Somethings not right-motor not working

Discussion in 'The Projects Forum' started by chimera, Nov 2, 2010.

  1. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    Well..here's another one. I made a proximity sensor for a robot project. Today i went a bought a toy car for $8. I opened it and took out the motor along with axle to drive four tires ( its dissembled at the moment). The objective was simple. To add a motor to the collector of the transistor and use the transistor as a switch. :cool:. When the proximity sensor senses an obstable, it drops to 0.34 V and causes the transistor to turn off---then led goes out. When there is no object the output of the photo transistor goes to 0.78 V and thus turning on the transistor

    I have already made the circuit on stripboard. When i temporarily connected the motor is series with the led, the led light went further bright. I also connected it in parallel to the led(to act as a snug diode configuration--check schematic named 'configurations') the led went out.:eek:--it didn't blow out..it still works..in case any of you jumped to that conclusion

    I wanted to test the motor to see if it working, just to be sure, i connected it directly to a 9v battery, Didn't work...Odd. I then made a quick voltage diving circuit to to vary 5 volts across it using a power supply. It spun. The motor works. I also made a very simple circuit to see why the motor was working (observe motor-test in attachment). That too didn't work.:mad:

    Now I have a empty structure of a car, a sensor (designed) and a motor..all ready to be used as a project prototype for initial development of a robot.

    What is going on with the motor..i don't know...:confused:

    help!!!!
     
    Last edited: Nov 2, 2010
  2. commandertom

    New Member

    Nov 2, 2010
    4
    0
    hey is a photo transitor capable of carring full voltage for the motor or is it just a switch therfor needing to hav a transistor probaly pnp to iniate the contact of the full powersuply voltage to the motor from the battery on the negative or positivr side switching turned on by the photo dector kindof like the photo dector is the igniton switch to the silinod on the starter motor in a car but try to get the ignition switch to carry all the juce for the starter it wold burn out the ignion switch so maybe the photo detector isnot rated to carry that much juce and is a simple switch
    ? not for shure try a transsitor linkied to photo on p (n) p and grounding the motor circut from positive direct to battery? good luck know how t step up voltage whout a transformer to a motor campacators or somthing wb
     
  3. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    Well the output of the phototransistor works to change the base voltage between 0.78 to 0.34 volts... hence making the transistor work like a switch. The motor shud be running when the output from the transistor is 0.78 and not running when the base votage becomes 0.34

    Hopefully tht explained it
     
  4. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,699
    905
    There may be several problems with your circuit.

    First, you have a 380 ohm resistor in series with the motor. I suspect the motor needs at least 100 mA, probably more, and that resistor is just too big. For starters, temporarily get rid of the LED. Get rid of the 380 ohm resistor.

    Second, a 9V battery cannot supply much current. Maybe your 9 V battery was old, which makes it even worse. How many batteries and what type were used in the car? I suspect 3 or 4 AA size. Were they rechargeable or alkaline? Anyway, use the same number and type of batteries for testing here. I realize your voltage divider worked, but that introduces another variable. At least show us the schematic that you used for it.

    Finally, the 2N3904 may get quite hot controlling that motor. It's current limit is 200 mA. A 2N2222 has a much higher current rating. You may want to use that transistor instead.

    Get that part working, then you can play with how to have the LED in the circuit too. You will not be able to have it in series, as it simply cannot pass enough current to run the motor.

    John
     
  5. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    John u always come to the rescue :D!!! Well after publishing the last post in this thread, I went and searched online about motors and their attributes. What i can conclude is that, the motor is not starting up because its not getting enough starter current.

    Here are my reading that I conducted on the motor:

    1- Current with no load at 9V = 0.15 [A]
    2- Starter current with no load at 9V = 0.32 [A]
    3- Stall current = 0.44 [A]
    4- Internal Resistance ≈ 10 [Ω]

    Clearly, with these real-time values, the sensor would work. Ive decided to redesign the proximity sensor circuit. This time Im going to have use a power BJT which will allow the Ic current to be max of 2+ amps. I will attach a current limiting resistor in series with that motor and a diode in paralllel.

    But before I go running to radio shack to get the power bjt (they are the closest electronic supplier I have)--- i want to confirm that the 2n2222 will be safe to run the motor.this transistor can handle up 0.6 amps MAX. And all i need is 0.48 amps for the worst case (stall current). And I must keep in mind that these values are with zero load. If there is load present, the values for the currents will be higher.

    Now all i gotta do is recalculate the values again for the proximity sensor. Shud be fun!!! *sarcastic sigh*
     
    Last edited: Nov 2, 2010
  6. jpanhalt

    AAC Fanatic!

    Jan 18, 2008
    5,699
    905
    Actually, I think it is fun to work through a problem one step at a time.

    As for transistor selection, the 2N2222 or 2N2222A type of transistors have about 800 mA current capacity. There are differences between vendors and exact part numbers, but for your purpose it will give you several times the current capacity of the 2N3904. You may not need a big power transistor. If you do need that much power, you may want to switch to a Darlington, which will give you more gain. (Radio Shack has a Darlington, but I don't recall its exact number. I think it begins TIP...)

    I mention Gain, which you need to consider. I don't know how much current your phototransistor can source, it will need to drive the base of the transistor to switch it fully on. A base resistor is also used to limit that current in the common emitter amplifier/switch. It seems almost every thread here includes its discussion. If you assume a gain of only 10 at high current for your transistor, then a starting point for the base resistor is: 10(Vcc-0.7V)/Ic

    Now, back to your motor. Stall current is the current the motor draws from a low-resistance supply when you physically stop the armature from turning. It is quite a bit higher than the running current. I suspect your number is a bit low, because of the limited ability of a 9V battery to supply current. But since your motor started at 0.3A, for the sake of discussion, you can calculate a base resistor for a 4.8V supply (i.e, 4X1.2 NiMH batteries) and a current of 0.4A -- i.e., 10(4.8-0.7)/0.4 = 102.5 ohm. So, you may want to try a 100 ohm base resistor.

    John
     
  7. SgtWookie

    Expert

    Jul 17, 2007
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    1,728
    I'll suggest that the 2N2222/PN2222 is good for up to about 500mA for continuous use, and 800mA pulsed.

    You'll need 1/10 the desired collector current to ensure that the transistor is saturated (ie: minimal Vce; this reduces power dissipation in the transistor).

    You may need to use a 2nd 2N2222 or 2N3904 or 2N4401 as a voltage follower to supply this extra current, as your CdS or photodiode or phototransistor light sensor won't supply that much current (50mA to 80mA).

    Either that, or you will need to go to a Darlington configuration; and Darlingtons are not particularly efficient due to their comparatively higher Vsat (0.7v to 1.6v typical)
     
  8. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    okay im a little confused. So I've attached a schematic (new one). Take a look at it. This is what I'm meaning to do.

    1- Run the photo transistor in the saturation stage.
    2- Connect the base of the 2n3904 with output from the photo transistor.
    3- Have a current limiting resistor in the collector of 2n3904 to make sure the current doesn't go above 50mA
    4- Connect the emitter of 2n3904 to the base of the 2n2222 (designed in the voltage follower).

    2n3904 and 2n2222 transistors will be designed in the ON stage. SO this is how it shud work. The photo transistor is in the saturation state. When the IR is not reflected back, the photo transistor remains in the saturation state. This provides the bases of the 2n3904 with 0.8 V to turn it on and hence conducts current. But this current is fed into the base of the 2n2222. This current will be magnified by the gain of the transistor (50mA x some number). This will be reflected in the emitter current of the 2n2222 and hence powering the motor at about 0.5A

    Simple. But a few unsure points: What will be voltage across the motor, do i need a resistor in the collector/emitter of the 2n2222, i only considered the stall current of the motor as that was the biggest value for the current. The running current and the start-up current can be handled.

    Hopefully after this, I can change the circuit.

    Critique on it. See if the set-up might not work
     
  9. wayneh

    Expert

    Sep 9, 2010
    12,139
    3,054
    That depends on the impedance of your power supply, and also on the relative impedances of the motor versus the transistor. If your 9v supply is low impedance (ie. NOT a 9v battery), it will hold at 9v regardless of the motor load. If the transistor can go fully on to a resistance much lower than the motor, then the motor will see a full 9v, less ~0.6v across the transistor. I think your motor is designed for that? If not, then a current limiting technique should be used.

    BTW, I think your circuit would be better if the motor was connected to Vcc, with the path to ground being regulated by the transistor.
     
  10. SgtWookie

    Expert

    Jul 17, 2007
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    That's not quite what I meant.

    Have a look at the attached. I used an optocoupler instead of an IR LED and a phototransistor; as it shows them in the same package - then I used a sine wave to power the IR emitter.

    The 2N3904 is the voltage follower that amplifies the current output of the phototransistor. The 2N2222 is used as a saturated switch.

    R3 and R5 make sure that Q1 and Q2 turn off.
     
  11. chimera

    Thread Starter Member

    Oct 21, 2010
    122
    2
    Okay, in my circuit, i believe im using the phototransistor as a switch. the 2n3904 works as a current amplier and the output of this is connected to the base of the 2n2222. If the 2n3904 shuts off, the 2n222 shuts of aswell. This way the motor would stop spinning. I will connect the motor to collecter via 9V and not to the emittter. I'll build the circuit and if it doesnt work, then i will try out ur schematic minus the sine wave and the opto coupler

    I really do appreciate the help. Everyone here had beeen really nice. The worst is that Im on my way to becoming an electronics engineer and wow.. theory is way different that practical work.. im experiecing it first hand
     
  12. wayneh

    Expert

    Sep 9, 2010
    12,139
    3,054
    True that. Wickedly satisfying when you can make them converge.
     
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