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Stepper motor

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Nigel Goodwin

Super Moderator
Most Helpful Member
Nigel Goodwin, I am not sure what you mean.

The output of the voltage regulator is 5V and the input is 9V. Bad idea feeding it from the output which is 5V??
Quite simple, feed the motor from the 9V before the regulator.

I't a bad idea feeding the motor from the regulator output.

I am trying to find a voltage regulator that can handle 2A if anyone find something let me know
It will still run red hot, and require a massive heatsink - there's no need to have a regulated feed to the motor.
 

uaefame

New Member
I am still cofused the motor requires 5V why i should feed it with 9V??

Can you show me a diagram how will you connect it??

Connecting a 9V to 5V motor sound not fun right!
 

yngndrw

New Member
Nigel: He is using the regulator to feed the motor because there is no current limiting setup.

uaefame: It doesn't sound right, but if you limit the current instead, you have either a PWM or chopper based system. This is the next step as it provides more torque and hence a greater top speed. It also puts you in a position to do microstepping.

I have a large 2V stepper motor which I run from 10V. When I get a proper power supply setup, I will be running it from 24V. It is current limited to ensure that it doesn't overheat.

There is a limit to the max voltage, because of "iron losses" from the high switching speed of the current limiter.

The highest advisable supply voltage is 32 * sqrt( Inductance ), where Inductance is in mH. Another method (Less accurate) is to run the motor upto the rated voltage * 10.
 

uaefame

New Member
Wow current limiting seems an interesting topic. I can save alot of money if i know more about it.

If i want to connect directly from 9V battery to Vcc2, it strange right the motor only handle 5V so 4V extra should i include a resistor for that drop? because i think the motor need 800mA per coil and i have 2 coil therefore 1.6A. V=RI therefore R=4/1.6=2.5ohms and the wattage P=VI=1.6*4=6.4W which i can't find easily.

Is my calculation about current limiting true or is there something wrong I did.

Any good website about current limiting is appericated

Thanks in advance
 
Last edited:

yngndrw

New Member
You can do it that way, but you effectivly have the same thing as you are doing now (With the regulator.) - All of the extra energy gets turned to heat which is a waste. The 6.4Watts you are talking about are currently turned to heat inside the regulator which is why it's getting hot.

You need to be using PWM to control the current.

You wanted a site: http://www.cs.uiowa.edu/~jones/step/
It's a big site.
 

Nigel Goodwin

Super Moderator
Most Helpful Member
Wow current limiting seems an interesting topic. I can save alot of money if i know more about it.

If i want to connect directly from 9V battery to Vcc2, it strange right the motor only handle 5V so 4V extra should i include a resistor for that drop? because i think the motor need 800mA per coil and i have 2 coil therefore 1.6A. V=RI therefore R=4/1.6=2.5ohms and the wattage P=VI=1.6*4=6.4W which i can't find easily.

Is my calculation about current limiting true or is there something wrong I did.
All you need for current limiting is some series resistors (PWM can be used, but makes life FAR more complicated for you).

But in your case I don't see as it's much of a concern, losses in the driver chip will probably only leave 5V or so for the motor.

Try it and see.
 

uaefame

New Member
I connect it directly to the 9V power supply battery amazing it work! I was suprised that it worked do you think sn754410 has anything to do with this?
 

yngndrw

New Member
Maybe, if you want it to last as long as you should, check the temperature of the motor against the maximum temperature in the datasheet. Also check the temperature of the driver chip.

For extra points you can check the current draw of each motor coil against the datasheet.

If you don't care about the life of the motor, you could probably just leave it. Just make sure it doesn't get too hot.
 

uaefame

New Member
Hello everyone,

I wanted to dervie a stepper motor from angle 22.5 degree till zero and then from zero back to 22.5 degree and do this again and again.

This codes i used for clockwise direction
Code:
movlw	b'00000101'
	movwf	PORTB
	call	Delay8ofsec
	movlw	b'00000110'
	movwf	PORTB
	call	Delay8ofsec
	movlw	b'00001010'
	movwf	PORTB
	call	Delay8ofsec
	movlw	b'00001001'
	movwf	PORTB
	call	Delay8ofsec
	goto	main
Q1>What code i should add before 'goto main' to let it move counter clock wise.

What i understand from the cw code
1-00000101
2-00000110
3-00001010
4-00001001

first one nothing will happen or maybe I think a lost in angle or something will be notice
second one 7.5degree will be moved cw
third 15degree will be moved cw measured from orgin
fourh 22.5degree will be moved cw measured from orgin

Q2> How can i prevent the first type of error introduce in case one of clock wise if there is an error?

Thanks in advance
 
Last edited:

yngndrw

New Member
Q1:

If the order for CW is:
1- 0101
2- 0110
3- 1010
4- 1001

Then just go through it backwards to go CCW.

For example, if it is currently at #2 (0110):
*) To step CW you would go to #3 (1010).
*) To step CCW you would go to #1 (0101).

Q2: I don't quite understand your question, I think you're asking what happens if the first step in the sequence (0101) does not correspond to the motor's current position.

This is a problem with stepper motors - There is no feedback so we must assume that it is in a known position and that when we tell it to step forwards one step that it is able to do so.

You could fit a rotary encoder onto the motor so that you always know where it is, or you could have a switch covering one part of the rotation - Rotate the motor in one direction untill you hit the switch and then you know where the motor is. You can then rotate the motor back so many steps and you can assume that the motor is in the position NumberOfSteps * DegreesPerStep away from the switch.

Does that make sense or am I talking nonsense ?
 
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