It's possile, but I'm not clear on why you need the double throws?  A DPST switch would work for connecting two voltages of electricity to the different light sources.

Or you could run both of the lights from a single 9v source - you'll need an appropriate dropping resistor in series with EACH of the 1.5v lights. Don't parallel the lamps and use a single dropping resistor. If one of the lamps burns out, the total current is reduced which reduces the voltage drop across the resistor, which increases the burnout chances of the remaining lamps. Oops.

btw, I' not sure on what a 9v led is. Led's are rated by current and it's up to you to provide an appropriate current limiting resistor in series with each LED (sound familiar).

A better circuit for those 1.5V lamps is to parallel them with a pair of series diodes.

A property of diodes is that they require a 'foward bias voltage' before they start conducting. For silicon rectifier diodes that voltage is about 0.7V. Once the diode starts conducting, the voltage across it grows no further. So the voltage across the 1.5V lamps is "clamped" at 1.4V by the diodes.

Here's the tricky part. You'll need to select the dropping resistor so that it will allow enough current to flow to illuminate the 1.5V lamps.  You'll need to know how much current that requires. Let's assume that these lamps each require 0.050amps (or 50 milliamps).

If the supply voltage is 12V then the voltage across the resistor will be

   12V - 1.4V = 10.6V

If there are 2 lamps, each needing 50mA we'll need to pick a resistor that will let at least 100mA (or 0.1amps) of current flow.  Use Ohms law to figure that out.  This law states that current (in Amps) is equal to voltage (in Volts) divided by resistance (in Ohms).  Current is abbreviated with a I (capital i), voltage with a V, and resistance with an R.  So

I = V / R

if we reaarange this equation to solve for resistance we get

R = V / I

Now plug in our values

R = 10.6V / 0.1A = 106 ohms

Lets pick 100 ohms which is a standard resistor value.

But we're not done yet. We need to figure out how many watts of power will be heating up the resistor and get one that's an appropriate size.

Power = Voltage x Current

plug in our values

Power = 10.6 x 0.1A = 1.06 watts

You'll probably want to use a 2watt 100 ohm resistor (although a 1 watt resistor) would probably work although it would get pretty hot to touch - don't let the resistors physically touch a wall of a plastic building or the plastic may melt. And touching the wall of a wood building could concievably be a fire hazard.

What if the power supply is delivering 9V?

The  resistance = (9 - 1.4) / 0.1 = 7.6 / 0.1 = 76 ohms

And  power - 7.6V * 0.1 = 0.76 watts (a 1 watt resistor should be ok here)

What if the lamps are 20mA each instead of 50mA?

the resistance = 7.6V / 0.040A = 180 ohms (which as it happens is a standard resistance value)

the power = 7.6A x 0.040A = 0.304 watts ( a half-watt resistor would work here)

(a good argument for bulbs needing less current)

OR

Use a 12V lamp (or lamps in parallel) instead of the dropping resistor. Make sure it's current rating is greater than the combined current requirements of the paralleled 1.5V lamps. I've done this. It's a bit tricky balancing the number of 1.5V bulbs with the current of the 12V bulbs. But you eliminate the waste of electicity in the dropping resistor.

Be sure to use rectifier diodes rated at greater than the current through the dropping resistor or 12V bulbs. If you have two 12V rated 100mA each then 1Amp rectifiers will have a good safety margin. Why a safety margin? Well if either of the two diodes fails then you lose you 1.4V clamping and all those 1.5V lamps may become flash bulbs! Personally I hate resplacing burned out lamps in buildings...

I hope this is at least somewhat helpful.

Cheers,

Charlie

I have SPDT also, so thats not a problem, and the info is very good,and will be very useful,  but one step further still using a DPDT, How would you wire the switch, We will label A and B, (middle) C and D, E and F.     Would I bring power in thru A and B after I have crossed A to F and B to E. and power out through C and D, or should I reverse my connections.  

If I understand what you're saying correctly, I think you're making it too hard.  You only need to break an electrical circuit path in one place to stop electrons from flowing. You don't need to have swith poles both before and after the lights. Even the following circuit is overkill - you don't really need to have separate switch poles for the LEDs and the 1.5V lamps.  Both could be powered through a SPST swtich (you can certainly use a single pole of a DPST switch if that's what you have on hand).

If you have 2 power supplies of different voltages, one of which you wanted to use to operate LEDs and the other the 1.5V (or 12V or whatever) lights then feeding each power pack through a separate pole of a DPST would be the way to do it.

If you can get your hands on a "How to Wire Your Model Railroad" book (maybe you can get it through your library?) it would answer lots of questions like this.

If this isn't addressing your question then let me know.

Best regards,

Charlie

Your info is very helpful, I was trying to get some info on if it could be done. you brought up a good point, although, I originally was planning incadesent lighting for inside of buildings, I have since purchased Clear LED's and will use inside of building, but for looks I would like to use 1.5 volt light on outside of building, above the doors. we are only talking about between 3-5 light per building per building. by using LED's I can come into building with low voltage, but I would still like to use a circuit for outside only and a circuit for inside. thats where my first question was coming from, but I forgot that LED's operate at low voltage.

on my layout I have a comercially purchased transformer that offers four pins(2 @12volts, and 2  @5 volts).  I realize I need one pin of 12 volts to handle the (10) Tortise switch machines, and was just trying to utilize the rest of power.  so I was going to use a resistor to reduce one of the 12 volt pins to 9 volts, and the other 5 volts down to 1.5 volts, leaving me with a 12,9,5,and 1.5 volts. in my control panel I need 9-12 volts for tortise's and 1.5 for lights. the rest is open for sugestions.     

an electrical question, tonight while soldering a 200 ohm resistor in on the 5 volt wire to try and reduce it to 1.5 volts,  my meter kept reading 5 volts, my question is did I cook this resistor by getting it too hot?    

If you use AC voltages then the circuit with the two diodes in series WILL NOT WORK. When the AC current is flowing the 'other' way the diodes don't conduct and you'll be depending on that dropping resistor to keep the lamps from frying. Not a safe thing.

To use AC you either need to rectify it before sending it to the lamps, or add a second pair of diodes in parallel with the first pair but aiming the other direction. You can use a bridge rectifier (4 pin package) to replace all these diodes with a single package.

Charlie

I think I see what you are trying to do with the DPDT switch.  I assume you do not want to be able to have all the lights, (indoor & outdoor) on at the same time.

It would help to know if the transformer you are using is outputting fully rectified DC voltage.  Knowing the specs of the LED's and light bulbs you will use helps too.

LED's are low voltage, typically requiring no more than 2 volts and only 10 - 30 milliamps of current to emit light.  Because you are using LED's and 1.5 volt lamps, why not just use the 5 volt tap on the transformer for your lighting?

I would run the negative side of the circuit as a buss from the transformer to the bulbs and LED's without switching it. The positive side needs to be switched and have voltage/current limiting in place to protect bulbs and LED's.

I would wire the +5 volt supply to the center of the toggle switch and then run the LED circuit off one side and the bulbs off the other side with the appropriate limiting diodes/resistors in place on each separate circuit.  The DPDT switch will still work for this, but one set of contacts will not be needed.

Remember that LED's have polarity, an anode and a cathode.  The anode must be connected to the positive side of the circuit.  Typically the Anode is the longer of the two leads on the LED.

If you run your light bulbs a little below rated voltage they will last longer and give off a nice yellow incandescent glow in the dark for night running.

Just my two cents.

yes you are correct. I want inside lights on with one side of switch and exterior with other side of switch.

the transformer converts 100-120 volts (wall current) to 2 EA pins of 5 volts DC @ 2 amp, and 2 EA pins to 12 volts DC @ 2 amp. I have each power lead attached to terminal bar and each wire attached to screw 

I thought I needed to reduce each of extra pins to a 9 volt and a 1.5-2 volts with applicable resistors. but after I checked the conumption I wont need to.

 I have lost the LED power Info, I just thought 5 volts was to much initially for the 8 LEDs and 32 1.5 bulbs.  but after I did the consumption rating of all the bulbs(17 mila amps per bulb).  it was relative close to a half an amp. well under the 2 amps, right now  I'm in the process of hooking up a control panel, and lights to layout.     I initially have been testing connections using 2 AA batterys. so I'm at least good with 3 volts, this was done with 5 bulbs.( they were bright but all lit up)  but after I add all the buildings together I should be OK with the 5 volts and a resister. 

Typically an LED might be rated at 3.3 volts @ 20 ma current.  Since you will be using 8 LED's you will have to wire a parallel circuit with dropping resistors because the 5V supply will not be able to drive 8 LED's in a series circuit.

So to be clear, you plan on using a total of 8 LED's for interior lights on one pole of the switch and 32 - 1.5V @17ma bulbs for exterior lighting on the other pole of the switch.  Also the LED's and light bulbs will never be on together.

never be on together.  I think I would recommend using two power supplies and two switches.  I think what Blue is getting at is that trying to turn on all of the LED's and all of the 1.5v bulbs will exceed the power output of your power supply that you have dedicated for lighting.

let me clear the air, some. I plan on using a DPDT or SPDT switch per building, a total of 8 buildings and all on a seperate switch, on each switch one pole will have an LED only, on other pole a max of 5 ea 1.5 volts bulb's.    all coming from a 5 volt source. now remember I have two 5 volt sources.

 I thought I was way under until I actually did power consumption test,  but have realized I actually am good on power. I know I could actually put 1/2 LED's on one of the 12 volt power source. and still be Ok, thats assumming they are rated at 3 volts apiece. but I have planned all along to use a resistor to at lease have a ceiling.

but talking among all of you, I have really gained an insight as to what is needed, but I'm nearing the end of this project, once complete will show pictures, if my small section of layout comes out half as good as my vision it will look good,

It has taken nearly two year's to go from nothing to making all new switches from jigs to making cornerstone kits to all come together as a cattle coral, large meat packing, leather processing facility, complete with sound. it has been a project.