Connect with us

How much of a Short circuit is too much?

Discussion in 'General Electronics Discussion' started by soma, Apr 7, 2016.

Scroll to continue with content
  1. soma

    soma

    8
    1
    Apr 7, 2016
    Hi
    I want to make an electronically heated wire (via small batteries), the simplest way to do this, is put some short, thin, high resistance wire in the circuit.
    I want to control it wirelessly via a relay and receiver.
    [​IMG]
    The lightbulb represents my wire/heating element.
    How do i know if i will blow the circuit? if so, what's the most efficient way to stop that happening?

    Thanks so much for reading.
     
  2. Bluejets

    Bluejets

    4,208
    893
    Oct 5, 2014
    Ohms law.
     
  3. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Use a proper heating element.
    A short-circuit is a 0Ω load in an ideal world... in the real world, a short is a very low resistance that allows excessive current that will damage the source or something else.
    Using your own wire, find the resistance and calculate using ohms law to determine the current draw. Is the draw too high? Increase the resistance or decrease the voltage.
    You can also use a high power resistor... using your own wire will be an interesting challenge I'm sure.
     
    soma likes this.
  4. soma

    soma

    8
    1
    Apr 7, 2016
    Thanks. its not my own wire. I bought some high resistance nichrome wire.
    If I put a really high power resistor in the circuit, this will solve the issue?
     
  5. Gryd3

    Gryd3

    4,098
    875
    Jun 25, 2014
    Yes, No, Maybe, so...
    Split your circuit into two parts:
    - Source
    - Load

    The source's capability is highly important. What voltage does it operate at, and how much current can it supply?
    The load will draw current based on the over-all sum of all the resistances of the circuit. The lower the resistance, the more current it will draw. If this current draw is more than the supply can handle then you must either decrease the voltage to compensate, or increase the resistance of the circuit. Please note that anything in-line with the nichrome wire will also have the same current through it. This means that the resistor will also heat up depending on the collective resistance and the source voltage.
    ** Special note here though... The amount of current through that nichrome wire directly affects how hot it gets. If you can't make it hot enough, you need a more capable power source.

    So... Break out a multi-meter, pencil, paper and a calculator.
    - Measure the resistance of the nichrome wire.
    - Determine the amount of current required to heat the wire up.
    - Use Ohms Law to determine if the voltage on your supply meets this requirement.

    Share the details and we can make suggestions.
    Your nichrome wire will have a resistance, and it will operate in a 'range' of currents that will allow it to get warm or red-hot. You choose how hot to make it. and adjust the circuit accordingly.
     
  6. Alec_t

    Alec_t

    2,882
    770
    Jul 7, 2015
    How small are your 'small batteries'? It takes a lot of current to heat a short length of nichrome wire to even a low red heat; maybe more than your batteries (which inevitably have internal resistance) can provide.
     
    Gryd3 likes this.
  7. AnalogKid

    AnalogKid

    2,386
    663
    Jun 10, 2015
    You control the answers to all of your questions.

    First, how much heat? In watts or joules or BTU or temperature rise above ambient, only you know how much heat you want.

    Second, how big are the batteries? Once you know how much energy you need, you can determine if your batteries can supply it. If they can't then no amount of circuit design or calculation can get you where you want to go.

    Energy load first.
    Energy source second.
    Characteristics of the resistance wire third.
    THEN you can calculate how much wire, and whether or not the relay contacts can handle the load, etc.

    ak
     
  8. ChosunOne

    ChosunOne

    372
    90
    Jun 20, 2010

    You can't really do that with nichrome heating wire. Its cold (room temp) resistance will increase dramatically when it heats up. All metals develop higher resistance when heated, so its resistance in use is a function of how hot it is, which in turn is a function of how much current it's carrying.

    Wikipedia's article on Nichrome gives tables on current as a function of the heat and wire gauge. You might find it easier to work from those instead of using cold resistance as a starting point. In any case, it will probably come down to trial-and-error at some point, but calculating what you need beforehand is a good idea. Researching the info on your batteries is a good idea too--the lithium family of batteries is very popular in electronics nowadays, but some lithium batteries are high energy but low power--long-lasting in low-current applications but not suitable for heavy current draw. Alkaline primary cells have a high internal resistance, which makes them not the best choice for electrical heating.
    Saying you are using "small batteries" doesn't tell us much. Do they fit the application, whatever it is?

    As the others have mentioned, electrical heating needs a lot of current. Most batteries can supply enough for a momentary pulse of heat, e.g. a rocket or fireworks igniter, but longer heating needs a hefty power supply. IF your application is to supply a momentary pulse of heating power, you might consider integrating a hefty capacitor in your power supply.
     
  9. soma

    soma

    8
    1
    Apr 7, 2016
    It is is for a remote igniter.
    2 AA batteries would be fine as the wire is 42 swg.
     
  10. Ratch

    Ratch

    1,079
    330
    Mar 10, 2013
    Last edited: Apr 8, 2016
  11. ChosunOne

    ChosunOne

    372
    90
    Jun 20, 2010
    Okay, I'm puzzled. :confused:

    Your schematic shows your circuit using ONE power source, presumably a battery, to power both the RF receiver-relay module and the heater/igniter nichrome element. But you mention planning to use 2 AA cells. If you're talking about alkaline batteries, 2 AA cells make a 3V battery, which won't power your RF receiver, and probably not the module's relay either.

    If you add a 2nd 3V battery (2 AA cells) to the assembly, it's going to make it bulkier and I think you're probably trying to hold down the size of your project....?

    I can't believe I didn't notice this before, but as the schematic is shown now, btw, it probably won't work. It can be made to work by adding a 2nd battery, or by adding a resistor in the right place--less efficient, but keeps the package smaller and simpler to do, and it's only a momentary pulse, so efficiency isn't that important.

    So which is it? Are you planning to add an extra 3V power supply to the assembly, or add the resistor to your schematic as it's shown?
     
  12. Bluejets

    Bluejets

    4,208
    893
    Oct 5, 2014
  13. ChosunOne

    ChosunOne

    372
    90
    Jun 20, 2010
    Soma, I'm still waiting for your answer:
    Do you want to modify the schematic as you posted it in the OP, with one 12VDC source (presumably a battery) powering everything? ( A 3V two-AA battery will not power your receiver and relay.)
    Or are you planning to change the diagram and use a separate 3V (two AA cells) battery for the nichrome igniter element?

    My advice for how to modify one will not be the same as how to design another. There is more than one problem with your design either way,
     
  14. soma

    soma

    8
    1
    Apr 7, 2016
  15. ChosunOne

    ChosunOne

    372
    90
    Jun 20, 2010
    Some of the guys on here are going to love you, Soma. The ones who just love to make solutions WAAAAY more complicated than they have to be.

    Using a boost module with input of 3V limits your output to 220mA Max, AND decreases the efficiency of your circuit. I don't know if that will even be enough to energize the receiver, relay, and igniter element together.
    The current draw of the receiver is NOT negligible--I work with quality RF receivers of that size and they typically draw around 40-60 mA--I have no experience with the cheap Chinese...stuff...off eBay. Remember, that's a steady current draw. I have to assume you mean to use the unit shortly after you power it up, because your little AA cells don't have enough energy to keep the receiver on standby for an extended period.

    The relay's current draw is a unknown. I don't even want to guess at what it draws, being a low-cost Chinese...part. Typically, cheaper relays will draw more power because they're made with less coil resistance. My guess is that it's going to draw more than the receiver, although it will be a momentary draw. The capacity of your battery to supply a momentary current surge might be a problem if you're limited to 220 mA through a booster.

    If you simplify the circuit by using (8) AA cells instead of a booster, in an 8-cell holder, the design will be more efficient and you'll have the full current output of the cells when it's needed to fire the igniter.
    http://www.ebay.com/sch/i.html?_fro...older.TRS0&_nkw=8X+AA+battery+holder&_sacat=0

    I used to work with larger alkaline batteries, but don't know the potential momentary output of those small AA's. I think it's safe to say it's at least 1 A, probably more, which will probably do the job.

    You will definitely need to put a resistor on the positive leg of that circuit, between the receiver power and the relay contact. Otherwise, when the relay is triggered, there will be a momentary short on that leg which will drop power to the receiver and the relay, causing the relay to de-energize, which will remove the short and allow the relay to energize again, which will short the input again and make the relay drop out again---the result will be a chattering relay which may or may not allow enough current to finally heat up the igniter element enough to do its job---I'd guess not.

    Yes, nichrome wire is high resistance for wire--but compared to other components, it may as well be a dead short when it's "cold", and the circuit as drawn won't allow it to heat up.
    Unfortunately, the ideal value of the resistor, low enough to give maximum current to the nichrome wire, but high enough to keep the receiver and relay energized, depends a lot of how much the receiver and relay--especially the relay--need to draw in order to work.
    You might shop around some more and find a seller who lists specifications better. The receivers I use always give the current draw at their operating voltage, with and without relays energized.
     
  16. soma

    soma

    8
    1
    Apr 7, 2016
    Hi, just thought I would update you guys on this. I made my remote controlled igniter and it works great!
    I used 9 AA Batteries (in series) to power the relay and heat the nicrome wire; I made my own battery holder using pvc pipe - I used tin foil to make 2 connectors for the + and - of the battery holder (molded tin foil discs around the connection/contact wires) all held together with electrical tape, it works great! I glued the relay device to the battery holder and put a simple on/off switch between the battery pack and relay so as not to drain the batteries. I made a bunch of E-matches using the nicrome wire which can easily be attached via clips. Its a very portable ignition device and only cost about £10 to make.
     
  17. Alec_t

    Alec_t

    2,882
    770
    Jul 7, 2015
    Good to hear a success story!
     
  18. CDRIVE

    CDRIVE Hauling 10' pipe on a Trek Shift3

    4,960
    648
    May 8, 2012
    I'm glad to hear that you solved your problem.... At least I think I am?

    Have you considered the dangers involved with wireless firing devices? If this was my project (doubt it), power to this board would be preceded with a timer circuit that would provide more than enough "Get the F!ck out of Dodge" time. Even then it's hard to beat the security of hard wired trigger circuits. ;)

    Chris
     
Ask a Question
Want to reply to this thread or ask your own question?
You'll need to choose a username for the site, which only take a couple of moments (here). After that, you can post your question and our members will help you out.
Electronics Point Logo
Continue to site
Quote of the day

-