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Tricky Dicky

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Posts posted by Tricky Dicky

  1. A quick way to find shorts without disconnecting everything is split the bus halfway. If power is restored in the first half your problem is in the second half and vice versa. Continue to split the problem section and repeat with those sections causing the fault. Eventually you will isolate your problem to a few connections at which point you might be best to disconnect all and test individually. This method works best if you suspect a single fault.

     

    Richard

    • Interesting/Thought-provoking 1
  2. Most insulated spade and bullet crimps come with either a standard red, blue or yellow sleeve so when going for a ratchet crimper look for one that has crimping jaws marked with those colours it takes out a lot error when it comes to making the crimp. I had one that a number of jaws marked in AWG which was a right PITA trying to remember which was for which crimp. Too small a jaw and the crimp would almost double over, whilst too large and the wire came out.

     

    Richard

     

  3. The electronics calculator I find most useful is Electronics Assistant found here;

     

    https://www.electronics2000.co.uk/

     

    It does a lot of things but as far as LEDs are concerned it can be as simple or detailed as you want. Simple is enter the number of LEDs in series, their colour and the supply voltage. The result tends to be the minimum resistor value so the LEDs tend to be quite bright. If you want dimmer use the starting value and increase the resistance until you are happy.

     

    Richard

  4. 15 hours ago, Furness Wagon said:

    Even as a mechanical engineer you know not to let the smoke out of the box.

    You learn that in beginners black masses and cross burning, sorry introduction to electrical/electronic engineering.

    Marc


    LOL

     

    Richard

  5. On 08/03/2021 at 23:24, d00m said:

    You might find it useful to type  'LED resistor calculator' in your search bar - it comes in handy when working out parallel or series resistors for led arrays. 

    A useful electronics calculator can be found FREE to download on the Electronics 2000 website, it is a Windows program so will only work on a Windows enabled machine. It has a very easy to use LED resistor calculator among other things. 
     

    www.electronics2000.co.uk

     

    Richard

     

  6. 3 hours ago, WhiteRose158 said:

    The problem is almost certainly the lack of agreement between the T-Tap and wire width. I've manually cut through the wire cover down to the strands, pushed them right down into the tap and hurrah. Lesson learned :yes: as my layout has been ten years in the building (with several of those years spent abroad so no progress made) I've got a mix of wire widths and types going on, which in the end will need standardising!

    I think you do need to look at your wiring and make sure it matches the specification of the T-taps you are using. If you are going to the lengths you describe above to make the splice then the long term prospects are that you will have the same issue again as a loose connection between wire core and displacement blade will simply produce a high resistance join similar to rails and rail joiners. IDC connectors are a very simple way of splicing wires but if you do not match your wires to the specs then they can be troublesome.

     

    Richard

    • Like 1
  7. 32 minutes ago, WIMorrison said:

    Sound like you have used the wrong size of IDC

    Got to agree, if you are using 7/0.2 wire  which is the equivalent of 24 AWG then even the smallest of the T-taps (Red) will struggle to displace the insulation as its minimum range is 22 AWG.

     

    Richard

  8. 18 hours ago, ColinK said:

    While all the pictures I’ve seen have the power being applied to the ends of LED strips, could I solder the power leads (3v in my case) to the solder pads in the middle of the strip?

    Yes you can, the sets of solder pads are simply sitting on two continuous power rails going the full length of the strip. The 3V I think you will find will be too low, these strips come in two varieties 24V and 12V the latter being the more common. As someone has already stated they will be quite bright and may need additional resistors to reduce the current to acceptable levels and only experimentation will reveal how much.

     

    Richard

     

  9. 16 hours ago, MOH said:

    Thank you,

     

    I quoted a 3v transformer as the leds were advertised as only to be used with 3v, I have test wired three leds to a mobile phone charger and those are working with no resistors added and no blow outs so far.

     

    Really no idea what voltage that mobile charger is rated as it's quite old and the details are blurred on the casing.

     

    I will plough on so and order that speaker cable and see how that goes.

    I do not know why people insist on trying to run LEDs without resistors, after all they are not the most expensive component neither do you need a vast collection of them. DC voltages associated with model railways tend to be 12V in the main. Other power sources often used such as ex-phone chargers are mostly 5V since all phone manufacturers and tablet makers have standardised on USB charging. Even the forward voltages of LEDs range from just below 2V to around 3.5V and since most modellers choose not to run their LEDs at full power the variety of resistors required is limited as long as the resistance exceeds the minimum calculated value.

     

    If you are not sure of the resistance required and cannot do Ohm’s Law calculations there are plenty of free calculators out there. One of the simplest to use is Electronics2000 it can work out values from the barest of information.

     

     www.electronics2000.co.uk

     

    Richard

     

     

    • Like 2
    • Agree 1
  10. 5 hours ago, mikeg said:

    Thanks Richard, I have put in a bridge rectifier across track supply and it’s output voltage is 12.5 so I can workout what I need from you answer. Having now checked all the coaches I have been wondering if it would be possible to connect them all together from the one set of pickups, I think I can get small 2 pin plug & socket connectors for between the coaches and as I can now get to the copper strips in each coach I could easily join them all together. 

    Can you see any problem with this approach and can I just add all the amps together and use one resister to the supply in the pickup coach?

     

    Hope I am not being a pest but electrics is not something that I fully understand so thanks for your help and time.

     

    Regards mike 

    No that would not work. All resistors dissipate heat the higher the current the more heat. The more heat needed to be dissipated physically the larger the surface area of the resistor needs to be. It is a fine balance not allowing a resistor to get too hot since that begins to alter the value of the resistor. This is the reason why it is recommended to provide each LED in a circuit with its own resistor thus each individual resistor has minimal heat to dissipate.

     

    Going of your figures for your coach lighting of 12.5V supply and 0.33A current draw you would require a resistor value of approx 38 Ohms, the nearest preferred value would probably be 39 Ohms. However, the total power that the resistor would need to cope with is just over 4W, you would have to go up to a 5W resistor. Resistors of that wattage are available round about 25mm long and in the region of 6mm round so doable. The more current you draw by linking all your coach lights together the smaller the ohmic value of the resistor will become but bigger physically the resistor will become  to manage the higher the wattage. As an example I recently had to install a power resistor in a circuit it’s Ohmic value was a mere 25 Ohms but it required a 50W resistor. This beast was about 60 mm long encased in an Aluminium heatsink approx. 16mm diam.
     

    Your requirement for a single resistor solution was always going to be less than ideal and I think you are going to have to experiment with the proposed resistor and see how much heat it generates before installing it into your coach.

     

    Richard

  11. Due to the DCC voltage being a hybrid AC most multimeters will not read the voltage correctly even on the AC setting, so your voltage numbers are highly likely suspect. Since you need a DC voltage for the LEDs anyway, connect a bridge rectifier to the coach pick ups, then measure the voltage on the output of the rectifier on the meters DC setting. Now a little maths, divide the voltage you have measured by the respective current readings you have for each set of coach lights that will be the value of the resistor you need to put in series with the LEDs. You will probably not get the exact value calculated since depending on which series of resistors you can get hold of, so simply use the nearest preferred value usually the nearest higher value. One final point is to multiply the voltage reading from the rectifier by the current readings which will give you the power requirement in Watts and make sure the resistors you use can sustain that power.

     

    Richard

  12. Testing as you go along is advisable, if you get something wrong it usually only means one step backwards to resolve it. Having to check out a whole layout to find where a problem has manifested can be a real PITA.

     

    Richard

    • Agree 1
  13. 21 hours ago, WIMorrison said:

    ScotchLoks are IDCs for splicing into wires without breaking the wire, whereas with Wago need you to split the wire which you cannot do in a loom or cable bundle.

     

    They are complementary to each other, not subsidies for each other

     

    I will concede that, but had already kind off hinted at that with the guitar string remark.

     

    The point I was making was prior to the availability of Wagos, ScotchLok was the only viable non-soldering screwless way of connecting/splicing wires that did not require a specialist tool. It is my opinion that wiring into a loom is probably better done with a ScotchLok but if wanting to take a solderless approach to wiring then I think Wago is superior in that they take a wide range of wire sizes unlike ScotchLok where you have to consider which connector to use in relation to the wires being joined.

     

    After all, there was a recent thread here where the same topic was discussed and I think you contributed to where the OP was about to splice wires together where the thinner wire he intended to use was outside the specifications of the particular connector he was going to use.

     

    As I have said before I have no particular objection to ScotchLok but there are pitfalls for the unwary whereas Wagos offer a more universal method of joining different sizes of wire which I think starters in this hobby would appreciate.

     

    Richard

  14. 2 hours ago, WIMorrison said:

    ScotchLok and Wago connectors have very different use cases and are not replacements for each other.

    Please explain how? You can make exactly the same kind of connection with the appropriate Wago as you can with a ScotchLok with the advantage that a Wago can cope with a greater range of wire sizes. The only circumstance I can see a ScotchLok would be better was if your wiring was strung out like guitar strings.

     

    Richard

  15. +1 for Wago connectors, they are far superior to Scotch Lok which are often recommended for quick screwless connections. Wagos will happily connect a wide range of wire from 4mm sq. down to 7/0.2. I can attest to the latter having tested to destruction, I pulled the wire apart before the Wago let go.

     

    Do go for the the genuine ones as there are a lot of clones on the market and their properties could be suspect.

     

    Richard

    • Agree 2
  16. I currently use a cheap Maplins 40W iron with supposed temperature control but I could not tell you what the temperature is set at since it is a knob control with a coloured scale going from green -cool, to red-very hot and I find setting it at 3/4 turn is right for me but i think if you can set temperature about 365C is about right.

     

    When you say the tips are burning out at high temps are you referring to the tip or the element? I suppose prolonged use at high temp will tend to take it out on an element over time but tips tend to get damaged more by abrasions. Most modern irons have copper tips coated by iron carbide. If you have cleaned the end of the tip with a file or emery cloth that coating abraded off and initially with the copper exposed you will find it solders very well but over time you will observe a cavity forming due to the action of the flux. Brass bristled brushes are about the toughest you can employ to clean tips. Of course the the secret is not to allow the tip to get cacked up with oxides which means frequent wipes on a damp sponge and re-tinning of the tip whilst soldering.

     

    Wattage and temperature do not necessarily equate. When soldering as soon as you touch the tip to the metal being soldered heat transfers from the tip to the metal, how quick the tip replenishes that heat is down to the wattage the higher the quicker and the quicker you solder. This is important when soldering to rails whose bulk has the effect of a heat sink drawing heat along the rail, this is what melts plastic sleepers. Low wattage irons need to be held on longer because they struggle to get the area of soldering up to fusing temperature thus allowing more heat to travel along the rail. A high wattage iron will allow you to deliver the heat quicker but more importantly localise the heat where you want because the whole process is quicker. Fusing of solder only takes place if the metal being soldered achieves fusing temperature in other words it is the heat in the metal that melts the solder. My maxim is you can have too little power but you can never have too much!

     

    Richard

     

    PS:  The bulk and diam. of the tip also have a bearing on the heat transfer - the wider the tip the more transfer. Mine is the fairly standard 2.3 wide tip

    • Thanks 1
  17. The spur should be completely isolated from the rest of the layout ie. both rails and needs its own feed. That feed should go to the centre terminals of a DPDT switch. The main feed should go to the two terminals on one side of the centre terminals and the programming feed to the other two. Just make sure the main feed is connected to the same rails as the rest of the layout otherwise you will cause a short.

     

    Richard

  18. Keith has outlined several possibilities, but at the end of the day it is all guess work and you really need to get an accurate reading of the current draw. If as you say you are having trouble finding a point where you can connect your meter in series, why not insert a thin piece of insulating material - plasticard say between one battery terminal and the contact in the battery holder then simply put one probe on the battery terminal and the other on the contact to complete the circuit.

     

    Richard

  19. Mike you have made the classic presumption that LEDs are like bulbs in that altering the voltage affects their brightness. LEDs are current driven devices and the series resistor that you get in LED circuits is a current limiting resistor. Hence as Keith has pointed out you need to establish the current draw of the LED circuit and as he has pointed out your meter needs to be in series with the battery and LED(s).

     

    As we are looking at 2 AA batteries which should be giving you 3V or 2.8V if they are rechargeable (2.4V seems odd unless the batteries are well used). The LEDs are probably wired in parallel with individual resistors already. A single white LED will draw about 30mA max. and probably less if as you say they are emitting a less than white light so with parallel LEDs you will be looking at a max of 30mA or less x the number of LEDs in the coach just to give some idea of the range of current you can expect.

     

    Now since voltage, current & resistance is related by Ohm’s Law ie V=IR. The calculation you need to make is:  R = V supply voltage - V led voltage/ I
     

    So in your case 13.6 - 2.4(?) divided by the measured current.

     

    Since your current value will in mA the answer you get will be in KOhms

     

    Richard

  20. Are you sure you have all the droppers connected correctly as you have indicated on the diagram. I would be tempted to as a quick fix to swap the two droppers on that siding and see what happens.

     

    Richard

     

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