RESISTOR MATH/ CHOOSING THE RIGHT RESISTOR

You don't mention if you have tested the flashing at different voltages for satisfactory operation. At 18 volts, they wil just flash faster... if operation is ideal at 14 volts, you may want to consider anti-parallel diode strings to power both canisters as you can more accurately determine the voltage drop, and the voltage drop is not dependent on the changing load of the bulbs flashing on & off.

Oman posted:

The correct resistor would be more like 20 ohms. Remember that you are trying to reduce the voltage across the lamps which means you are also reducing the current. 4 volts across 16 ohms is still 0.25 A. First calculate the resistance of the lamp at 18 volts (18/0.25 = 72). Then calculate the current of the lamp at 14 volts (14/72 = 0.194). Finally calculate the resistor (4/0.194 = 20.6)

Hmm...  He measured the lamp current at 14 volts, I think his calculations were correct. 

Desert Railer posted:

When this car is operating at 14 volts, my ZW transformer tells me they are pulling .25 amps. 

One thing you're not considering here. You are going to dissipate 1 watt, so you'll want at least a 2W resistor, and I'd probably use a 5 watt resistor.  A 2W resistor will get pretty toasty after a spell, that's why I go a bit higher.

http://www.digikey.com/product...B-16R/16W-5-ND/18660

gunrunnerjohn posted:

Oman posted:

The correct resistor would be more like 20 ohms. Remember that you are trying to reduce the voltage across the lamps which means you are also reducing the current. 4 volts across 16 ohms is still 0.25 A. First calculate the resistance of the lamp at 18 volts (18/0.25 = 72). Then calculate the current of the lamp at 14 volts (14/72 = 0.194). Finally calculate the resistor (4/0.194 = 20.6)

Hmm...  He measured the lamp current at 14 volts, I think his calculations were correct. 

Desert Railer posted:

When this car is operating at 14 volts, my ZW transformer tells me they are pulling .25 amps. 

One thing you're not considering here. You are going to dissipate 1 watt, so you'll want at least a 2W resistor, and I'd probably use a 5 watt resistor.  A 2W resistor will get pretty toasty after a spell, that's why I go a bit higher.

http://www.digikey.com/product...B-16R/16W-5-ND/18660

 

Oops, I missed that.

gunrunnerjohn posted:

Hmm...  He measured the lamp current at 14 volts, I think his calculations were correct.

But is that per lamp, or both? What about figuring for the flashing?  The current draw is steadily changing from both lamps, to one, sometimes none, randomly.

Valid point, and three diode bridges wired as diode pairs would be a better solution, I agree.

A little further clarification:

• The measured amperage of .25amps was for both lights/canister operating at the same time. The lights flash randomly but this was the peak amperage I noted on the ZW transformer. (The lights are illuminated once track power is applied, they flash independently (lamp #402-300)).  On this particular flatcar, the canister slide onto two rails on the topside of the flatcar. ( #6805 flatcar circa 1958) These rails get their power directly from the track the flatcar is on. The positive through a pick-up attached to a truck set and the ground/ return also attached to the other truck. I would rather control both canisters together vice independently since there is little room in the canister to add anything. The underside of the flatcar has plenty of room to add either a resistor or diode. ( I added some pictures from a previous posting... on this forum " 6805 Atomic Energy Disposal Flatcar Truck Removal")
• How do I  make a diode bridge? Would I be using a Zener diode? If so what size diodes would be best? (voltage  and wattage wise) How would I link/wire them together? (keeping in mind they have an orientation)
• Thanks again

I'd use a few 1A diode bridge rectifiers, they're connected up as follows.  Connect the + & - terminals together, and daisy-chain the bridges using the AC terminals, it doesn't matter which AC terminal goes in which direction.  You can simply stack them like this to get the desired AC voltage drop, each bridge rectifier drops about 1.3 to 1.4 volts, to the two shown would drop 2.6 to 2.8.  In your case, you'd wire up a 3rd one to get roughly 4V.  The 1A bridges are pretty small, about 3/8" square and less than 1/4" thick.

If using diode drop what about your initial idea to use back-to-back (in series) 1 Watt Zener diodes?  Not sure why you propose 5.6V and 6.2V diodes.  A pair of 3.3V 1 Watt Zeners (10 for $1, free shippiing on eBay) would drop 4V which I thought was your objective.  A pair of 1W diodes effectively becomes 2 Watts as heat is split evenly between the two when dropping AC.  Just seems this is far less wiring/assembly than multiple bridge rectifiers.

Stan, I never actually thought about using Zener Diodes, but that's a good idea.

Thanks Stan...From a post about a year ago, you mentioned the Zener Diode as an option to a bridge rectifier, you even gave an example of various diodes and how they could be used together to drop voltage. That is where I got the idea from. (something small and compact...and easy to add into the circuit)  I choose a 5.6 volt diode for two reasons: I thought that it would provide a bit of under voltage (the lamp is rated at 14 volts) and still give me the same lighting effect. (The flatcar uses what I would consider a  special lamp and I wanted to maintain its lifespan by not powering it to the maximum voltage.) Secondly,at the time I wrote the post, I was unaware of what sizes Zener diode come in. ( I just used your example from the previous post looking for a jumping off place to start the conversation)

After doing some internet searches on  Zener Diodes, it appears  that these type of diodes can be used to either "clip" off a voltage or can be used to regulate a circuit. From a oscilloscope/ sine curve perspective, what is happening in your example when you just add the two Zener diodes in series to the "hot" leg of the AC circuit?

It would appear, I could build a simple constant voltage 12 or 13 volt circuit (common Zener sizes) (providing my input voltage is at least two volts higher) by adding a resistor on the hot leg then having the diodes oppose one another and span both the hot and return leg. On an oscilloscope, this type of circuit flattens both the top and bottom of the sine curve at the chosen voltage. I am not sure how much heat would be produced this way. Would I "beef up" the resistor as GRJ suggests to handle the voltage drop as well as"beef up " the wattage on the Zener diode to 3 or 5 watts vice a 1 watt which is what I calculated I would be using up?

Thanks again for your help.

If you were dropping say 5V across your two Zener diodes at .25 amps average current, that's 1.25 watts dissipated in the two Zener diodes.  Using my basic rule of thumb that the power rating should be at least double the expected load, I'd be looking at the 2 watt Zener diodes to avoid them getting toasty.

That's some remarkable preparation you before posing your questions! 

So to respond in kind, I'll give you a more thorough answer.  Here's a simulation of what's going on for the "drop" method vs. the "clamp" method.  Since the lamp is a resistive load, what we're after is the RMS voltage at the terminals of the bulb.

The blue line is the 18V AC from your transformer.  This is 18V RMS so it peaks at 25V in both polarities. 

Inserting a pair of 6.2V Zeners inserts a 6.9V drop and yields the orange line.  If the input voltage at any time drops below 6.9V in either polarity, the output is 0 since dropping the voltage cannot change its polarity.  So you get the flat-region around the zero-crossings.  As it turns out, the RMS output is 12V.  Yes, I stuffed-the-ballot-box to make it come out this way!

Inserting a pair of 13V Zeners inserts a 13.7V clamp and yields the yellow line.  If the output voltage at any time attempts to exceed 13.7V in either polarity, the output is clamped to 13.7V.  So you get the flat-region around the peaks of the incoming voltage.  As it turns out, the RMS output is also 12V - again I stuffed-the-ballot box to force the outcome.

So in the case of the drop method, simply subtracting the 6.9V drop from the 18V input under-estimates (11.1V) the actual RMS output voltage of 12V.  And in the clamp method, simply using the 13V clamp over-estimates the actual RMS output voltage.

Note that I chose 6.2V and 13V Zeners since these are common values.  Nearby values would be 5.6V or 6.8V, and 12V or 14V respectively.  In each case the effective drop or clamp voltage for back-to-back Zeners in series is the sum of the Zener voltage and 0.7V.

Cutting to the chase.  I suggest the drop method for one simple reason.  With the clamp method if the bulb fails and stops drawing current, the diode clamp will have to absorb all of the power that previously went to the bulb.  So if, for example, the bulb draws 1/4 Amp, that same 1/4 Amp would now flow into the clamp...in round numbers 14V x 1/4 Amp = 3.5 Watts.

I have no beef with GRJ's suggestion to 2x the power ratings of components.  But I'll suggest that if you lower the voltage applied to the bulb, the 0.25 Amp you saw at 14V will also drop.

Using the Zener diodes in the clamp-mode configuration also requires you to drop all that voltage in the resistor, so that's cookin' as well.

While we're on the subject of using Zener diodes to drop voltage. How much current can they handle, 1/2A, 1A, 20A, 50A as much as 75A? Or, is the rectifier method better for higher amperage? Just curious.

The amount of current they handle is dependent on the voltage they're dropping.  If you are dropping 10 volts and you have half an amp of current, you are dissipating 5 watts.  OTOH, if you are dropping 2 volts with half an amp of current, you only have 1 watt of dissipation.

Truthfully, if you have high currents and need to drop a lot of voltage, a switching regulator would probably be a better choice if you can use DC on the load.  For AC, if you want to drop 20 amps, use a transformer.