Advice Needed

Discussion in 'General Electronics Chat' started by r_james14, May 9, 2010.

  1. r_james14

    Thread Starter New Member

    Apr 23, 2010
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    Hi there,

    Attached i have a circuit diagram and it's PCB, it is designed to control a stepper motor, its input is 5v.

    On the outputs the middle lines into the motors are desigend to be fixed 5v with the outside four controlling it,

    In simulation this works perfectly however when i have constructed it on PCB i keep getting 10volts on the middle pins and very low voltage on the outside (meant to be 5v), basically i have made two pcb's now and experiencing exactly the same problem, is it my soldering (which is awful) or is there a fundamental issue with the design.

    Thanks

    James
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    1N400x diodes are not suitable; they are not fast enough. You need fast-recovery diodes, like FR303's, 1N4934's, 1N5817 thru 1N5819, MBR320-MBR350, MUR105-MUR160, etc.

    What are you using to control the transistor bases with?
     
  3. r_james14

    Thread Starter New Member

    Apr 23, 2010
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    a microcontroller, atmega32 to be precise change pulses about every 20 milliseconds

    james
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    What are the motor requirements as far as current, what is their rated voltage and winding resistance?
     
  5. r_james14

    Thread Starter New Member

    Apr 23, 2010
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    Hi there
    Ive attached the main datasheet for the motor, i didnt want to misread it and give incorrect information

    thanks

    James
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    Having datasheets available is great. It usually eliminates a lot of guesswork.

    OK, so 5v, 860mA, 5.8 Ohms +/-10%, 6.5mH@1kHz +/-20%.

    The transistors you used (ZTX869) have an unusually high gain; which is a good thing.

    If the Vcc of your uC can output 5v into a 1.1k Ohm load, then:
    Ib = (Vin-Vbe)/1.1k = (5v-0.7v)/1100 = 4.3/1100 = 3.91 mA.
    You need Ic=860
    We'll use the forced beta of 100 as shown in the plots. Ib=Ic/100 = 8.6mA base current required. You need 2.2 times the base current that you're giving it.

    So, let's calculate some new base resistors. Normally, the forced beta is 10, but the datasheet shows that a forced beta of 100 is quite acceptable.
    Rb = (Vin-Vbe)/(Ic/100) = 4.3/8.6mA = 500 Ohms.

    I suggest that you go even lower than that. Use 330 Ohms for the base resistors.

    Currently, you are using 120k resistors for the base pull-down resistors. Change them to 1.1k instead. They will only take 0.64mA away from the base current when the transistor is on - and besides, now you have something you can use those 1.1k resistors for. ;)

    Add a much larger cap or two across your motor supply. 470uF to 2200uF.
     
    kingdano likes this.
  7. r_james14

    Thread Starter New Member

    Apr 23, 2010
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    in the solution you refer to the beta value, would the same method be carried out if the transistor were a ZTX689 instead of the ZTX869, sorry to sound daft

    james
     
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