# An interesting problem for all the electronically inclined



## Dave Crocker (Jan 2, 2008)

I was reading the post on the Massoth pulsed smoke unit and found out why I thought mine had failed.
The 5 vdc unit needed almost 7 v to operate.
I put together the circuit below to put out 7.1v and the smoke unit doesn't work.
So I connected the smoke unit directly to my variable supply and it works even at 6.7-6.8v.
The question is WHY?









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## astrayelmgod (Jan 2, 2008)

I get 6.5V, not 7.1. How much heat sink do you have on it? You need at least 4W, whiich is pretty big.


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## Dave Crocker (Jan 2, 2008)

The 7.1 v input does not make the unit run. 
It maybe 6.5v, my power supply gets itchy around there, but it does run on raw DC and not on the higher regulator output.


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## GaryR (Feb 6, 2010)

Maybe it has something to do with how many amps it takes to run the smoke unit. 


GaryR


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## Dave Crocker (Jan 2, 2008)

The smoke unit pulls about 625ma and the regulator is good for 1.5 amps. 
I measure the same current running it with each supply. 
I have both removed the capacitors and changed the output cap to 50uf, neither helped.


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## krs (Feb 29, 2008)

Did you measure the voltage at the output of the regulator while the Massoth smoke unit is connected? 
Do you get 7.1 volts or something close? 

The schematic looks OK. 
If the regulator can't supply the current because your heat sink is too small it will shut down and your output voltage will drop. 

Knut


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## JPCaputo (Jul 26, 2009)

There are a few things. Check the cooling of the 317. 

Also I would suggest a 7805ct 1amp regulator, 5 volt fixed, no external parts, simpler. 

Since your dropping roughly 10 volts at .6 amp. That's 6 watts of heat, and the regulators, either 317 or 7805 has thermal protection and will cut out when overheated. so add a sizable heat sink. 

Also you can parallel the regulators with a .1ohm resistor on the output of each. Splitting the load and heat. 

JP 
The lm780x ct series has multiple voltages available. Also NTE has equivalents for all the range. They have 5, 6, 8, 9, 12, 15 volts.


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## Dave Crocker (Jan 2, 2008)

I measured 7.1v at the output of the regulator and 625ma. It never starts so it is not the regulator shutting down.
I have a large heatsink as I was expecting 4+watts.
I can't use a 5v regulator as the unit will not work at that voltage, even if run from a power supply.
????


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## Mike Reilley (Jan 2, 2008)

Dave....hook it to the smoke generator...turn it on...THEN measure the voltage at the input to the smoke generator with a high impedance meter. You might find that the generator loads the voltage regulator down.


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## mbendebba (Jan 22, 2011)

Dave: I am not electronically inclined, but you may want to check this manual for a clue: http://www.massoth.com/dlbereich/down.php?action=en&kategorie=5&gruppe=56


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## astrayelmgod (Jan 2, 2008)

When those regulators don't start at all, it's because they are seeing too much load, which could be too low a load resistance, or else the load capacitance is too high. 1uF is the recommended value, but I don't know what the input capacitance of the smoke unit is. Take the smoke unit off, and see if you get any output. The 240 ohm resistor is all the load you need to keep the LM317 in regulation. If the output is 6-7V, then you have a problem in the load circuit. Either a short, or too much capacitance. If the smoke unit works when you apply battery voltage directly, then I would say it is too much capacitance.


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## Dave Crocker (Jan 2, 2008)

Mike,
I have three meters connected when I am testing this.
One monitoring the input voltage to the bridge a second measuring the input to the Massoth unit and a third measuring the current the smoke unit is pulling.
The 7.1v from a p/s works, the 7.1v from the regulator doesn't.


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## toddalin (Jan 4, 2008)

The 0.1 uf cap is used "if more than 6" from filter capacitors." Also the 1 uF can be as high as 1,000 mf. Maybe try increasing these values or putting a filter cap in parallel. Also the caps should be aluminum or tantalum.

BTW, I've never had much luck with this regulator either when trying to drive any kind of a load. (Same problems you seem to be having.)


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## Mike Reilley (Jan 2, 2008)

Posted By Dave Crocker on 02 Oct 2011 08:13 PM 
,,,, The 7.1v from a p/s works, the 7.1v from the regulator doesn't. 
What's that mean? What is the voltage at the input to the connected smoke unit?


If the voltage at the input to the connected smoke unit is LOW...it means the smoke unit is loading the regulator down...and it ain't regulating because it can't. If the circuit the regulator is looking into is a very low resistance from the smoke unit, the regulator may NOT be able to get going. 


Think of it this way...what if the smoke unit was a near short circuit. In that case, the regulator can't put out enough current to get the smoke unit input up to 7.1v no matter how much current it passes. If it can't pass enough current...it can't regulate. That's what I mean about not being able to get going.

If you have an output current and voltage meter hooked up, you should be able to compute the input resistance of the smoke unit...which I believe might change as it heats up. If that resistance is low...your regulator can't provide enough current and therefore, you see low voltages at the input...and it dunna work. Does that make sense?


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## JPCaputo (Jul 26, 2009)

If it poses almost a dead short then increases resistance and lowers current when warm to an operating 625ma. You can do a pass transistor. Such as a 3055. Connect the collector to the input, base with a 100 ohm resistor to the output of the regulator, and the emitter to the load. 

This is roughly what is used in most linear power supplies.


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## Dave Crocker (Jan 2, 2008)

Thanks for the input Mike,
The regulator is regulating OK it is putting out 7.1v and supplying 625ma to the smoke unit (the same pull as when it is being fed by my bench supply)
The bench supply makes the smoke unit work but the 7.1v out of the regulator will not.
I wish I had a scope to see what the output waveform looks like.
Something about it is not liked by the Massoth unit.
Todd,
I'll try playing with larger output caps in the morning, I tried 50uf but I'll try some larger.
My input cap is not a tantalum so I'll change that as well.


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## Semper Vaporo (Jan 2, 2008)

Sumpthin screwy here. 

The smoke unit works on WATTS. Which is Voltage times Amperage. If the Voltage and the Amperage is the same in both setups then both setups are producing the same Watts and the same Watts will produce the same result in the smoke unit. 

Maybe your meter is measuing PEAK voltage and PEAK current and you have severe ripple on the regulated supply that is "not working" such that the Wattage is not averaging high enough. The type of filter capacitor (electrolytic, Tantalum, etc.) will make no difference except the physical size and possibly internal leakage causing more power to be drawn from the regulator and causing it to run hotter.


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## krs (Feb 29, 2008)

Posted By Dave Crocker on 02 Oct 2011 10:13 PM 
Thanks for the input Mike, The regulator is regulating OK it is putting out 7.1v and supplying 625ma to the smoke unit (the same pull as when it is being fed by my bench supply) The bench supply makes the smoke unit work but the 7.1v out of the regulator will not. I wish I had a scope to see what the output waveform looks like. Something about it is not liked by the Massoth unit. Todd, I'll try playing with larger output caps in the morning, I tried 50uf but I'll try some larger. My input cap is not a tantalum so I'll change that as well. 
The problem has to be something simpler than the wrong value of capacitors.

Are you measuring the 7.1 volt at the output of the regulator *with the Massoth smoke unit connected*?
I assume the answer is yes. 


Also - take the ammeter out of the circuit completely - that might give you too much of a voltage drop so that the smoke unit won't work.
I have seen that before.

Knut


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## astrayelmgod (Jan 2, 2008)

Take the power supply off the smoke generator, and measure the resistance of the smoke generator. It should be around 7.5 ohms, but plus or minus up to about 2 ohms wouldn't be a problem. 

I second Knut in taking the ampmeter out of the circuit. There have been some extraordinarily badly designed current meters in the past, and not all from shady manufacturers, either. If you have a very small value resistor, like 0.1 ohm, or even lower, you can put that in instead, measure the voltage drop across iit, and calculate the current. I keep 0.1 ohm, 10 W resistors just for that purpose. 

In my experience, LM317s are very reliable, and not fussy about capacitance. I should say, though, that most of my circuits are on proper PC boards, with good grounds.


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## krs (Feb 29, 2008)

If none of the above suggestons get you anywhere, I would take a second careful look at the wiring. 

First of all, get the manufacturers spec sheet of the actual unit you are using to make sure they follow the industry standard pin-outs of this regulator. 
It's different than the other industry standard three terminal regulators like the 7800, 340 and 323 types. 

and then make sure that you have wired it correctly. 

Knut


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## toddalin (Jan 4, 2008)

Posted By Dave Crocker on 02 Oct 2011 10:13 PM 
Thanks for the input Mike, The regulator is regulating OK it is putting out 7.1v and supplying 625ma to the smoke unit (the same pull as when it is being fed by my bench supply) The bench supply makes the smoke unit work but the 7.1v out of the regulator will not. I wish I had a scope to see what the output waveform looks like. Something about it is not liked by the Massoth unit. Todd, I'll try playing with larger output caps in the morning, I tried 50uf but I'll try some larger. My input cap is not a tantalum so I'll change that as well. 

I've always thought that the wiring for this regulator was "screwy."

Look at any voltage regulator that outputs a fixed voltage, and the ratio of C1 to C2 (C1 before the regulator and C2 after the regulator) is typically 10:1 (i.e., C1 is 1,000 mfd and C2 is 100 mfd). The ratio is reported to avoid oscillation and other nasties.

On these things, I continually see the schematic that you posted with a C1:C2 ratio of 1:10 and this always seemed screwy to me.


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## krs (Feb 29, 2008)

For the external capacitors it's best to follow the manufacturers recommendations.
The values shown of 0.1 uf and 1.0 uf are what the manufacturer recommends for a typical application which is what this circuit is.

The type of capacitor can make a difference, this is from the National datasheet:

EXTERNAL CAPACITORS
An input bypass capacitor is recommended. A 0.1μF disc or 1μF solid tantalum on the input is suitable input bypassing for almost all applications. The device is more sensitive to the absence of input bypassing when adjustment or output capacitors are used but the above values will eliminate the possibility of problems.
The adjustment terminal can be bypassed to ground on the LM117 to improve ripple rejection. This bypass capacitor pre- vents ripple from being amplified as the output voltage is increased. With a 10 μF bypass capacitor 80dB ripple rejec- tion is obtainable at any output level. Increases over 10 μF do not appreciably improve the ripple rejection at frequencies above 120Hz. If the bypass capacitor is used, it is sometimes necessary to include protection diodes to prevent the capacitor from discharging through internal low current paths and damaging the device.
In general, the best type of capacitors to use is solid tantalum. Solid tantalum capacitors have low impedance even at high frequencies. Depending upon capacitor construction, it takes about 25 μF in aluminum electrolytic to equal 1μF solid tan- talum at high frequencies. Ceramic capacitors are also good at high frequencies; but some types have a large decrease in capacitance at frequencies around 0.5 MHz. For this reason, 0.01 μF disc may seem to work better than a 0.1 μF disc as a bypass.
Although the LM117 is stable with no output capacitors, like any feedback circuit, certain values of external capacitance can cause excessive ringing. This occurs with values be- tween 500 pF and 5000 pF. A 1 μF solid tantalum (or 25 μF aluminum electrolytic) on the output swamps this effect and insures stability. Any increase of the load capacitance larger than 10 μF will merely improve the loop stability and output impedance.


If all else fails, maybe a post on the Massoth forum will bring some suggestions.

Knut


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## Greg Elmassian (Jan 3, 2008)

There seems to be a little confusion about bypass capacitors, filtering capacitors, and what kind of noise you are trying to filter. 

If you are running your little circuit from 60 Hz ac lines, I don't think there will be very much 500,000 Hz noise !!! 

There's something fundamentally wrong here, I doubt seriously that the regulator is in oscillation, or there is some component causing the meter to not read the output right. 

But, there is no way that the 2 outputs are the same, since the smoke unit is not acting the same. 

So either there is something radically different in the 2 outputs, or there is something not being reported as different in the 2 configurations. 

Greg


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## krs (Feb 29, 2008)

I'm taking the schematic posted at face value:










14.4 VDC at the input to the bridge - so I don't expect much ripple at the input.
Pin numbers for input, output and adjustment terminal are correct for the LM 317
Capacitor values are as recommended by manufacturer
Types of capacitors used I don't know, but with DC as an input I doubt that makes much of a difference

That circuit should work just fine the way it's drawn.

Only possibilities I see is that
a. The 7.1 voltage is measured at the output of the regulator, not the input to the smoke unit, and
b. There is a multimeter connected between the output of the regulator and the input of the smoke unit that drops the voltage below the operating voltage of the smoke unit.

I posted this schematic and the question on the Massoth forum a few minutes ago to see if they come up with anything.

Knut


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## Greg Elmassian (Jan 3, 2008)

I agree completely Knut... 

there is something not "visible" or communicated in the current situation... 

If the regulator was oscillating, it would be obvious. 

I see all this stuff about the 5v unit working better on 7v, but I would want to hear from the manufacturer directly on what is the REAL working voltage range. 

Regards, Greg


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## Trains West (Oct 4, 2008)

what i see math wize with a 240 r1 and a 1 k r2 is 6.46 volts 

looks like you need 214 and 1 k or 240 and 1123 to hit 7.1


even if your meter says you are at 7.1 it does not mean you are.....as some ripple may be read by the meter but not seen as power to the smoke unit


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## krs (Feb 29, 2008)

Posted By Trains West on 03 Oct 2011 04:39 PM 
what i see math wize with a 240 r1 and a 1 k r2 is 6.46 volts 

looks like you need 214 and 1 k or 240 and 1123 to hit 7.1


even if your meter says you are at 7.1 it does not mean you are.....as some ripple may be read by the meter but not seen as power to the smoke unit 




True enough, the calculated value is 6.46 volts.

But if the meter reads high at 7.1 volts at the output of the regulator which seems to be the case, it would also read high when the voltage of the bench supply is measured at 6.8 volts.


Unless of course different meters are used for each of the measurements which may well be the case reading the thread.


Knut

BTW - the ripple at the output of these regulators is negligible.


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## Greg Elmassian (Jan 3, 2008)

Where is this 14.4 volts DC coming from? 

Is it from an unfiltered transformer, just rectified? 

That would explain ripple. 

Greg


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## Trains West (Oct 4, 2008)

depends on if it is just a power pack on both tests or if one is battery and the other is power pack but the fact that calulated is not the same as seen on meter shows a problem


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## krs (Feb 29, 2008)

Posted By Trains West on 04 Oct 2011 06:25 AM 
........but the fact that calulated is not the same as seen on meter shows a problem 

Yes, there is a problem but probably more with the meter than the actual circuit.

I have had these inexpensive digital multimeters myself where I only noticed months after I bought it that it read high on the DC scale.
And that only came up because I was checking an AA battery and the meter read over 2 volts.

Probably something Dave should do - check each of the meters to make sure they read correctly and consistently.
But theoretically the circuit should work the way the schematic is drawn.


Massoth, btw, came back with a reply.
They claim one needs a much bigger capacitor at the output - they suggest 2000uf because the smoke unit doesn't like residual ripple.
I don't agree with that assessment. There is no residual ripple to speak of with these regulators unless the input voltage drops below the minimum required for regulation.
All a large capacitor like that will do is change the source impedance of the supply.
But I would still try a much larger capacitor at the output.


In any case, I think we need to hear back from Dave to see what he tried so far. 
I still think the problem wil be something very simple that was overlooked, something that, as Greg said, is "invisible" from what was posted so far.
The circuit the way it is drawn has been used millions of times before without an issue ....but maybe not with a Massoth smoke unit.

Knut


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## Greg Elmassian (Jan 3, 2008)

I agree... waiting to hear what the source of the 14.4 volt DC supply is. If it's a filtered supply already, then no additional filtering is needed. If not, then it could be. 

Dave can you answer this? 

Greg


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## Semper Vaporo (Jan 2, 2008)

Why the bridge rectifier if the supply is DC?


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## Greg Elmassian (Jan 3, 2008)

If you are running a loco on DC track power, then the polarity can change (so the loco can go in the other direction). 

A bridge as a "front end" to track power makes sense. Just a guess, but probably right. 

Greg


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## Dave Crocker (Jan 2, 2008)

Sorry I didn't get back on this, I spent the whole day yesterday getting my wife's drivers license renewed. 
She has a little short term memory loss and the DMV really wants to verify you are OK to drive. 

The output meter is my old Simpson. I get the same reading with two other DVMs 
The input 14.4v supply is a battery that will run the circuit when I get it working. 
The 6.8v supply is my desk 1A LGB supply 
The board holding the bridge and regulator is a commercial piece I removed from something earlier. 
It didn't have to capacitors, they were added after the original board wouldn't activate the smoke unit. 
I'll get back to it this morning and let you know the results. 
Dave


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## Dave Crocker (Jan 2, 2008)

I just realized what Greg said. 
I am going to run this on DC battery so I will bypass the bridge. 
Dumb me. 
Dave


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## Greg Elmassian (Jan 3, 2008)

Well, that answers the question of ripple, unless a battery charger was attached when testing. No ripple, no need for large filter caps. 

The bridge is unnecessary, but it could not have been causing the problem/difference. 

Down to slight differences in voltage making the unit not run. 

If I was in your situation, I'd put the unit on a variable DC power supply (filtered DC, not a "transformer") and see the voltage range it works with. 

I'll bet there is something strange in it's behavior at these voltages. 

Too bad you aren't close by, we could check it out at my place. 

Greg


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## krs (Feb 29, 2008)

Posted By Dave Crocker on 04 Oct 2011 11:46 AM 
Sorry I didn't get back on this, I spent the whole day yesterday getting my wife's drivers license renewed. 
She has a little short term memory loss and the DMV really wants to verify you are OK to drive. 

The output meter is my old Simpson. I get the same reading with two other DVMs 
The input 14.4v supply is a battery that will run the circuit when I get it working. 
The 6.8v supply is my desk 1A LGB supply 
The board holding the bridge and regulator is a commercial piece I removed from something earlier. 
It didn't have to capacitors, they were added after the original board wouldn't activate the smoke unit. 
I'll get back to it this morning and let you know the results. 
Dave 
Dave -

What you just posted may explain at least some of what we are seeing/measuring.

The 1 amp LGB power packs I have will not provide anything close to pure DC, especially at the lower output voltage there is a fair amount of rectified AC in the output which is part of the design.

So when the meter shows 6.8 volt on the DC scale the peak voltage is actually higher.

And if the Massoth smoke unit has some filtering capacitor on the input, that peak voltage will be stored and seen as a higher voltage by the unit.
So the actual output voltage of the regulator may still be the issue. 


One way to check if that is the issue with the LGB supply you have, or at least part of it, is to connect a large filter capacitor at the output of the 1 amp LGB supply and measure the output voltage with and without a load.

Make sure the polarity of the capacitor is correct.
If the LGB power pack output includes some rectified AC, the capacitor will charge to the peak AC voltage at the output and that's what the meter will read (on DC!) when there is no load.
It's also what the smoke unit would see. 


Knut


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## Dave Crocker (Jan 2, 2008)

Well here's how it stands.
I put a variable resistor in place of the 1K (measured at 1250 ohms which was why the output was high for that resistor pair)
I removed the bridge. 

The smoke unit activated at 7.4v input.
I rebuilt the whole circuit on a breadboard and got the same results.
I ran it directly from a 7.2v battery pack (output was 7.6v) and it ran.
Input and output caps made no change.


I tried a second LGB supply directly to the smoke unit and it activated at 6.6v

I filtered the LGB with a large cap and things changed, the smoke unit would not activate until the voltage was increased to around 7.4v.

OK Greg and Knut, I think you got it right. The LGB supply output was skewing the results.

A nice filtered supply will replace my existing units.

I also tried to get it to work with 4 AA cells as this is supposed to be a 5v unit, no luck.

My 5v unit seems to need 7.4v, astrayelmgod said his unit needed 6.5v.

I'll run it that way and we'll see how long it lasts.
Thanks for all the help.
Dave


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## Greg Elmassian (Jan 3, 2008)

Interesting results. You would think that the Massoth documentation and what we have heard from Mohammed that it needs 5 volts. 

The reality seems to be different than the documentation. 

Greg


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## Dwight Ennis (Jan 2, 2008)

I tried a second LGB supply directly to the smoke unit and it activated at 6.6vI suspect that you were measuring average voltage her, not peak voltage of the ripple.
I filtered the LGB with a large cap and things changed, the smoke unit would not activate until the voltage was increased to around 7.4v.The cap would smooth out the ripple, making the meter read a value closer to the average value (less variance). Anyway, that would be my amateurish guess (that and a buck might get you a cup of coffee).


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## krs (Feb 29, 2008)

Greg,

This idea that LGB 5 volt equipment actually runs with 5 volts is a bit of a myth.


I just didn't want to say too much about it with the Massoth smoke unit because I wasn't sure if it was really designed to operate on 5 volts or actually required the "LGB 5 volts" which is closer to 8 volts.

These smoke units aren't exactly cheap and I didn't want to be responsible for Dave burning one out by applying too high a voltage.

Way back when LGB first came out with constant voltage lighting people in Germany were complaining about 3rd party 5 volt bulbs burning out rather quickly and they asked me to take a look at what was happening.

Turns out that the original regulator that LGB used in their RhB passenger cars with their 5 volt constant lighting actually put out something like 8 volts and this was confirmed by the actual circuit in the car.

Here is the schematic and the changes I proposed at the time - it's in German but the schematic is pretty well universal.

The Zener used at the time was an 9.2 volt Zener and the output voltage of that regulator was just over 8 volts.


I found out later that the 5 volt bulbs that LGB was selling probably had a nominal rating closer to 6 volts - at least that's what some German dealers advertised.

But the 8.1 volt from that regulator would still cause those bulbs to burn out relatively quickly so I changed the Zener in my cars to a 5.6 volt Zener.
That provided pretty close to 5.0 volts to the lights which now were a bit dim but still looked good and at that voltage will last a long time. 













A few years back Massoth developed a switching regulator to replace the old linear one in the LGB cars with lighting.
Quite a complex circuit.
I just went out instead and bought a bunch of these three terminal 5 volt switching regulators for 5 bucks each to eventually replace the old linear ones in my LGB cars.

It would be interesting what the actual output voltage of the regulator is that Massoth sells as a "6 volt regulator"

http://www.massoth.com/en/produkte/8242050.en.php

Knut


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## krs (Feb 29, 2008)

Posted By Dwight Ennis on 04 Oct 2011 04:26 PM 
I tried a second LGB supply directly to the smoke unit and it activated at 6.6vI suspect that you were measuring average voltage her, not peak voltage of the ripple.
I filtered the LGB with a large cap and things changed, the smoke unit would not activate until the voltage was increased to around 7.4v.The cap would smooth out the ripple, making the meter read a value closer to the average value (less variance). Anyway, that would be my amateurish guess (that and a buck might get you a cup of coffee).








In this case, the voltage a DC meter will indicate depends on the load.

Yes, with a DC meter you will measure the average voltage if you just measure the output as is with ripple of some sort, but if you add a capacitor at the output with *no typical load*, only the load of a high impedance meter, you would normally measure the peak voltage.

I say normally because it also depends on the design of the supply, if there is some path within the supply for the capacitor to discharge then you would be back to measure more of an average value - so to be sure one measures the peak voltage that way one needs to add a diode between the output of the supply and the capacitor. 


I think it becomes obvious in this thread that sometimes it's not that easy to measure accurately or that measurements may not be what they seem.
A scope is really very useful in those situations.

Knut


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## Dave Crocker (Jan 2, 2008)

Thank you Knut for your incite on LGB, 
It seems to help show why things were working, or weren't working they way they did. 
It has been interesting, but a scope would have shown the problem immediately. 
thanks for all the help 
Dave


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## Dwight Ennis (Jan 2, 2008)

Which is exactly why I kept my mouth shut until the trouble was resolved. No point in my mucking up the conversation with my limited understanding.







I'm more a mechanical guy, not electrical.


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## mbendebba (Jan 22, 2011)

I am like Dwight, I do not know much about electronics either. 

I have tried a couple of things and I found out that the output of the Massoth 6V voltage regulator does activate the smoke unit. I measured the voltage regulator's output, using a Fluke digital meter and it read at 5.89V, which is within the regulator's design specs ( 5.8-6.2)


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## Dave Crocker (Jan 2, 2008)

No comment Deeewight


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## Greg Elmassian (Jan 3, 2008)

Crazy... Massoth smoke unit 5 volt input does not work at 5v... sounds like the manual should be updated, or the information they are giving customers. 

Does make sense it works (or is designed to work with) the 6v output on Massoth decoders... although some experience here indicates that more than 6 volts may be required. 

It does underscore the feeling I got that Massoth is not really looking at general aftermarket sales, just the configurations they show in the manual. 

That's ok, just interesting.. 

Greg


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## krs (Feb 29, 2008)

Posted By Dave Crocker on 04 Oct 2011 02:42 PM 
Well here's how it stands.
I put a variable resistor in place of the 1K (measured at 1250 ohms which was why the output was high for that resistor pair)
I removed the bridge. 

The smoke unit activated at 7.4v input.
I rebuilt the whole circuit on a breadboard and got the same results.
I ran it directly from a 7.2v battery pack (output was 7.6v) and it ran.
Input and output caps made no change.


I tried a second LGB supply directly to the smoke unit and it activated at 6.6v

I filtered the LGB with a large cap and things changed, the smoke unit would not activate until the voltage was increased to around 7.4v.

OK Greg and Knut, I think you got it right. The LGB supply output was skewing the results.

A nice filtered supply will replace my existing units.

I also tried to get it to work with 4 AA cells as this is supposed to be a 5v unit, no luck.

My 5v unit seems to need 7.4v, astrayelmgod said his unit needed 6.5v.

I'll run it that way and we'll see how long it lasts.
Thanks for all the help.
Dave 




Looks like astrayelmgod is using some sort of power pack with some half-wave rectified AC on it, similar to the LGB power packs.
With those the DC voltage meter reading can be lower than what the Massoth smoke unit actually "sees" as a voltage.

The ML317 will put out a DC voltage with negligible ripple as long as the input is above the minimum for regulation.

Dave - just to close this out, what are the final componenst you used in your circuit?

I assume the two capacitors are as shown at 0.1 and 1 uf - I would have been really surprised if a large capacitor at the output of the rergulator had made any difference.

But you must have changed at least one of the resistor values to get a slightly higher output voltage.

I want to go back to the Massoth forum and post what the problem was, the solution and the final component values.

Knut


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## krs (Feb 29, 2008)

Posted By mbendebba on 04 Oct 2011 08:46 PM 
I am like Dwight, I do not know much about electronics either. 

I have tried a couple of things and I found out that the output of the Massoth 6V voltage regulator does activate the smoke unit. I measured the voltage regulator's output, using a Fluke digital meter and it read at 5.89V, which is within the regulator's design specs ( 5.8-6.2) 
Could be tolerance issue with the smoke unit - or still a measurement issue somewhere.
Fluke makes nice meters that are usually pretty accurate but if the waveform is complex, ie DC and AC superimposed, you won't necessarily get meaningful readings unless you take some additional steps when measuring to separate the AC and DC components.

If Dave posts the two resistor values in his final circuit, one can calculate the nominal and max and min output voltage of that regulator and those values will be dead on for all practical purposes.


Mohammed - what you could is measure the output of the Massoth 6 volt regulator on the Fluke meter with the meter set to the AC range.
Start at the 10 volt or higher range and then go to lower AC ranges to see if you get any reading (or if the meter has auto-ranging just read the value of the AC portion which should be in the millivolt range)


Knut


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## mbendebba (Jan 22, 2011)

Knut: the meter is auto-ranging, it read 16.3mv; can you please explain why the wave form could be complex with both ac and dc components involved? 

BTW: according to the smoke unit's manual, the operating voltage is 5-7V DC.


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## krs (Feb 29, 2008)

Posted By mbendebba on 04 Oct 2011 09:42 PM 
Knut: the meter is auto-ranging, it read 16.3mv; can you please explain why the wave form could be complex with both ac and dc components involved? 

BTW: according to the smoke unit's manual, the operating voltage is 5-7V DC.

16.3mv is good - no problem there.
At least not any obvious problem but I'm sure Massoth has checked the output on a scope during the design cycle.

As to a complex waveform with both AC and DC....yo just measured one.

The 16,3 mv is the residual AC from the switching regulator.
When you use the AC range on a meter, a series blocking capacitor is switched into the measuring circuit to block any DC component and all one measures is the AC component of the output that is left.
However, what this AC actually looks like and what the true values are is another story. AC by definition has a frequebcy component associated with it - typical multimeters "assume" that the AC frequency measured is 60 Hz (cycles/second) and are calibrated to indicate the proper rms value if the AC waveform is a 60 Hz sine wave.
True RMS meters work a bit different, they actually analyse the waveform to provide a correct reading even if the waveform is not a sine wave but they still have some limited frequency range - often only 400 Hz or maybe 1000 Hz. Depends on the meter.

16 millivolts of AC on a 6 volt DC output izs a very small percentage so that is fine. 
But a typical moder railroad power supply gives you an output wave form like this:










This is the output from an MRC Tech 4 power pack I found on the net.
It shows a trapezoidal pulse with DC between the pulses.
The various horizontal lines are different DC levels as the throttle is cranked up - the traces are stored on the scope to show them but in reality there would only be one horizontal line at any time.
The horizontal line is the DC level - it has some ripple on it if you look closely.

This is a good example of a complex waveform with DC and AC components. What do you think a meter would read on the DC range if you measured a waveform like that?
One sure wouldn't have an inkling what the waveform actually looks like without a scope.


Pure DC btw can only be provided y a battery - any DC supply that is derived from AC will have some DC component in it. But if the supply is designed properly, the AC componet will be too smal to have any effect on the connecting circuit.

In the Tech 4 power pach shown in the scope trace above, the pulses are intended and are part of the design - they help to provide good slow speed operation of typical model train motors.


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Back to the smoke unit - now that you tested one that runs well with just under 6 volts but Dave needs well over 7 volts to get his to even work, I wonder what the difference is.

Are there any date codes or lot numbers on these units?
I wonder if these units come from different manufacturing lots or even different manufacturers.


Knut


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## mbendebba (Jan 22, 2011)

Knut: 

I do not see a date code or a lot number on my unit. I have sold a few of thes units along with voltage regulators for use in analog operations and I have not had of any complaints; so I can safely assume they are working as expected. Could the element in Dave's unit have anything to do with the need for 7V? 

BTW- Thanks for the explanation, and I understand the jist of it.


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## krs (Feb 29, 2008)

Posted By mbendebba on 05 Oct 2011 07:48 AM 
Knut: 

I do not see a date code or a lot number on my unit. I have sold a few of thes units along with voltage regulators for use in analog operations and I have not had of any complaints; so I can safely assume they are working as expected. Could the element in Dave's unit have anything to do with the need for 7V?


Mohammed -

What element?
The heating element?


I don't know, can't even come up with an educated guess since I have no clue what the smoke unit circuit looks like.

From other posts here and on other forums it seems the smoke unit circuit is more than just a heating element and a fan.

The Massoth switching regulator will deliver 700ma at 6 volts for a very wide input range. Since it is a switching regulator the heat generated is relatively small - doesn't look like it's intended to use an external heat sink.

The LM 317 that Dave uses is a linear regulator, those typically require heat sinks since the voltage drop and current drawn are dissipated as heat.
If the heat sink is not large enough and the internal temperature of the regulator therfore gets too high, the regulator will protect itself by limiting the current and/or voltage to a safe value.
That is one possibility of what MIGHT be happening.
In other words, the output voltage in Dave's circuit maybe set to 7.4 volts and the meter may read 7.4 volts when the output is measured with just the multimeter, but it's possible that the output voltage drops when the smoke unit is connected because the heat sink is not large enough and the regulator goes into its protection mode.
I'm just guessing here at a possible scenario to explain the "supposed"difference in the voltage requirements of the smoke unit.

But it could be something totally different - hard to determine what the actual situation is via posts on a discussion forum.


Knut


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## Dave Crocker (Jan 2, 2008)

Knut, I'll go and measure the value that I have the pot set at. 
In the last tests, I tried input and output caps but did most of the testing without either as they did not seem to have any effect 
I ordered this unit from Germany but I'll see if I still have the packaging. 
FYI the point at which the internal fan started spinning was when I said it was operating.


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## mbendebba (Jan 22, 2011)

Knut: I was thinking about the heating element in the smoke generator itself (I am just guessing), but your hypothesis about the regulator going into protection mode seems to be more likely.


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## Dave Crocker (Jan 2, 2008)

The 2 resistors measured as 235 & 1100 which says the output of the LM317 should be 7.1v but I measured 7.4 with 3 different meters.
The regulator output stays the same with and without the smoke unit as a load.
Here's the unit.


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## krs (Feb 29, 2008)

Posted By Dave Crocker on 05 Oct 2011 12:11 PM 
The 2 resistors measured as 235 & 1100 which says the output of the LM317 should be 7.1v but I measured 7.4 with 3 different meters.
The regulator output stays the same with and without the smoke unit as a load.





A measured output voltage of 7.4 vots is fine with the resistor values you posted.


I don't know how you derived the 7.1 volts, but that is just the nominal output voltage - if you calculate the output voltage using the posted equation for that regulator and the take the tolerances of the reference voltage into account, and also the small voltage offset due to the adjustment current, you can get a calculated value of up to 7.495 volts and that is still not allowing for any tolerances of the resistor values (measured or otherwise).


Still begs the question why the units that Mohammed has work on 6 volts and yours clearly doesn't.
The difference is way beyond what one might be able to explain with measurement accuracy.


Knut


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