If I am powering an atmega328 with a 7805 voltage regulator...is it okay to also power the 4 servos I am controlling with that power? Or do I need to power them with something separate? I am not so familiar with how servos work exactly. I know each one takes 160mA or 60mA when it is idle...the atmega takes 25mA...the 7805 gives 500mA. So this should be okay? (I wouldn't ever be controlling all four at once...just one or two at a time max)
The servos I am using are here: http://users.ece.utexas.edu/~valvano/Datasheets/ServoHS311.pdf
Thanks for any feedback
The standard 7805 is rated at 1A, so it should be more than adequate for the task, although it may need a heatsink depending on the input voltage.
If you look at your data sheet, you will see the servos use 700-800ma at stall conditions.
thanks for your help...so since they take 700mA at stall current...one 7805 is not enough? what is the difference between stall current and running current at no load? why would stall current take so much more amperage than running current? if a 7805 is not enough to power these do i just need an adapter that is 4.8-6v and gives out like 3.5A? do those even exist? or would i need to power each servo separately?
The servo stall current happens when the servo is trying to exert maximum force. If there is any load on the servo, a 7805 chip will get hot. Below is a 2A regulator chip that might be worth considering. Keep the servo and arduino power supplies seperate to aviod low voltage resets of the arduino when the servos draw a lot of current. You can put a small diode on the ground of the 7805 chip to raise its output voltage from 5v to 5.7v for better servo performance (bottom). The 7805 chips are inexpensive, so one for eaxh servo might be good.
Use external source to power up your servos..and connect that source's ground to arduino ground. So now reference voltage level is same!!
you can use 7806 for better performance.
I would recommend capacitors direct from the input and output to ground on the 7805 swell as what you have. Putting that decoupling through a diode is not very effective.