Project Help/Direction

Discussion in 'The Projects Forum' started by maz719, Mar 16, 2009.

  1. maz719

    Thread Starter New Member

    Feb 23, 2009
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    Hey guys, I've googled a lot of broad topics and I'm here to ask for a little advice, or direction - as I am a tad bit confused about how to implement certain things.

    I am currently in my last year for my aerospace engineering degree at ucf, and for my senior design project the last piece that we need in order for it to work is a circuit that can amplify a signal. Now the signal is coming from a separate board that uses PWM to send voltage from 0 to 3.3V, and 0 to 132mA (varying by duty cycle 0~100%). This board was premade by Rabbit, so we are avoiding modifying it. The power is output to a magentic torque coil (just a wire wrapped around a square) so we are adjusting the power through them regularly from the Rabbit board, and we need to be able to range from 0W to the maximum power (currently 436mW).

    Basically I'm asking for the best way to take the power output of the Rabbit boardv(436mW) and amplify it. Keep in mind that the entire contents of the Rabbit Board, and this circuit must fit in a 10cm x 10cm x 10cm cube, so components that would all fit on a breadboard are preferred.

    Also, for now we can only run everything off of a 9V battery. This may change later.

    I have tried to use a NPN transistor, but I couldn't find the best way to wire it so that in the end there was more than 436mW. I also tried to use an op-amp (TL082CP) but it looks like it requires 15VDC, which I don't know how to get from 9V (please forgive me if this is noobish, it's been a while since I took an EE class).

    If anyone has any ideas of how to take the power from the Rabbit board and generate a larger power somehow (even it it's larger by 50mW - its still acceptable) please let me know. I am lost in reading many forums and chapters of ebooks and any help from you guys would be greatly appreciated.
     
  2. SgtWookie

    Expert

    Jul 17, 2007
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    Do you have a datasheet for the Rabbit board? It would be most helpful if you could provide a link or upload it to the Forum from your computer (use the "Go Advanced" button below, then "Manage Attachments").

    9v "transistor" batteries are very limited in their current output. Internally, they usually contain six cells that look like "AAA" batteries, but much smaller in diameter. The more current you try to draw from them, the more power is dissipated across the battery's internal resistance, quickly heating them up instead of doing useful work.

    If you are using a linear regulator to drop the 9v down to 3.3v for your board, most of the battery power is being dissipated in the regulator and the battery.
     
  3. maz719

    Thread Starter New Member

    Feb 23, 2009
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    0
    here is the rabbit board:
    http://rabbit.com/products/RCM5700/index.shtml

    The battery isn't much of a concern (we have the rabbit board running off a power supply plugged in the wall) as much as just a way to amplify the power coming from the board and then running that through the magnetic torque coil. If ya need a further explanation please let me know.
     
  4. SgtWookie

    Expert

    Jul 17, 2007
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    You might use something like this:

    [​IMG]

    R1 limits the current through the base of the transistor.
    R2 is optional; if you don't want current sensing capability then just connect the emitter to ground.
    L1 represents your coil.
    D1 is for reverse-EMF suppresion for when Q1 turns off; otherwise there will be a high reverse voltage spike that will likely kill Q1. C1 "buys time" for D1 to turn on.

    You can substitute a PN2222 for the 2N2222A. Note that 2N3904 and 2N4401 transistors are much more limited in current sinking ability.
     
  5. maz719

    Thread Starter New Member

    Feb 23, 2009
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    Awesome! Thank you very much, I will try this in the morning. Just for reference - I-sense is the output?
     
  6. SgtWookie

    Expert

    Jul 17, 2007
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    I-sense gives you a voltage representing the current flow through the inductor, but only while Q1 is conducting. The recirculated current via the "flywheel" diode D1 has no sense output.

    Since I=E/R, and R=0.5, if I-sense = 0.5v then 0.5 Volts/0.5 Ohms = 1 Ampere.
    It would be quite a bit more efficient to use a linear Hall-effect sensor instead of a resistor, but you'd have to order it somewhere. I don't know of a local supplier for them. Skycraft Surplus might, but that's iffy. They're over on Fairbanks Ave, just this side of I-4 on the north side of the street. It's about a 20 minute drive from the campus. You would have to come armed with a long list of candidate sensors, and do a lot of digging around. They ARE a great source for wire, transistors, and various TTL/CMOS logic IC's. They have so much stuff there it's not possible to describe.
    Website: http://www.skycraftsurplus.com/

    You could increase R2 up to 1 Ohm, but more than that and you'll start consuming a good bit of power in the resistor.

    There are a couple of Radio Shacks not far from the campus, but the selection is pretty limited. One is on Colonial a few blocks east of Alafaya Trail, another is in the big shopping plaza just north of the 408 on the east side of Alafaya.
     
    Last edited: Mar 16, 2009
  7. Audioguru

    New Member

    Dec 20, 2007
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    They are AAAA cells. Their capacity is 1/3rd the capacity of AAA cells.

    The rabbit needs 200mA when it has the ethernet. without amplifying the signal a 9V battery with a 3.3V linear regulator will power it for only 2 hours then the little 9V alkaline battery will be dead.
     
  8. maz719

    Thread Starter New Member

    Feb 23, 2009
    7
    0
    Our project is to do an ADCS (attitude and determination) subsystem for a small CubeSAT. The way we will show how it works is by having a very strong magnetic field (by having a large power supple wrapped many times around a cylinder) with the adcs inside, and with the power we currently have we would only need about 15 minutes of battery anyway.

    The main question was how we can acheive a higher power output (from an input of 0 to 3.3V, and 0 to 120~ mA) through the use of either transistors or opamps, if possible.
     
  9. SgtWookie

    Expert

    Jul 17, 2007
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    Well, you won't get there with a TL082. You might with an L272, but those are mighty slow power opamps. You'll be better off with just a 2N2222; it's cutoff frequency is hundreds of MHz.

    Your 3.3v PWM output pretty much limits you to using transistors. You can get very low-threshold MOSFETs, but you'd almost have to order them.

    How are you winding your inductor? eg: diameter of the tube, wire gauge, insulation thickness, number of turns, single or multilayer, total width of windings?

    Or are you just shooting for a target inductance, and need some help calculating it?
     
  10. gregdevid

    Member

    Feb 4, 2009
    18
    0
    Hi,

    Sorry ,
    but i also concern this problem not same but this type of.
    So, if u find any solution please tell me.
     
  11. SgtWookie

    Expert

    Jul 17, 2007
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    Gregdavid,
    Even though your project might be related, if it's not exactly the same as the original poster's, you should start a new topic of your own.

    What you have done is called "hijacking a topic", and it is frowned upon here.
     
  12. maz719

    Thread Starter New Member

    Feb 23, 2009
    7
    0
    From my understanding the equation for inductance is :

    [SIZE=-1]H = (4 * Pi * #Turns * #Turns * coil Area * mu) / (coil Length * 10,000,000)[/SIZE]

    and mu is dependant on the core. I'm not too sure how to approach making my own inductor since I don't know where to get the right material. It sounds like a run-of-the-mill iron core will work, and I'm sure the mu for that can be found.

    I also looked into maybe buying one, all I found was this one from radioshack:

    http://www.radioshack.com/product/index.jsp?productId=2103978

    I don't suppose I could have two of those in series, with a set of five in parallel to create a complete inductance of 220uH? Would it be better to have just one inductor instead of the seven it would take for that one to create the desired inductance?
     
  13. SgtWookie

    Expert

    Jul 17, 2007
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    I thought you wanted a cylindrical inductor that would surround your project sensor?

    I just threw 220uH in that schematic somewhat pseudo-randomly. I don't know what the PWM capablities of your Rabbit unit is. The larger the value of inductance, the less ripple you'll have for a given PWM frequency. By the same token, increasing the value of inductance allows you to operate at a lower PWM frequency with the ripple current staying relatively constant.

    One thing you'll need to avoid is saturation; that will make current soar without doing any useful work.
     
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