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Resqueline
- Jul 31, 2009
- 2,848
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- Jul 31, 2009
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Hi,
Sorry if i disturb u guys.
i made simulation about this way but i have questions.
1- why you add this resistance and why you use this value of resistance ( 47Ohm ) ?
I use BC238BP and the output current was 0.074A .
I use BC337 and the output current was 0.071A .
so i decide to try using uln2803 and the current was 0.089A.
here the circuit =>
http://img822.imageshack.us/img822/628/34902715.png
2-so what about using uln2803 , this ic will save alot space in pcb BECAUSE i need 3*4 pin ( i will run 3 stepper so i need 12pin from micro controller ).
Sorry if i disturb u guys.
i made simulation about this way but i have questions.
1- why you add this resistance and why you use this value of resistance ( 47Ohm ) ?
I use BC238BP and the output current was 0.074A .
I use BC337 and the output current was 0.071A .
so i decide to try using uln2803 and the current was 0.089A.
here the circuit =>
http://img822.imageshack.us/img822/628/34902715.png
2-so what about using uln2803 , this ic will save alot space in pcb BECAUSE i need 3*4 pin ( i will run 3 stepper so i need 12pin from micro controller ).
Resqueline
- Jul 31, 2009
- 2,848
- Joined
- Jul 31, 2009
- Messages
- 2,848
The resistor is neccessary for current limitation, otherwise the TIP base would receive 4V directly which could burn it out, or at least waste power.
Let's say the PIC outputs 4.6V high. The b-e drop being 0.6V leaves 4V out. The TIP might need as much as 2.5V base voltage in. This leaves 1.5V to be dropped.
(2.5V and 32mA is taken from the datasheet.) Thus 1.5V / 0.032A = 47Ω.
I'm surprised the simulation says the TIP base voltage is ( 5V - ( 0.089A * 47Ω ) ) = 0.817V only. I'd expect a lot higher voltage there (1-2V).
There's nothing wrong with using an ULN2803 as you suggest, only you'll have to invert the outputs in your programming, since the IC inverts the base drive.
Let's say the PIC outputs 4.6V high. The b-e drop being 0.6V leaves 4V out. The TIP might need as much as 2.5V base voltage in. This leaves 1.5V to be dropped.
(2.5V and 32mA is taken from the datasheet.) Thus 1.5V / 0.032A = 47Ω.
I'm surprised the simulation says the TIP base voltage is ( 5V - ( 0.089A * 47Ω ) ) = 0.817V only. I'd expect a lot higher voltage there (1-2V).
There's nothing wrong with using an ULN2803 as you suggest, only you'll have to invert the outputs in your programming, since the IC inverts the base drive.
I used orcade and it give me other results .
using 50ohm the volt on bjt base is 2.56v and ib = 48.66mA
so i will try to find 56ohm (1/2w or 1/4w) resistance or any one between this number and i will try it live
, and tell you what happen.
using 50ohm the volt on bjt base is 2.56v and ib = 48.66mA
so i will try to find 56ohm (1/2w or 1/4w) resistance or any one between this number and i will try it live
, and tell you what happen.hi,
i do some test.
i run the circuit with uln2803, but when i connect the circuit , the microcontroller stop working and when i remove the uln2803 the microcontroller start work without any problem.
i try to put diode at pin10 but the problem still.
note that the circuit is the same as last pic in my replay.
i do some test.
i run the circuit with uln2803, but when i connect the circuit , the microcontroller stop working and when i remove the uln2803 the microcontroller start work without any problem
.i try to put diode at pin10 but the problem still.
note that the circuit is the same as last pic in my replay.
Resqueline
- Jul 31, 2009
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Maybe RB0/INT and/or RB3/PGM influences the PIC mode at startup? Did you remember to connect ULN pin 9 to GND?
Did you check the power supply? If all 4 PortB outputs happen to be low at power-on then a very high motor current sum will flow.
Did you check the power supply? If all 4 PortB outputs happen to be low at power-on then a very high motor current sum will flow.
here the problem in details.
I connect the unl2803
1- pin1 direct to GND or 5+ ( so this ic don't have any connection with pic )
2- pin9 to GND
3- pin10 to 5+
4- pin18 is the out but it connect to 57 ohm 1/4 watt and it connect to pin10
5- pin18 is the out and it connect to bjt
when i connect pin10 i found that the volt on power supply is ~ 0.9 !
when i remove connection at pin10 i found that the volt is 5
so i think that there are problem in connection of uln2803 ( short circuit ) ?
here the circuit but just remove the input of unl2803 and make it manual ( GND, 5+ )
http://img822.imageshack.us/img822/628/34902715.png
EDIT : when i remove the resistance the problem solve !
I connect the unl2803
1- pin1 direct to GND or 5+ ( so this ic don't have any connection with pic )
2- pin9 to GND
3- pin10 to 5+
4- pin18 is the out but it connect to 57 ohm 1/4 watt and it connect to pin10
5- pin18 is the out and it connect to bjt
when i connect pin10 i found that the volt on power supply is ~ 0.9 !
when i remove connection at pin10 i found that the volt is 5
so i think that there are problem in connection of uln2803 ( short circuit ) ?
here the circuit but just remove the input of unl2803 and make it manual ( GND, 5+ )
http://img822.imageshack.us/img822/628/34902715.png
EDIT : when i remove the resistance the problem solve !
Last edited:
You wrongly placed 47 ohm as pull up resistor. No wonder the transistor was turn on to saturation because of +5V directly applied to the base. You should placed 47 ohm resistor in series between uln2803 pin 18 to base of transistor as base current limit resistor.
Question:
What is the DC supply voltage of your PIC?
Question:
What is the DC supply voltage of your PIC?
Last edited:
I using pc power supply.
so i must add 10k as a pull up resistance and 47 as current limitation
I will try and tell what happen.
I'm about to re-edit my post because I remember that ULN2803 was a open collector device. So, you need a 10k pull up resistor.
And also. Maybe you have to place inverters to the inputs of uln2803. So that don't have to re edit your program. Or the other way around.
I've never tried using PIC controller. So that why I still asking.
Most of my microcontroller project were Z80 and Intel 8080.
But I'm planning to use PIC16F84 using Assembler for my Multiple Spark CDIS.
i love using this microcontroller pic16f877a , try it and u will love this controller ( contain all option i need to make different project ).
8080 need alot space to make it run , you must suffer to make it run .
i use pc power supply because it give me 5v and 12v with alot power and it sheep.
i will fix the problem and post what happen , but now i going to sleep.
( contain all option i need to make different project ).8080 need alot space to make it run , you must suffer to make it run
.i use pc power supply because it give me 5v and 12v with alot power and it sheep.
i will fix the problem and post what happen , but now i going to sleep.
Resqueline
- Jul 31, 2009
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Forget the 10k... The 52 ohm pull-up resistor is needed to deliver the neccessary 32mA base drive to the TIP...
You don't say if you've only connected one TIP or all 4, nor if you have connected the motor, or what voltage you apply to the motor.
Unless all ULN inputs are high at turn-on you're probably overloading your PSU with all 4 TIP's turned on hard.
I suspect your driver setup is now working well, but that you've run into another problem due to your inverted driver setup.
You don't say if you've only connected one TIP or all 4, nor if you have connected the motor, or what voltage you apply to the motor.
Unless all ULN inputs are high at turn-on you're probably overloading your PSU with all 4 TIP's turned on hard.
I suspect your driver setup is now working well, but that you've run into another problem due to your inverted driver setup.
I connect 3 TIP to GND and the last one connect it with ULN and the microcontroller not connect to the ULN so i put the input manual.
so when i put 56 ohm as pull-up ( connect it to betwen pin10 and pin18 ) and connect the input of bjt to pin18 => the total voltage at circuit became ~0.1
note that i use regulator with 5+ to supply the microcontroller .
i try to put 10k between pin10 and pin18 and connect pin18 to 56ohm and this is the output => i get the same problem .
When i remove the resistance between pin10 and 18 => the problem gone !
so when i put 56 ohm as pull-up ( connect it to betwen pin10 and pin18 ) and connect the input of bjt to pin18 => the total voltage at circuit became ~0.1
note that i use regulator with 5+ to supply the microcontroller .
i try to put 10k between pin10 and pin18 and connect pin18 to 56ohm and this is the output => i get the same problem .
When i remove the resistance between pin10 and 18 => the problem gone !
Resqueline
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I'm still missing a lot of vital information, so let's take it one question at a time then..
First; what regulator part # are you using?
Second; what is the TIP collector connected to?
Third; what does the +12V become when the +5V drops to 0.1V?
First; what regulator part # are you using?
Second; what is the TIP collector connected to?
Third; what does the +12V become when the +5V drops to 0.1V?
I draw the circuit that i use.
http://img856.imageshack.us/img856/4272/28756564.png
First : lm7805
second : connect to direct 5+ ( not taking from regulator ) or 12+ from power supply
Third :
{ this what happen when using 10k as pull up with 56 ohm as current limiter
#when I connect the the motor and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.5V and the volt on regelator become 1.08V
and the volt on power supply (pin that give me 12+ ) became 11.57v
#when I connect the the motor 12+ and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.66V and the volt on regelator become 2.7V
and the volt on power supply (pin that give me 12+ ) became 10.23v
and the circuit run but the bjt become hot fast !
}
{ this what happen when using 56 ohm as pull up and don't use 10k ohm
#when I connect the the motor and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.1V and the volt on regelator become 0.56V
and the volt on power supply (pin that give me 12+ ) became 11.58v
#when I connect the the motor 12+ and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.4V and the volt on regelator become 1.1V
and the volt on power supply (pin that give me 12+ ) became 9.35v
and the circuit don't run
}
http://img856.imageshack.us/img856/4272/28756564.png
First : lm7805
second : connect to direct 5+ ( not taking from regulator ) or 12+ from power supply
Third :
{ this what happen when using 10k as pull up with 56 ohm as current limiter
#when I connect the the motor and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.5V and the volt on regelator become 1.08V
and the volt on power supply (pin that give me 12+ ) became 11.57v
#when I connect the the motor 12+ and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.66V and the volt on regelator become 2.7V
and the volt on power supply (pin that give me 12+ ) became 10.23v
and the circuit run but the bjt become hot fast !
}
{ this what happen when using 56 ohm as pull up and don't use 10k ohm
#when I connect the the motor and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.1V and the volt on regelator become 0.56V
and the volt on power supply (pin that give me 12+ ) became 11.58v
#when I connect the the motor 12+ and ULN with 5+ direct from power supply and run the circuit
i found that the volt on power supply (pin that give me 5+ ) became 4.4V and the volt on regelator become 1.1V
and the volt on power supply (pin that give me 12+ ) became 9.35v
and the circuit don't run
}
Are you sure your using +5V from pc power supply to supply input of LM7805? You should connect Vin of LM7805 to +12V not +5V.
And if your using your pc power supply as dc power source. Place additional load on dc supply connector such as hardisk to have well regulated output from psu.
And if your using your pc power supply as dc power source. Place additional load on dc supply connector such as hardisk to have well regulated output from psu.
Last edited:
If your using +5 to Vin of LM7805 it will not work. It needs at least 10V Vin to function properly. See datasheet of LM7805.
Or just remove the LM7805 and directly connect +5V from psu supply to your PIC and other circuit.
Place fuse to +5V and +12V for safety measure.
Or just remove the LM7805 and directly connect +5V from psu supply to your PIC and other circuit.
Place fuse to +5V and +12V for safety measure.
Last edited:
what about the reason that make the volt down also the pic is the only ic that use the V out of regulator and the pic don't connect to any other ic.
so why the volt down when connect resistance with ULN ! ( did u get what i mean ) .
i agrea with you about that " when i put 12v on regelator the drop when be small and the out will be 5 so the pic will run " but i just went to know the reason of the drop
so why the volt down when connect resistance with ULN ! ( did u get what i mean ) .
i agrea with you about that " when i put 12v on regelator the drop when be small and the out will be 5 so the pic will run " but i just went to know the reason of the drop
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