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pwm to control a dc motor

Discussion in 'Sensors and Actuators' started by mariomoskis, Mar 14, 2012.

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  1. mariomoskis

    mariomoskis

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    Mar 13, 2012
    ok,but remeber that i need 4.5V to supply the 555
    so you said that i start with duty cycle 0% and supply 5V
    and check duty cycle until the motor consumes 0.3A,yes?
     
  2. mariomoskis

    mariomoskis

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    Mar 13, 2012
    i just try about you said with my new transistor BD135:

    when i have supply voltage 3V,motor consumes 0.3A with 100%duty cycle and for 50%duty cycle consume 0.26A. and of course the speed change too
    it seems work well,but the 555 don´t have 4.5V of supply voltage as it need

    so i need supply 5V to the 555,how could i do it with only one power supply and the motor can´t have across him 3V?
    maybe about fix 2V at the emitter with a diode or something like that?

    will i get with this in saturation: 2V emitter,2V colector
    and in cut-off : 0V emitter,5V colector
    do you know what i mean?
     
  3. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Not quite sure what the problem is. The 555 can easily handle 5V. It's rated for up to 15V.
     
  4. jackorocko

    jackorocko

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    Apr 4, 2010
    I thought we already went through this? If you need a load to see 3V and you are using PWM then you change the duty cycle. But you also need to figure in the voltage drop of the transistor and emitter resistor(if using one).

    Voltage is what changes the speed of the motor. I don't know why you are so concerned about the amperage. I don't think amperage changes linearly in a motor when using PWM.
     
  5. mariomoskis

    mariomoskis

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    Mar 13, 2012
    555 need at least 5V,but the motor only can have 3V maxium across it, so if supply is 5V, i will have 2 V more across my motor when the duty cycle is 100%
    but i would like to have 3V when duty cycle is 100%,and supply 5V(because of the 555)
    do you understand now?

    the amperer was to see what was happening with the current consumed by the motor.

    is there a way to get about i said before??
     
  6. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    Go back and read the entire thread again.

    Please don't post again until you understand that it can.
     
  7. mariomoskis

    mariomoskis

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    Mar 13, 2012
    yes,if i reduce the duty cycle to 60% and supply is 5V,the motor will have across it 3V
    so that is the best i can do to get 3V,yes?
     
  8. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    It achieves the ends required with the minimum of complexity and the maximum of efficiency.

    Note that because of the voltage drop across the transistor, you may require a little more than 60% duty cycle.

    Also note that 5V is getting toward the lower range at which the 555 operates.
     
  9. mariomoskis

    mariomoskis

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    Mar 13, 2012
    the problem was that i was confused about this
    i thought that if for example i had supply 5V, my motor could have 3V across him with a duty cycle 100%, if i fixed a voltage 2V with a diode at the emittor
     
  10. mariomoskis

    mariomoskis

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    Mar 13, 2012
    but for example for the transistor BD135 the problem is that when i have dudy cycle 60% and supply 3V,i already have the maxium current consumes by the motor
    and i can´t get the supply 5V
     
  11. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    If 60% duty cycle is sufficient to make the motor run at full speed then make that your maximum duty cycle.

    I really don't understand the rest of your post.

    And to answer an earlier question, it's not quite as simple as saying that 50% duty cycle is the same as half the voltage. There are a number of other variables.

    At 50% (say) you motor would receive 5V for 50% of the time and 0V for 50% of the time. This results in an average of 2.5V, however while the motor has 5V across it, the current through it will be doubled, so the power will actually be 4 times the normal power. Then it will be zero for half the time. So the average power will actually be twice the normal amount. However, the motor is an inductor and that will limit the rate at which current can change. This may result in the current not actually reaching the theoretical value. It will also result in current continuing to flow (through the diode) for at least part of the 50% of the time when the power is off. This will result in a lower average current.
     
  12. mariomoskis

    mariomoskis

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    Mar 13, 2012
    hello,i have been a time which i couldn´t do nothing about this topic
    finally i did the next:

    http://imageshack.us/photo/my-images/703/darbueno.jpg/

    i used a darling transistor because i can´t have more than 2mA at the base of the transistor,and i got the next results

    for duty cycle 50% the voltage across the motor is 1.5V
    for duty cycle 75% the voltage across the motor is 2.1V
    for duty cycle 100% the voltage across the motor is 2.5V

    why i can´t get 3V when the duty cycle is 100%?? what is hapening?
     
  13. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    I don't know if you're supplying enough base current.

    I don't know if the difference between your supply voltage and 2.5V is simply Vce(sat).

    That image is too small to read anything on. It looks like your supply voltage is 3V which means you're pretty lucky to be getting 2.5V to the motor, especially through a darlington.

    You really don't seem to have caught on yet that you need a higher supply voltage, allowing you to control the effective maximum voltage supplied (if necessary) by limiting the maximum duty cycle.

    We haven't dwelt on the Vce characteristics of the transistors because we have been encouraging you to use 5V (at a minimum) as your supply voltage. We've also been gently encouraging you to use a mosfet (although at 5V, a BJT should work OK.
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    OK, I've squinted and made a huge number of assumptions. That circuit you have will not go up to 100% duty cycle.

    I can't see the values of the components, so I don't know how close it will go, or what frequency it will operate at.

    The very low voltage and the steering diodes will strongly affect operation in any case.

    Do you have a 555 that is rated to as low as 3V operation?

    Are you using schottky diodes to get a minimum voltage drop across them? (Not normally needed, but where 1/3Vcc is so low, the voltage drop across the diode becomes a real issue.)
     
  15. Wabajig

    Wabajig

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    Apr 14, 2012
    Change the R1 to a 3.3V zener and put the motor in the R spot and tie collector to the supply. Also put a 10uf / 16V cap across the motor (watch polarity). John
     
    Last edited: Apr 16, 2012
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    That seems like very poor advice.

    Why on earth would you want to do that?

    It sounds like you want to place the transistor across the supply rails, and drive the motor using the current flowing between the 555s output and the base of the transistor. Is that what you really mean?
     
  17. mariomoskis

    mariomoskis

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    Mar 13, 2012
    This is the circuit:
    http://imageshack.us/photo/my-images/35/darl.jpg/

    forget about the CINE555,i generate the PWM signal with a digital mode of CompactRIO,then the at the output of CRIO i have 5V and the current must be between 100microA and 2mA,so i use a darling transistor( BD135),the resistor at the base is 2.2K and supply is 3V,i don´t know if this image is more clear. frequency will be between 50Hz and 500Hz, it won´t be higher than 500Hz.diode are 1N4004,and i don´t understand about u said about it.

    is it more clear now?

    thanks
     
    Last edited: Apr 14, 2012
  18. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

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    That diagram still shows two 2N2222's. Can you tell me what the actual transistors are that you're using so I can calculate the base current required.
     
  19. mariomoskis

    mariomoskis

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    Mar 13, 2012
    BD135,sorry
     
  20. Harald Kapp

    Harald Kapp Moderator Moderator

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    Leaving aside the specific kind of transistor: In this application a Darlington circuit can never be as efficient as a two-stage amplifier. The voltage across collector-emitter of the end stage in a darlington configuration can never reach saturation. Let's call the first stage Q1 and the second stage Q2. Then Vce(Q2)= VBE(Q2)+Vce(Q1) which is at least 0.6 V + 0.1 V = 0.7 V ore more, depoending on the transistors and the currents. If you want to drive Q2 into saturation (Vce ~ 0.1 V), then change the darlington configuration to a 2-stage amplifier:
    - remove the connection of collector (Q1) - collector (Q2).
    - add a current limiting resistor from collector(Q1) to VCC

    Now Q1 can drive a base current into Q2 such that Q2 is saturated with Vce(Q2) ~ 0.1 V.

    Unfortunately that requires that you use another transistor, not an integrated Darlington.

    Harald
     
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