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Going Surface Mount from Through Hole — Parallax Forums

Going Surface Mount from Through Hole

eagletalontimeagletalontim Posts: 1,399
edited 2011-12-12 18:46 in General Discussion
I am worried about switching to surface mount components due to the current of my circuit on the high side 12v rail. I was hoping to convert everything I could to surface mount making more room for additional features. I know some components will not be able to be converted, but I want to convert as much as possible. One question I have is, the current ratings between surface mount resistors vs through hole. I currently have a 10k 1/2 watt resistor that pulls down a TIP42 transistor's base. The resistor has 12 volts DC to it at all times which is connected to a 10 amp supply. The transistors are the only thing that handle high current. Can I replace the 10k 1/2 watt resistor with this and will it handle the current?

http://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_2078944_-1

Comments

  • Duane DegnDuane Degn Posts: 10,588
    edited 2011-12-11 08:39
    0805 resistors are usually 1/8 Watt. At 12V it depends on the current whether or not 1/8W is enough.

    P = IV or I = P/V so the current through the reistor needs to be limited to 10.4mA. I doubt that's enough current for your puposes.

    EDIT: I answered too quickly. I see it's a 10K ohm resistor. At 10K ohm the current may be low enough. Time to drink some caffeine.
    EDIT (again): With 10K ohm at 12 V your current is only 1.2mA (Ohm's Law). So you should be okay.
  • eagletalontimeagletalontim Posts: 1,399
    edited 2011-12-11 08:45
    Is there a surface mount resistor that will work? Do they make 1/2 watt surface mount resistors?
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-11 08:47
    Yageo has some literature on their chip resistors. Perhaps you can find the power rating somewhere in this:

    http://www.yageo.com/portal/product/open_literature.jsp?SWITCH_CATEGORY=/product/open_literature

    I'm guessing that surface chip is probably rated to only 1/8 watt, so it makes sense to figure out exactly what sort of power it is actually handling. Of course you could always get a bigger chip resistor if you think it's a possible problem.
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-11 08:49
    eagletalontim wrote: »
    ...Do they make 1/2 watt surface mount resistors?

    Try looking at this product selection guide:

    http://www.yageo.com/exep/pages/download/literatures/ProductSelectionGuide2011.pdf

    EDIT: I think Duane is right: at 10K, you are probably okay, at least so far as power dissipation goes.
  • Duane DegnDuane Degn Posts: 10,588
    edited 2011-12-11 09:15
    Okay, I've had a couple of sips of Dr. Pepper (my current favorite source of caffeine).

    This is the electronic theory stuff I actually still remember from my physics classes. There are two equations that are important to us here. One is Ohm's law (I'll use the notation on my Inventor's wand) E = I*R and the power equation P = I*E. (R is resistance, P is power, I is current and E is voltage (IIRC it stands for electromotive (or something like that) force).

    So we want to know the power through your resistor. We start with P = I*E. P is our unknown. E is 12V. We don't yet know I but we can get it from Ohm's Law.

    I = E/R or 12V/10000ohm = 0.0012A or 1.2mA

    We plug 1.2mA into P = I*E for

    P = 0.0012A * 12V = 0.0144W

    This is about 1/69th of a Watt much less than 1/8 or even 1/10 (another common resistor size).

    Thanks to my Inventor's Wand I could also tell you about drill and tap sizes or the Three Laws of Robotics.

    Just don't expect me to answer a question about a LRC circuit.
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-11 13:16
    Duane Degn wrote: »
    ...

    Just don't expect me to answer a question about a LRC circuit.

    What's LRC stand for?
  • PublisonPublison Posts: 12,366
    edited 2011-12-11 13:24
    ElectricAye wrote: »
    What's LRC stand for?

    I've remember it as LRC, but I guess RLC is the norm now:

    http://en.wikipedia.org/wiki/RLC_circuit
  • $WMc%$WMc% Posts: 1,884
    edited 2011-12-11 13:28
    Inductor/resister/capacitor
    '
    '
    EDIT:...Sorry Publison. We posted @ around the same time....
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-11 13:36
    Duane Degn wrote: »
    ...

    Just don't expect me to answer a question about a LRC circuit.
    ElectricAye wrote: »
    What's LRC stand for?


    $WMc% wrote: »
    Inductor/resister/capacitor....


    Geeeez, guys, you gave him the answer!
  • kwinnkwinn Posts: 8,697
    edited 2011-12-11 13:57
    Try these formulas.
  • Duane DegnDuane Degn Posts: 10,588
    edited 2011-12-11 14:07
    Publison wrote: »
    I've remember it as LRC, but I guess RLC is the norm now:

    http://en.wikipedia.org/wiki/RLC_circuit



    Yes, we had to calculated the various parts of such a circuit in one of my physics classes. I like to think I could learn to use those equations again but for now they're a bit of a fog.

    I also remember something about ELI the ICE man. (A memory aid for AC circuit stuff, I think.)
  • eagletalontimeagletalontim Posts: 1,399
    edited 2011-12-11 16:08
    Not sure about the math coming out to less than 1/8 watt for the 10K. I have tried a 1/4 watt and it got hot. Thanks for the feedback! The PDF is quite useful!
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-11 16:22
    eagletalontim wrote: »
    Not sure about the math coming out to less than 1/8 watt for the 10K. I have tried a 1/4 watt and it got hot. Thanks for the feedback! The PDF is quite useful!

    Duane's math looks reasonable. Maybe you should post your circuit schematic so it can be checked for other possible problems.
  • Duane DegnDuane Degn Posts: 10,588
    edited 2011-12-11 18:17
    I imagine the information I've read here on the forum about voltage regulator also applies to resistors.

    Just because a regulator (and I assume resistor) can handle a stated voltage and current (power) doesn't mean it won't get too hot. You still have to worry about moving heat away from a component. (Mike Green comes to mind as I write this. I'm pretty sure I learned the above from his posts.)
  • eagletalontimeagletalontim Posts: 1,399
    edited 2011-12-11 18:36
    The components are in a completely sealed box so it is hard to pull heat from the components. My goal is to not produce any heat which has been working for 2 years now on the same circuit...problem is now, I need to redesign to fit smaller components on it to be able to switch to the Prop eventually. I am currently using the SX28, but since it is at an EOL, I need to get things switched around a bit.

    Right now, I don't have a schematic, but I can explain it very well :) I have a TIP42 220 case which is the power feed for the device that is being activated. Since the device needs 12vdc, I cannot use an NPN. What I currently use is a MPSA06 NPN transistor which has a 1k 1/2W resistor between the base and the SX28, then there is a 1k 1/2W resistor between the collector of the NPN and the base of the TIP42. The TIP42 has at 10k 1/2W resistor between the positive 12V rail and the base. I know it seems complicated, but I will try to get a schematic up as soon as possible.
  • eagletalontimeagletalontim Posts: 1,399
    edited 2011-12-11 18:48
    Ok, here is a schematic. Hopefully I did it right.


    Also.. The current that goes through the TIP42 is about 2 amps
    12v-control-sx28.jpg
    775 x 514 - 29K
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-11 19:02
    Did you ever monitor the heat situation in your old design? Based on your schematic, it doesn't look like these resistors would generate very much heat, but if the box is "completely sealed" then you might have a problem. Keep in mind that when components are rated for a certain wattage, the manufacturers are making some broad assumptions about the situation in which the component is being used. The manufacturers assume a certain amount of convective or conductive heat transfer is able to happen, for example. So if you have a component generating even a small amount of thermal energy but it has no way to conduct that heat to the environment, or if the environment is too hot to accept any of that heat, then the component temperature could eventually rise to a point where it fails. So if your box is sealed too well to allow the components to dump their heat, you might need to mount your heat-generating components perhaps to the wall of the box, etc. so there is at least some chance of getting rid of the heat.
  • kwinnkwinn Posts: 8,697
    edited 2011-12-11 20:22
    Your schematic looks good. Here are the worst case power calculations.
  • eagletalontimeagletalontim Posts: 1,399
    edited 2011-12-12 02:17
    I have checked the temp of the box inside when running for 8 hours under full load. The temp difference from ambient temp to the inside of the box was only 2 degrees. I was wondering if I could change the TIP42 to something different like a MOSFET to eliminate some of the cost of components and free up some space by using surface mount components. I am not very familiar with using MOSFETs so I worry about changing something I know that works. I would rather have overkill for current ratings than barely have enough.
  • eagletalontimeagletalontim Posts: 1,399
    edited 2011-12-12 16:15
    Would this MOSFET be suitable for activating a 12v 2A circuit for long periods of time? I see it has a Max Drain Current of 3A and a Max drain voltage of 30V. It seems like it would work, but I would need to verify before I move on. Would the SX or PROP be able to activate this MOSFET without a complete other circuit from the SX or Prop like I am using now?

    https://www.jameco.com/webapp/wcs/stores/servlet/Product_10001_10001_1563690_-1
  • ElectricAyeElectricAye Posts: 4,561
    edited 2011-12-12 18:46
    eagletalontim wrote: »
    Would this MOSFET be suitable for activating a 12v 2A circuit for long periods of time? ...



    One thing you want to look at is the mosfet's "Maximum Drain Source Resistance", which looks to be approximately 0.098 ohms at 10 volts (but you should check the charts in the spec to see if you can get a better number based on your exact operating situation). "Drain Source Resistance" is how much ohmic resistance the mosfet appears to have given a certain amount of current racing through it at a certain voltage.

    From V = IR, you can see that the voltage drop across the mosfet is going to be about V = (2amps)(0.098 ohms) = about 0.2 volts.

    To determine the amount of heat produced across the mosfet, you use Power = VI, which gives you (0.2 volt drop)(2 amps) = about 0.4 watts. The spec shows a "Maximum Power Dissipation" of about 1.2 watts (but you should check the charts in the spec to see if you can get a better number based on your exact operating situation), so you should be careful about how the circuit will be packaged and whether or not it's got some ventilation, etc. so it can dissipate this kind of heat. That much heat on a tiny package could easily make it hot fairly fast if it had no way to dump the heat.
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