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problem with tip122 at high volt

xmen33

Aug 28, 2011
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Hi,
i Design a Unipolar stepper motor drive with tip122.
the motor run without any problem with 5v but when i try to run it with 12v it stop at the step and don't move and the tip122 became hot !.
note that the motor at 5v it use 3.5A from power supply and at 12v it use 7.5A from power supply.
is this problem because the high ampere ( so i need a current limiter ) or it because tip 122.
the circuit here
http://dl.dropbox.com/u/2193931/test1.JPG
 

Rleo6965

Jan 22, 2012
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You need to replaced TIP122 with higher current and place heatsink for that transistor.
 

xmen33

Aug 28, 2011
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TIP122 => ic = 5A and ic = 8A at Pulse .
i Think the reason of this problem is that the motor is 3A so when the circuit draw 7.5A , the motor stop run and hold the step .
so i think that i should put current limiter.
is this right or ?
Waiting :)
 
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jackorocko

Apr 4, 2010
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You should be running the motor at its rated current. Otherwise at low voltages your max speed will suffer. AFAIK steppers are constant current devices or as close as possible to constant. They will work hard to try and maintain a constant current.
 

xmen33

Aug 28, 2011
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so i should put a current limiter by 3A to make stepper run at 12 V ?

# by the way , the stepper is 3A
 

jackorocko

Apr 4, 2010
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I am finding it hard to understand why you need to limit the current? If you run the motor at its max current then you won't need a limit.
 

xmen33

Aug 28, 2011
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the motor is 3A and this pic ok it
http://dl.dropbox.com/u/2193931/m1.JPG

when i run the circuit with 5V , the circuit draw 3.5A and the motor work fine .
But when i run the circuit with 12V , the circuit draw 7.5A and the motor stop work and hold the step .

so i think that the reason that the motor don't work at 12v because it draw 7.5A and the motor should run at 3A.

is this right or ?
 

jackorocko

Apr 4, 2010
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3A per phase. If I understand you it seems like both phases are energized at the same time. Do you have a schematic of the driver?
 

jackorocko

Apr 4, 2010
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I don't know what is going on, what mode are you using to drive the motor? Is the code something you wrote?
 

xmen33

Aug 28, 2011
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i using 16f877a to send 0011,0110,1100,1001 to tip122 but between them i use buffer 4050
to improve the signal that come from micro controller .
tip122 act like 5 => GND , 0 => float .
do you understand now ?
(sorry for my bad language)
 

Resqueline

Jul 31, 2009
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Rleo has the solution. You're pushing the TIP122 to its limits. The 8A rating is only good for 5ms btw..
Also how much base current are you able to provide for it? The 4050 is only able to provide around 1.5mA..

Have a good look at the datasheets and ask yourself if you don't want to use Logic-Level MOSFET's instead.
TIP122 extract:
Vce(sat) = 2.0V max @ Ic = 3A, Ib = 12mA
Vbe(on) = 2.5V max @ Vce = 3V, Ic = 3A
Vce(sat) = 1.0V typ @ Ic = 3A, Ib = 12mA
Vbe(on) = 1.85V typ @ Vce = 1V, Ic = 3A

Vce(sat) = 4.0V max @ Ic = 5A, Ib = 20mA
Vce(sat) = 1.3V typ @ Ic = 5A, Ib = 20mA
Vbe(on) = 2.25V typ @ Vce = 1.3V, IC = 5A

Vce(sat) = 2.0V typ @ Ic = 8A, Ib = 32mA
Vbe(on) = 2.8V typ @ Vce = 2V, Ic = 8A

The motor being a Sanyo Denki Step-Syn type 103H6707-0741 FH6-1176 rated at 3A seems to have 6 wires with a resistance of 1.2 ohms from the centre tap(s).
Going by the currents it seems that you're running it in single-mode, which as jackorocko points out is important to know when you're only quoting the supply current.
Going by your drive sequence however you're running it in double-mode, which should have resulted in a supply current of 20A (had the drive circuit been up to it).
You may want to ask yourself how a 3A motor is supposed to survive this kind of abuse, and why you want to engage two windings at once when you're not half-stepping it..?

I also want to point to your other recent thread about the very same stepper, where there was already dug up some pertinent information..
 
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xmen33

Aug 28, 2011
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the other recent thread is my old topic, i run the the motor in bipolar mode and i decide to run it at Unipolar mode.
i remember that at bipolar mode the first phase draw 3.2A and the second phase draw 0.1A and it was running at 12V !.

I don't understand the last section of your post.

are you say that the motor should draw more than 3A ? but i went to tell you that when i running the motor by bipolar mode i use 2 => l298 and it have limiter inside it and when i increase the limiter, the motor became crazy and when i minimize the limiter, the motor became crazy.
so i was change the limiter untill the phase draw 3A=>3.5A.

I understant that i should change the buffer 4050 and change tip122.
So what you suggest ?
 

Resqueline

Jul 31, 2009
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Ok, different mode.. It would still be useful for knowing the motor spec's.
I don't understand what you mean by the second phase drawing 0.1A?
Of course it was drawing the prescribed current, regardless of voltage, it was after all set up to be a constant-current driver (L297+L298).

Driving it with a "dumb" transistor driver is a different matter and requires 3A * 1.2 ohms = 3.6V per winding, as I pointed out in the other thread.

The math is simple. 12V / 1.2 ohms = 10A. That is for each winding, and you're attempting to always drive two windings simultaneously, hence 20A.
While double-driving (I called it double-mode; my own term) can increase the available torque but it also doubles the dissipation which could burn the motor.
Driving two windings at once is only used during half-stepping, and is thus applied only half the time, only increasing dissipation by 50%.
Single-stepping (one winding at a time) is the ordinary step sequence, giving a dissipation for which the motor current is rated.
Look here for information about step sequences.

We were never informed of your results and problems in the other thread. It could have been useful.

The current available from the PIC877A outputs is probably much higher than the current available from the 4050 outputs. Check the datasheets.
You can either try to remove the 4050, or replace it with a better driver. The TIP's would like at least 8mA in for 5A out, preferably 20mA.
If you replace the TIP122's with Logic-Level MOSFET's it doesn't matter about the driver current, but then you will fry your motor running it on 12V.
It's only the poor driver setup you have now that has saved it.
 

Rleo6965

Jan 22, 2012
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It's also much better if you have followed the required pulse width to drive the stepper motor. Use oscilloscope to confirm it.
 

xmen33

Aug 28, 2011
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I understand now.

Logic-Level MOSFET's good but i buy all i need of tip and Logic-Level MOSFET's expensive in my country .

so i try to change my old step sequence and make it ( 1000,0100,0010,0001 ).
When i try 5V , the circuit draw just 1.8A but torque not good,
so i try 12V , the circuit draw just 3.6A and the motor run very good.
so in this case tip122 can Handel 3.6A and 12V , right ?
If yes , so are there any comment about the circuit that not right.
Other thing , what the useful for using this diode D1,D2,D3,D4 .
I know that D5,D6,D7,D8 using to protect the tip.

http://img14.imageshack.us/img14/8161/32683895.png
( I Will remove the 4050 )
by the way the old method run without any problem and it was very good.
 

Resqueline

Jul 31, 2009
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The TIP's are presently just not getting enough base current to conduct fully, that's all. Improve that and you'll get higher winding currents & cooler transistors.

D1-4 are strictly not needed, and besides the TIP122 already has a built-in diode in that position. D5-8 are doing an important protective job though.

Rleo is probably thinking about using PWM (within each step) to effectively reduce the applied voltage and get 3A even with 12V applied.
That's what the L297+L298 combo did.
 

xmen33

Aug 28, 2011
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all buffer i found has low Ioh,Iol .
So are there any way to improve Ioh,Iol of micro controller?
 
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