Voltage drop for Dispatch Board

When you say 1.2 Volts left to right, did you apply the "higher" Accessory voltage on the left, measure the voltage on the left with a meter, then measure the voltage on the right with a meter, and subtract the two readings?

I assume the Dispatch Board was NOT attached?

If so, this is to be expected as you need to run a "reasonable" amount of current thru a diode to get the ~3/4 Volt drop.   Reasonable means, say, 0.1 Amps to 1 Amp which I figure is what a Dispatch Board would draw.

With just a typical digital meter attached to the right side, the current thru the diodes would be extremely low... in the neighborhood of maybe 0.000001 Amps... where the voltage drop thru a diode would be maybe ~1/4 Volt.  So the total drop across 5 diodes (with just a meter attached) would be about 1.2 Volts!

Hope this explains what you are seeing...or else I've got some explaining to do!

Yes, that's exactly what I did, without the Dispatch Board attached.

After I attach the Board, I can always eliminate a pair or 2 of diodes if the reduction turns out to be excessive.

Your guidance is appreciated.

This being a discussion forum and inquiring minds just want to know, Stan's drawing shows 3 sets of diodes wired in parallel. Those pairs are wired  in series with the hot wire.  The picture shows 5 diodes (x2) wired in series then in parallel with the hot wire.  Is this electronically the same?

Why diodes in a series parallel circuit and not a resistor to drop the voltage?

This will no doubt be Too Much Information, but here's are some photos and diagrams from earlier OGR threads that may peel a few more layers of the onion...

For the purposes of dropping voltage, components like resistors and diodes have a well defined relationship between current (Amps) and voltage (Volts).  A resistor is simplest to explain...the voltage drop is exactly proportional to the current flowing thru the resistor.  Double the current, double the drop.  Triple the current, triple the drop.  This "straight-line" relationship is shown as the Red line.

A diode has what is often referred to as a hockey-stick relationship (Orange curve).  In a relatively small zone around 3/4 Volts, you can double, triple, quadruple the current and the voltage drop stays relatively constant.  To be sure, it changes, but nothing like a resistor.  I attempted to show this in the diagram... with the same change in current, the voltage drop thru a diode is relatively small vs. the voltage drop thru a resistor.

A specific example helps.  As I understand it, the Dispatch Board has a light that is always on.  Then you trigger the accessory and some motor or whatever moves the figure.  So the current changes from lights-only to lights + motor.  With a resistor as the dropping device, the additional current when triggered would effect a larger drop and the accessory would see less voltage than with just the light on.  The lights would dim...and I'd think quite noticeably.  To be sure, with the diode method, the voltage drop will change a bit...and you may even see some dimming, but nothing like that with a resistor.  Additionally, with the diode method, you don't have to fuss with "stocking" a variety of resistors or using somewhat spendy variable resistors (aka rheostats).  And in case not obvious, it's the relatively recent (in the long history of O-gauge) availability of 5 or 10 cent diodes that makes the diode dropping method practical vs. the variable resistor method.

As an aside, suppose the OP had used a resistor or rheostat (instead of diodes) and attempted to adjust the voltage drop with JUST a meter attached.  This would have been quite the head-scratcher since there would have been zero measurable drop.  No current, no drop!

Enough about that.  On to implementation.

Good catch on the wiring.  In my diagram I show each pair of diodes joined back-to-back.  The OP shows the 5 upper diodes in a string (in series) only joined at the ends to the 5 lower diodes in a string.  This will behave/perform the same.   Note that in the OP's wiring photo, with the terminal strip method used for interconnections, I believe his method always results in less wire cutting/stripping.  So what's the back story?

Again, here I go again opening the information floodgates :

As it turns out, you can buy so-called bridge-rectifiers which are 4 diodes packaged in a way that makes it useful for AC to DC voltage conversion.  But with a simple "trick" a bridge rectifier can be put into service as an O-gauge voltage-dropper!

Above shows a recycled OGR photo where I assembled 4 bridge rectifiers (8 pairs of diodes) in a voltage dropping configuration.  In this case I used relatively high current bridge rectifiers (10 Amps) to show how the diode dropping method can be used to simultaneously power multiple accessories that have differing voltage needs.  This multiple tap method requires each pair of diodes wired back-to-back rather than only once on each end of the overall string; I hope this makes sense!

I now cease and desist.

Stan, thank you for the explanation about using diodes to drop the voltage.  In the spirit of discussion, even though Dusty asked about using one resistor to drop the voltage, maybe he was thinking of a voltage divider circuit.

My understanding of the voltage divider is that, the tapped voltage (in between two series resistors) stays consistent regardless of current flow.  Even though there's a little more to it, the math to determine the tapped voltage, as I understand it, basically boils down to making the resistor values proportional to the ratios of voltages dropped across each of the 2 resistors.  Granted the resistors would need to have a high enough wattage rating to supply the current required by what ever load is being fed by them (plus the total load the resistors impose).  They would likely produce a fair amount of heat (depending on the load) and perhaps not work as efficiently as using diodes.  So in essence, probably not as practical as using discrete diodes or bridge rectifiers.

Link to Voltage Dividers Explained

Your link to Voltage Dividers is to instructions on how to save a webpage:

...how-to-save-all-the-webpages-linked-from-one...

Whatever.

You are correct that a 2-resistor divider is a technique to create a reduced voltage with excellent precision based on the ratio of the resistor values.

So for example, starting from a 14V AC Accessory voltage, a 2-resistor divider with resistor ratio of 10:4 as shown above would divide/reduce the 14V to 10V AC at the junction of the 2 resistors.

If measured with a voltmeter (which is a negligible load) the meter would read 10 Volts.  And the  contrived "goal" of reducing the voltage by 4 Volts would be met.

BUT.  As soon as the accessory draws any current to operate as an accessory, the output voltage would drop below 10 Volts.  It becomes a "do the math" scenario depending on how much current the 2 resistors draw relative to how much additional current the accessory draws.  In round numbers, what they teach you in electrical engineering class is to set the divider's current draw to, say, 10 times the current draw of the accessory.  Again, you have to do the math but what this does is make the effect of the load 10 times smaller than without the resistor divider (i.e., with just 1 resistor).  The penalty, if you can call it that, is the divider circuit is constantly burning power...and lots of it...just to demote the effect of the incremental current when the accessory itself is triggered.

I'm not satisfied with my explanation but, if I understand what you mean by resistor divider, I don't think it's a practical solution to the Dispatch Board.

Stan, once again you've provided an excellent explanation, I followed it just fine.  Thanks for pointing out my linking error, I've corrected that in my earlier reply.

I understand what you mean now about the load's current draw pulling down the voltage at the tap. I typically see voltage dividers used in circuits with smaller loads and was thinking about it in the context of the signalling circuit implied here in this thread, not so much to run an accessory.  Seems like it might work ok for LED signalling, but not for running a model train size motor or accessory or coil.

If one were to use a voltage divider to power a single typical  3mm 2V 18mA LED, would something like this work?

As you may have guessed I'm not an electrical engineer, so thanks for being patient and taking the time to explain this.

I'm impressed that you applied the voltage divider circuit as instructed...10x the divider current relative to the load current, and proper ratio to divide the incoming voltage to match the 2V load.

However, an LED would not be a good practical use of a voltage divider circuit.

1. Burning 0.2 Amps as the setup or bias current is awfully wasteful of power - that's almost 3 Watts warming up the room (14V x 0.2 A = 2.8 Watts).  And that's for the "privilege" of driving a 2V 18mA LED which is a measly 0.036 Watts (=2V x 0.018A).  Kind of defeats the purpose of using LEDs to save power!  As you correctly observe, the technique is most suitable for signaling (as opposed to power transfer) where currents are small.  And, BTW, a couple of 1/4 Watt resistors will soon be toast if dissipating 2.8 Watts!

2. Note that with 14V AC input, the divider output is ~2V AC.  LEDs are DC devices and don't like to be driven with the wrong polarity.  In this case, since it's "only" 2V AC, the so-called reverse voltage is small and won't damage the LED but a consideration nonetheless.

3. But backing up a bit, I refer to the orange diode "curve" I showed earlier which shows the interesting property where the voltage changes relatively a small amount for a large change in current.  Of course this behavior makes it useful for dropping a relatively constant voltage across a range of currents.  The LED is of course a diode and it too has a hockey-stick like orange curve where the steep vertical line is around 2V (for a red LED).  But brightness in an LED is exactly proportional to current (not voltage).  Double the current, double the "photons" going out.  Triple the current, triple the light output; it's actually fairly phenomenal how exact this relationship.  Anyway, the point is you control an LED by adjusting the current, NOT the voltage.  And since a voltage divider is a technique to adjust/set voltage, it stands to reason that it might not be the best way to control current.  I don't know if this makes any sense but I'm sticking with it!

So if we exit the voltage-divider train, if addressing the specific issue of driving a DC LED from 14V AC, we can turn those off-the-shelf plug-and-play LED bulb replacements used in passenger cars or turnout lamps.  These have been discussed at length on OGR and the "design" is basically the LED itself, a resistor to limit the current, and a diode to convert the AC to DC so that the LED only sees the correct voltage polarity.

In the interest of completeness, if your eyes have not yet glazed over, there's a principle in circuit design which I'm loosely calling reciprocity but with reason.  Since the thread has focused on using back-to-back diodes to drop and AC voltage, how about this:

Why not take a pair of back-to-back LEDs and use the diode voltage dropping method as per the OP's original application?!    So the 2 LEDs become a ~2V AC drop.  Again, the voltage drop will be zero if just measuring the output with a meter...but hook up a load that draws about 18 mA and you'll get the ~2V drop across the LEDs...and they'll light up as a bonus!  And by being back-to-back in opposite polarity, each diode protects the other from seeing more than 2V of reverse voltage.

Stan, thank you for thinking this through and your explanations.  I blame my wattage mistake on midnight math.  I scribbled the current flow for 70 Ohms @ 14V = 0.2 but didn't write the unit Amps.  Later when noting the resistor wattages as an afterthought, I looked at that number (0.2) and forgot to convert it to Watts.   Yes, indeed 2.8 Watts is a lot to burn just to power an LED.

Also, this morning thinking about the over simplicity of my proposed circuit, I realized (as you pointed out) it would also need a diode for reverse current protection in this AC circuit.

Thank you for the alternate circuit suggestion and your continued willingness to help educate the rest of us in our quest for electronics adventures. 

Hopefully CPASAM and Dusty's questions have been answered.

Way above my pay scale!

Thanks gentlemen.