Model train turn-out problem

Discussion in 'The Projects Forum' started by kbirecki, Dec 11, 2009.

  1. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    I cannot figure out what I'm doing wrong. I have a model train turn-out that has a wired remote to control the switch. To activate the switch, I can apply 9-19 VDC momentarily to flip the switch and I'm trying to use a transistor to control the switch. I already have a circuit that will apply a momentary voltage, for instance, to a base of a transistor. I was thinking I would have this other circuit apply a small current to the base of a transistor and the emitter would be connected to the switch to apply the voltage and trigger the switch on the turn-out. But it doesn't work. I've tried various configurations with NPN (2N4401) and PNP (2N4403) transistors, and even a TIP130. But the switch on the turn-out will only trigger if I directly apply the voltage to the leads directly to the switch. Can I be missing something? If I measure the coil resistance it is 20 ohms. Are there any suggestions of what I should try?
    Thanks,
    Ken
     
  2. hgmjr

    Moderator

    Jan 28, 2005
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    Post your schematic.

    hgmjr
     
  3. SgtWookie

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    Jul 17, 2007
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    Just for some numbers:
    9v across 20 Ohms is 0.45A, or 450mA.
    19V across 20 Ohms is 0.95A, or 950mA.

    While 2N4401 and 2N4403 transistors are rated for up to 600mA collector current, they're pretty much out of steam by 500mA - and that's if they have 50mA base current.

    A TIP130 is an NPN Darlington. It can sink quite a bit more current, and do so with a lot less base current.

    However, it would help us help you a lot more quickly if you post a schematic.
     
  4. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    Thanks for the quick replies. It took me until this morning to document my ckt. I was only working on a breadboard. I'll attach the ckt layout I have so far. It is intended to control switching the turnout one direction. Another copy of a portion of this ckt will switch the turnout in the other direction. I am having trouble with section #5, which seems like it should be very straight forward. I've tried various values for R9 and R10, and various other configurations around Q5 to try to engage the coil in the turnout. The only way I've been able to make the switch respond is to directly apply 12 Vdc (I found it will accept anything from 9-19 Vdc momentarily across either A-B or B-C to make the switch change directions.)
    The coil in the turn out has a resistance of ~20 Ohms.
    Any help would be appreciated on the proper way to make this work.
    Thanks,
    Ken
     
  5. SgtWookie

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    Nice job at the documentation - it's definitely appreciated, and will speed things up considerably.

    1) Power Supply
    You are using a 12v wall adapter plug connected to a full wave bridge, which is constructed from 1N400n diodes.

    These 1N400n series diodes have a Vf (forward voltage) of about .8v at 100mA current; at 1A, it's about 1v. Since a full wave bridge is basically adding two diodes in series with the load, you will lose roughly 1.6v to 2v in the bridge alone.

    Using 1N540n series diodes will help somewhat; they are rated for 3A, and will have a lower Vf at 1A.
    Using Schottky diodes, like 1N5817-1N5819 would be even better, as they have a very low Vf. They have a lower maximum reverse voltage, however.

    Also, you have no filter capacitor(s) across the output supply rails; those would aid quite a bit in taking care of transient loads. 3,000uF per 1A of output load for a FW rectifier and 60Hz AC is a "rule of thumb" number. If it were halfwave rectification, the capacitance would need to be doubled.

    Wall warts may have small caps in them, but they are tiny at best.

    Your wall wart will need to be rated for 1A or more.

    2) 5v regulator
    It'll work for very light loads, but a 1N4401 can only handle 625mW power dissipation due to the packaging. As it is, since your input is 12v (or somewhere near there) and you're dropping at least half that across Q1, most of the power in the 5v rail is being dissipated in Q1. Basically, you're limited to 89mA output current before Q1's power rating gets exceeded. That's not very much. A 7805 would handle up to 1A output current, and dissipate more power than a 2N4401 without a heat sink.

    3) Signal acquisition
    A bit more detail is needed here.

    4) Signal processing
    You might consider replacing this whole block with a 555 timer wired as a monostable multivibrator. It'll cut your parts count and simplify things quite a bit. You would no longer require the 5v supply, either.

    You have a 15k resistor (R6) in your output signal path. Your output signal will measure in the microampere region.

    5) Solenoid controls
    You have a TIP130 wired as an emitter follower with R9 limiting the base current even further; now you're into nanoamps for base current. You'll lose about 0.7v with a standard bjt or about 1.3v across a Darlington transistor that way. R10 shouldn't be there.

    Connect point B (center tap) of the solenoid to +12v. Connect a transistor's collector to point A. Connect a transistor's emitter to ground.

    If you can't connect point B to +12v instead of ground, then you will need to use PNP transistors to source current to point A.

    With a standard BJT, you'll need about 50mA current through the base to saturate it to get 500mA through the collector to actuate the switch solenoid. If you use a Darlington, you will need much less base current; probably 1mA-2mA would be sufficient.
     
    Last edited: Dec 12, 2009
  6. SgtWookie

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    Here's a proposed solution to your 1st problem; that is ensuring that the polarity is correct when you connect a "strange" wall wart to your system, without the drop in voltage you get across the diodes:

    [​IMG]

    V1 on the left simply represents your wallwart.
    S1a and S1b are two poles of a DPDT toggle switch.

    R1 is a current limiter resistor for D1 and D2, which are LEDs.

    If the polarity is not correct, the red LED will be lit. The relay coil will not get power, because D3 will block current flow. The red LED informs you the polarity is not correct, and that you need to flip the switch.

    If the polarity is correct, the green LED will be lit, and the relay will close, supplying power to the load.

    D4 takes care of the reverse-EMF voltage spike from the relay coil when the power is turned off.

    The penalty is having a relay energized the entire time you're operating the system, but there is no voltage drop across a bridge, and you won't accidentally apply power in reverse polarity. I suggest using a relay rated for at least three times the expected current. An automotive relay would be a good choice.

    If you add capacitors to the supply, they should be after the relay contacts.

    [eta]
    After thinking about it for a bit, this circuit would not protect absolutely against a faulty wall wart that was putting out AC instead of DC. You would get an indication of AC by both the red and green LEDs being lit, but the relay could still engage on 1/2 of the cycle. A single Schottky diode, like a 1N5818-1N5819, in the load path would prevent damage.
     
    Last edited: Dec 13, 2009
  7. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    Thank you SgtWookie! Those are some great suggestions. I appreciate your prompt replies and *very* thorough analysis, much more than I expected. It is nice to have found a site with some very dedicated and helpful people.

    I have a couple questions about your suggestions. Regarding:
    1) Power Supply
    a. I’ll change the diodes as suggested, but I’m thinking I don’t have to go to far to get 12 Vdc precisely because the turnout will operate as low as 9 Vdc, so some drop in voltage would be OK. The other ckt suggestion you made looks like a good one too, and I have almost everything except the relay. I may try that just to see how well it works for me.
    b. Question 1: You mentioned I should have a capacitor across the output supply rails. Since the wallwart already gives me DC voltage, why do I need a filter capacitor(s)? Is this because we can’t know for sure how reliable the output from the wallwart is? That would make sense because I have about a half dozen wallwarts rated at 12 Vdc, but when I measure their output with no load I get anywhere between 14-16.5 Vdc, so there is some variability.
    c. Question 2: Did you suggest my wallwart would need to be rated at 1A or better because this ckt needs it? I haven’t calculated the current draw of this ckt (or measured it because I don’t think I’m using my multimeter correctly when I try). I am planning on controlling 4 similar turnouts, so portions of this ckt will need to be replicated. Does it look like this ckt alone needs 1A?
    2) 5V Regulator
    a. I found a datasheet for a 7805 that you recommend (http://www.datasheetcatalog.org/datasheet/fairchild/KA7805.pdf). Question 3: Do you recommend putting it in place of Q1 with R1 and D5 as-is? Or should something else change as well?
    3) Signal Acquisition
    a. This is going to be IR Emitter/Phototransistor pairs placed on either side of the train track. When the train passes through, it breaks the connection and the signal from the Phototransistor goes low. Then I do what I need to based on that “go-low” signal. I planned to use CAT3 cable (I think it is 22 AWG wire) for the sensors. There will be a maximum of ~12 feet from the Signal Acquisition sensors to the Signal Processing ckt. I was trying to keep the components at the remote location (around the track) to a minimum, so all I should need is one resistor for the emitter and one for the phototransistor.
    Question 4: Does this make sense? I’ve tested that portion and it works well. I can easily place them 1.5 inches apart and get a very good signal response. (In the future, I’ll disguise them as something that could be near the tracks.
    4) Signal processing
    a. Replacing the whole block with a 555 sounds like a very good suggestion. I originally started with considering that, and then I took apart the remote switch that came with the turnout to see how it worked. It uses something like the block I have, but it only uses two transistors. I tried reproducing it, but I must have missed something while following the traces because I couldn’t get it to work. I was thinking they did it that way so I should too, but I think you are right. I’ll convert that to use a 555 timer. And it makes sense to do that and keep the processing at 12 V. Simplifies things greatly.
    b. My understanding is that R6 is actually in the path from 5Vdc Rail-R8-R6-base of Q3. I do now see that Visio didn’t hop that line over the connection from C1 to Q4. So my understanding of the path to the output is essentially 5 Vdc rail to the collector of Q4, which is triggered on/off by the rest of that block, and in turn provides the output to the relay control ckt. Question 5: Am I understanding it correctly or missing something?
    Having resistor R9 in that path may be a problem, as you point out in your next section.
    5) Solenoid Controls
    a. I should have no problem connecting as you recommend. I do also have a TIP120 (Darlington), so I can try that. Between the change in the Signal processing portion and this, maybe that will work, and simplify the whole ckt as well.

    I also noticed your edit on your follow-up comment about the polarity issue. It’s a good point about a faulty wallwart. Question 6: I assume you mean that the Schottky diode should be placed reversed biased?

    SgtWookie, thank you for your evaluation of this ckt. I really appreciate it.
    Ken
     
  8. SgtWookie

    Expert

    Jul 17, 2007
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    If you just change the diodes from 1N400x to 1N540x (3A silicone diodes) ro 1N5820 (3A Schottky diodes) and add a 3,300uF cap to the output of the bridge, you should be fine. The cap is important.

    Wallwarts don't have a voltage regulated output. When the current load is what the rating of the wall wart is, the voltage will measure within about 10% of the rated voltage. When the load is less, the voltage is higher. A large cap will store the higher voltage for when a sudden and temporary (transient) load is applied.
    The current rating is to re-charge the cap quickly after partial discharge by the switch. You want to get the voltage back up quickly, ready for the next actuation.
    If you used a fixed regulator like the 7805, the need for R1 and D5 goes away.
    OK, you may run into a problem with ambient light triggering it. That could be "a bit fiddly" as they say in the UK.
    CAT 3 at that short of a run should be OK. I think it's smaller gauge wire, though.
    How about a young boy with a fishing pole on one side, and his dog on the other side? ;)
    It's not like you can't make a monostable multivibrator out of a couple transistors, a few resistors and a capacitor - but the more parts you have in a circuit, the more things you have that can fail. If you can get an inexpensive IC that replaces a bunch of parts, you're time, money and reliability ahead.

    Q6 is an emitter follower. This means that the voltage on the emitter will "follow" the voltage that is placed on the base (Vb), but it will be about 0.7v less than Vb under a fairly light load. The problem here is that Vb can't exceed 5v due to your regulator, so Ve can't exceed 4.3v even if the current through the base were unlimited. However, the base current is very limited by R6. I would be surprised if you were getting any voltage reading at all at the emitter of Q1 with the load connected.

    R10 is limiting the current through Q6, which you don't want at all. But, that whole section needs to be re-done anyway.

    You'll also need a diode connected in reverse across the solenoid coil, otherwise it'll blast the transistor when the current gets cut off.

    Based on some further thinking on this, I'm going to suggest a complete revision; just changing the diodes to the 1N540x type would be a big help and cheap, 1N582x diodes would be better but perhaps $0.20 more per diode. A low-dropout 15v regulator would ensure that the voltage couldn't go too high.

    If you want a deal on a 4A 12v unregulated supply, look at this for $4: http://www.mpja.com/email/12-08-09a.asp?r=%%ref%%&s=1

    Or if you want to just eliminate all the external regulation stuff that you'll need, have a look at this:
    http://www.mpja.com/prodinfo.asp?number=16007+PS

    That last one should be able to power a pretty large layout, as long as you weren't operating more than 6 or so switches simultaneously.
     
  9. SgtWookie

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    OK, have a look at the attached.

    The upper left area is the power supply. After the output of the regulator, U1, all connections to the + power rail are referred to as Vcc. The LT1086-12 is a low-dropout fixed 12v regulator. There are others similar on the market.

    U2 down on the lower left represents your IR emitter and detector pair.
    The voltage symbol marked Vcc that R4 is connected to is actually putting out a pulse, but that was one of the few ways I could simulate the train passing.

    U4 is a NAND gate, but it's more than that; it's a Schmitt trigger. It "squares up" a slowly transitioning signal. That will make tripping the timer more reliable.

    U3 is a standard bjt (transistorized) 555 timer; it can't be a CMOS model due to the output current required. If you want to use a CMOS version, you'll have to use a Darlington transistor or power MOSFET. It's set up to output a pulse about 110mS long (slightly over 1/10 of a second). I don't know how long your switches take to switch. If you want a shorter time, decrease R1. 470k Ohms would give you about 50mS.

    R5 limits the base current into Q1. D6 and D7 are for reverse-EMF suppression; D7 isn't necessary at the moment, just showing how it would be connected when it's necessary.
     
  10. SgtWookie

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    Just a slight revision; only the connection to the IR receiver was changed to invert the logic. I realized I had the logic for the IR emitter reversed. :rolleyes:

    One thing that would help a great deal in eliminating inadvertent or repeat switching - do your turnout switches have a position indicator that works electrically? Such as; if the switch is in mainline, there is a voltage available, and in turnout, an absence of voltage?
     
  11. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    On the topic of the power supply, I tried to post a question this morn, but I must not have hit submit - What about using a computer power supply? They put out 12 V and 5 V and I think they would be regulated, and plenty of power, albeit somewhat noisy. I have spares of those.

    However, the power supplies you mentioned may be easier to deal with, and reusable for other projects. The wall warts I have are all 1A or less. And it looks like I'm going to need more than that for everything I am planning on as I read your feedback.

    If I understood your last question correctly, the turnout has a single connector with three contacts, and requires a momentary pulse of voltage to be applied to either A & B (shown in my original schematic) to turn one way or B & C to turn the other way. So absence of voltage has no control.

    I knew I was going to have to deal with power requirements at some point, but thought I'd start with small components of my bigger picture. I may be at that point because it seems to possibly have an impact on my overall implementation. Before I go through the revised ckt, let me ask this question about the power supply. Are you getting a sense that the (apparently) small wallwarts I'm thinking of using would only have enough power to supply this one ckt? Would you say I may not have enough power for my planned usage to include everything I'm thinking of? Briefly, I'm planning on:
    - 4 IR Emitter/phototransistor pairs.
    - 4 turnout control ckts, so if this one we've been talking about is half of one turnout, so I'll need 8 of the solenoid control ckt portions.
    - 2 DPDT relays to control polarity of two small sections of track.
    - 2 DPST relays to control power of two small sections of track.

    Maybe I should just switch to a beefier power supply now. Your thoughts? (I'm not looking to have you work out the details, I'm just wondering if you have gut feeling. If it needs some calcs, I'll do the math.)

    OK, so I'm going through the revised ckt diagram.
    Power Supply: I guess this depends on whether or not putting my own full wave bridge together with a voltage regulator is "better" than one of the commercial units you mentioned in an earlier post. If I won't have enough power for my whole project with a wall wart, maybe I just keep what I have on the breadboard for testing/development, but as I build up the real project, maybe I should get something with higher amperage. Is that what you'd recommend?

    NAND Gate/Schmitt trigger: I have some of those, so I can easily do that. I expected I needed to invert my signal somewhere along the way, but I was going to use a hex inverter. I can use the Schmitt trigger NAND. Thanks for the suggestion.

    Timer: I already have some NE556N timers (http://search.digikey.com/scripts/DkSearch/dksus.dll?Detail&name=296-6504-5-ND), which I assume are CMOS; not sure how to tell and it doesn't list it on the spec page. And I have TIP120's, so I'd rather try that combo. I assume when you referenced needing a Darlington for the output when using a CMOS 555 timer, you meant in place of Q1, correct?

    And I think ~.1s would probably work well. If anything, maybe longer, but I can adjust that.

    Question on the caps: Can you refresh my memory: I know the straight line caps are non-polarized, and the curved line caps are polarized. Under what conditions do you use polarized or electrolytic as opposed to non-polarized? Is there a rule of thumb? Are there very different results for differing types even in a small project like mine?

    Thanks again for your time!
    Ken
     
    Last edited: Dec 15, 2009
  12. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    OK, Let me try this again. I've posted this morning and again this evening and neither post is showing. This is a test of the emergency posting system. This is only a test.
     
  13. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    Now I see what is going on. My morning post did not work, and I see my evening post is on a second page.
     
  14. SgtWookie

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    This happens to me frequently. Sorry about that. It's best to compose your post in Notepad or or other text editor. Just make certain that you paste and then copy it from a "plain text only" type editor like Notepad before you post it, or it will make it very difficult on me to address your questions step-by-step, as I am wont to do.

    Perfect solution.
    Efficient and extremely cheap. I have converted a salvaged ATX form factor PC power supply to be a bench supply. Such a supply should eliminate much of the foolery that I posted - about the power supply, at least.

    Google "ATX bench supply" for lots of ideas.

    You can use the low voltages to power LEDs, but you'll need the 12v to power the switches.

    You really do need to keep track of your power requirements. If you do not, you will have problems; big ones. Even professional electronic engineers frequently exceed their power budget; ask me how I know this.

    OK. Too bad. It would help save power, and extend the life of your turnout solenoids, if you can figure out how to interface a SPDT switch with the points, many "false switch" requests could be avoided.

    You'll ALWAYS have to deal with power requirements. Best to "get green" as quickly as possible. Use an efficient supply. Scads of "wall warts" do not qualify. You need efficient power supplies that have excess capacity. Switching power supplies fall into that category.

    Two, maximum. Simultaneously, that is.
    You're hopelessly over-budget. Are you a Congressperson? :)

    I've already stated that you need a beefier and more efficient supply.

    I suggest that Marlin P. Jones & Associates who are doing business at http://www.mpja.com are a very competitive and satisfactory resource for you to check out for pre-made supplies. I have no relationship with them other than being a satisfied customer.

    Keep it simple. Trains seem to demand more trackage, sideshows, accessories, and only the Engineer knows what else. ;)

    You need to start figuring out what the power demands might be, how to best manage them, and then multiply by at least four.

    With the efficiency of the switching power supplies available nowadays, you will never be sorry that you bought more capacity than you need.

    OK, there are very few CMOS IC's that have Schmitt-trigger inputs.
    The 4093 is one of them.
    The 4106 inverter is another.
    For the purpose of my schematic, consider the two interchangeable.

    They are both very useful IC's. However, you'll probably find that the 4093 is less expensive. It is also far more useful than the 4106, as it can be used to construct any other logic gate function, whereas the 4106 cannot.

    If they say NE556, they are dual bjt timers, and will be OK for the project.

    Those will work. I love swatting flies with battleships ;)
    You will need to increase the value of the base transistor to perhaps 2k-5k Ohms though.

    Yep. CMOS 555 timers have very wimpy source current capabilities. They can source about 10mA when Vcc/Vdd=15V, but sink 100mA at the same voltage. If the Vcc/Vdd is less, you'll have to extrapolate.

    If your switch won't actuate within 100mS, there's something wrong. I padded the time deliberately.

    OK, I'm trying to interpolate your question via convention.
    I haven't come up with a good answer.

    However, polarized caps should never be used in a situation where the polarity across the capacitor would reverse.

    With non-polarized capacitors, it does not matter.

    Polarized electrolytic caps are very popular, due to their small size and high capacitance. However, you can quickly destroy them if you reverse the polarity across them.

    You can connect two polarized electrolytic caps back-to-back (ie: + to + or - to -), however that will reduce their uF rating by 1/2 (assuming they have identical ratings) while no improvement of their maximum voltage rating will be noted.

    You are better off to choose the most suitable cap to begin with.

    Don't use polarized caps in the schematic I presented unless they show a polarity. Capacitors with parallel plates indicate polarity insensitive. Caps with one plate curved indicate that the caps can be sensitive to polarity.

    I hope you're having fun ;)

    You realize that you're obligating yourself to post photos of your railroad, right?
     
  15. kbirecki

    Thread Starter New Member

    Oct 22, 2009
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    SgtWookie, I *AM* having fun with this. My wife can't understand what I'm doing with all this, but I'm OK with that. Maybe my daughter will get it once she see's it in action, but she's only marginally interested right now (mostly when the lights blink as she passes her had through the sensors). And I absolutely plan on posting final schematics, pictures of the track, and I'll post a link to a video if I can get that far.

    I do have some questions about your last post, but I'll have to come back to that tonight. I'm sure you know your assistance has been invaluable.
    Ken
     
    Last edited: Dec 16, 2009
  16. SgtWookie

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    Just a thought...

    Placing the IR emitter/detector pair at an angle across the track would help reduce multiple activations of the switch solenoid due to the gaps between the cars. Otherwise, you'll get the solenoid activating once (or more) each time a car passes.

    I've been idly thinking about adding a flip-flop to the circuit. The idea with the flip-flop is that once the solenoid has been actuated in one direction and the 555 timer has expired, the flip-flop is set or reset, and the output of the flip-flop is connected to the RESET input of the 555 timer. This would prevent the same solenoid from actuating again until the other solenoid was actuated, which would reset or set the flip-flop.

    Clear as mud?
     
  17. SgtWookie

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    Onwards through the fog! :D

    Here's a newer version (you were sort of expecting this, right? ;))

    This will handle both frog-side approaches to the turnout.

    No provisions for manual switching have been added - yet!

    The upper half of the schematic is basically a clone of the lower half. I haven't changed Q1 & Q2 to TIP120s because I don't have them in my library yet, which means that R5 and R10 will also need to be changed to higher values.

    Note that because you're using TIP120s, you'll lose about 0.7v to 1.1v across the base-emitter junction; you wouldn't lose that with the 2N2222's. This is because the 2n2222's saturate; the TIP120's don't.

    New stuff: over on the right, a CD4013 D-type flip-flop has been added.
    Note that the Clock and Data inputs are grounded.

    When the active timer's output goes high, it turns on it's accompanying transistor, which saturates and sinks current from it's solenoid. This also pulls C9 or C10 low; it discharges through D5 or D6.

    When the timer's output goes low, the transistor turns off, and the coil pulls the cap high again, which triggers either the SET or RESET input of the 4013.

    This brings the RESET input to the timer that triggered it low, so that it won't trigger again - there is no need for it to pulse the solenoid again until the switch points are flipped the other way.

    [eta]
    Off to the far right, you'll see a couple of transistors with 6.2k base resistors, LEDs, and 680 Ohm current limiting resistors. If you make those red and green LEDs, you'll have switch point indicators for your approaches. ;)
     
    Last edited: Dec 16, 2009
  18. SgtWookie

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    Yet another revision :rolleyes:

    I've updated to include the TIP120 change and the base resistor values; they went to 2.2k, which should be about right. Note that the TIP120's Vce will be around 0.6v to 0.7v with a 500mA load (your solenoids), but that will probably be OK.

    Note that to the left of the TIP120 base resistors, diodes have been added - and input connector flags marked "A" and "C". If you apply 12v momentarily at that point using manual electrical switches, it will change the track points, regardless of the state of the 4013 or anything else. If there is a train blocking the IR detector, the points will switch after a gap between the cars is seen.

    Due to the 4013 addition, the installation of the IR detector/emitter pair at an angle across the track is no longer suggested; as a matter of fact, it would be best for them to be just barely above the top of the rails. This will ensure that it's tripped a bunch of times due to the passing of the wheels; in case busy little fingers happen to be switching the points the wrong way, the movement of the train past the IR emitter/detector pair will continually switch the points to the correct position, if they are not there already.
     
  19. SgtWookie

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    One thing I forgot about is the frog polarity.

    I'm not quite certain how to handle that at the moment.

    Adding a relay to the mix would be rather ugly in terms of power consumption.
     
  20. SgtWookie

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    Jul 17, 2007
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    Same schematic; realized I'd forgotten to add the part numbers to D9 & D10. 1N4148/1N914 would work fine, so would any 1N400x series.

    "Tidied up" area around C9/C10/D5/D6/R11/R12. No real electrical changes; just a bit easier to look at.

    Still not sure what to do about powering the frog. If your frogs are insulated, no point in worrying about it.
    [eta]
    A bit more tidy.
     
    Last edited: Dec 16, 2009
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