This may seem a very silly question, but does a servo draw more current when it's moving? I need to be able to log when a servo flips, and being able to log the current drawn from the battery pack seems to be the most obvious and simple solution
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\$\begingroup\$ You might try it? :-) \$\endgroup\$– user17592Commented Jan 9, 2013 at 17:35
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\$\begingroup\$ umm... I don't have access to anything at the moment, and my (very limited) budget will be sanctioned only after I have finalized what I require, hence the question :) \$\endgroup\$– phoCommented Jan 9, 2013 at 17:37
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\$\begingroup\$ Okay, then it's okay :-) \$\endgroup\$– user17592Commented Jan 9, 2013 at 17:51
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\$\begingroup\$ @PranavHosangadi If you're still looking for alternative approaches, it may be worth joining the Electrical Engineering Stack Exchange chat room and seeking views from those present there. \$\endgroup\$– Anindo GhoshCommented Jan 9, 2013 at 19:23
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3 Answers
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Assumption: Servo motor in question is a hobby-type servo, not an industrial high torque or heavy duty servo.
The bulk of the current through a (hobby-type) servo motor is the current through the coils of the DC motor typically used within the servo housing. Some current is required by the servo controller circuitry, but that is negligible in comparison.
Current through the servo will thus be high when it is changing position, as the DC motor must turn till target angle is achieved.
Current is then low after it reaches the new position and stops, as long as there is no opposing torque that the servo must fight to retain the new position.
Current will be high again if a torque is applied to the servo arm... this last can actually be higher than current while moving, up to the stall current of the internal DC motor. This is because the servo controller drives the motor hard to force it to hold the target position.
Thus, depending on current as a means to detect servo traversal is unreliable at best. Instead, a mechanical solution such as a homemade rotary limit switch made using a conventional limit switch levered by one of the servo arms, would be a recommended approach.
answered Jan 9, 2013 at 18:34
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1\$\begingroup\$ The others seems to forget to mention your 3rd bullet, which is important to realise. Worth a +1 \$\endgroup\$– jippieCommented Jan 9, 2013 at 18:41
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Like a motor, if it isn't moving, there's no current in the windings (or insufficient current).
(Though technically a motor can be under too heavy a load and there's a lot of current despite its inability to move, hastening its failure, but that's another subject.)
Simply put, a servo that isn't moving is not powered, and therefore is not drawing current.
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\$\begingroup\$ The servo will still be on, just in the -45 state. When I flip it, it'll go to the +45 state. From what I understand, a servo that is off will move when the torque on its arm exceeds the gear friction (not what needs to be done). Am I right? \$\endgroup\$– phoCommented Jan 9, 2013 at 17:41
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\$\begingroup\$ If it is being held at -45 degrees with power, it will have a current draw. That current may be more or less than that which is required to move it to +45. It depends on the physical load. \$\endgroup\$– JYeltonCommented Jan 9, 2013 at 17:44
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\$\begingroup\$ If I'm logging a graph of the current draw, will it be obvious when the servo flipped? \$\endgroup\$– phoCommented Jan 9, 2013 at 17:46
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\$\begingroup\$ @PranavHosangadi Explain "flipped"? \$\endgroup\$ Commented Jan 9, 2013 at 17:50
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\$\begingroup\$ @AnindoGhosh, flipped = went from state 1 to state 2 (e.g. far left to far right) \$\endgroup\$– phoCommented Jan 9, 2013 at 17:51
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The amount of current a servo draws will be related to the load on the servo (and consequently the servo's torque. There also may be additional torque required (/current drawn) when moving a load from a stand still.
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\$\begingroup\$ also, if its position you're after - you may be better off opening up the servo and soldering a wire to the internal potentiometer. that can probably give you a decent estimate of position to track 'flips' \$\endgroup\$– JonCommented Jan 9, 2013 at 18:50