problem with tip122 at high volt

[Rleo6965]

LM7805 was a +5V fixed output regulator. Higher Vin was needed to be able to have +5V regulated output. Remember that you need to put heatsink if you use +12V as your Vin. LM7805.

To avoid all this problem. Just connect your PIC to +5V psu. No need to use LM7805 and heatsink.

Read this link. http://www.learningaboutelectronics.com/Articles/What-is-a-LM7805-voltage-regulator

 

[xmen33]

I put 12+ on regulator and the circuit run fine but there is problem ( i think that it not big one ) .
when using 56 ohm as pull up and don't use 10k ohm.
with 5+ on motor i found that current in bjt base = 38mA and volt between base and ground = 2.48v ( very good )

with 12+ on motor i found that current in bjt base = 22mA and volt between base and ground = 3.6 v AND the bjt heating fast ! ( not good )

so what i should do to decrease 3.6v and increase the current when i run the motor with 12+ ( because it must be 2.5v and greater then 33ma as the datasheet and u )

 

[Rleo6965]

I told you.
I think your BJT need heatsink.

 

[xmen33]

i will use heatsink when i make the pcb , but i think that the VBE must be ~2.3V and Ib ~50mA or more so what i must do ? ( i think that i should change the resistance ? )

 

[(*steve*)]

Do the 12V motors require a higher current than the 5V ones?

 

[Resqueline]

xmen33: did you actually connect the TIP collector directly to +5V?
In that case: disconnect it, and the "short" it creates (when turned on) will magically disappear.

 

[xmen33]

Do the 12V motors require a higher current than the 5V ones?

it's the same motor , but when i run it by 12V it draw more current.

xmen33: did you actually connect the TIP collector directly to +5V?
In that case: disconnect it, and the "short" it creates (when turned on) will magically disappear.

no i add a diode then connect it with 12+ or 5+
here the circuit
http://img856.imageshack.us/img856/4272/28756564.png

do you mean that i should add 5+ not 12+ at diode after collector and add 12+ at the wire 1,6 of motor ?

 

[Resqueline]

Steve: it's a 3.6V 3A/winding (1.2 ohm) stepper motor.

Ok, so you still have the motor connected to the TIP collectors, and you have flyback diodes aross the motor, good.
Now, what power supplies do you have for the PIC - and for the motor, and what are their capacities?

 

[xmen33]

I use pc power supply 350w.
for pic i use regulator ( the input is 12+ ),the motor i use 5+or 12+ from power supply ( but i will use 12+ because it give me more torque ).

 

[Resqueline]

Ok, so what are the actual current ratings of the +5V and the +12V?
Do you get the short condition when connecting the motor to PC PSU +5V too?
Have you checked the diodes for shorts?
It's good to use the 12V and a 7805 for the PIC, but you'll have to use the PC PSU 5V for the motor. I don't expect the 12V to have a 40A rating.
And like I've said many times before you'll burn the motor running it on 12V with a good driver. Poor driving is what has limited the torque before.
Also, it's not possible to load a PC PSU +12V fully without loading the +5V somewhat. PC PSU's only regulate the +5V, and the +12V only follow suit.

 

[xmen33]

the short circuit gone when Rleo tell me that i should put 12+ on regulator not 5+.

And like I've said many times before you'll burn the motor running it on 12V with a good driver. Poor driving is what has limited the torque before.

5+ give me low torque.
let say that i need more torque without using 12+ us u say , how i should do that ?

 

[Resqueline]

....
If you can now just measure the c-e voltages we can determine if you're getting as much torque from the motor as it's specified to deliver...
And forget about using the 10k resistor..

You do realize that the motor is only specified to dissipate 10.8W? And that if you apply 10.8V it will dissipate 97.2W?
That's nine times too much, and what can you do about that? Apply liquid nitrogen cooling? Or run it only 1 second every 8 seconds?

 

[xmen33]

with 5+ => 1.44v , with 12+ => 2.34v ( now i remove the uln and use 56 ohm with pic, the reason is down ).

Tell me what about this idea => .
First ULN provide me about 33mA and pic provide me about 28mA so i remove ULN and put the 56 ohm on input pic and run the motor.
I found some thing very good,
with 5+ no bjt heat ! and with 12+ no bjt get very hot and the motor doesn't get hot ! ( i run it 1 min without using any heat sink ! AND the motor is running ) .

the heat appear when the motor stop on step using 12+ ( so the heat appear ) but no heat appear using 5+ when the motor stop on step.
so what about to run the motor on 12+ supply and when i went to hold the motor i give it 5+ supply ?
with this way
1- solve the problem with heat
2- have a higher torque .
if you agree with this idea , what about using relay to switch between two volt ( 5+,12+ ) but the relay slow so if you have any thing better then relay ?

 

[Resqueline]

All this is to be expected, and is the very reason I previously suggested using a constant-current PWM driver..
I suggest you resurrect your previous thread instead of trying to reinvent the wheel with one full of compromises.

 

[xmen33]

All this is to be expected, and is the very reason I previously suggested using a constant-current PWM driver..
I suggest you resurrect your previous thread instead of trying to reinvent the wheel with one full of compromises.

ok, i will search about this driver.
Thanks alot Resqueline.