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DaleSchultz		#51 Posted : 08 May 2019 21:36:49(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		We are talking about the digital track current, which is not AC sinusoidal but similar in that it alternates but with square 'bits' as I understand it. I am keen to get to the bottom of this myself.... A visualization: https://cabin-layout.mixmox.com/2007/07/voltage-visualization.html
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Minok		#52 Posted : 08 May 2019 21:42:18(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		Originally Posted by: DaleSchultz We are talking about the digital track current, which is not AC sinusoidal but similar in that it alternates but with square 'bits' as I understand it. I am keen to get to the bottom of this myself.... A visualization: https://cabin-layout.mixmox.com/2007/07/voltage-visualization.html Thats what I thought, in which case the transformer part of those rectifier circuits is even more irrelevant; center tap doesn't come into play, its a pure square wave signal. If the low voltage of that doesn't reverse the current, then its not even AC, and then a full wave bridge isn't needed and a half-wave is just fine, since there isn't a negative side of the wave at all. I haven't hooked up my oscilloscope to the track to see what my MS2 puts out. Its an old school analog tube scope, so no digital storage to easily capture the signal (they had to use Polaroid film camera attachments on that old Tektronix unit back in the 1980s)
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DaleSchultz		#53 Posted : 08 May 2019 22:46:24(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		yes the center tap portion of the diagram does not apply, but I assumed it was just a standard graphic that also showed a center tap. I don't have an oscilloscope, so physically trying it is my only option.
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tiono		#54 Posted : 09 May 2019 02:33:03(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Originally Posted by: Minok So are we talking about a purely AC sinusoidal driven system here, where the mains is stepped down to a train-level AC signal? Otherwise the presence of the transformer and where its tapped doesn't matter. With a full-wave bridge rectifier, yes where the slider is one pole, and the rails is the other pole, the output of that is a +ve power signal and the rectified ground, so that produces a ground/return signal inside the post-rectifier space. But that would work just fine. Whichever rectifier circuit one uses it has to be wired up correctly. Using a full wave bridge gets you more of the power signal to work with, so you need a smaller capacitor for the same smoothing, so why not use the full-wave bridge solution? Everything attached to the right side +V/GND side is a complete circuit just inside the wagon/coach/car, without contact to the slider or rails. Yes I agree. Whenever the DC circuit is within a wagon/coach, then I use bridge rectifier. However, if the rectified DC must be passed between coaches, then I prefer half-wave because I do not want to use 3-pole connector. For those who don't mind using 3-pole connector, the full-wave bridge should be good. The downside of half-wave rectification is its high ripple. But this will not be a concern if there is a regulator on the DC line.
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DaleSchultz		#55 Posted : 09 May 2019 03:02:31(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		I am struggling to understand why there is a difference when you move to a second coach. I understand by the 3 pole connector that you are taking three wires to the second coach, what would those three be? Why can the rail ground not be used as Minok suggests?
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tiono		#56 Posted : 09 May 2019 03:39:10(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Originally Posted by: DaleSchultz I am struggling to understand why there is a difference when you move to a second coach. I understand by the 3 pole connector that you are taking three wires to the second coach, what would those three be? Why can the rail ground not be used as Minok suggests? ahhh... now I understand. Sorry I misunderstood Minok's message. Below is the modified diagram. Still using 2-pole connector between coaches, but with bride rectifier, plus optional feature; back up power to decoder (this must be connected to the DC supply in the decoder, after decoder's rectifier)
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DaleSchultz		#57 Posted : 09 May 2019 03:55:52(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		perhaps this helps explain the problem I having having (and perhaps Minok also) I don't understand why the ground that goes into the rectifier cannot be the same as the ground for the LED strips. The connection shown arrowed in purple. Is it because the DC-DC regulator is using N-MOSFETs and so is in fact switching the negative?
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Minok		#58 Posted : 09 May 2019 04:09:25(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		Dale that would be ok solong as the right input is not an AC signal where the current reverses. But then the rectifier can be vastly simplified as you really just need a one way gate with a diode so the half wave. Your design is connecting the purple back to one of the AC inputs (wheel pickup) and effectively taking the lower left diode is the FWB rectifier out is the circuit and then with the two parallel diodes you only need one so the half wave bridge is all you need.
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DaleSchultz		#59 Posted : 09 May 2019 04:13:07(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		ah yes, I think I get it now... yes I am not expecting AC or reverse impulses but I see it is the same as half wave now.... (will look at it again in the morning)
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rbw993		#60 Posted : 09 May 2019 21:21:58(UTC) Retweet
Joined: 19/08/2008(UTC) Posts: 1,055		That is pretty much how I do it except the resistor is on the output side of the capacitor.
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DaleSchultz		#61 Posted : 09 May 2019 21:39:14(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Originally Posted by: rbw993 That is pretty much how I do it except the resistor is on the output side of the capacitor. I think the resistor is there to slow the charging up of the capacitor, otherwise the inrush current trips the digital controller.
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Minok		#62 Posted : 09 May 2019 22:42:03(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		Originally Posted by: DaleSchultz Originally Posted by: rbw993 That is pretty much how I do it except the resistor is on the output side of the capacitor. I think the resistor is there to slow the charging up of the capacitor, otherwise the inrush current trips the digital controller. Exactly, it limits the current draw when you turn on the power (with a bunch of capacitors all needing to be charged at once ). Similar circuits designed as keep-alives for motors and decoders in locos would also have a diode parallel to the resistor so that the discharge from the capacitor doesn't have to take the path across the resistor, but for this in-wagon stuff, all the current goes out of the right side DC connections.
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rbw993		#63 Posted : 10 May 2019 17:08:57(UTC) Retweet
Joined: 19/08/2008(UTC) Posts: 1,055		So far there has been no apparent issue with in rush of current. That is with ~35 lit cars on the layout (also about 20 locos). I use the resistors to adjust LED brightness. Without them the light strips are too bright.
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rbw993		#64 Posted : 10 May 2019 17:12:28(UTC) Retweet
Joined: 19/08/2008(UTC) Posts: 1,055		It may be that the way I segmented the layout in rush isn't an issue. The track is split into 3 sections w/boosters. The CS2 only powers turnouts on one side of the room. Turnouts on the other side are powered by another booster.
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DaleSchultz		#65 Posted : 10 May 2019 17:18:08(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		I see. Another factor will be the size of your capacitors.... and the threshold for your system to trip on sudden large current draws... 35 lit cars is a fair number!
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rbw993		#66 Posted : 10 May 2019 19:06:15(UTC) Retweet
Joined: 19/08/2008(UTC) Posts: 1,055		330 microfarad
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DaleSchultz		#67 Posted : 28 May 2019 22:22:09(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		my tiny DC-DC converters finally arrived today! So experimenting can begin soon...
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DaleSchultz		#68 Posted : 29 May 2019 02:52:28(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		so far so good... I tested some LEDs cut to length for a tinplate coach - 4 x 3 LEDs and I like their brightness at 8 Volts. I adjusted the output of these tiny regulators to 8V I ran digital track power into the regulator, with a single diode on the positive input. That all works, is simple, and looks good. Now I come to add a capacitor, and it strikes me that there are two places where it could be added. a. between the diode and the regulator input. b. between the regulator output and the LED strip Which would be better? The voltage after the regulator is lower (8 V) The circuit is so cheap and compact, it may be viable to have these components in every coach and simply feed track power from coach to coach as opposed to feeding the 8 V DC output from coach to coach. That way I don't have to worry about how many I chain together form a single pickup.
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Minok		#69 Posted : 29 May 2019 02:59:13(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		Since the cap is there to sustain the lighting I’d think at the lighting LEDs but check the regulator sheet to be sure that having the output voltage held up by that cap isn’t a problem.
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DaleSchultz		#70 Posted : 29 May 2019 03:14:20(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Originally Posted by: Minok Since the cap is there to sustain the lighting I’d think at the lighting LEDs but check the regulator sheet to be sure that having the output voltage held up by that cap isn’t a problem. Ah yes, good point. Finding a data sheet may be the big task though!
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DaleSchultz		#71 Posted : 29 May 2019 03:20:02(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		And tiono places it before the regulator... so I bet that is best! Originally Posted by: tiono I do add a capacitor at the output of bridge rectifier (the input of PWM regulator).
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DaleSchultz		#72 Posted : 29 May 2019 03:26:31(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Originally Posted by: Minok Since the cap is there to sustain the lighting I’d think at the lighting LEDs but check the regulator sheet to be sure that having the output voltage held up by that cap isn’t a problem. If that value is called Feedback Voltage, it has a max of 0.95V ! https://www.digchip.com/...heet/1006/MP2307-pdf.php
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Minok		#73 Posted : 29 May 2019 04:46:15(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		Originally Posted by: DaleSchultz And tiono places it before the regulator... so I bet that is best! Originally Posted by: tiono I do add a capacitor at the output of bridge rectifier (the input of PWM regulator). Yes in that circuit the brightness is controlled by a waveform and the cap supports keeping the circuit running. But if the cap directly feeds a resistor and LED then it’s ok, the resistor controls the brightness. It all depends how the circuit is designed.
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tiono		#74 Posted : 29 May 2019 04:55:19(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Originally Posted by: DaleSchultz And tiono places it before the regulator... so I bet that is best! Originally Posted by: tiono I do add a capacitor at the output of bridge rectifier (the input of PWM regulator). I prefer to put the capacitor before the regulator, the reason; - to reduce voltage spike due to intermittent contact of the wheel or slider. - to reduce the on-off cycle of regulator if the voltage fluctuate due to intermittent contact. A switching regulator need time to start the switching cycle, and it also has under-voltage protection which will shutdown the regulator when the input voltage drop below the output.
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DaleSchultz		#75 Posted : 29 May 2019 15:10:22(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Thanks tiono, good reasons! My next decision is where to place a resistor to reduce inrush to the capacitor. As I see it there are three possibilities: a. on the output of the diode on the positive line, before the junction with the capacitor b. in series with the capacitor, on the positive side c. between the capacitor and the regulator positive I have read that thermistors are ideal for preventing inrush so that the resistance is only present during start up. Resistors not as good. But to keep costs low I think I should look at resistors first. My feeling is that option (b) above is the correct place to pace it. Placing it there will reduce the speed at which it charges up and also releases current, and will not waste energy for when the pickup is getting contact and current is flowing to the regulator. If it also causes the dissipation of charge to be slowed, it might mean that the lights would be a bit dimmer, but last longer. (So long as the resistor is small enough to still provide enough forward voltage for the LEDs.) Which leads to the last issue, what size resistor to use. It needs to balance the inrush against the outgoing supply when power is lost.... perhaps a diode in parallel to the resistor would be the solution. Like this: D1 cuts digital power to half wave R1 inrush resistor D2 inrush bypass C1 Capacitor (big as space allows) Last night I tested with a 470uF capacitor and it works really nicely, in that I can disconnect for a short moment and the lights stay on. I also tried a 220uF one I had on hand and it worked too but not as well as one would expect. Of course I cannot tell with a single coach if I will have an inrush problem with my system. Perhaps I have to connect 30 or 40 to the track to see if that many coaches would have enough inrush to trigger the short circuit threshold of my Intellibox.
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tiono		#76 Posted : 29 May 2019 17:55:59(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Your circuit is good. Rough calculation of the resistor value would be (assuming all coaches use lighting and big capacitors): 18V divide by {maximum output current of your controller, divide by the number of coaches}. So if the controller is 2A max, and there are 10 coaches, then the resistor should be 90 ohm. But since the in-rush current happen only in very short period, then usually the controller able to handle more than the specified maximum (this need to consult the controller's spec), therefore most likely you can put resistor with much lower value.
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DaleSchultz		#77 Posted : 29 May 2019 18:27:07(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Thanks tiono, your help has been very valuable on this project! The lower the value the faster it will charge up, so I was thinking of putting in a large value resistor so it charges slowly. With my Intellibox which can deliver 3A, your formula would suggest 180Ω for 30 coaches, but of course there are other consumers as well, so I might use 240Ω as a starting point and do some tests. Edit: My Märklin boosters can supply 2.5A, so 30 coaches would call for 216Ω My hidden station that would hold most of the illuminated coaches is running off one of the boosters so I may even go higher than 240Ω
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DaleSchultz		#78 Posted : 29 May 2019 20:35:47(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Note for those who use an Intellibox, there is a special option (SO 36) that allows one to control the time during which short circuit faults from boosters are to be ignored. The default is 20 = 200ms - so if boosters report a short circuit during the first 200ms of starting up due to a high inrush, the IB will ignore that. (other controllers may have similar settings too) This could remove the controller shutting down but of course the current draw from a large inrush would still exist and should not exceed 2.5 A
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Minok		#79 Posted : 29 May 2019 21:00:15(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		The design with the diode in parallel to the resistor is the way to go, thus the inrush current (when the capacitor is empty) runs through the resistor, but when the current runs out of the capacitor, it bypasses the resistor over the diode. This design also relieves your DC:DC converter from having to handle the inrush current, but makes the capacitor have to be the higher voltage of the track-side, thus maybe larger but I doubt that is an issue. If you compute the time it takes to charge the capacitor, if you use 470uF with an approx track side voltage or 18v, and a 500 ohm resistor it takes 0.24 seconds to charge the capacitor. t = c * r ; c is in Farads (so 470 / 1,000,000); r is in Ohms (so 500 ) = result in seconds. You can make the resistor larger without any really significant issue, but reducing the inrush load on the layout substantially.
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DaleSchultz		#80 Posted : 29 May 2019 22:36:42(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Based on all this fact finding, I have now drawn a better diagram and arranged the circuit so that it can be extended to multiple coaches in a modular fashion. The light strip and DC-DC regulator can be built as 'modules' with connectors at each end, allowing them to be connected to adjacent coaches. A single pickup shoe, with its own single diode, can supply power to multiple coaches, by connecting to any of the light strip modules. Optionally, flicker preventer modules can be connected within each coach, or just one with a hefty capacitor anywhere in the set of illuminated coaches. Components: D1 1N4001 D2 1N4001 R1 TBD but my initial feeling is 240 Ohm C1 470uF (or the largest one that can be hidden away) LED strip - 12 V strips cut to length as needed Mini DC-DC regulator using PWM MP2307 chip. I am setting mine to 8V output. Connectors: not yet decided but perhaps JST style so that polarity can be maintained Limitations: Digital track power only - a reverse impulse of an analog system may destroy the DC-DC regulator Only one pickup per string if you have multiple power districts with boosters Advantages Very low current consumption Very little power wastage Brightness can be adjusted on the PWM regulator Inrush prevention Each coach can be connected to the ground line ensuring that wheel contact loss will not result in a break in the circuit. No batteries. If desired, remote switching can be added by adding a relay at D1 Plugs allow coaches to be attached or detached with minimal effort. Coaches can be connected in any direction Unknowns Will the PWM nature of the regulator output produce any visible flicker at 340 MHz when the coach is in motion? The optimal size of R1 TBD Critical review invited!
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DaleSchultz		#81 Posted : 29 May 2019 22:45:40(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Originally Posted by: Minok The design with the diode in parallel to the resistor is the way to go, thus the inrush current (when the capacitor is empty) runs through the resistor, but when the current runs out of the capacitor, it bypasses the resistor over the diode. Exactly my thinking. Originally Posted by: Minok This design also relieves your DC:DC converter from having to handle the inrush current, but makes the capacitor have to be the higher voltage of the track-side, thus maybe larger but I doubt that is an issue. Most seem to be 25V anyway... so yes I don't think a problem. Originally Posted by: Minok If you compute the time it takes to charge the capacitor, if you use 470uF with an approx track side voltage or 18v, and a 500 ohm resistor it takes 0.24 seconds to charge the capacitor. t = c * r ; c is in Farads (so 470 / 1,000,000); r is in Ohms (so 500 ) = result in seconds. Interestingly that is just more than the default time to ignore shorts by the Intellibox (0.2s) Originally Posted by: Minok You can make the resistor larger without any really significant issue, but reducing the inrush load on the layout substantially. Yes, and if it takes 2s to charge up, that sounds fine suggesting I could go to 2K Ohms!
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DaleSchultz		#82 Posted : 30 May 2019 00:21:13(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Initial tests using 1K Ohm inrush resistor on the workbench suggest that it may in fact take a bit too long to charge the cap. I need to build it into a an actual coach to see if the power interruptions are frequent enough to warrant faster charging.
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DaleSchultz		#83 Posted : 31 May 2019 04:00:18(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		First test results are interesting: I initially tried a 1K Ohm inrush resistor. It seemed to take too long to charge up if I manipulated the contact so it only had a short period of power, so I tried a 430 Ohm resistor and built it into a coach so I could actually run it. The lights stays on very well and does not seem too bright. In real life they do not look as yellow as in the picture. The coach I found already had two red LED tail lights. I measured the mA consumed which also includes the existing LEDs and it uses about 80 to 85 mA. At least 20mA of that is the two tail lights so I estimate the coach lighting to be using about 65 mA Edit: LEDs are using about 15mA When track power is applied, I see an initial draw of 115mA, which suggests that 22 such coaches would draw 2.5 Amps initially. I may need to move to a higher impedance for the inrush resistor. Edit: that is 22mA not 115mA I also hear a very feint buzzing sound when the coach gets power, I am wondering what is causing that? Edited by user 06 June 2019 20:16:47(UTC)  | Reason: Not specified
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tiono		#84 Posted : 31 May 2019 04:25:24(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Originally Posted by: DaleSchultz I measured the mA consumed which also includes the existing LEDs and it uses about 80 to 85 mA. At least 20mA of that is the two tail lights so I estimate the coach lighting to be using about 65 mA When track power is applied, I see an initial draw of 115mA, which suggests that 22 such coaches would draw 2.5 Amps initially. I may need to move to a higher impedance for the inrush resistor. I also hear a very feint buzzing sound when the coach gets power, I am wondering what is causing that? 80mA seems too high. How did you measure the current? My Silberlinge coach consume less than 10mA of track current, and the light is bright enough. I measured the DC current after rectifier. No tail light. The faint buzzing sound is the switching regulator. Although the switching frequency is ultrasonic, but usually there will be audible harmonics across the inductor.
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DaleSchultz		#85 Posted : 31 May 2019 04:34:12(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		I only got around to measuring after I put it all together so I simply connected my meter in series with the whole coach, on the digital track current. I can't recall if the Märklin circuit board for the two tail lights is running them in series or in parallel. They have a 690 Ohm resistor and a diode so I presume they are using 20mA in series.
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tiono		#86 Posted : 31 May 2019 05:02:26(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Originally Posted by: DaleSchultz I only got around to measuring after I put it all together so I simply connected my meter in series with the whole coach, on the digital track current. I can't recall if the Märklin circuit board for the two tail lights is running them in series or in parallel. They have a 690 Ohm resistor and a diode so I presume they are using 20mA in series. The track is AC. Am I right that your ampere-meter is DC ? If your meter is DC, obviously it can not get correct reading if the measured current is AC. If the tail-light is the old LED bulb, instead of SMD LED, then most likely it will draw 20mA each.
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DaleSchultz		#87 Posted : 31 May 2019 15:24:28(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		The meter is very top end and even tells me the frequency of the current (3.3 kHz) so it probably is handling AC current OK. I will connect up another string without additional tail lights and make more measurements. If the two LEDs are in series then together they will draw 20mA. If they are in parallel, then 40mA for both. (in an 18V circuit with 1 diode and a 690 Ohms resistor)
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DaleSchultz		#88 Posted : 31 May 2019 20:42:26(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		I connected a length of LED strip to another regulator, fed it pure DC (around 11V) and did some more measurements: With an output voltage of 8.5V the device consumes 311mA With an output voltage of 7.35V the device consumes 270mA The strip of LEDs contain 4 triplets, 12 LEDs in all. I cant recall what the spec of the LEDs are, perhaps these draw too much? I presume the even higher draw is because it is getting pure DC, whereas the one in the coach is getting half wave rectified digital current. I'll connect up a diode and run it off digital current next.
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mario54i		#89 Posted : 31 May 2019 22:02:25(UTC) Retweet
Joined: 28/03/2007(UTC) Views messages in topic : 303 Location: Torino,		All this looks quite strange. Usually this kind of LED strip gives a sensible brightness for a coach drawing 2-3 mA per triplet. If there are really three white LEDs in series with a 240 Ohm resistor (seen in the picture) each triplet should draw ( 8.5 - 7.5 ) / 240 = ~4 mA Beware that a stepdown converter works as a transformer: high voltage low current at input, low voltage high current at output. If you set it at 8 V output and test it at 11 V input there is not much advantage. Test it at 18 V input and input current will be nearly halved. This is what I measured on a similar stepdown set at 8 V output for different input DC voltages. Regards
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DaleSchultz		#90 Posted : 31 May 2019 22:04:10(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		using track digital power with Ammeter in series with track power - it draws 80mA (regulator set to 7.5V) - like the one in the coach. Next week I will put the ammeter between the regulator and light strip.
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DaleSchultz		#91 Posted : 31 May 2019 22:08:40(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Perhaps my fancy meter is giving me crazy readings, will try with another.
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DaleSchultz		#92 Posted : 31 May 2019 22:12:25(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		I agree Mario, I would also expect 4mA
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tiono		#93 Posted : 01 June 2019 02:17:23(UTC) Retweet
Joined: 09/02/2010(UTC) Posts: 234		Originally Posted by: DaleSchultz Perhaps my fancy meter is giving me crazy readings, will try with another. I suspect your meter is giving incorrect reading. A triplet of LED in series with 240 ohm resistor, should draw less than 5mA in 8.5V. To verify your meter; measure the current thru resistor and battery. A 100 ohm resistor with 1.5V battery, should generate current of 15mA. If your meter failed to give correct reading, then you have to calibrate.
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DaleSchultz		#94 Posted : 01 June 2019 13:26:50(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		thanks yes, it is also likely I have it in some incorrect mode for the non pure-DC reading
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GaryTrooper		#95 Posted : 01 June 2019 16:03:52(UTC) Retweet
Joined: 26/01/2018(UTC) Posts: 406 Location: Hailey, Idaho		Originally Posted by: tiono Originally Posted by: ktsolias Originally Posted by: tiono This circuit is not working because you mix rectified DC current and current from the rails Costas Could you explain why it does not work? Please note: the rectifier is not a bridge, but only half wave. See the diagram below: The reason of using single diode (half wave rectifier) is to allow the usage of 2-pole connector. If using bridge rectifier, then yes, the ground can not be mixed with the rail, and that means; must use 3-pole connector. Very interesting concept. First time I've seen it done this way.
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DaleSchultz		#96 Posted : 03 June 2019 03:41:58(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		OK, so I found out that if I leave my meter in auto mode it thinks the digital signal is AC and so it measures the current in AC mode. If I force it into DC mode, I get much better results. All these measurements were made after the diode from the pickup shoe - so half wave rectified. The regulator was adjusted to provide 7.5 Volts. Some interesting things turn up... If I feed in 18 V DC, the regulator and LEDs draw 17.4mA If I feed in 9 V DC, the regulator and LEDs draw 4.15mA So, it appears that the smaller the conversion, the less power is used. If I feed in digital track power, the regulator and LEDs draw 13.6 mA (The voltage after the diode is 16V) I then added the anti-flicker module, this time with a 560 Ohm inrush resistor. Here is a plot of the mA being drawn showing an initial switch-on, and some drop outs where I removed the power for small moments. Generally the inrush seems to not exceed 22.06 mA Yet, I did manage to capture an inrush as high as about 55 mA - not sure if that is a sampling error or some condition I can't reproduce. Unfortunately, I did not have the range mode on to show the peak reading. So, overall good news, a coach set up like this will draw about 15 mA and around 22 mA initially. That will allow more than 100 such coaches on a 2.5 A system.
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DaleSchultz		#97 Posted : 03 June 2019 04:08:26(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Now, here is something I saw a few times.... The lights are running, and I break the contact with the track power. The current drops, but then stabilizes at around 9mA for about 1 second, before dropping to zero. I then connect track power again, and this time I don't even see any inrush, it jumps up to normal 15mA draw. I don't understand why there was no inrush visible. The second time I break contact, make it again and break it. It drops to about 8mA, spikes to 20mA and then we see the little shelf again around 9 mA before dropping to zero. I believe the I am seeing the period in which the capacitor supplies power to the lights for about 1 second - supplying around 9 mA
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mario54i		#98 Posted : 03 June 2019 11:09:34(UTC) Retweet
Joined: 28/03/2007(UTC) Views messages in topic : 303 Location: Torino,		Originally Posted by: DaleSchultz Some interesting things turn up... If I feed in 18 V DC, the regulator and LEDs draw 17.4mA If I feed in 9 V DC, the regulator and LEDs draw 4.15mA Does LED brightness change ?
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DaleSchultz		#99 Posted : 03 June 2019 14:09:15(UTC) Retweet
Joined: 10/02/2006(UTC) Posts: 3,997		Originally Posted by: mario54i Originally Posted by: DaleSchultz Some interesting things turn up... If I feed in 18 V DC, the regulator and LEDs draw 17.4mA If I feed in 9 V DC, the regulator and LEDs draw 4.15mA Does LED brightness change ? I did this on my workbench under good light, and I did not notice any difference. I will try and see if there is any. Inside a coach it might.
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Minok		#100 Posted : 03 June 2019 20:22:06(UTC) Retweet
Joined: 15/10/2006(UTC) Posts: 2,319 Location: Washington, Pacific Northwest		Originally Posted by: DaleSchultz Now, here is something I saw a few times.... The lights are running, and I break the contact with the track power. The current drops, but then stabilizes at around 9mA for about 1 second, before dropping to zero. I then connect track power again, and this time I don't even see any inrush, it jumps up to normal 15mA draw. I don't understand why there was no inrush visible. The second time I break contact, make it again and break it. It drops to about 8mA, spikes to 20mA and then we see the little shelf again around 9 mA before dropping to zero. I believe the I am seeing the period in which the capacitor supplies power to the lights for about 1 second - supplying around 9 mA As the capacitor buffer is feeding input to the voltage regulator, is it the case that as the capacitor discharges to a certain point, the voltage regulator no longer can provide its requested output and shuts off the load effectively, thus not further discharging the capacitor, so when the power comes on again, the capacitor still has a some charge in it and thus needs much less current to top its charge off?
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