Transistor/LM317/current/dc motor help :)

Discussion in 'General Electronics Chat' started by AlmightyJu, Oct 23, 2011.

  1. AlmightyJu

    Thread Starter New Member

    Oct 14, 2011
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    Hi guys,

    a bit of a long post im afraid!

    I've made a little 555 timer circuit to run a LED but I want it to run a motor instead (i think its one of these: http://www.ebay.co.uk/itm/2x-New-Miniature-DC-Electric-Motor-1-5-4-5V-13000rpm-/130474122559), so I was looking at doing what it says here http://www.kpsec.freeuk.com/trancirc.htm#ic and use a transistor as a switch

    Considering the worst case scenario that the motor takes 1A I figure i need a transistor which meets:
    Ic(max) > 1A
    hfe(min) > 25 ( 5 x (1000mA/200mA)) - 555 timer Ic max output is 200mA

    the only transistor I have over 1A is a BD135 (http://www.fairchildsemi.com/ds/BD/BD135.pdf) but it seems overkill to be honest so not sure if its quite right but even though the motor will only be pulsing on/off (PWM?) it will still take 1A theoretically?

    so assuming I use that transistor my next question is to do with powering the motor, my power input is 5V so my LM317 is set up to output roughly 3.125V, 2 questions come to mind:

    1. if the Vout of the 317 goes to my motor and the 555 will there be 1A current into the 555 and the motor? - i.e. will my entire circuit have a 1A current or does the current "branch" and go to wherever it is needed, so the 555 will only take 200mA say (because of its output) and the motor takes 1A so the 317 will output 1.2A (hope that makes sense, I'm not quite 100% on the basics yet :) although im fairly certain it branches)

    2. the Vout goes to my 555 and the current is quite low with an LED but with it going to a transistor will the current to the base pin be 200mA? - im worried about the load on the 317 and power dissipation (does this also effect the resistors that regulate the voltage??). I know the formula for dissipation is:
    PD = ((VIN − VOUT) × IL) + (VIN × IG)
    PD = ((5V - 3.125V) x 1.2A) + (5V x ??)
    PD = 2.25? + ??

    but that doesn't help me so im stuck.

    The alternative is to power the motor straight from the 5V but is there an easy way to bring the voltage down to at max 4.5V for a 1A load?

    many thanks
     
  2. Adjuster

    Well-Known Member

    Dec 26, 2010
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    The length of your post is not such a big issue as its clarity. This would be much improved if you would provide a schematic of what you want to build. Questions about what current goes where will then be much more easily understood.

    My first reaction is that using a logic-level FET instead of a bipolar could help you to avoid sweating over gain and the maximum current drive available from the 555.
     
  3. Audioguru

    New Member

    Dec 20, 2007
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    hFE of a transistor is used when it is a linear amplifier with plenty of collector to emitter voltage. But a switching transistor needs its base current at 1/10th the collector current (or more) because its collector to emitter voltage is very low. The datasheet of every transistor shows its Max Saturation Voltage Loss when its base current is much more than is calulated by its hFE.
     
  4. AlmightyJu

    Thread Starter New Member

    Oct 14, 2011
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    I've never used a transistor before so its a bit new so I'm going to read that again tomorrow after a good sleep and hope it sinks in! :)

    I've attached the circuit of what I want, and I'm at the point where I've got the output from the 555 to come on and off (dont worry about the timing of it) so its just R5 and Q1 im stuck at.

    the attachment "from lm317" seems the best way because the voltage is at 3V, but what would the current be for all the resistors? would it blow up 1/4 watt ones or would the 1A current go straight to the motor and no where else?

    the other option is "from 5v" but I would somehow need to bring the voltage down.

    hope that makes things clearer :)

    edit: just realised the names of the files dont show up - the left one is "from lm317" and the right is "from 5v"
     
  5. Audioguru

    New Member

    Dec 20, 2007
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    The cheap Chinese motor has no detailed specs. Its starting and stalled current is not stated. It might be 5A or 10A. Then the voltage regulator will limit the current and the motor will not start running.

    The minimum supply voltage of an ordinary 555 is 4.5V so it won't work in your circuit that has a supply of only 3.1V.
    A Cmos 555 works at 3.1V but its output current is too low for driving only a single transistor.

    Why do you need a voltage regulator? An ordinary 555 works from an unregulated supply that is 4.5V to 15V.
     
  6. Adjuster

    Well-Known Member

    Dec 26, 2010
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    If the motor is fed via the LM317, then its current will come through the IC.: current will not pass into the small resistors around the regulator.

    Note that the IC must in a package suitable for fitting to a heatsink capable of dissipating a few watts safely. (That is, the wattage given by your motor current, times the voltage dropped accross the regulator).

    LM317 is also made in small packages with a lesser current rating. These will not do.

    http://www.national.com/ds/LM/LM117.pdf

    The transistor or FET driving the motor also requires cooling.
     
  7. Adjuster

    Well-Known Member

    Dec 26, 2010
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    Noting Audioguru's comment about CMOS 555s, if you have got one of those you will either need to add an extra transistor, or use a MOSFET*. If the latter, you need to be sure to use one that will be quite well turned on by the timer output voltage.

    *Or get a bipolar 555, but that will need more supply voltage.
     
  8. #12

    Expert

    Nov 30, 2010
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    6,797
    Here are some ideas. The second drawing will deliver close to 3 volts to the motor due to losses in the transistors. You could add a resistor to from the base of the first transistor to ground to lower the voltage more, if necessary.
     
  9. AlmightyJu

    Thread Starter New Member

    Oct 14, 2011
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    The same conclusion I came to :/

    and thats what happens when you dont read the data sheet fully :rolleyes: its a NE555, but it works on 3V ok lol!

    I only used the regulator for the motor since its 3V, so i'll take a look at #12's idea and how that works and take the regulator out :)

    Hmm, going to have to do some reading as to how that works I think! what transistor's would you use for that and the resistors?

    thanks :)
     
  10. #12

    Expert

    Nov 30, 2010
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    The transistors are BD135.
    The motor uses 1.07 amps at full load, so the second transistor will have a gain of about 55 when 2 volts is left over from its collector to emitter.
    It will need base current of at least 19.45 milliamps and its Vbe will be about .9V
    The first transistor sending 20ma will have a gain of about 90 with a Vbe of .7V
    So, 3+.9+.7 is the voltage at the base of Q1. If the 555 gives a perfect 5.00 volts out, and this is not likely, you have .2 volts to waste in resistors.

    For Q1, .2V/216 ua = 925 ohms. Try 1k
    That leaves .184V for the second resistor which carries 19.45 milliamps.
    9.46 ohms
    Try 10 ohms

    That's a start, but it depends on your exact voltages. Do some measuring and see what happens.

    Now that you've named your 555 as an NE555, you could run it from 5 volts and just make a voltage divider on the base of ONE transistor so it has at least 20 ma of drive current and the base voltage is set at 3.7 volts. That will lock the emitter at 3.0 volts.
     
  11. AlmightyJu

    Thread Starter New Member

    Oct 14, 2011
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    Sorry its been a while since I got back, been a bit busy!

    does the attached work? I'm still working on what R3 needs to be, and I worked out the voltage drop from http://www.raltron.com/cust/tools/voltage_divider.asp, not sure if its quite right or if the 16k res is avaliable so dont worry about that too much ;)

    also do you think there will be quite a bit of heat in the transistor which would need a heatsink at 1A ?

    thanks :D
     
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  12. Audioguru

    New Member

    Dec 20, 2007
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    The "maximum load current" of the cheap motor is 1.07A when it is powered from 3V. Then it is 4.75/3 times more (1.7A) at 4.75V. But the motor draws up to 10 times more when it starts, maybe 17A!

    Since the transistor is driven from a 555 then a voltage divider at its base is not needed. Your voltage divider supplied hardly any current anyway. Only a single current-limiting resistor is needed.
    I calculated its value for a 1A load on the transistor. It will not get hot unless its current is much more than 1A (its max allowed current is only 1.5A to 3A).
     
  13. AlmightyJu

    Thread Starter New Member

    Oct 14, 2011
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    17A :O! would a FET be better than a transistor then?

    Everything makes sense how you calculated that, but how do you know a base voltage of 0.95V allows the motor to have 4.75V?

    The 555 output goes to the base yes. then the 5V goes stright to the motor and the transistor.

    I don't have any 1/2W resistors but if I was to run the motor at 3V would I be able to get around it? or do I just go get a 1/2W one?
     
  14. Audioguru

    New Member

    Dec 20, 2007
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    Since the cheap motor does not have a detailed datasheet then we must guess that its "maximum load current" is when it is running, not when it has a much higher current to start. Its starting current is the same as when it is stalled. Then it is simply a very low resistance that uses a very high current.

    A graph on the datasheet for the BD135 shows that with a collector current of 1A then its base-emitter voltage when it is saturated is typically 0.95V. Another graph shows that its collector to emitter saturation voltage is typically 0.25V. Your power supply is 5.0V so the motor gets 5.0 - 0.25= 4.75V.

    The ratings on the datasheet show maximum voltages which are much higher than typical so they must be calculated separately from the typical ones. The temperature also affects the voltages as shown on the graphs.

    The calculations for the amount of current and heating in a resistor are simple.
     
  15. AlmightyJu

    Thread Starter New Member

    Oct 14, 2011
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    I know, what i'm trying to work out is that the motor is max 4.5V so I'm working out how I can drop that down to 3V using resistors or the BD135 with the current things I have without blowing things up :)
     
  16. Audioguru

    New Member

    Dec 20, 2007
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    If you try to drop the voltage to a motor with a resistor then the motor might not start running because the much higher starting current will cause a much higher voltage drop in the resistor then the motor will get a very low voltage and won't start.

    You can try using the BD135 as an emitter-follower and its emitter voltage will be typically 0.85V less than its base voltage when its emitter current is 1A. Make its base +3.85V then its emitter will be typically 3.0V. It will heat with 2W so it will need a little heatsink. The graph on the datasheet shows that its base-emitter voltage is less when it is not saturated.
     
  17. sheldons

    Active Member

    Oct 26, 2011
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    try this as a starter i pulled from the net
     
  18. Audioguru

    New Member

    Dec 20, 2007
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    If the motor does not start running with the LM317 variable voltage regulator circuit then its starting current is more than 2.2A.
     
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