We are using this stepper motor driver (https://www.sparkfun.com/products/12779). In its datasheet, there is no mention of efficiency? How much is the efficiency in motor drivers like these?
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This question is a little problematic - old drivers using things like power resistors had low efficiency, but modern *drivers* are quite efficient. The overlooked problem is that stepper *motors* are generally quite inefficient. So your question can be interpreted *literally* about efficiency of the *driver*, or the question you haven't quite asked but probably should have about efficiency of the *system* can be answered instead. So far you have one of each type of response... – Chris Stratton Jul 19 '17 at 16:21
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Does neither of the answers proposed, answer your question? – Harry Svensson Jul 22 '17 at 17:18
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A stepper motor driver is more or less a DC/DC converter with variable voltage output. So it's probably something around 90%. The same efficiency as a regular DC/DC converter.
Harry Svensson
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It may actually be better than that, arguably the only loss in the *driver* is the FET RDSon. However, efficiency of the *system* is going to be low, with most of the losses in the motor. – Chris Stratton Jul 19 '17 at 16:19
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Hmmm, 90% is still kind of a good **approximation**, in order for a stepper to get any current, the current has to pass through 2 FETS, rather than 1 as a DC/DC would do. And then the current will rotate back into the stepper motor through the body diode of one of the FET's (or just a regular diode component), so there's some losses in the diode. And on top of this you got the switching losses, higher frequency => more losses in switching. If you want to get rid of the audible noise you need 20kHz+ => quite a bit of switching losses. 95+% maybe if you got some high quality top notch stuff. – Harry Svensson Jul 19 '17 at 16:31
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We did our own experiment by measuring the input and output power of the motor driver and the efficiency came out to be around 75%. – Soumil07 Jul 24 '17 at 06:04
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In motion control efficiency is usually mechanical output power (w*T) divided by electrical input power (U/I). Obviously it is different for different working points. For a standing stepper it's zero. On higher speeds sometimes nice figures may be reached, but bot very nice. Because for stepper you always take spares, otherwise it slips.
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This is the stepper motor linked in the question: -
How much is the efficiency in motor drivers like these?
It's not great. Consider the saturation voltage of the output transistors in the A3967 chip that drives the stepper: -
If you are taking the full 750 mA into your stepper then a total volt drop of about 3 volts can be expected. If you are running from a supply of (say) 15 volts, you are going to only see about 12 volts across the stepper motor coils. This means that peak input power is 15 x 0.75 amps = 11.25 watts whereas peak output power is only 9 watts. That means the peak efficiency is about 80%. If the supply voltage is (say) only 6 volts then nearly half that voltage is lost in the driver rendering an efficiency of only 50%.
There are better devices that only drop about 0.5 volts. See this for a critique of other similar basic stepper drives / motor drives for a comparison.
Andy aka
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Thanks for the downvote whoever it was (and I have my suspicions given that 4 downvotes in rapid succession have just happened to answers I have given and, this is very much reminiscent of last year when a certain person had a mini vendetta against me). – Andy aka Jul 19 '17 at 18:47
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Thanks Andy for the answer. Did you actually mean a voltage drop of 2V instead of 3 since the output voltage is saturating at around 2V in the graph? – Soumil07 Jul 19 '17 at 19:09
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It's 3 volts given that there are two active elements in a H bridge: a source and a sink and the sink drops about a volt. Both elements are in series with the motor winding. – Andy aka Jul 19 '17 at 19:10