# Powering a Futaba S3003 servo

I'm working on a project and I need to use a Futaba S3003 servo, I'm about to buy a transformer that gives 6v, the thing I'm not sure about is the amperage.

Is this transformer ok? I can choose 3V, 4.5V, 6V, 7.5V, 9V and 12V, but its 800mA and I don't know if it is OK with the servo.

EDIT: Link to servo datasheet: http://www.es.co.th/schemetic/pdf/et-servo-s3003.pdf

The transformer is only for powering the servo

• How much current does servo need? Is the transformer only for servo? Commented Jul 2, 2015 at 14:19
• How about a hyperlink to the servo's data sheet (I don't mean some ebay page but a proper data sheet) plus, have you read the servo's data sheet? Commented Jul 2, 2015 at 14:20
• @Umar the datasheet says: Current Drain (6.0V): 8mA/idle. But I have no idea what it means. Yes is for powering only the servo. Commented Jul 2, 2015 at 14:27
• @Andyaka I've read the datasheet I found (I've just pasted the link in the question) and ti says: Current Drain (6.0V): 8mA/idle. But I don't know what it means. Commented Jul 2, 2015 at 14:28
• The data sheet do not have the current requirement. OR there is a surprise for me. But, 800 mA is definitely over kill. It will work for sure Commented Jul 2, 2015 at 14:37

I'm going to do a bit of math on this and make some outlandish assumptions because the data sheet does not give enough information. Stall torque is reckoned to be about ~3.5 kg.cm and full speed is reckoned to be about ~0.2 seconds per 60 degrees of rotation.

Many thanks to Brian Drummond for correcting my stupid math mistake so here's the corrected answer with a bit more theory. First, it is a fair assumption that the servo will contain a small dc motor and, generally, this is the torque-speed characteristic for such a motor: -

Note the blue line - at zero speed, torque is maximum (stall torque) and with no load (zero mechanical torque other than losses in the bearings), speed is maximum. This is just a graph of any old motor I stole from the internet BUT it is largely representative of all DC motors.

Now look at the red curve - this is mechanical power out and equates to: -

Mechanical power = $2\pi n T$ where n is revs per second and T is torque in newton metres (N.m)

So, the 3.5 kg.cm translates to approximately 35 N.cm or about 0.35 N.m

A "speed" of 0.2 seconds per 60 degrees is upside down but, rearranging, it basically means 0.166 revs per fifth of a second and this translates to 0.833 revs per second.

Go back to the graph and note that max power is roughly when both torque and speed are at their respective half values therefore,

Power is $2 \times\pi \times (0.833/2) \times (0.35/2)$ = 0.458 watts

This is the peak mechanical output power I have estimated from the limited information in the data sheet. Of course it might be a bit higher or lower.

Next, a motor this small may not be very efficient at converting electrical power to mechancial power so we need to apply a "frig" factor. Let's say it's 50% efficient. This now means the input electical power might be about 0.9 watts.

There is a lot of hand waving here and to play a little more safely you might assume that the power supply needs to supply maybe double this value. I suspect there will be a little H bridge controller inside the servo and this might only be 50% efficient so maybe 2 watts should cover most eventualities.

Please insert your own numbers and frig factors into the equations if you think I may have over-egged the omelette.

• With all due respect, can you help me in understanding deriving power from torque and speed?, approximately. Commented Jul 2, 2015 at 15:17
• power = torque * speed. In SI units, that's Nm * radians/second. 1kg force is approx 10N so approximately 0.35Nm * 5 radians/second or 1.65W. Andy, you might want to double check.
– user16324
Commented Jul 2, 2015 at 15:18
• @brian oops I shall correct my math later when I get off this android and on my pc. Looks like any old power supply should be good enough. Commented Jul 2, 2015 at 15:49
• I was about to comment that a 20 watts power supply was to much, is it? Commented Jul 2, 2015 at 16:08
• @elunicotomas - please see my answer amendments - I estimate you need 2 watts of power. That's 6V at 333mA. Commented Jul 2, 2015 at 17:16