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Newb needs help with a 2 second delay for milsim

Discussion in 'General Electronics Discussion' started by Blacker360, Apr 10, 2013.

  1. Blacker360

    Blacker360

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    Apr 10, 2013
    So I made an account here to ask for help. Im a complete noob when it comes to electronics so I could use some help...well maybe alot. First off ill explain my situation. Me and 2 friends are milsim players. For those of you who don't know what that is, we do airsoft and paintball military simulations. Its kind of our hobby. But anyways, I came up with the idea of making our own smoke grenades. but I want to make them cool, not the light a fuse and run kind. Ive got some mini push buttons, coin size 3v batteries, and electronic fuses (need 6v so 2 batteries to ignite). It all works but I don't like having the things ignite in our hands. So to improve the design we need a 2 second delay. So, my questions are:

    How can I easily build a 2 second delay for my 6v electronic igniter (small enough to fit in cardboard tube with other crap?
    Actually, do I even have to build it or can I buy them complete some place?
    If not, where can I buy the parts (at a cheap, affordable price that is)?

    Any help would be much appreciated lol
     
  2. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    OK, take a look at this circuit diagram (schematic) and voltage and current graphs.

    [​IMG]

    This is probably the simplest solution. Q1 is an N-channel MOSFET. It works like an electronically controlled switch. When the voltage between its gate terminal (the left hand connection) and the source terminal (the bottom connection) reaches a certain voltage, called the gate threshold voltage, it begins to conduct current between its drain terminal (top) and its source (bottom). This completes the circuit between the battery (shown at the left, as "V1") and the igniter.

    Resistor RT and capacitor CT produce a rising voltage at Q1's gate terminal. This is because CT acts a bit like a rechargeable battery; current flows from the positive supply rail, through RT, into CT, and causes the voltage across CT to increase. After a certain amount of time, this voltage reaches the gate threshold voltage of Q1 and Q1 turns on and fires the igniter.

    The graph above the schematic shows the voltage on Q1's gate (the blue trace), and the current through the igniter (the green trace, which uses the markings on the right side of the graph area), plotted against time on the horizontal axis. The start of the graph is when the switch is turned on. (The switch isn't shown in the schematic, but the place where it goes is marked.)

    You can see how the gate voltage increases, starting from zero when power is first applied to the circuit. It's not a straight line. And you can see that when the gate voltage reaches about 4.3 volts, Q1 turns ON (starts conducting) and the igniter current ramps up quite quickly to the maximum current.

    RD is there to ensure that CT discharges (eventually) when power is removed from the circuit.

    That's the basic explanation, but there's a lot more to it.

    1. Battery current capacity and igniter current

    The battery needs to be able to comfortably supply the current required by the igniter, without its voltage dropping too much. You mentioned you're using coin cells; these usually have a high internal resistance and a fairly low maximum current output. They would only be suitable if the igniter is designed for low operating current, i.e. it has a fairly high resistance.

    I've assumed that the igniter has a resistance of 60 ohms so it will draw 100 mA (0.1 amps) at 6 volts (this is calculated using Ohm's Law, I=V/R, where I is current in amps, V is voltage in volts and R is resistance in ohms; if you're not familiar with Ohm's Law, you definitely should be - use Wikipedia or Google).

    I've also assumed that the battery's internal resistance is 6 ohms. This means that the battery voltage will drop to about 5.4V when the igniter is activated, and the igniter current will be only about 90 mA.

    You should look up the specifications for the cells you're using and the igniter you're using, to see whether you need to use bigger cells. If you don't, you may not get reliable ignition, and your cells may wear out quickly.

    2. Q1 gate threshold differences and variations, and suitable part numbers

    In the simulation, Q1 starts to conduct at a gate-source voltage of about 4.3V. The gate threshold voltage of a MOSFET that you buy is not well-defined; it varies between units and between production batches, and it can be quite different with different part numbers.

    The MOSFET I used in the simulation, BSC something, is just one that was available with the simulation software I use (this software is called LTSpice). I don't recommend that part for this application.

    I've suggested a MOSFET called NTD4906N, which is cheap (USD 0.57 from Digikey) and small and has a very low ON-resistance, so very little voltage is dropped across it when it's ON. But its gate threshold voltage is roughly about 2.4V, not 4.3V. This means it will turn on earlier, at about 0.8 seconds. To adjust for this, you need to adjust the time constant, by changing RT and/or CT. See later for details.

    You can actually use a wide variety of MOSFET devices in this circuit. The only important specifications are that it needs to be an N-channel MOSFET, with appropriate voltage and current capabilities. That is, Vds (maximum voltage across the top and bottom terminals) should be at least 10V, Id (maximum current through it) should be at least a few amps, Rds(ON) should be low to minimise voltage dropped across it when it's ON, and the gate threshold voltage, Vgs(th), should be between roughly 1V and 4.5V. If the gate threshold voltage is much higher than 4.5V there may not be enough voltage available to turn it on fully.

    You can search on Digikey (http://www.digikey.com) or Mouser (http://www.mouser.com) if you're in the U.S., or Farnell/Element14 in the UK, or Jaycar in Australia. Digikey and Mouser will ship internationally, of course, but they charge an arm and a leg!

    Also, MOSFETs are static-sensitive and should be handled carefully until they're connected into your circuit. Google those keywords for specific advice.

    3. MOSFET heating

    MOSFETs are normally used as switches. That is, they're either OFF (not conducting current through their drain-source path) or ON (conducting). This means that they dissipate relatively little power, and therefore don't get very hot.

    The formula for power dissipation is P = I V, where P is power in watts, I is current in amps (that is, current through the MOSFET) and V is voltage in volts (that is, voltage across the MOSFET, i.e. between its drain and its source).

    When the MOSFET is fully OFF, there may be voltage across it, but there's no current through it, so I*V is zero (or close to it). When the MOSFET is fully ON, there may be current through it, but there's no voltage across it (or very little voltage), so I*V again is close to zero.

    The problem occurs when the MOSFET is not FULLY conducting. This happens during the angled ramp-up on the green trace in the graph. During this ramp-up period, the MOSFET is partly conducting (it's said to be "in its linear region"), and it has voltage across it, AND current through it. This causes it to dissipate power during this time, and this power is dissipated as heat.

    As long as the time spent in the linear region is short, the thermal energy will not accumulate, and the MOSFET will not be damaged. I am assuming (hoping) that the time spent in the linear region will be short enough, and the MOSFET current will be low enough, that the MOSFET will not be damaged. I can't really know, until I know more about the igniter's characteristics. A part number for the igniter could really help here.

    This is a shortcoming of this circuit. To keep it simple, I connected the R-C timing circuit straight to the MOSFET's gate, without any kind of "cleaning-up" circuit to make the MOSFET switch quickly from fully OFF to fully ON. This forces the MOSFET to run in its linear region for a short time. I could have added more components to produce a clean control signal for the MOSFET. If you like, I can show you a modified circuit that does have this feature. Let me know if you want to see it.

    4. Delay time period

    The delay time is determined by the MOSFET's gate threshold voltage, and the values of RT and CT. An R-C (resistor-capacitor) circuit like this can be summarised by its "time constant", measured in seconds, which is the product of the resistance and the capacitance, i.e. T=RC. In this formula, R is in ohms, and C is in FARADS, not microfarads. So the time constant of a 47k resistor and a 33 uF (microfarad) capacitor is:
    T = 47,000 ohms * 0.000033
    = 1.55 seconds.

    This figure, 1.55 seconds, is not simply the time delay before the MOSFET turns on. That time depends on the gate threshold voltage of the MOSFET. If the MOSFET has a low gate threshold voltage, it will turn on sooner, because the threshold voltage will be reached sooner as the capacitor charges up. If the MOSFET has a high gate threshold voltage, it will turn on later. If the MOSFET gate threshold voltage is 63% of the power supply voltage (that is, the voltage at the top end of RT), then the delay will be 1.55 seconds, as calculated from the time constant formula, T=RC. This 63% figure is one of many numerical constants, like pi and sqrt(2), that are used in engineering.

    The point of all of that is that given a fixed gate voltage threshold, the delay time is proportional to the time constant of RT and CT, which is T=RC, so increasing either of them will increase the delay, and vice versa.

    Once you choose a MOSFET and build up the circuit, you can measure the ACTUAL delay time, and if it's too short, you can increase the time constant (by increasing RT and/or CT) proportionally, to get the delay you want.

    Sorry this is so long. You said you were a noob so I've tried to cover all the bases. You'll probably want to read it a few times!
     

    Attached Files:

  3. Blacker360

    Blacker360

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    Apr 10, 2013
    Response:

    I was looking at a smaller, simpler circuit that could possibly do the exact same thing. You see, the discharging of the capacitor is relatively unimportant as each circuit is a one time use, then the resistor/fuse (I will call it the resistor from now on) will have to be replaced anyways, so we can just manually discharge the capacitor. So the particular circuit i was looking at is at http://www.instructables.com/answers/Simple-onoff-delay-switch-circuit/, but of course some of the numbers will be different. I am looking for an approximate 2-4 (preferably around 3) second delay in the time it would take the resistor to ignite. I haven't actually tested anything yet, mostly because I am unsure of what numbers to start with. I was thinking perhaps using about the same resistor and capacitor rates as you suggested in your diagram from the middle circuit loop, but if you could help with the math because it has been a year since I took Electricity and Magnatism, I dont really remember too much of what i was SUPPOSED to have learned. My reasoning behind the circuit i mentioned is because it is small (needs to fit in a tight space and I am making it by hand), it needs to be simple (cost is extremely important), and I believe the charging of the capacitor will draw enough current not to ignite the resistor until the cap is relatively charged and most of the amperage goes through the resistor anyways, hence the delay b/c of the time to charge. If you could help with that, that would be fantastic.
    P.S. I am blacker's brother, hence the "not knowing anything" vs. taking a Physics course. Just felt it was somewhat important to clarify.
     
    Last edited: Apr 12, 2013
  4. Blacker360

    Blacker360

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    0
    Apr 10, 2013
    Also, I have a general understanding of everything you stated, except I have no idea what a MOSFET is.
     
  5. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

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    Nov 28, 2011
    I don't think you'll find a simpler or smaller circuit that will do what you want. That thread on instructables.com is not relevant to your needs, and nor is the idea of using a large capacitor to draw current from the battery for a short time to delay the firing of the igniter. In any case, a 1F capacitor will blow your size and cost budgets.

    Re discharging the capacitor, if you don't care about that, you can omit RD. Personally, I would keep it. But the rest of the circuit is still needed in any case.

    A MOSFET is a three-terminal semiconductor device that acts as an electrically controlled switch. It's a type of transistor (that's what the "T" at the end stands for).

    If you want a three second delay with the NTD4906N MOSFET I specified, RT should be about 180 kilohms. RD can be the same value, for convenience. (It doesn't have to be 10k; that was just a suggestion.)

    Here are some links to the Digikey pages for the components I've specified in that circuit diagram.

    RD,RT: 180k SMT resistor:
    http://www.digikey.com/product-detail/en/RMCF0402JT180K/RMCF0402JT180KCT-ND/2417839 (0402 - tiny)
    http://www.digikey.com/product-detail/en/RMCF0603JT180K/RMCF0603JT180KCT-ND/1943205 (0603 - very small)
    http://www.digikey.com/product-detail/en/RMCF0805JT180K/RMCF0805JT180KCT-ND/1942594 (0805 - small)
    http://www.digikey.com/product-detail/en/RMCF1206JT180K/RMCF1206JT180KCT-ND/1942822 (1206 - biggest of these options)
    Have a look at the size options available and choose the size that you think you'll be able to work with. The tiniest options are quite difficult to handle. The other sizes aren't so bad, but you'll still need tweezers.

    CT: 33 uF, 10V SMT tantalum capacitor:
    http://www.digikey.com/product-detail/en/F931A336MBA/493-2359-1-ND/677827

    Q1: NTD4906N MOSFET:
    http://www.digikey.com/product-detail/en/NTD4906N-35G/NTD4906N-35GOS-ND/2194521

    Your cost budget is about USD 1.30 for 1-up of those components. If you buy them in larger quantities, the price goes down.

    These components are designed to be surface-mounted ("SMT" = surface-mount technology) onto a small printed circuit board. With this simple circuit, you can just solder them together in what's called "skeleton wiring", where the larger components support the smaller ones, like the bones in a skeleton. You might want to encapsulate the final assembled circuit so it doesn't get damaged when the igniter ignites.

    Apart from those components, all you'll need is a battery, a switch, and an igniter.

    Although I believe this is probably the best circuit for you to use, I can't be sure it will work properly in your application. I need more information from you.

    1. What kind of cells are you using?

    2. What are the details of the igniter? If it's a resistor, what value is it?

    If you can answer those questions, I can be more confident in telling you whether that circuit will work or not.
     
    Last edited: Apr 12, 2013
  6. KrisBlueNZ

    KrisBlueNZ Sadly passed away in 2015

    8,393
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    Nov 28, 2011
    Neither of those igniter companies provides even the most basic specifications on the igniters! That's disappointing. I guess there's a standard voltage that all ignition systems use. Is that where you got your 6V figure from?

    Can you measure the resistance of one of your igniters?

    The batteries you want to use are completely unsuitable. These are CR2025 lithium coin cells which are designed for clock backup in motherboards and for powering calculators with LCD displays. They have a high internal resistance and are not able to supply the significant amount of current needed to fire an igniter. Try one and see.

    You need a much larger cell with low internal resistance. Have a look at http://www.digikey.com/product-detail/en/CR-123A/P151-ND/431596 - this MIGHT be suitable, but you'd have to test it first.

    I'd say your best bet would be four 1.5V Alkaline cells. AAs would probably do it; AAAs might not. There are other options available but they tend to be very expensive.

    You might also be able to find lower current options for the igniter. For example, a very small resistor of a suitable value will get pretty hot and might be able to ignite some chemical mixture if its ignition temperature is low enough.
    http://www.digikey.com/product-detail/en/CFR-12JB-52-33R/33EBK-ND/4030
    This 33 ohm resistor will dissipate about a watt at 6V and will draw about 200 mA. That's still far too much current for lithium button cells but it would be OK with four Alkaline AAA cells.
     
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