Another newbie needs help with 555 timer for a motor.

Discussion in 'The Projects Forum' started by 6octagon, Nov 24, 2009.

  1. 6octagon

    Thread Starter New Member

    Nov 9, 2009
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    I searched for day but dont understand how to use the formulas to calculate the on off times of a 555 timer. I have been playing with the Yenka software and i almost have what i want but im not sure if it is ok. I really only have limited knowledge of this stuff. I know alittle about resisters and capacitors but nothing about how they work together.
    I have a 6v 4mA battery running a motor continuously at an unknown RPM. I want to use a timer to run it for about 5 seconds on and 5 seconds off. My simulation shows that the motor winds up and down at these times but dosnt turn on and off. I have seen a npn transister used in a few schmatics but cant get it to work right in the yenka software. I have been just guessing up to this point. I also noticed that the resister on th output leg of the timer seems to control RPM of the motor. Is that needed? Here is what I have come up with. Dont laugh to hard.
     
  2. ke5nnt

    Active Member

    Mar 1, 2009
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    A charging capacitor needs to reach \frac{2}{3} the input voltage to trigger the 555 timer to "on". When voltage falls below \frac{1}{3} supply voltage, the output (pin 3) changes state to +VCC (supply voltage) and the discharge transistor (pin 7) is turned off.

    In your example, V_{cc} is 6V, so the charging capacitor would need to charge to 4V to trigger "on" and discharge to 2V to trigger "off". The way it does this is to charge the capacitor through a resistor. For the charging capacitor, I'm not sure you should use anything but a polarized electrolytic type capacitor rated for a voltage greater than or equal to V_{cc}, in your case probably 12 or 15 Volts.

    As far as choosing the value of the capacitor and resistor, use this formula: T=1.1*R*C, where T is Time in seconds, R is expressed in Ohms, and C is expressed in Farads. Your desire for 5 second intervals can be accomplished using a Resistor Value of 100K Ohms (100,000 Ohms) and a capacitor value of 47μF (0.000047 Farads). This will give you:

    1.1 x 100000 x 0.000047 = 5.17 seconds.
    [​IMG]

    What kind of motor are you talking about exactly? 4mA seems like a very small current for a motor unless its really small, maybe it's just me. This could be your problem, if the motor requires more current to operate than the battery can supply.

    Not necessarily, that resistor is probably there from the original schematic to limit current to the motor, which in the case of a 4mA supply current, probably wont be a problem, if it works at all for the reason I stated above.

    AAC (All About Circuits) members don't laugh, they just help. I may not have answered all your questions, but I'm not an expert. Hope this helps you. Welcome to the forums.

    Sources, and a couple links to check out:

    Play-Hookey.com
    Bill Marsden's Blog
     
  3. 6octagon

    Thread Starter New Member

    Nov 9, 2009
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    Thanks for the help. I will check out the links. I have looked at so much stuff that i dont know up from down any more. I think i made a mistake on the battery info. I think it is 4.0 aH. Here is a pic of one that is used in a similar application. The motor i dont know much about. I will keep playing.
    Here is a link to what im playing with minus the remote and reciever. It is a motion duck decoy. http://www.mojooutdoors.com/wiring-diagram-for-mojo-decoys.htm
     
  4. Wendy

    Moderator

    Mar 24, 2008
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  5. ke5nnt

    Active Member

    Mar 1, 2009
    384
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    OOOHHHH KAYYYY... yes, 4.5 AH (amp hours) means your battery can supply 1 full amp of current for 4.5 hours (assuming full charge at start). Anything less than 1 amp will be available over 4.5 hours, anything over 1 amp will be less than 4.5 hours. Get it?

    Bill's reply post has the schematic set-up you need at the very bottom using an NPN transistor.
     
  6. lmartinez

    Active Member

    Mar 8, 2009
    224
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    Impressive how a 50% percent duty cycle can be achieved by the above given circuit. Is it a fact? :rolleyes:
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    Yes.

    The diode is the key; it provides a charge path for C1 around R2. The discharge path is strictly via R2 to pin 7.

    Your mileage may vary somewhat. It depends upon the Vf of the diode that's used, the current being sunk via pin 7, the tolerance of the R1/R2 resistors and the internal tolerance of the voltage divider network.

    Without the diode, you can't get down to a 50% duty cycle with the standard R1/R2/C1 timing configuration, even with very small values of R1, C1, and large values of R2.

    R1 values should not be lower than 100 Ohms per volt of Vcc, or the Vce of pin 7 will become excessive, and you risk blowing the lid off the 555 timer due to high power dissipation.
     
  8. lmartinez

    Active Member

    Mar 8, 2009
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    Which schematic are you referring to?
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    The very first one.

    OK, if you are referring to Bill Marsden's schematic where C1 is being charged/discharged by a resistor connected to pin 3;

    That will work pretty well for a CMOS 555 timer that's under light load.

    It can also be made to work quite well with a BJT 555, but the control pin needs to be pulled down towards ground a bit, using a pot to parallel the internal 15k (nominal) divider used for the threshold and trigger levels.

    That's because the bjt 555's output will get down within .2v or so of ground, but won't get much higher than Vcc-1.3v even under light load due to the Darlington output.
     
  10. Wendy

    Moderator

    Mar 24, 2008
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    The transistor could be replaced with a digital MOSFET with no other change. Note, I said digital MOSFET, these parts take less voltage to turn on than a conventional MOSFET. MOSFETs are generally better than BJTs for this kind of application.
     
  11. 6octagon

    Thread Starter New Member

    Nov 9, 2009
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    Wow! Thank you for all the input but you guys are getting to technical for me to follow. Do you think if i do what i created in this diagram will it blow up my battery and motor. Also I like the idea of using a variable resistor to change the seconds on and off.
     
  12. Wendy

    Moderator

    Mar 24, 2008
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    Why the 330Ω resistor? I suspect it is a project killer all by itself.

    We were discussing components that would use less power. Heat in a project like this is the enemy, it is waste. It shortens battery life, and has no real benefits (unless you need to warm the battery). What is you're motor voltage and current?
     
  13. beenthere

    Retired Moderator

    Apr 20, 2004
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    One other problem - the diode in parallel with the motor is forward biased. That means only .7 volts left across the motor. Change the orientation of the diode before removing the 330 ohm resistor.
     
  14. 6octagon

    Thread Starter New Member

    Nov 9, 2009
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    OK - I flipped the diode and removed the resistor and it exploded the npn transistor in my simulator. How do you know what transister you need. I picked some random one that worked (TIP120). I have no info about the motor at all. From the factory the setup is 6V 4.0aH battery - on/off switch - motor. So i guess i need to keep the output the same as what the battery would ouput right? Will all these components create resistance? I can handle the rpm of the motor dropping alittle if thats all that would happen. Thanks again for all the help.
     
  15. SgtWookie

    Expert

    Jul 17, 2007
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    OK, a few things:
    1) You are using the TIP120 as an emitter follower. It would be better if you used it as a current sink. Move the motor and diode from the emitter - ground connection, and move them above the TIP120's collector (motor + & diodes cathod connected to +6v, motor - and diodes' anode connected to the TIP120's collector).

    2) The base of the TIP120 needs current limiting; so does the output of the 555 timer.
    The datasheet for the TIP120 says that the maximum base current is 120mA.

    The minimum gain of a TIP120 is 1000. The maximum current for a TIP120 is 5A. So, since 5A/1000 = 5mA, you shouldn't really need more than 5mA current through the base. However, the datasheet shows 20mA for a 5A current; let's go with that.

    Since the TIP120 is a Darlington there are two BE junctions between ground and the base. So roughly, that's 0.65v x 2=1.3v.

    If your 555 timer is a standard BJT (transistor) timer, the highest the output can go is Vcc-1.3v, so 6v-1.3v = 4.7v. Then we subtract the base-emitter voltage of the TIP120, so that's 4.7v-1.3v = 3.4v
    Now what resistance do you need to get 20mA current from a 3.4v drop?
    R = E/I (Resistance = Voltage / Current)
    R = 3.4/20mA
    R = 3.4/0.02
    R = 170 Ohms.
    So, place a 170 Ohm resistor between the 555 output (pin 3) and the base of the TIP120.
    It could actually be anywhere from 170 Ohms to 340 Ohms. 340 Ohms would allow 10mA current; that should still be enough.

    3) You have R1 as a 75k Ohm potentiometer. That's OK, but 75k isn't a standard value. 10k, 20k 50k, 100k, 200k, 500k are standard values.
    4) Something important - R1 should never be set to less than 100 Ohms per volt of Vcc. Since you have a Vcc of 6v, R1 must never be set to less than 600 Ohms. To make sure that doesn't happen, you can put another resistor in series with R1.

    The reason for this, is when the timer is discharging C1, pin 7 is shorted to ground (it's an open-collector transistor) - if R1 is set to too low of a value, you will be shorting out the power supply, and probably burn up the 555 timer.

    [eta]
    Here's what I'm talking about; I moved the motor & diode, and added the 600 Ohm resistor to R1, and the 170 Ohm resistor between the 555 output and the TIP120's base.

    [​IMG]
     
    Last edited: Nov 25, 2009
  16. 6octagon

    Thread Starter New Member

    Nov 9, 2009
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    Sgtwookie I think i have it. But If i change R1 to 600Ω wont that make my motor on time super long. I only want it to run 5-10 seconds.
     
  17. SgtWookie

    Expert

    Jul 17, 2007
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    You have a 75k pot in series with a 600 Ohm resistor.
    75k + 600 Ohms = 75.6k Ohms. That only increased the maximum resistance of R1.

    What it does is prevent you from accidentally setting the 75k Ohm pot to 0 Ohms, which would destroy the 555 timer.

    Note that the minimum value is 600 Ohms for a 6v supply. You can use more resistance, but not less.
    A 620 Ohm, 680 Ohm, and 750 Ohm resistor is a standard value. They would all be fine to use.
     
    Last edited: Nov 25, 2009
  18. Audioguru

    New Member

    Dec 20, 2007
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    Your battery is rated for 4Ah, which is 4 ampere-hours. One charge will last for 0.4A for 10 hours or 0.2A for 20 hours. It might not be able to provide 4A but maybe only for a couple of minutes.
     
  19. 6octagon

    Thread Starter New Member

    Nov 9, 2009
    8
    0
    ok the 680Ω is only a safety. I originally put a 680kΩ in that made the time to long. Hows this one look? Do my capacitors look ok? I think the hardest thing is going to buy this stuff. There are millions of different diodes alone. Is it as simple as buying components with the values i need within the voltage and amp range as the battery? I see the resistors go by watts. I attached a BOM that the software spit out. You guys are amazing! I dont know how you can keep all this straight. If i was to think way ahead and want to use this on other things with different batteries what would need to change? The R1 and the resistor before the transistor? Im not going to worry about that now, but since i have your attention. :)
     
  20. 6octagon

    Thread Starter New Member

    Nov 9, 2009
    8
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    Oh no I flipped the switch to off in the simulator and the timer exploded. It said the pin voltage was taken below the negative supply voltage by 2.92V. The Max is 1V.
     
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