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I took this out of a locomotive I was working on and failed to note where the leads went.? Would someone who knows more electronics than I do (It won't take much) please advise where the leads to the motor and to the bulbs should go?
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 Stan Stokrocki
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Hi Stan --
This is an easy diagram to understand.? ?Thanks for posting.? ?I seem to recall that the motor is an essential ingredient?in the circuit, or you have to use a "ballast" resistor.? ?Am I correct, or just blowing smoke?
Bill Winans? --------------------------
 Stan Stokrocki
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Back in my straight DC days, I
installed a constant lighting board that permitted the
numberboards on my S2 switcher to be on for both forward and
reverse along with the headlight and backup? for the correct
direction.? I thought it was a -wizard of sorts but I had made it
about a 1/8 too large to fit the narrow hood.? So I tried it again
compressing the size a bit.? That board is still in the hood of
that engine scans lights unfortunately as I know I'd be blowing
smoke!?
At the time PFM sold a series of tiny
rivets (sockets), so by using just some phospher bronze wire as
plugs, everything was removeable.?
It worked really well--found out later
that PFM actually published a booklet with that and other circuits
for their sound/lighting installations.?
I sort of miss the PFM system but
sometimes we throw what was great for something that this old man
forgets!
Bob Werre
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Hi Stan --
This is an easy diagram to understand.? ?Thanks for
posting.? ?I seem to recall that the motor is an essential
ingredient?in the circuit, or you have to use a "ballast"
resistor.? ?Am I correct, or just blowing smoke?
Bill Winans?
--------------------------
Stan Stokrocki
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Thanks- a little searching found the diagram. Yes, you need the motor or a ballast lamp/resistor to pull a limited range of current through two diodes. A pair in series drops .7v +.7v, which will light a 1.5v mini lamp. Stan Stokrocki
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I noticed that an LED wired directly across DCC rails will light no matter which lead goes to which rail.? Is this because of the split phase version of DC which sort of mimics AC?? So the diode is only responding to the half wave...does that mean it is at half intensity as compared to pure 12v DC?
?
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AC isn't being mimicked.? It IS AC.? It just happens to be a square wave.? There is a lot of confusion about this due to some other naming conventions.? Most notably "Bi-polar DC" which is a term also used elsewhere in electronics.? By definition, DC cannot be bipolar, so it is just a descriptive term.
I'm pretty sure that the actual light output would cut in half as the waveform has a 50% duty cycle.? Not necessarily compared to 12 VDC, but whatever the RMS voltage of the AC waveform is (which is the same as peak voltage since it is square wave) which is usually a bit higher than 12 volts for DCC unless for N scale.
I don't know if the human eye, or brain, would perceive it as half brightness, however.
To get way down in the weeds, LEDs are typically specified with a maximum reverse voltage of 5 VDC.? What saves them from serious damage is the series resistor.? However, damage can accumulate over time from a sort of micro damage to areas of the junction.? It is probably highly unlikely anyone would see this get to the point of a failure.? People do it all the time.? But it is a circuit that violates a specified limit, and an electronics engineer is likely to avoid a design applying excessive AC to LEDs.
Charles E. "Chuck" Kinzer
?
On Wednesday, February 26, 2025 at 10:10:16 PM PST, Kim Hartshorn via groups.io <w.kim.hartshorn@...> wrote:
I noticed that an LED wired directly across DCC rails will light no matter which lead goes to which rail.? Is this because of the split phase version of DC which sort of mimics AC?? So the diode is only responding to the half wave...does that mean it is at half intensity as compared to pure 12v DC?
?
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It is AC. And the rails act as a transmitting antenna. If you tried to turn on your AM radio while your DCC equipment is on, the noise from the DCC overrides the strongest AM station! Regards,
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On Feb 27, 2025, at 2:16?AM, Charles Kinzer <ckinzer@...> wrote:
?
AC isn't being mimicked.? It IS AC.? It just happens to be a square wave.? There is a lot of confusion about this due to some other naming conventions.? Most notably "Bi-polar DC" which is a term also used elsewhere in electronics.? By definition, DC cannot be bipolar, so it is just a descriptive term.
I'm pretty sure that the actual light output would cut in half as the waveform has a 50% duty cycle.? Not necessarily compared to 12 VDC, but whatever the RMS voltage of the AC waveform is (which is the same as peak voltage since it is square wave) which is usually a bit higher than 12 volts for DCC unless for N scale.
I don't know if the human eye, or brain, would perceive it as half brightness, however.
To get way down in the weeds, LEDs are typically specified with a maximum reverse voltage of 5 VDC.? What saves them from serious damage is the series resistor.? However, damage can accumulate over time from a sort of micro damage to areas of the junction.? It is probably highly unlikely anyone would see this get to the point of a failure.? People do it all the time.? But it is a circuit that violates a specified limit, and an electronics engineer is likely to avoid a design applying excessive AC to LEDs.
Charles E. "Chuck" Kinzer
?
On Wednesday, February 26, 2025 at 10:10:16 PM PST, Kim Hartshorn via groups.io <w.kim.hartshorn@...> wrote:
I noticed that an LED wired directly across DCC rails will light no matter which lead goes to which rail.? Is this because of the split phase version of DC which sort of mimics AC?? So the diode is only responding to the half wave...does that mean it is at half intensity as compared to pure 12v DC?
?
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So this is where,?i think, there is confusion.? My understanding...and I could be totally wrong, is that DCC puts a half wave on each rail so that one rail, for the sake of argument, is a 6v square wave from the 'top' of the wave to neutral.? In other words 30 pulses and the other rail is the bottom half of the wave -6v from neutral to the 'bottom' of the wave.Each 180 degrees out of phase with each other.? And wiring across the waves results in 12v AC.? In other words there is no 'neutral' rail with a full square +6 to -6 wave on the other rail.? Each rail by itself represents a square wave DC current to neutral, only when bridged do we get AC.
Does this sound right?
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Hey there
GREAT QUESTION?
Looking forward to the answer!!!!!
Roger Haag. ? ? ? S scale man
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On Feb 27, 2025, at 9:47?AM, Kim Hartshorn via groups.io <w.kim.hartshorn@...> wrote:
?
So this is where,?i think, there is confusion.? My understanding...and I could be totally wrong, is that DCC puts a half wave on each rail so that one rail, for the sake of argument, is a 6v square wave from the 'top' of the wave to neutral.?
In other words 30 pulses and the other rail is the bottom half of the wave -6v from neutral to the 'bottom' of the wave.Each 180 degrees out of phase with each other.? And wiring across the waves results in 12v AC.? In other words there is no 'neutral' rail
with a full square +6 to -6 wave on the other rail.? Each rail by itself represents a square wave DC current to neutral, only when bridged do we get AC.
Does this sound right?
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This is where some of the confusion comes in.? The way Digitrax creates the waveform (and it is not the only possible way) is to hold one "Rail A" at common and apply +14 VDC (approx) to "Rail B."? Then to hold "Rail B" at common and apply +14 VDC (approx) to "Rail A."? The result ACROSS THE RAILS is an AC voltage that is 28 volts peak-to-peak, 14 volts peak, and 14 volts rms.
Some argue that there can be no AC "because there is no negative voltage anywhere."? This is true relative to the power supply common, but the DCC voltage is NOT relative to that common, but from rail to rail.
Charles E. "Chuck" Kinzer
On Thursday, February 27, 2025 at 04:07:29 AM PST, Anthony Salvate via groups.io <n1tks@...> wrote:
It is AC. And the rails act as a transmitting antenna. If you tried to turn on your AM radio while your DCC equipment is on, the noise from the DCC overrides the strongest AM station! Regards,
On Feb 27, 2025, at 2:16?AM, Charles Kinzer <ckinzer@...> wrote:
?
AC isn't being mimicked.? It IS AC.? It just happens to be a square wave.? There is a lot of confusion about this due to some other naming conventions.? Most notably "Bi-polar DC" which is a term also used elsewhere in electronics.? By definition, DC cannot be bipolar, so it is just a descriptive term.
I'm pretty sure that the actual light output would cut in half as the waveform has a 50% duty cycle.? Not necessarily compared to 12 VDC, but whatever the RMS voltage of the AC waveform is (which is the same as peak voltage since it is square wave) which is usually a bit higher than 12 volts for DCC unless for N scale.
I don't know if the human eye, or brain, would perceive it as half brightness, however.
To get way down in the weeds, LEDs are typically specified with a maximum reverse voltage of 5 VDC.? What saves them from serious damage is the series resistor.? However, damage can accumulate over time from a sort of micro damage to areas of the junction.? It is probably highly unlikely anyone would see this get to the point of a failure.? People do it all the time.? But it is a circuit that violates a specified limit, and an electronics engineer is likely to avoid a design applying excessive AC to LEDs.
Charles E. "Chuck" Kinzer
?
On Wednesday, February 26, 2025 at 10:10:16 PM PST, Kim Hartshorn via groups.io <w.kim.hartshorn@...> wrote:
I noticed that an LED wired directly across DCC rails will light no matter which lead goes to which rail.? Is this because of the split phase version of DC which sort of mimics AC?? So the diode is only responding to the half wave...does that mean it is at half intensity as compared to pure 12v DC?
?
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On Wed, Feb 26, 2025 at 11:16 PM, Charles Kinzer wrote:
To get way down in the weeds, LEDs are typically specified with a maximum reverse voltage of 5 VDC.? What saves them from serious damage is the series resistor.? However, damage can accumulate over time from a sort of micro damage to areas of the junction.
Seems pretty easy to add a diode to the circuit to protect the LED.
-Michael Eldridge
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Bob Werre
Roger Haag