 terminals(one common ground, 3.7v , 7.4v and 11.1v) .. so, two cells will do the work. correct?

- - - Updated - - -

A Lithium rechargeable battery cell voltage AVERAGES at 3.7V. It is 4.2V when fully charged (a 3-cell battery will be 12.6V) and a discharged cell needing a charge is 3.2V (a 2-cell battery is only 6.4V which is too low to feed an ordinary 5V regulator).

each cell will provide 3.7V .. 3.7*2 = 7.4 V ... will that not be enough for the 7805?

 

[crutschow]

To insure operation at the end of the battery's discharge voltage of 6.4V you could use a low-dropout regulator such as this.

 

[bajji93]

To insure operation at the end of the battery's discharge voltage of 6.4V you could use a low-dropout regulator such as this.

Okay.. thank you! I'll do that !!