Brew Man Posted January 30, 2023 Share Posted January 30, 2023 I will change the leaf to volt to see how much voltage the led is using.My mistake, I thought to read resistance reading I needed to check this way.Thanks all for correcting me and hopefully we will get to bottom of this eventually.The red lead plugs into the right hand socket for everything except the 10A current range, so it's not just voltage, it's voltage, current, resistance and the continuity buzzer. Therefore just change the red lead socket, switch to the 200mA range and measure current as intended. Link to comment Share on other sites More sharing options...
Deem Posted January 30, 2023 Author Share Posted January 30, 2023 Thanks Brew Man. Will update later. Link to comment Share on other sites More sharing options...
Deem Posted January 30, 2023 Author Share Posted January 30, 2023 It is weird or am I missing something, please see attached pictures which will explain. Once they are approved. Link to comment Share on other sites More sharing options...
96RAF Posted January 30, 2023 Share Posted January 30, 2023 You are measuring resistance i.e. Ohms ΩPut it on the 200μA scale i.e. current A Link to comment Share on other sites More sharing options...
Deem Posted January 30, 2023 Author Share Posted January 30, 2023 Thanks RobI have tried again and I go the reading but to avoid confusion I am uploading pictures, so you all can advice me accordingly. Link to comment Share on other sites More sharing options...
Deem Posted January 30, 2023 Author Share Posted January 30, 2023 Link to comment Share on other sites More sharing options...
Fishmanoz Posted January 31, 2023 Share Posted January 31, 2023 6.2mA Link to comment Share on other sites More sharing options...
P-Henny Posted January 31, 2023 Share Posted January 31, 2023 You can't use the ohms scale (in the first set of images on previous page) to directly measure the resistor value because you have the LED diode still in circuit. The LED will affect the ohms reading and give inaccurate results. The LED is also a diode, which means that you will get widly different ohms readings when the red and black meter leads are reversed.The second set of images on this page show a nominal current reading of 6mA (6.1mA / 6.2mA actual).Now you haven't stated the voltage of your power supply which means an accurate resistance reading cannot be calculated. But for the sake of an example lets assume 12 volts. You can then substitute the 12 in the calculations below with the voltage supply you were using and recalculate the formula.The resistor ohms resistance value is:supply voltage minus forward bias voltage divided by current.The LED was white, so assume a LED forward bias voltage of 3 volts. This gives (assuming a 12 volt supply):12 volts - 3 volts = 9 volts divided by 6mA (0.006 Amps)9 divided by 0.006 = 1,500 ohmsA value of 1.5K for the resistor is in a range of values that would be expected.P.S. The 0.76 is a false reading because you have the rotary switch on the 10 Amp setting but the red lead is not in the 10 Amp socket. Link to comment Share on other sites More sharing options...
Deem Posted January 31, 2023 Author Share Posted January 31, 2023 Thanks everyone, little bit complicated for me at my age but I am positive if I read it couple of times and I will get to grip the idea on how to do this correctly, so I can understand fully. I am going to delete the last page pictures to avoid the confusion.As for what power supply I am using, I had Hornby elite connected to the track and had 2 crocodile clips from track to, one to meter probe and other to led one cable, using the third cable with crocodile clips, I had second probe connected to led second cable so I have steady current and have led on all the time to test on various reading. Link to comment Share on other sites More sharing options...
Deem Posted January 31, 2023 Author Share Posted January 31, 2023 Can I send 30 white, 30 red and 30 yellow to forum head office from where someone can share 10 of each cable to brew man,fishmonaz and p- Henry, so I don't have there details but they can get it free of charge. So they can try them tensor themselves or use them.I have installed same led's on buffer stops and soldered on to track.Mod note - there is no ‘forum head office’ to send them to. The moderators do not have sight of members emails or other contact details. Your best bet at making contact is via another forum that they also frequent which allows direct messaging. They can let you know on this thread if they want to accept your offer and post their other forum name and which forum. Unfortunately until this forum gets direct messaging that is the most practical way of making contact with others. Link to comment Share on other sites More sharing options...
Fishmanoz Posted January 31, 2023 Share Posted January 31, 2023 A little more complicated than we thought Paul. What’s the rms value of half wave rectified DCC?It’s interesting that a cheap 12V LED with modest series resistor doesn’t get its knickers in a knot when hit with the bipolar DCC volts. PS. Deem, apologies for the techno-talk. Link to comment Share on other sites More sharing options...
P-Henny Posted January 31, 2023 Share Posted January 31, 2023 Regarding Deem's offer. No thank you as I have boxes of component LED's and resistors already and don't need any more. Plus, I rarely use pre-wired LEDs, preferring to make my own with custom resistor values chosen for the specific task I want to use them for.The formula used to calculate the resistor value is Ohm's Law and only really applicable to DC. DCC track voltage is not DC, thus using DCC as a testing power supply is not really appropriate for these type of calculations.If Deem really wants to know the resistor value in his pre-wired LEDs then the most accurate way would be to use a fully charged 9 volt DC battery with the wired LED. Then use the 9 volt DC voltage rather than track DCC supply to measure a current. Then the formula becomes:Resistance = 6 volts (from 9 - 3) divided by new measured current in Amps.@Deem, remember with a 9 volt DC battery, the LED will only light up when connected to the battery the right way round. Link to comment Share on other sites More sharing options...
Deem Posted January 31, 2023 Author Share Posted January 31, 2023 PS. Deem, apologies for the techno-talk.No need to apologise, unless more knowledgeable people or person won't talk - techs talk - how a person like me will learn.Thanks for your contribution. Link to comment Share on other sites More sharing options...
Deem Posted January 31, 2023 Author Share Posted January 31, 2023 Resistance = 6 volts (from 9 - 3) divided by new measured current in Amps.@Deem, remember with a 9 volt DC battery, the LED will only light up when connected to the battery the right way round.Thanks Paul for detailed explanation, I will try to use the 9V battery, send you reading again and hopefully will learn a thing or 2 in process. Regards Link to comment Share on other sites More sharing options...
Deem Posted January 31, 2023 Author Share Posted January 31, 2023 Mod note - there is no ‘forum head office’ to send them to. The moderators do not have sight of members emails or other contact details. Your best bet at making contact is via another forum that they also frequent which allows direct messaging. They can let you know on this thread if they want to accept your offer and post their other forum name and which forum. Unfortunately until this forum gets direct messaging that is the most practical way of making contact with others.Not to worry. Link to comment Share on other sites More sharing options...
Rog RJ Posted January 31, 2023 Share Posted January 31, 2023 Simply cut away the black insulation from the base of an LED and measure the resistance of the resistor that is revealed. Link to comment Share on other sites More sharing options...
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