What equivalent DC voltage should I use in calculations?

[chaz]

System is a Lenz with a L150 power supply which is marked as outputting 15V AC.

 

I am about to put indicator LEDs on some cassettes, which will take their power from the DCC bus, and I need to do some calculating to decide on resistor values.

Current consumption will not be a problem as there will never be more than two LEDs lit at any one time.

 

I will be putting a 1N4001 in series with the LED to protect it.

 

My question is what value should I use for an equivalent DC voltage - is 15V a safe bet?

[34theletterbetweenB&D]

It is what the system is set to that determines the peak voltage of the track supply. This is adjustable in the range 11 - 22V, but I haven't found a statement on what the factory setting is; I estimated my set 100 outputting 14V as supplied. Perhaps worth contacting A&H for advice, as unless you have a meter with the capability for square wave you cannot get an accurate measurement.

[chaz]

It is what the system is set to that determines the peak voltage of the track supply. This is adjustable in the range 11 - 22V, but I haven't found a statement on what the factory setting is; I estimated my set 100 outputting 14V as supplied. Perhaps worth contacting A&H for advice, as unless you have a meter with the capability for square wave you cannot get an accurate measurement.

 

Thanks for that.  I had forgotten that the peak voltage can be set by the user. I haven't ever changed the factory setting so I assume that 14V may be a reasonable value, yes?  I am inclined to go with this, do some maths and give it a try. The worst that can happen is one fried LED and I have plenty. I am looking for a bright output as I can't be sure of ambient light levels when the layout goes to shows.

[Ken Anderson]

Why not add a simple 5V regulator for all the LED supply?

[chaz]

Why not add a simple 5V regulator for all the LED supply?

 

Not possible economic Ken. The LEDs will be fixed to the loco cassettes and fed from the track. When they are lit they tell the operator that an empty cassette is powered ready to receive a loco and will prove the chain (usually two wagon cassettes and a loco cassette) is live all the way along. This is to avoid an embarrassing stall when the tail of a departing train stops suddenly in full view.

 

I could fit a VR to all seven loco cassettes but this seems a touch OTT.

[Crosland]

The 15V AC will be rectified to about 20V being available to the command station. If the track voltage can be adjusted (some systems cannot be) then use the value you set if you trust the adjustment is accurate.

 

DCC is nominally a square wave so peak and rms do not need to be considered.

 

The LED will only light on each half cycle (50% of the time) but I doubt you'll notice.

 

Andrew

[chaz]

The 15V AC will be rectified to about 20V being available to the command station. If the track voltage can be adjusted (some systems cannot be) then use the value you set if you trust the adjustment is accurate.

 

DCC is nominally a square wave so peak and rms do not need to be considered.

 

The LED will only light on each half cycle (50% of the time) but I doubt you'll notice.

 

Andrew

 

"The 15V AC will be rectified to about 20V being available to the command station. If the track voltage can be adjusted (some systems cannot be) then use the value you set if you trust the adjustment is accurate."   There's the rub - I have never made any such adjustment - I don't know if it's possible with the Lenz kit (although I seem to remember reading some such reference in a manual). So I don't know what voltage to assume, and no way to measure it  - I have no access to a 'scope  - hence my OP.

 

"DCC is nominally a square wave so peak and rms do not need to be considered."   Yes, I knew that.

 

"The LED will only light on each half cycle (50% of the time) but I doubt you'll notice."   Yes, I knew that too, from previous experience. In any case a slight flicker (at mains frequency?) is not important.

 

Now here's a rule of thumb calculation, based on a value of 15V for the supply.

 

Supply - 15V

voltage drop across protecting diode 0.7V

voltage drop across LED (typically) 2V

so voltage across the resistor will be 12.3V

max' current for a 5mm green LED typically 20mA

using ohms law

R = V / I

R= 12.3 x 1000/20

 

R = 615R

Although one or two of the values I have assumed might be arguable the final value looks credible to me.

 

moving to the "safe side" of 615 I could use the preferred value 680R - I am tempted to smack in a 1K0 and if the LED proves bright enough it's a working solution.

 

Anybody see anything wrong with the above.....?

[Ken Anderson]

Little difference between adding a VR to adding a diode.

[Crosland]

"The 15V AC will be rectified to about 20V being available to the command station. If the track voltage can be adjusted (some systems cannot be) then use the value you set if you trust the adjustment is accurate."   There's the rub - I have never made any such adjustment - I don't know if it's possible with the Lenz kit (although I seem to remember reading some such reference in a manual). So I don't know what voltage to assume, and no way to measure it  - I have no access to a 'scope  - hence my OP.

No need for a 'scope. Ther are two ways to measure track voltage reasonably accurately:

Use a normal DVM between the blue wire and a function wire of a decoder with the function on. Add a couple of volts to account for the rectifier and function output transistor in the decoder.

 

Or,

 

Connect a bridge rectifier to the track (make one from some of the diodes used to protect the LEDs). Add a 100uF, say, capacitor to the DC side to smooth any noise (the DCC will not be a perfect square wave) and measure the voltage with a DVM, again adding a couple of volts.

 

"The LED will only light on each half cycle (50% of the time) but I doubt you'll notice."   Yes, I knew that too, from previous experience. In any case a slight flicker (at mains frequency?) is not important.

DCC is more like 9KHz.

 

moving to the "safe side" of 615 I could use the preferred value 680R - I am tempted to smack in a 1K0 and if the LED proves bright enough it's a working solution.

I would go to at least 1K. Modern LEDs are usually bright enough with only a few mA but it does depend on the ambiemt lighting conditions.

 

Andrew

[chaz]

Little difference between adding a VR to adding a diode.

 

Does not a VR require a couple of discrete components? If so a diode strikes me as simpler (although I agree there is not much to choose between them).

 

NO NEED FOR ANY MORE POSTINGS ON THIS TOPIC FOR ME. THANKS GUYS.

[Ken Anderson]

It was only a suggestion. :-(

[chaz]

Don suggested using a breadboard, I must admit it hadn't ocurred to me (!), but it was a quick way of trying out the LED indicator idea.

 

 

A green LED, a 1N4001 diode and a 1K0 resistor in series. Baseboard is up on end and the circuit model is connected to the DCC bus. The LED is bright enough but not obviously overbright! A quick check confirms that the circuit works the same which ever way it's connected to the bus. It's probably the simplest circuit I have ever modelled on a breadboard.

 

I will try some higher value resistors to see how much the brightness is affected with the current further reduced.

 

It has been suggested that there is no need for the diode, that in practice the LEDs life will not be affected if it is omitted. The cost saving is insignificant but it would make the indicators smaller. Any comments?

 

Chaz

[Phil S]

Re: Protective diode eg 1N4148

When bought in bulk - and that is how I also use them! - these can cost less than 1/3rd of a penny each.

 

'Standard LED's frequently have a REVERSE_BREAKDOWN voltage of about 5volts ... which is considerably less than the TOTAL track voltage available eg 12-24V 0-peak

[As I also work in G Scale, I usually work on the 'worst-case' values for dcc - it is easier and safer to do so]

 

So, 'in theory' when say 20V reverse voltage is applied, and NO CURRENT is flowing; there is no voltage drop across the series resistor, and so ALL the voltage is available as a reverse-voltage acrtoss the LED .... of course, as soon as any flows, the Resistor drops voltage, and the LED reverse voltage would fall ... unless it had just died !

 

So, a 1N4148 diode, with 50V reverse-breakdown voltage (or similar) in series with the rsisitor and LED, WILL provide protection against the reverse failure.

 

IF HOWEVER, 2 LEDs are 'back-to-back, Inverse-Parallel .... even with a White LED which might need almost 5V forward voltage  to operate, an adjacent, reversed diode will ONLY get <5V reverse voltage, and may therefore be considered as 'protected'.  [Obviously, the same applies to a 2-wire Bi-Colour LED used for a Red/Green Searchlight Signal head]

 

HOWEVER: You MIGHT also be using some '12V FLASHING LEDs' - these contain an Integrated Circuit producing the flashing timing ...  and the REVERSE BREAKDOWN is often as LITTLE AS 0.5V in their specs !! .... in this case, I would always use a protective diode in series.

 

WHEN SEVERAL LEDS are used in series, then it might be reasonable to assume that the breakdown voltages can be added ... and therefore gaining a greater protection than the 5V .... but also bare in mind, that with old fashioned christmas light strings ... when 1 bulb fails, the full voltage is 'available' on either side of the failed bulb - a shocking thought 8-)

I spend the extra 1/3p --- even sometimes with  DCC-controlled outputs (which are DC only, of course) - because I construct them in 'modules' and they can be risk of mis-connection - especially between carriages.

 

Finally: Having a stock of Diodes to hand allows them to be used to drop known volts and reduce brightness as required - but don't use 1N4148's in parallel to get a higher current limit - use  higher rating diodes.

[Phil S]

Re: Protective diode eg 1N4148

When bought in bulk - and that is how I also use them! - these can cost less than 1/3rd of a penny each.

 

'Standard LED's frequently have a REVERSE_BREAKDOWN voltage of about 5volts ... which is considerably less than the TOTAL track voltage available eg 12-24V 0-peak [As I also work in G Scale, I usually work on the 'worst-case' values for dcc - it is easier and safer to do so]

 

So, 'in theory'; when say 20V reverse voltage is applied, and NO CURRENT is flowing; there is no voltage drop across the series resistor, and so ALL the voltage is available as a reverse-voltage across the LED .... of course, as soon as any flows, the Resistor drops voltage, and the LED reverse voltage would fall ... unless it had just died !

 

So, a 1N4148 diode, with 50V reverse-breakdown voltage (or similar) in series with the resistor and LED, WILL provide protection against the reverse failure.

 

IF HOWEVER, 2 LEDs are 'back-to-back, Inverse-Parallel' .... even with a White LED which might need almost 5V forward voltage  to operate, an adjacent, reversed diode will ONLY get <5V reverse voltage, and may therefore be considered as 'protected'.  [Obviously, the same applies to a 2-wire Bi-Colour LED used for a Red/Green Searchlight Signal head]

 

HOWEVER: You MIGHT also be using some '12V FLASHING LEDs' - these contain an Integrated Circuit producing the flashing timing ...  and the REVERSE BREAKDOWN is often as LITTLE AS 0.5V in their specs !! .... in this case, I would always use a protective diode in series.

 

WHEN SEVERAL LEDS are used in series, then it might be reasonable to assume that the breakdown voltages can be added ... and therefore gaining a greater protection than the 5V .... but also bare in mind, that with old fashioned christmas light strings ... when 1 bulb fails, the full voltage is 'available' on either side of the failed bulb - a shocking thought 8-)

I spend the extra 1/3p --- even sometimes with  DCC-controlled outputs (which are DC only, of course) - because I construct them in 'modules' and they can be risk of mis-connection - especially between carriages.

 

Finally: Having a stock of Diodes to hand allows them to be used to drop known volts and reduce brightness as required - but don't use 1N4148's in parallel to get a higher current limit - use  higher rating diode

[chaz]

Re: Protective diode eg 1N4148

When bought in bulk - and that is how I also use them! - these can cost less than 1/3rd of a penny each.

 

'Standard LED's frequently have a REVERSE_BREAKDOWN voltage of about 5volts ... which is considerably less than the TOTAL track voltage available eg 12-24V 0-peak [As I also work in G Scale, I usually work on the 'worst-case' values for dcc - it is easier and safer to do so]

 

So, 'in theory'; when say 20V reverse voltage is applied, and NO CURRENT is flowing; there is no voltage drop across the series resistor, and so ALL the voltage is available as a reverse-voltage across the LED .... of course, as soon as any flows, the Resistor drops voltage, and the LED reverse voltage would fall ... unless it had just died !

 

So, a 1N4148 diode, with 50V reverse-breakdown voltage (or similar) in series with the resistor and LED, WILL provide protection against the reverse failure.

 

IF HOWEVER, 2 LEDs are 'back-to-back, Inverse-Parallel' .... even with a White LED which might need almost 5V forward voltage  to operate, an adjacent, reversed diode will ONLY get <5V reverse voltage, and may therefore be considered as 'protected'.  [Obviously, the same applies to a 2-wire Bi-Colour LED used for a Red/Green Searchlight Signal head]

 

HOWEVER: You MIGHT also be using some '12V FLASHING LEDs' - these contain an Integrated Circuit producing the flashing timing ...  and the REVERSE BREAKDOWN is often as LITTLE AS 0.5V in their specs !! .... in this case, I would always use a protective diode in series.

 

WHEN SEVERAL LEDS are used in series, then it might be reasonable to assume that the breakdown voltages can be added ... and therefore gaining a greater protection than the 5V .... but also bare in mind, that with old fashioned christmas light strings ... when 1 bulb fails, the full voltage is 'available' on either side of the failed bulb - a shocking thought 8-)

I spend the extra 1/3p --- even sometimes with  DCC-controlled outputs (which are DC only, of course) - because I construct them in 'modules' and they can be risk of mis-connection - especially between carriages.

 

Finally: Having a stock of Diodes to hand allows them to be used to drop known volts and reduce brightness as required - but don't use 1N4148's in parallel to get a higher current limit - use  higher rating diode

 

"So, 'in theory'; when say 20V reverse voltage is applied, and NO CURRENT is flowing; there is no voltage drop across the series resistor, and so ALL the voltage is available as a reverse-voltage across the LED .... of course, as soon as any flows, the Resistor drops voltage, and the LED reverse voltage would fall ... unless it had just died !"

 

The last part of this quote explains the difference between your 'in theory' and what people have found in practice. As you say "as soon as any [current] flows, the Resistor drops voltage" so the LED will not fail because the protection is simultaneous. Current flow and voltage drop are really expressions of the same physical effect so there can be no question of the LED failing before that lazy old resistor gets off its backside and limits the current......each time the DCC square wave produces a reverse bias the current flow and the voltage levels are instant.

 

I did consider using flashing LEDs but then I would have to add a voltage regulator (and also a rectifier?).

[Crosland]

The last part of this quote explains the difference between your 'in theory' and what people have found in practice. As you say "as soon as any [current] flows, the Resistor drops voltage" so the LED will not fail because the protection is simultaneous. Current flow and voltage drop are really expressions of the same physical effect so there can be no question of the LED failing before that lazy old resistor gets off its backside and limits the current......each time the DCC square wave produces a reverse bias the current flow and the voltage levels are instant.

 

The people that "find in practice" tend to be hobbyists who don't really appreciate the subtleties of data sheets, nor why things are done that might appear uneccessary at first glance.

 

If you choose the resistor to operate the LED with a forward currebt of 20mA, what current flows when the voltage is reversed? Is this current within the LEDs specification for reverse current? Only the data sheet will tell you for sure.

 

Just fit the diodes and be done.

 

Andrew