Stepping mode of unipolar driver

[same79]

This is the Unipolar driver I am going to build. But I want to confirm that can it provide facility of half stepping. Please explain about its stepping mode.

[doorknob]

I'm not an expert, but I do not see how that circuit can provide half stepping.

The motor has four coils (two center-tapped windings). Each winding is driven by the Q and not-Q outputs of a D-type flip flop. Therefore, when one half of a winding is turned on, the other half will always be turned off. So, looking at the entire motor, two coils are always on and two coils are always off.

My understanding of a half stepping sequence for a unipolar motor would require that the driver be able to alternate between one coil on and two coils on. I do not see how this circuit can do that.

[john_100]

full steps only

the two D type flip flops in the 4013 ic switch on alternate step pulses

the 4030/4070 ic provides the steering logic to control the direction of the count

note for one direction D input on flipflop 1 is connected to the Q output of flipflop 2

and D input on flipflop 2 is connected to the -Q output on flipflop 1

for reverse D i/p f/f 1 connected to -Q o/p f/f2
D i/p f/f 2 connected to Q o/p f/f1


each flip flop swiches the two fet transistors
and hence the power to one end of coil or other
one end of other is always on

if you want to avoid using a PIC chip or other microcontroller

you could use a up/down counter like the 4029 counter

decoding the count with a 4017 to a 1 of 8 output
and then via a few diodes you can
generate the correct pattern for the half step drive

John

[same79]

Thanks for your reply. Is there any way to change it to full and also half stepping mode?

[john_100]

Hi

this a modified circuit I'm thinking of

John

[same79]

Thanks for your big effort. Pleas also explain it some bit more that how to use it.

[john_100]

Hi Same79 ,

tonight when I'm home
I'l try to draw the complete diagram showing the connection to the FET's and power supply

John

[john_100]

Hi Same79 ,

for half step mode

only 3 of the 4029 binary / decimal up down counter outputs are used ,
count 000 to 111 ( logic 1 on pin 9 selects binary count)
logic on pin 10 controls the direction of the count

the step signal on pin 15 increments the count

the first eight 4028 outputs are used ( selected output is high )

Y0 to Y2 switch coil 1 end A to ground via a 3 i/p OR gate and Q1
Y2 to Y4 switch coil 2 end B to ground via a 3 i/p OR gate and Q2
Y4 to Y6 switch coil 1 end B to ground via a 3 i/p OR gate and Q3
Y6 to Y0 switch coil 2 end A to ground via a 3 i/p OR gate and Q4

for full step

pin 10 (LSB) on the 4028 is wired low

and only the first 2 counter output pins 6 &11
drive the 4028 input pins 13 and 12

now only only the decoded outputs Y0 ,Y2, Y4 and Y6
go high to give the full step sequence

the two 4075 IC's contain 3 x 3 input OR gates
only 4 gates are used

the IRFZ44 FET's are "logic level " FET's that only require 0 to + 5V
to switch them on and off


John

[same79]

Thank, that is really very smart. There are switches between 4029 & 4028 to switch full or half step. I think these have to change manually not electronically. How to control the stepping mode electronically automatically same as Step & Dir?
And how about its frequency rate (steps per second) capability. I think it should be minimum 1500 steps per second. If it can not then how it can?

Thanks,

[john_100]

Hi Same79 ,

if you replace the mechanical switch / jumpers with two 4016 /4066 quad bilateral switch ic's or 74244?

you could change from full to half step with a logic input

i'll add another diagram tonight when I'm home

now I had a look at some ic data , a better solution is a 40258 data selector ( 4 electronic toggle switches ) (cmos version of the TTL 74157 )

John

[same79]

That are great and very much appreciated but also please tell me about its frequency rate (steps per second) capability.

[john_100]

Hi Same79 ,

since the cmos logic gates can be clocked at several megahertz

I'd expect the limiting factor to be the motor inductance
and how you drive the motor.

a 12v motor run from a 12v supply having a lower step rate
than a 3v motor run from 35v via a dropper resistor.
as this will give you a higher initial current through the coil

I've not converted my lathe to cnc ,its still on my to do list!
so I have no practical experiance at the moment

one thought , is in the half step mode
on every other step only one coil is energised and
produce less torque than the full step mode .

John