MP Posted May 8, 2006 Report Share Posted May 8, 2006 It sounds like you are looking for a short cut. Of course, taking short cuts will allow the chance for mistake. You should compare all parameters of the data sheets. That is why they are written. Mostly, compare voltages, current ratings, and timing. Please note that timing (speed and delay) of the device will be very important when swapping digital devices in some applications.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted May 8, 2006 Author Report Share Posted May 8, 2006 Hi again,I collect some detalis about the two ICs as belowM74LS244:Vcc= -0.5v >> +7vVi= -0.5v >> +15 vVo= -0.5 >> +5.5vtPHL/ tPLH propagation delay1An to 1Yn;2An to 2Yntyp. value = 9ns for tPHL & 8ns for tPLHtyp. value = 15ns for tPZH & 12ns for tPZLtyp. value = 11ns for tPLZ & 12ns for tPHZ74HCT244:Vcc = 5vAC waveform s -- Vi= 0v >> 3vtPHL/ tPLH propagation delay1An to 1Yn;2An to 2Yntyp. value >>13ns @ +25 Quote Link to comment Share on other sites More sharing options...
MP Posted May 9, 2006 Report Share Posted May 9, 2006 From what you have posted, I see no reason why the substitution would not work in a stepper application.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted May 13, 2006 Author Report Share Posted May 13, 2006 Hi again MP,please, can you explain for me what did they mean by the J3 Connections in the following link:http://home.att.net/~wzmicro/l298.htmlplease, if you can give me description about them and what is the value must entered to them until the motor will run :-\ Quote Link to comment Share on other sites More sharing options...
MP Posted May 14, 2006 Report Share Posted May 14, 2006 Hi Queen,J3 in your link are digital inputs. These should be connected to a micro or computer. A low would be 0 volts and a high would be 5VDC.J3-1 is self explanatory. Ground.J3-2 is a connection for the pulse stream that provides steps to the stepper windings.J3-3 will cause the motor to turn in a clockwise direction if it is high and a counter-clockwise direction if it is low.J3-4 will cause the controller to provide half steps if the signal is high and full steps if the input is low.J3-5 should be left open. It is connected to a high. When it is set low, your board is reset.J3-6 is Normal winding current when high or reduced winding current when low. I have never used this feature. Try normal to start.J3-7 Winding current on when high and off when low. It is the "Enable" for the controller chip.J3-8 is a step oscillator output. The 555 provides a data stream that you can connect to the step input (J3-2) to provide a continuous motor run.J3-9 Provides a function so that you can home your stepper.J3-10 As with J3-1, is self explanatory. It is a ground connection.Note that U1 is just a buffer chip. 1A1 goes out to 1Y1, 1A2 goes to 1Y2, etc. It cleans up your signals. Pins 1 and 19 enable the chip when connected to ground.Hope this helps. MP Quote Link to comment Share on other sites More sharing options...
Queen Posted May 16, 2006 Author Report Share Posted May 16, 2006 Thank you very much MP ;DPlease, I want to use the circuit in the previous link, can you give me your notices on it and if it really will work properly :-\And can you give me the sequence that I must write it to the (J3-2 ..... J3-9) to get the motor work :-[am very sorry since I asking all these question but I have a project and I must finish it quickly :-[ Quote Link to comment Share on other sites More sharing options...
MP Posted May 17, 2006 Report Share Posted May 17, 2006 I see no reason that this would not work. You also do not need the part of the schematic with the 555 if you will send the pulses from a micro instead of using the on-board pulse stream. In other words, you can eliminate U1, R1, R2, R3, C1, C2. You only need to write a sequence to J3-2. The stepper will make one step for each pulse. I recommend that you use a 50% duty cycle and a low frequency to start. Then you can speed up until your stepper starts to skip steps and back off. Change the direction of the motor with J3-3. When you connect 5 volts DC to it, the stepper will go in one direction and when you apply 0 vdc to it, the motor will go in the other direction. Connect pins 1 and 10 to ground. Tie pin 4 low (0 VDC). Leave pins 5, 6, 7, 8, and 9 open.This should get you started. Hopefully, if I have missed something someone else will jump in to assist.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted May 19, 2006 Author Report Share Posted May 19, 2006 Hello MPThank you ::)But I have suggestions as follows:Pins 1, 10 will be groundedPin 2 will run the motorPin 3 will make the motor CW or CCW Pin 4 will be at 5VPin 5 will be openPin 6 >> in this pin I will give the motor reduced current to ensure that the motor will become run, I will create something as "Idle state" for the motor Pin 7 will be at 5VPin 8 will be openPin 9 will be open Can you give me your opinion, please?? 8) Quote Link to comment Share on other sites More sharing options...
MP Posted May 20, 2006 Report Share Posted May 20, 2006 I have never used the feature for pin 6. You will not hurt anything for trying it out. Perhaps you could add a shunt so that it can be high or low if you decide to change.Pins 4 and 7 are already at 5V because of RP1.Have fun!MP Quote Link to comment Share on other sites More sharing options...
Queen Posted May 22, 2006 Author Report Share Posted May 22, 2006 Thank you very much MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 3, 2006 Author Report Share Posted June 3, 2006 HelloWhat is the meaning of winding current??and if I put J3-7 to GND what will happen??please reply Quote Link to comment Share on other sites More sharing options...
MP Posted June 4, 2006 Report Share Posted June 4, 2006 This refers to the current flow in the windings of the motor. If you ground this pin, then the motors will be turned OFF. This option is available so that you can connect a safety switch if needed.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 4, 2006 Author Report Share Posted June 4, 2006 Hi If J3-7 remains open what will happen?? Quote Link to comment Share on other sites More sharing options...
MP Posted June 5, 2006 Report Share Posted June 5, 2006 RP1 is pulling it to logic high through a resistor. Leaving it open is essentailly a high. You only need to connect a switch to ground if you ever want to pull it low.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 10, 2006 Author Report Share Posted June 10, 2006 Hi MPPlease, can you give me a hint of how I can use the battery of 7Ah and 12V with three motors and a PIC18f4550, each motor need 2.5A ??? Quote Link to comment Share on other sites More sharing options...
Queen Posted June 10, 2006 Author Report Share Posted June 10, 2006 hi again :)i have one more question which i hope you have an answer forits about setting a delay for stepper motor in the PIC18F4550 ???in other way... what is the maximum frequency could the circuit handle??? Quote Link to comment Share on other sites More sharing options...
MP Posted June 11, 2006 Report Share Posted June 11, 2006 Queen,Frequency that the motor can handle is pretty much different from motor manufacturer to motor manufacturer. You will have to make some adjustments when you make this circuit. You should make the frequency adjustable anyway, so that you can use different speeds. I would recommend that you start with some obviously too slow frequency, like 500HZ and work your way up until the stepper starts missing steps. Then, you need to back down to a slower frequency.In regards to the battery question, 2.5A per motor is sure a lot to feed from a small battery. Are you sure you need this much current? If you are using 3 motors @ 2.5A each, that is 7.5A consumption from your 7Ah battery. Keeping in mind that the battery will not work well at the 50% level or lower, this gives you less than a half hour of usage. I would recommend a larger battery, such as a deep cycle RV type or even a car battery for the initial design. I assume you do not have AC power at the site. Perhaps other members will have better ideas.I can certainly give you suggestions and guidance, but I do not have the time available to help with the design work to program the PIC. www.avrprojects.net has some sample basic code to run steppers with AVR micros. Perhaps you could port this to PIC or find something similar on the web in the language that you will be using. It sounds like an interesting undertaking and I wish I did have a little more time to help further.Good luck!MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 15, 2006 Author Report Share Posted June 15, 2006 Hello MP,I have a big problem and I dont know what is the solution for it.Look, am building three circuits for three motors using the L297 & L298, and am using a battery of 12V & 7AH, two of the motors are running but the third motor didnt work and the L298 burn.Please give me the reasons why L298 burn.Best Regards. Quote Link to comment Share on other sites More sharing options...
MP Posted June 16, 2006 Report Share Posted June 16, 2006 I would look to see if there is anything different on this circuit than the other two. Perhaps you ran a wire wrong or a component is different. This type of error is the most common.Can you post pictures and diagrams? It is hard to guess what might be the problem. Did you heat sink the driver chips? Your L298 might have gotten too hot.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 16, 2006 Author Report Share Posted June 16, 2006 Look, when I took a look at the circuit I found that a pin for L298 didnt conect in an appropriate way, Is this may be the reason why it get heat then burned?? ??? Quote Link to comment Share on other sites More sharing options...
MP Posted June 17, 2006 Report Share Posted June 17, 2006 Possible. Which pin? Was it shorted to another pin or left open?MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 17, 2006 Author Report Share Posted June 17, 2006 The pin was 8, and it was open ??? Quote Link to comment Share on other sites More sharing options...
MP Posted June 18, 2006 Report Share Posted June 18, 2006 Leaving pin 8 open should not have caused a problem. There must be something else also wrong in the circuit or it is possible something touched a pin that caused a power surge to the chip. Perhaps you had one component that was wrong in this one circuit that was different than in the other two. There are many possibilities, including that the chip just got too hot. You are fortunate to have two other duplicate circuits which are working ok. My only suggestion is to rebuild the circuit, make sure it is the same as the other two which are working correctly, try all three circuits one at a time so that you know they are working ok individually, then connect them together again as a complete unit.MP Quote Link to comment Share on other sites More sharing options...
Queen Posted June 19, 2006 Author Report Share Posted June 19, 2006 Thank you MP, I think the problem was occured since the tempreture of L298 increased, but please can you give me suggestion of how I can know that l298 become destroyed?? Quote Link to comment Share on other sites More sharing options...
MP Posted June 20, 2006 Report Share Posted June 20, 2006 You should use heat sinks on all of your L298 chips. Please refer to the data sheet for information regarding this. They run hot. In fact, I was just looking through the data sheet and found that you should not exceed 2 amps unless you use a bridge configuration (figure 7). You should read it very carefully so that you fully understand all that is required for this chip.I have attached the data sheet.MPL298_Data_Sheet.pdf Quote Link to comment Share on other sites More sharing options...
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