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  1. @PSi If the LEDs had significantly different brightness characteristics I would suggest my method, but as Dhjgreen, who posted a few seconds before me assures us they are close enough that they can all use 1k, Steve W's approach with less resistors and less soldering is suitable. As for a protection diode, unless you're particularly clumsy with connecting things up, or don't use polarised connections, I wouldn't worry about that.
  2. As DJHGreen suggests 1k gives an appropriate brightness for each colour, that approach should be fine.
  3. I haven't gone through all the switch permutations but one thing is certain: you shouldn't have a situation where you can have LEDs in parallel (particularly if they are different colours) - the forward voltage at which an LED starts to conduct can vary slightly from LED to LED of the same type, and considerably for LEDs of different colours - when this happens the LED that has the lowest forward voltage will grab all, or at least, most, of the current and will be brightly lit, while the others will be dimly lit, or not lit at all. The way to prevent this is to give each LED its own series resistor (if I've counted right, that's 13 resistors) - the easiest place to put them might be to solder one end of each directly to the tabs on the rotary switches, and the other end to the wires leading to the signals (put a bit of sleeving over them to prevent anything shorting on the back of the switches). As for balancing the brightness with different resistor values, I can't suggest anything more than trial and error to get a brightness you are happy with - I'd temporarily use a set of resistors and pots in series, adjust them until I was happy and then measure the combined resistances for each colour - e.g. 1K 1K +12V o-----/\/\/\/-------/\/\/\/ ^ ^^ | // RED +---------|>|------o OV 1K 1K +12V o-----/\/\/\/-------/\/\/\/ ^ ^^ | // GREEN +---------|>|------o OV 1K 1K +12V o-----/\/\/\/-------/\/\/\/ ^ ^^ | // YELLOW +---------|>|------o OV with the values shown, you'll be able to adjust the LED current between approximately 5-10mA (using the fact that you've been supplied with a 1K ohm resistor as a guide for the maximum current).
  4. https://www.maximintegrated.com/en/products/analog/data-converters/digital-potentiometers.html Digital pots generally contain a chain of, typically, 100 or 256 (sometimes less) resistors internally between a Vmax and Vmin (or some similarity named pins) which correspond to the two ends of a traditional pot. A multiplexor controlled by a digital register selects a point in the chain to tap. Some digital pots let you write the position register directly, others have an up/down counter which you pulse to get you to where you want to be. Usually, because of the way the technology works the ends of the resistor chain must be within 0V to the logic power rail (typically 5V) - i.e. at least with the ones I've seen you can't use a digital pot directly to generate a 0-12V output. To a circuit they are connected to they do look much like a traditional pot.
  5. I thought you were the one suggesting a PWM driver... I read it as either feeding the PWM into an H-bridge chip (which has the advantage that with the right choice it will work straight from logic levels so you don't need any level shifting), or using the PWM output as the input to an emitter follower (in which case your digital ~0V and ~3-5V depending on your PIC power requirements would need boosting to switch between 0-12V) - apologies if I misunderstood and you meant to use the PIC DAC module to drive an emitter follower as a true DC controller - you'd still need a gain stage, probably easiest with a non-inverting op-amp circuit to boost your DAC output (I can't remember the analogue voltage range for a PIC) to 0-12V for a DC controller at the base of a emitter-follower - actually I'd suggest a power supply of about 15V and gain to max out the base voltage at a little over 12V to take into account op-amp head-room and the 0.6V Vbe (or more for a darlington) drop on the emitter follower if you want to go to full 12V output.
  6. Are you sure the plain boarding isn't just the other end? I would have thought one end or the other would have steps and handrail to the roof.
  7. You could, but it would be hard to beat Andy's PNP switching circuit on page 1 for simplicity. I think driving the output with an emitter follower rather than as a hard driven switch will increase the power dissipation in the output transistor too.
  8. Not the display cabinet, but today I offered SWMBO that I would do the Lidl shopping... Not entirely out of altruism- I thought it was cheap soldering iron and rotary tools week! Anyway, apparently not, but my Lidl (Wimbledon) was doing furniture and I noticed they had a Livarno sewing machine table for ~£50 that might equally well function as a cheap modelling bench. http://offers.kd2.org/en/gb/lidl/pdXdi/ Note: this usually seems to be an October offer so I don't know if my Lidl was just clearing old stock.
  9. I wouldn't be the first to try to control motors like this... after a bit of googling I found: https://www.depotbassam.com/2012/02/analog-dc-motor-driver-push-pull-power.html although as a bench-top experiment rather than for moving model trains. H-drives are a kind of push-pull configuration but with PWM usually used with the transistors driven hard on or hard off to act as switches which minimises the power dissipated in each transistor. With a class B or AB amplifier the transistors are not driven into saturation, so are working as amplifiers rather than switches, which means that a lot of power can be dumped in the transistors (or for a DC controller, one transistor or the other depending on the polarity) - I suppose a hybrid might be possible a push-pull configuration driven hard with +ve or -ve pulses to create a bipolar PWM output without the same power losses in the transistors. The principle achievements would be: a single control knob - reverse one way, forward the other (as per the Morley as David mentioned) a no-fuss common return without any need to isolate the supplies of each control channel to prevent shorts. Edit: Andrew: Thanks for that example - I thought I most likely hadn't had an original idea! Nearholmer: as someone who grew up with H&M controllers I'm quite happy with a centre-off controller.
  10. No - not the type of passenger train with a driving trailer! I was thinking about amplifiers (otherwise known as class B or class AB) and train controllers, and wondered if anyone had ever put a push-pull output stage on a model railway controller. My thinking was - if the output is symmetrical about ground to +/-12V, then it removes the need for the traditional DPDT polarity reversing switch between the power output and track, and makes common return easy with multiple controllers powered from the same source. The slight downside- your power source has to provide smooth +12V and -12V (and a bit) rails instead of a single 12V rail. To be honest, I've only seen them used in the past for audio and RF where the signal is approximately equally divided between +ve and -ve, and hence both output transistors see approximately the same power. Are there any problems when driving DC and one transistor will be driven continuously and dissipating power while the other is off? (drift possibly as they warm at different rates?)
  11. I notice there are two versions a GM500 and a GM500D - the latter is 'DCC friendly', not sure if that will make a difference for you.
  12. It's not obvious from the original sketches, but the LEDs should be wired with both anodes or both cathodes pointing towards the frog.
  13. I'm guessing it's for a DCC set-up so there will always be power to the track - half the time it will be the correct polarity to light the LED.
  14. One thing to consider is the type of controller the PSU will be connected to. A computer power supply will only output smooth DC, which may be fine with appropriate protection for many controllers. However, if the controller is a PWM type, or measures back EMF by sampling, it may well not be suitable- most (all?) commercial controllers (and many published d-i-y designs) of that type require an AC input as it (or the unsmoothed full-wave rectified version) is used as a timebase for the PWM frequency and pulse start point - e.g. via a triac circuit, and the zero-crossing as a point to sample the emf.
  15. How did I misquote you? I used the 'quote' button on the site (unless you edited your post after). You said 'the earth pin could take 115 watts' - your exact words. What flows through the earth pin is amps, not watts. All your lights may, combined, have a total of 115 watts, but those watts are not flowing through the earth pin - 10A or so of current is flowing through that earth pin. Ideally, that earth pin should be contributing no watts to the scenario. If your battery was somehow to be 24V, and the caravan lights took 230W, there would still be exactly the same current going through your earth pin - the same strain on it, or your battery was 6V, and the lights took 57.5W, the same current again - it just doesn't make sense to say how many watts are passing through the connector, it's the current it's rated for. In your post you appear to be liberally exchanging watts and amps. Particularly when the currents involved are sufficient to do some damage, some rigour should be applied to the correct use of units.
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