Model Railway Forum banner
1 - 10 of 10 Posts

·
Ian Wigglesworth
Joined
·
750 Posts
Hi folks,

I was messing about today, couldn't really get focused on anything, and so measured the rail voltage on my little "Thomas&Friends" layout.

This is being powered by the Bachmann EZ-command, the power supply states an output of 16vAC.
I measured this at 17.8 volts, so a little out there then!

The rail voltage was 20v!

It's taken me ages to find the post that Richard made about using diodes to lower the voltage, these are now on order!

I reckon that rail voltage maybe a little on the high side for N-gauge models!

Cheers
 

·
Registered
Joined
·
1,647 Posts
QUOTE (wiggy25 @ 8 Jul 2008, 18:54) <{POST_SNAPBACK}>Hi folks,

I was messing about today, couldn't really get focused on anything, and so measured the rail voltage on my little "Thomas&Friends" layout.

This is being powered by the Bachmann EZ-command, the power supply states an output of 16vAC.
I measured this at 17.8 volts, so a little out there then!

The rail voltage was 20v!

It's taken me ages to find the post that Richard made about using diodes to lower the voltage, these are now on order!

I reckon that rail voltage maybe a little on the high side for N-gauge models!

I too found the EZ PSU kicks out quite a high voltage, and a crude AC meter on the track was 18.6v.

I found the EZ works a fine on a lower voltage AC PSU. I've tried 15v, 12v and 9v AC. I was happiest with a 12v supply, but it seemed OK on all of them (except for my internal mods which are voltage dependent, but doubt you've been hacking inside!). You can get a 12v AC supply power brick fairly cheaply.

I don't know what you were planning with diodes, but my EZ is very unhappy on DC (either just rectified AC or from a switched mode PSU). The EZ either refuses to work (one polarity), or goes haywire (the other, sending locos off at random high speeds). It must be the one exception to the "any DCC controller will be OK with a DC PSU" rule. I doubt my internal mod could cause the behaviour (its only a relay which swaps the standard control pot to a remote one on a wander lead when the wander lead is connected).

- Nigel
 

·
Ian Wigglesworth
Joined
·
750 Posts
The diodes are to reduce the voltage.

A Mr Johnson of this parish posted a very useful bit of information.

If you solder some diodes together in series, say 6 in a line, then repeat, you will have two lines of 6 diodes.
Now turn one of the line of diodes round to face the opposite way then solder the lines together in parallel at each end.

This diode string is then placed in one line of the power supply lead, the diodes give a volt drop across them which lower the voltage.

Just bought 100 of them for £1.90 free P&P off ebay, so cheapest easiest way of doing it, will hopefully be playing about with that lot when it turns up!

Cheers
 

·
Paul Hamilton aka &quot;Lancashire Fusilier&quot;
Joined
·
844 Posts
My understanding is that the DCC signal is a square wave form not a sine wave like AC and as such there will be some discrepancy when using an AC voltage meter like a multimeter to measure it. Osciloscopes I believe are the best diagnostic tool and look quite nifty too!
 

·
Registered
Joined
·
825 Posts
QUOTE (Lancashire Fusilier @ 9 Jul 2008, 00:59) <{POST_SNAPBACK}>My understanding is that the DCC signal is a square wave form not a sine wave like AC and as such there will be some discrepancy when using an AC voltage meter like a multimeter to measure it. Osciloscopes I believe are the best diagnostic tool and look quite nifty too!
The frequency is also very different. Hobbyist multimeters are only calibrated for 50Hz sine waves on the AC range. A meter needs to be "true-RMS" reading.

Cheaper than a scope or a true-RMS meter is to connect a loco decoder to the track and measure (with a normal DC multimeter) the voltage from the blue wire to a function output with the output on. This will be only a little lower than the track voltage. Voltage drop through the decoder's rectifier will be quite small as you are drawing virtually no current.

Andrew Crosland
 

·
Ian Wigglesworth
Joined
·
750 Posts
Thanks Andrew for that!
I will check that when my decoder arrives.

My IN4004 diodes have arrived!
Just as a test I only used 1 at first, so basically two diodes connected in parallel with one end going to the PSU the other end going to the EZ-command. In one lead only! the other lead runs straight from PSU to EZ-command.

With one in circuit it's only dropping by 0.3volts!
It means I need 18 diodes in each string to drop the voltage down to 12.5volts.
I know that there will be a difference in electronic components, but I thought the drop would be about 0.7 volts, to have it less than 50% of that seems high.

It's no problem I bought 100 of them and at £1.90 it's not broken the bank, I just wanted to make sure that this seems ok.
 

·
Just another modeller
Joined
·
9,983 Posts
QUOTE (wiggy25 @ 12 Jul 2008, 21:57) <{POST_SNAPBACK}>Thanks Andrew for that!
I will check that when my decoder arrives.

My IN4004 diodes have arrived!
Just as a test I only used 1 at first, so basically two diodes connected in parallel with one end going to the PSU the other end going to the EZ-command. In one lead only! the other lead runs straight from PSU to EZ-command.

With one in circuit it's only dropping by 0.3volts!
It means I need 18 diodes in each string to drop the voltage down to 12.5volts.
I know that there will be a difference in electronic components, but I thought the drop would be about 0.7 volts, to have it less than 50% of that seems high.

It's no problem I bought 100 of them and at £1.90 it's not broken the bank, I just wanted to make sure that this seems ok.

Hi Wiggy/Ian

* on the track, use Andrews measurement - the output at blue and white and meter set to DC to check the real rail voltage. I could be wrong but it won't be as high as 20v measured that way I think.

* for the tranny, the voltage will be a no load voltage if its not connected to a working EZ command. Either add a resistor f appx 500 ohms across it and measureit with meter set to AC OR connect it up and then measure it with a loco running - also with meter set to AC.

This will be more realistic as "no load" on a low cost AC transformer will always be a couple of volts higher than a the voltage with some load. That "Loaded Voltage" is the one you need to drop a little..... so its probably less diodes than U think.

Richard
 

·
Registered
Joined
·
825 Posts
QUOTE (wiggy25 @ 12 Jul 2008, 13:57) <{POST_SNAPBACK}>My IN4004 diodes have arrived!
Just as a test I only used 1 at first, so basically two diodes connected in parallel with one end going to the PSU the other end going to the EZ-command. In one lead only! the other lead runs straight from PSU to EZ-command.

With one in circuit it's only dropping by 0.3volts!
It means I need 18 diodes in each string to drop the voltage down to 12.5volts.
I know that there will be a difference in electronic components, but I thought the drop would be about 0.7 volts, to have it less than 50% of that seems high.
How did you measure the voltage, what kind of load?

The figure of 0.7V is often used for silicon diodes but it's only a typical value. At their rated maximum current of 1A, a 1N4001 will actually drop 1V.

Try it with the layout running a representative set of locos.

Andrew
 

·
Ian Wigglesworth
Joined
·
750 Posts
Richard & Andrew,

Thanks for the advice.
I was not measuring the output under load.

I have now connected it all up and running two locos on the continuous loops, the power supply was reading 17.7volts AC going in to the EZ-command.

After connecting the 10 diodes in each string from the PSU to the EZ-command the voltage has dropped to 12v while running the two locos.

Problem is nothing will run, well it will but very slow and jerky and then stop!

I will take a couple of the diodes out which will leave 8 diodes in each string this should leave the voltage at about 13v.
I will then have to clean the track again and reset CV2 and CV5 so the locos all run nicely.

Thanks again.
 

·
Just another modeller
Joined
·
9,983 Posts
QUOTE (wiggy25 @ 14 Jul 2008, 18:54) <{POST_SNAPBACK}>I will take a couple of the diodes out which will leave 8 diodes in each string this should leave the voltage at about 13v.

Thanks again.

***8 .....from memory the number I recommended day one in an earlier thread


I'm pleased you went through the process though - there's a bit of good learning & info created by it for others.

* Diode voltage drop is like all other voltage drops - it depends on the current loading
* AC transformer voltages always read high with no load so need measuring under load
* Track voltages are the same - you cannot see any voltage drop without a reasonable load on the track
* Voltages are often too high as bought but for stability, there will be a lower limit too.
* Measuring the output voltage of a function with the meter set to DC is the easiest and most accurate method

Keep experimenting and asking.... its good for the whole forums knowledge

kindest regards

Richard
DCCconcepts
 
1 - 10 of 10 Posts
Top