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[General] Doubt on stepper motor ratings

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ismu

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hi am using SY57STH76-2804A stepper motor , which datasheet rating is
3.2v , 2.8A and Coil resistance 1.13Ohms .

1-so can drive thios motor with 12v/24V ?
2-When i drive with 12V then current will be =12/1.13=10.6A :shock:
3- what is the impact of coil inductance? what should i take care on designing?
 

The usual way to do this is to drive it with 12v/24v but with a current limited supply set to 2.8A (or less).

The higher supply voltage will produce a very fast ramp up of current in the inductive motor windings, right up to the current limit value.
This makes much faster stepping rates possible, and with higher torque when running fast.

If you do not need the speed just run the motor at 3.0 volts.
 
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    ismu

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what is the meaning of rated voltage 3.2v ? then how can give >3.2v?
 

hi am using SY57STH76-2804A stepper motor , which datasheet rating is
3.2v , 2.8A and Coil resistance 1.13Ohms .

Those are the CONTINUOUS ratings for drive with direct switched dc current.

You can run it like that but it will be slow, and torque will fall off with increasing speed.
 
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    ismu

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The usual way to operate a stepper motor at higher voltage is to apply PWM with current regulation.
 
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    ismu

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if i use 12v supply instead of rated 3.2v [datasheet rating is 3.2v , 2.8A and Coil resistance 1.13Ohms .]
then how much should be the Driver chip H bridge driver current capability ?
i have seen in some case rated volt is less still people are driving with high voltage, any reason?
 

The motor driver should implement constant current operation, maximum current is rated 2.8 A.
 

A more simple and very crude way to do this is to fit external resistors to each winding.

If its driven with 12v and you want 2.8 amps, then total resistance needs to be 4.285 ohms.

The stepper already has 1.13 ohms, so add a 3.15 ohm resistor.
3R3 would be about right.
That's a huge 26 watts of power dissipation which really makes it impractical, but it is one possible solution.
 
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    ismu

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The motor driver should implement constant current operation, maximum current is rated 2.8 A.

you know any example circuit to get constant current to motor? and should be cheapest way

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Just use fixed PWM duty cycle to get proper current.

for driving bipolar stepper mototr ,we wrer using pulsed signed on A, A+ ,B, B+ terminal , then how can i make again PWM on this ? can you please draw a simple wave form for half drive signal with PWM ?

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A more simple and very crude way to do this is to fit external resistors to each winding.

If its driven with 12v and you want 2.8 amps, then total resistance needs to be 4.285 ohms.

The stepper already has 1.13 ohms, so add a 3.15 ohm resistor.
3R3 would be about right.
That's a huge 26 watts of power dissipation which really makes it impractical, but it is one possible solution.

agree.... ! But am driving with 12v Source and motor rating is 3.2v ,,, will it damage?
 

agree.... ! But am driving with 12v Source and motor rating is 3.2v ,,, will it damage?
Not if you use a 3R3 resistor.

But the resistor is going to get mighty hot.
 
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    ismu

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A more simple and very crude way to do this is to fit external resistors to each winding.
That's not funny, man! You will reduce efficency 5 times. 2.8A * 20V will give 56W power lost on each!

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for driving bipolar stepper mototr ,we wrer using pulsed signed on A, A+ ,B, B+ terminal , then how can i make again PWM on this ? can you please draw a simple wave form for half drive signal with PWM ?
Just take a PWM frequency 10 times higher than expected step time. For example, if you planing to use 1ms step time, 1kHz per step, use 10kHz or higher PWM frequency. I suppose, 1/10 duty cycle will be good enought. But this is actual only for 'on hold' static mode. For movement PWM duty cycle have to be calculated according constand current law. But if you not planing to use variable frequency, just find a good value for duty cycle that gives you expected force and current. It can be lower than maximum rating shown on sticker.
 
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Hi,

External resistors make no sense because of their huge power loss = heating. But for sure it prevents the stepper from overload.

A fixed pwm does not give the same benefit as a constant current source regarding stepper speed. But for sure it is a way to drice the stepper with 12V.

There are a lot of stepper driver ICs with pwm controlled current source. All of them have different price .. and availability.

You say "the cheapest way".
With the resistor solution..the huge resistors may not be cheap. You need a power supply able to give the motor power and the resistor power, maybe heatsink, baybe fans. Is it cheap?

The constant pwm solution...has good efficiency, so you need a smaller power supply, smaller heatsink, maybe no fan. Maybe this is the cheapest way.

The constant current pwm IC solution is the most flexible. Good efficiency, high motor speed. Power supply is somewhere inbetween the above solutions, depending on motor speed. Small heatsink, maybe no fan.

So it's on the OP to select the "best solution for him". I recommend to go to a distributor site for stepper motor driver ICs and find "cheap" IC with all the specifications, read through datasheets and compare them...
(Try farnell, digikey, mouser ...)

Hope this helps, although you need to spend some time.
Good luck

Klaus

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Added to easyrider's post:
I suppose, 1/10 duty cycle will be good enought.
Just to safe the motor from overcurrent the max duty cycle is: (raw estimation)
Motor_voltage / Supply_voltage = 3.2V /12V = 0.266 = 27%.

For a true constant current it will move from 100% down to about 27% after each step.

Klaus
 
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I agree fixed resistors are an ugly solution.

But some low impedance automotive fuel injectors still use series ballast resistors to achieve pretty much the same thing.
 

Constant pwm works like lower constant voltage and doesn't bring the advantage of constant current: faster phase current rise and thus higher speed. At best you get faster current decay in free wheeling which is restricted in a low voltage bridge driver.
 

Sure, for real device purpose there are a lot of difficult calculations have to be performed for proper accelaration and perfomance.
 

That's not funny, man! You will reduce efficency 5 times. 2.8A * 20V will give 56W power lost on each!

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Just take a PWM frequency 10 times higher than expected step time. For example, if you planing to use 1ms step time, 1kHz per step, use 10kHz or higher PWM frequency. I suppose, 1/10 duty cycle will be good enought. But this is actual only for 'on hold' static mode. For movement PWM duty cycle have to be calculated according constand current law. But if you not planing to use variable frequency, just find a good value for duty cycle that gives you expected force and current. It can be lower than maximum rating shown on sticker.

thanks ,, understood
 

Jut try running the stepper motor at reduced current and see how it performs.
It may have sufficient speed and power for your purpose at only half an amp.
You will never know unless you try it.

In that case, some 22 ohm resistors running off 12v might be all you need.
Power would be about six watts which should be tolerable.
 

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