# Smoke units



## Madman (Jan 5, 2008)

This has been covered numerous times in the past. Stepping the voltage down, in an 18 volt system, to 5 volts to power a 5 volt smoke generator. While we are here, does anyone have the diagram and parts need ed to do so? 
Now, on the other hand, can voltage be stepped up to an 18 volt smoke generator, so that it will produce smoke without making the loco travel at warp speed? I am talking mostly about LGB Stainz locos.


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## Greg Elmassian (Jan 3, 2008)

5 volt regulator, but you need to know the current you need. 1 amp regulator circuit and regulator and parts and circuit all available at Radio Shack. LM7805 is the regulator part number. 

2 caps, one regulator, input, ground output... use a heat sink (again need to know the amperage you are drawing). 

Yes, voltage CAN be stepped up, but you need a DC to DC inverter, not cheap, better and I believe cheaper to change out to lower voltage smoke unit. 

Regards, Greg


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## Madman (Jan 5, 2008)

Thanks Greg. Would you have the wiring diagram for the voltage reducing system? Would you please label the components. I'm not familiar with the symbols.


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## Greg Elmassian (Jan 3, 2008)

You can leave off the transformer (the squiggly lines) the diodes are necessary, but you can buy the 4 diodes all in one package at RS, called a "full wave bridge" comes with 4 leads, just like the 4 connections.


The first component is a 25 microfarad 25 volt electrolytic capacitor... the next component is the 7805 5 volt regulator itself, and the last component is a 0.1 microfarad disc capacitor, any voltage over 5 volts is fine.

RS will have all your components.

NOW, you have to determine the current rating to make sure all will work. You will need to google and research how much current your 5 volt unit draws, or if you have the model number, possibly some people will have the knowledge here on the forum.

Radio shack has this schematic on the back of the regulator ic, and you should google "7805 5 volt regulator", there's tons of stuff online.

Regards, Greg


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## Madman (Jan 5, 2008)

Thanks Greg. I'll let you know how I make out.


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## K27_463 (Jan 2, 2008)

Even with a heat sink a single 7805 will run at or beyond its design limits when driving a current load such as a smoke unit The combination of high drop voltage ( 18 to 5 is too much) and significant current load will cause the regulator to shut down in a short time. I have tried parallel 7805, works sort of OK but not really well due to imbalance. I have tried the same circuit with a 7812 (12 v out) then driving a 7805 in cascade , this seems to work generally, but takes a lot of space( both regs need heat sink). I have also tried numerous other solutions, all have drawbacks. If you do this, be ready to spend some time and effort getting it all to work. 78 series regs are 1 amp devices on a good day. The RS stuff varies wildly in performance. 

Jonathan/EMW


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## Greg Elmassian (Jan 3, 2008)

Yep, thanks for underscoring my request for the current load. I had the same issue trying to run a 12 sweeper motor from my 24v DCC track.. I would up putting an appropriate 20 watt resistor in the open air and did it the easy way... of course you can burn your fingers if you touch the resistor. 

The basic law of physics is the regulator is going to dissipate power in watts that equal the voltage dropped times the current. 

Dan, you REALLY have to let us know how much current the smoke unit draws. 

Regards, Greg


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## K27_463 (Jan 2, 2008)

Well, the typical Seuthe (LGB) style 5 volt smoke unit consumes 250- 350 ma under normal temp conditions. They can go higher. So, we get 13 volts drop times 300 ma on average. So, you can see the problem... 

Jonathan/EMW


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## Greg Elmassian (Jan 3, 2008)

Maybe George S. will chime in here, but I believe if you set a lm317 up as a current regulator, the power dissipation is much less. 

Using the regulator to get down to 5 volts gives you 3.9 watts, could use a big heat sink and forced air flow. (read: not practical). 

That's why, for example, Aristo uses a switching regulator to keep heat down. A series regulator has the above problem. 

Dan, you have not indicated track or battery power. If battery power, select a smoke unit to be close to the supply voltage. 

If you want to run a 5 volt smoke unit from DC track power, then, a big heat sink and a little fan is one option. A more compact but expensive option would be a switching regulator, most likely impractical in terms of cost. 

I'd try the regulator, heat sink and fan.... or maybe this: http://www.powerstream.com/dc6.htm 

This one takes 30v input and is apparently a switcher... 2.2 amp output you will have to look around, most car type units don't have such a high input voltage.. 

Regards, Greg


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## ConrailRay (Jan 2, 2008)

These guys have a few versions of switching regulators. Believe I put one in my Spectrum Radio at one point to save some juice/heat. 
http://www.dimensionengineering.com/DE-SW050.htm 

-Ray


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## Madman (Jan 5, 2008)

How about this item, http://www.allelectronics.com/make-a-store/item/DC-49/DC-DC-CONVERTER-TYCO-QW010A0A1-H/1.html

I just noticed the minimum voltage input seems to be 36 volts. 


Or perhaps this one. http://cgi.ebay.com/DC-DC-Converter...5adb01ec68

Wouldn't it do what I want to do without all the pats needed in Greg's diagram?


Here is another one with adjustable input voltage http://cgi.ebay.com/DC-5-30V-Out-DC...1c1458d165 

I run track power, about 10-14 volts, depending on the locomotive and consist.


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## Tom Bray (Jan 20, 2009)

I haven't implemented the smoke units in my F7 conversion to DCC. The MTH uses a 5V smoke unit but used more current than the Zimo would deliver at 5V. I found PT78ST105H switching regulator at DigiKey. It is rated at 1.5A and handles the typical track voltages. My plan was to power it off of the track level DC output of the Zimo decoder so I didn't have to use another set of diodes. I bought 3 regulators but shipping was still a considerable chunk of the cost. 

If you use a linear regulator the current draw gets directly translated up to the track voltage. So at 18V it needs to drop 13V across the regulator to make 5V, assuming 0.5A for the smoke generator (being sort of optimistic here) that means the regulator will have 6.5 watts to dissipate, more than twice what the smoke generator will use to make smoke. The total for the linear regulator and the smoke unit works out to be 9W total. The switching regulator will only heat up relative to its 10 to 20% loss due to its efficiency. With a switcher, if the smoke unit uses the same .5A at 5V, the 2.5W will get translated up to the 18V as 2.5W plus the switcher's losses - only about 150mA. 

Tom


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## Tom Bray (Jan 20, 2009)

By the way, the EBay unit listed at the top in Madman's post is only rated to 16V. Most do top out around 36V unless it is rated for both automotive and telecom. Automotive is usually 9 to 30V or so, telecom goes over 40 but usually doesn't get much below 24V.

Regardless of which regulator you pick, you will need a bit of capacitance on the input and the output of the regulator. Pull down the spec sheets on any that you are considering and look at what the manufacturer recommends. 

Tom


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## Madman (Jan 5, 2008)

Posted By Tom Bray on 22 Oct 2010 10:09 PM 
By the way, the EBay unit listed at the top in Madman's post is only rated to 16V. Most do top out around 36V unless it is rated for both automotive and telecom. Automotive is usually 9 to 30V or so, telecom goes over 40 but usually doesn't get much below 24V.

Regardless of which regulator you pick, you will need a bit of capacitance on the input and the output of the regulator. Pull down the spec sheets on any that you are considering and look at what the manufacturer recommends. 

Tom 



Tom,

First, what is capacitance? Second, why do we need it in this instance?


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## Greg Elmassian (Jan 3, 2008)

The two components on either side of the regulator 7805 were referred to as "capacitors" in my post. 

Capacitors have "capacitance" which is the CAPACITY (see the wood origin) to store charge (electrons). 

Depending on the value they can "smooth out" DC that is "irregular"... this is called "filtering"... that is the electrolytic (has polarity) capacitor to the left of the regulator... right off the diode bridge that converted AC to DC... 

"smaller" capacitors are also used for filtering, but much higher frequencies... have you ever seen the little capacitors on electric motors? They can help filter electrical noise from bugging your TV for example. 

That is the function (basically) of the small capacitor on the right hand side of the 7805. 

Take that explanation for why you "need" them. Further explanation would really not help, you need them. 

Greg


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## Tom Bray (Jan 20, 2009)

Greg is right ... here is a simple explanation of why you need them (the next level of detail will end up looking like Circuits 200): 

What the capacitors (i.e. capacitance) do for the voltage regulators is help calm down the circuit and keep it from going unstable, the regulators have a tendency to want to oscillate which will cause them to self destruct (like what uncontrolled feedback in a PA system does to your ears and the speakers). In this case, the capacitors work sort of like shock absorbers in your car. 

The neat thing about Integrated Circuits is that you can build a fairly complex circuit using just a few parts. But, the ICs need a certain amount of help to make them work correctly, especially related to how they are powered and what kind of load they are working with. 

You want to have some small capacitors, .1uF or so on either side of the regulator ICs to control the higher frequencies. The larger capacitor on either the input or output, say 100uF, makes sure the part sees a the right kind of supply input and output load, again to help keep it from wanting oscillate and self destruct. The capacitors need to be rated for 50V for anything on the track power supply side of the circuit, the smoke generator side needs to be 10V or higher (50V works here also). Radio Shack parts are fine for this. Note that Greg's circuit has one capacitor on each side of the voltage regulator. My "standard" design is to put both kinds of capacitors on both sides of the regulator. 

Greg probably has a picture of the circuit on his website. If not, I'll try to come up with something. 

Tom


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## Tom Bray (Jan 20, 2009)

One thing that was asked about originally was being able to boost the power up for the smoke generator. The problem with simple DC - DC converters, the IC variety at least, is that they are designed to either boost the voltage or lower the voltage. The design gets far more complex if it needs to do both and there are very few affordable modules that will do that, especially ones that will fit inside a locomotive. 

Also 18V is not a particularly popular voltage to provide high current drive for, so finding something that just does that is going to be hard to find. Also something that boosts the voltage up to 18V is not going to be happy over about 15V or so. 

The simplest most economical route would be to swap the smoke units for 5V types and use a switching regulator to convert the track voltage down to 5V. 

You could also put some 10W resistors in series with the motor leads so that you can run a higher input voltage to achieve the same speed. If the motor draws 2A, a 2 ohm resistor will allow you to set the voltage 4V higher on power pack. Of course your lights will be brighter also and may not last quite as long compared to running everything at a lower voltage. 


Tom


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## Madman (Jan 5, 2008)

Posted By Tom Bray on 23 Oct 2010 01:21 PM 
One thing that was asked about originally was being able to boost the power up for the smoke generator. The problem with simple DC - DC converters, the IC variety at least, is that they are designed to either boost the voltage or lower the voltage. The design gets far more complex if it needs to do both and there are very few affordable modules that will do that, especially ones that will fit inside a locomotive. 

Also 18V is not a particularly popular voltage to provide high current drive for, so finding something that just does that is going to be hard to find. Also something that boosts the voltage up to 18V is not going to be happy over about 15V or so. 

The simplest most economical route would be to swap the smoke units for 5V types and use a switching regulator to convert the track voltage down to 5V. 

You could also put some 10W resistors in series with the motor leads so that you can run a higher input voltage to achieve the same speed. If the motor draws 2A, a 2 ohm resistor will allow you to set the voltage 4V higher on power pack. Of course your lights will be brighter also and may not last quite as long compared to running everything at a lower voltage. 




Tom 






Tom, Would this unit be considered a switching regulator? I am in contact with the seller who seems to be very helpful. It would help me if I could ask questions that sort of sound like I know what I am talking about. Both you and Greg have been helpful so far. I am looking for the simplest way out. To buy a ready made product would be the simplest way. 

http://cgi.ebay.com/DC-5-30V-Out-DC-1-28-26V-adjustable-DC-DC-Converter-/120600449381?pt=LH_DefaultDomain_0&hash=item1c1458d165


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## Greg Elmassian (Jan 3, 2008)

In DC 5-30V Out DC 1.28-26V adjustable DC-DC Converter 

Yep, it's a switcher, adjustable... notice my first post refers to a DC to DC inverter. 

Will work fine except it a little big... that's mostly because of the current it handles. 

Greg


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## Tom Bray (Jan 20, 2009)

The module is pretty big, being 1.5"W x 2"L x 1"H but yes it will work. That seems to be much larger than most of the decoders I've worked with. 
Don't forget about the heat sink - probably at 1A or less you won't need it. 

That module should work pretty much as is if there is room for it. You will want to use the full wave bridge rectifier that Greg talked about above so that you can run the engine backwards. If you ever want to run it on DCC you will need 100uF of capacitance after the rectifiers going into the module.

I noticed that one link has expired but he has several more for sale. That module is a good deal ... if you want one, jump on it because they won't last long.

Tom


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## Tom Bray (Jan 20, 2009)

One word of caution about the module. It is going to work all the way down to 5V. If your smoke unit need 1A at 18V for example, the module is going to draw about 3.5A when the track voltage is at 5V but the smoke unit will still be running at full speed. 

Switching power supplies work backward from ohms law. Normally as you reduce the voltage, a device will draw less current. A switching regulator does just the opposite, the lower the voltage, the more current it will draw. 

Tom


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## Greg Elmassian (Jan 3, 2008)

The easy way to say that is that on each "side" the volts times the amps is the same. 

If you "drew" 2 amps at 5 volts on the output (2 x 5 = 10) and you had 10 volts on the input then the current draw would be 1 amp (1 x 10 = 10)... 

voltage times current is wattage, and for the same wattage load on the output, the lower the input voltage the higher the current, to keep the wattage the same. 

Tom's warning is a good one, because, you could really start taxing the power pickups. (I'm assuming track power based on your first post)

Regards, Greg


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## Tom Bray (Jan 20, 2009)

Either description works. It still comes down to the point that the behavior is backwards from what is considered normal. 

If I were to use that switcher module, I would add a relay or something that would not turn on until the voltage reached some reasonable level, maybe 10V. It would also drop out if the voltage was below 10V or so. Otherwise, it might be difficult getting the engine in motion smoothly without the possibility of having the circuit breaker trip or having the engine slow down as soon as the regulator started up and turned on the smoke generator full blast. 

It may also require a larger capacitor to hold up the power while the switcher turns on. Also, it may have to be a special type of capacitor with what is called Low "ESR" which stands for Equivalent Series Resistance. 

Tom


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## George Schreyer (Jan 16, 2009)

If you want to step down high track voltage to 5 volts for a smoke unit, then the easiest way is the 5 volt regulator with an initial voltage dropping resistor. This is similar to the 7812/7805 combination mentioned, but cheaper. 

DO NOT use a current regluator like a LM317 to drive a smoke unit. The LGB units are "impedance protected" to some extent. When they run dry and get hotter, the resistance increases and reduces their power consumption somewhat. This helps them from burning out. If you use a current regulator, it will just shove the wet operating current into a dry smoke unit and defeat the impedance protection. 

Raising the voltage on any smoke unit will materially increase the smoke output, but at the risk to toasting the thing when it runs dry. 

see my smoke tips page for information on various units and ways to power them. 

http://www.girr.org/girr/tips/tips3/smoke_tips.html


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## George Schreyer (Jan 16, 2009)

oops, double post


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## Madman (Jan 5, 2008)

How would one test a resistor? I experimented with a small resistor, an LED, and a voltage tester. An LED that will burn out with more than 4 volts applied to it, to it runs perfectly on twelve or more volts, when a resistor is spliced into the circuit. Yet when I replace the LED with a voltage testor, after the resistor, the voltage does not drop.


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## Greg Elmassian (Jan 3, 2008)

Hmmm... electronics 101... 

Dan, first you need to realize that the controlling factor is current in leds... yes, there is a NOMINAL voltage for leds, but what allows them to burn out is excessive current. You need to calculate the resistor needed to supply the correct current. 

There are many calculators online. 

But, this is not what you have asked, you are apparently trying to measure the voltage drop, but you cannot substitute a meter for the led, because the meter does not draw the same current as the led. 

Since the circuit is "tuned" for a certain current through the led, and the led is not there, and the voltmeter does not draw the same current, you cannot do this to measure anything. Also, having a meter in series is only used when measuring current, and you are probably on volts. 

1. get the specs on the led... current and nominal voltage. 
2. based on the supply voltage (select the max if track powered), and the current for the led (usually 20 ma) and the nominal voltage (varies all over the place), calculate the resistor needed. 
3. connect the circuit... the led and the resistor in series and apply power. 
4. now, using the volt meter, in volts mode, put the test leads on each side of the resistor, you can see the voltage drop. That really means nothing though, because it will appear to be the same as the supply voltage. 

So why is this? Because you need to understand that you want to measure and control current. 

1. Switch you meter to measure current... usually requires changing one of the socket connections on the meter. 
2. Now put all THREE devices in series, the resistor, the led, and the meter.. the order does not matter, the SAME current must flow through each device no matter what the order. 
3. now you can measure the current. 

Try to remember to focus on the current, it seems weird, but a LED is basically a diode, a short circuit in one direction, so trying to measure the voltage drop across a shot circuit will be almost impossible to do, and really means nothing when using a LED. 

This is all the opposite of an incandescent lamp, which operate on voltage. 

It's weird at first, but this is the way stuff works. 

Regards, Greg


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## George Schreyer (Jan 16, 2009)

You don't want to JUST use a resistor in series with a smoke unit. You want a resistor to drop about half of your track voltage. The resistor will get hot but take the heat load off the regulator that follows the resistor. To calculate the resistor value to use with the regulator circuit that Greg supplied above, you need to know the current that your smoke unit wants to draw at it's rated voltage. Then take half your track voltage, divide it by this current and you'll get the resistor value in ohms. Then multiply half the track voltage by the smoke unit current and you'll get the power dissipation of the resistor. Pick a resistor with a higher power dissipation capability.


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## rayjturner (Feb 20, 2008)

Most engines have a large metal weight in them. This would make an ideal heat sink for the 5V regulator IC. The metal tab on the IC is NOT insulated, so you would be well advised to get an insulating mica washer assembly so you can screw the IC to the weight. You'll need to find a flat spot on the weight and drill a hole either through it or tap it.


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## Greg Elmassian (Jan 3, 2008)

Consider the metal weight, while it has a large thermal mass, it is probably not the best radiator for getting rid of heat. Indeed, in USAT locos, where the regulator is bolted to a large lead weight, there is a finned heat sink used. 

It takes longer to heat up the weight, but once hot, it does not cool well. It merely delays a situation if you cannot dissipate enough heat on a continuous basis. 

Also, as shown earlier, you can easily generate too much heat using a 5v regulator from 18 volts... 

This thread is getting a little mixed up, I interpreted the last post by Dan as a new question about a LED, having nothing to do with the support of the smoke unit. 

So, there seems to be 3 questions in this thread: 

1. How to run a 5v unit from 18v 
2. How to run an 18v unit from a lower voltage / track power 
3. How the voltage drop across a led - resistor combination is observed 

Greg


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## Madman (Jan 5, 2008)

Greg, you interpreted my question correctly. I probably should have posted the question in a new thread. However, in my mind, I considered the LED/testing meter experiment as part of a learning curve involving the smoke unit question. Having never really delved into electronics, I find it as difficult, at my age, as trying to learn another language. And boy, is *that* taking forever. 
I would suspect that my meter will not measure amps, or current flow. It is set up for volts, both AC and DC, and also for measuring OHMS, which is resistance in a conductor, I believe. I haven't had the time to test the meter on a completed circuit, suggested by George.


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## Greg Elmassian (Jan 3, 2008)

No problem by me Dan, if I get confused, then I ask the question like I did, no harm no foul! 

If you meter has amps capability, there are usually at least 2 clues: there will be a dial position that says "A" or "mA" and there is usually at least another place to plug in the positive probe lead. 

If you don't have a meter that will do this, I'd buy another meter, they are inexpensive nowadays. There's a digital one at harbor freight for about $5, well worth spending. 

Greg


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## Madman (Jan 5, 2008)

We have smoke! I was going through my inventory of spare parts, and parts I have on hand for future projects. It seems that some time ago, I collected the neccessary parts to power a five volt LGB smoke unit via track power. I always say that I have forgotten more than I ever new. Anyway, here is a photo of the wiring diagram, from one of our members, not sure who, and another of my bench test of the assembly. The second photo got butcherd somehow, and the smoke unit is not visible. On the bench, there are no capacitors in the circuit. I take it then that they should be installed when I decide to put the system into a locomotive?? The end result was good. I was even able to check the voltage at the two leads of the smoke unit. It read five volts plus and minus. It seemed to fluctuate a tenth of a volt or so while in operation.


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## Madman (Jan 5, 2008)

I am not sure why the second photo is so large. I reduced it's size, along with the first photo before I uploaded them to my web space here.


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## Greg Elmassian (Jan 3, 2008)

Hmm... the diode to ground? Won't that raise the voltage a bit from 5 volts? Have not seen that in a 7805 circuit. 

Greg


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## Madman (Jan 5, 2008)

Greg, your asking me







. All I know is that it works on the bench. By the way, my meter doesn't have the DC current capability. I snatched one from Ebay last night for a penny. Where else but China. $7.99 shipping.


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## Greg Elmassian (Jan 3, 2008)

Dan, if you have a link to the site where you got that diagram, I would be interested. Always ready to learn something. 

Yep, good price, we can talk you through some basic measurement techniques. Once it "clicks" you will be on your way. 

Regards, Greg


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## Madman (Jan 5, 2008)

Greg, I just did a search of past topics that had anything to do with smoke. For the life of me I could not find anything with that diagram. But I do know it came from one of our members.


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## Greg Elmassian (Jan 3, 2008)

Yeah, no problem. That center pin is normally connected to ground. If it is "above" ground, then it should raise the output voltage a bit. Normally this is done with a resistor. A diode in that position I have never seen. 

I would not do it, I would follow the manufacturer's circuit. 

In your second picture, I don't see the diode connected, but it's hard to see. Also I see no evidence of the 2 capacitors in the circuit. Be apprised that the regulator can act screwy in some circumstances without those in the circuit. 

Regards, Greg


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## George Schreyer (Jan 16, 2009)

the diode to ground trick is often used to tweak a fixed regulator upward by 0.7 volts/diode. The current in the "ground" is about 10 mA. 

- gws


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## Tom Bray (Jan 20, 2009)

The diode is probably OK but you really should move the capacitors to go between the two power terminals and ground lug on the regulator. This will keep regulator from potentially going into oscillation or doing something else unintended. The diode pretty much negates the capacitors in the circuit the way it is currently wired. 

Tom


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## Madman (Jan 5, 2008)

Posted By Greg Elmassian on 26 Oct 2010 01:05 PM 
Yeah, no problem. That center pin is normally connected to ground. If it is "above" ground, then it should raise the output voltage a bit. Normally this is done with a resistor. A diode in that position I have never seen. 

I would not do it, I would follow the manufacturer's circuit. 

In your second picture, I don't see the diode connected, but it's hard to see. Also I see no evidence of the 2 capacitors in the circuit. Be apprised that the regulator can act screwy in some circumstances without those in the circuit. 

Regards, Greg 


I don't have any of the items you mentioned in the circuit. I have the capacitors to install in the circuit if they are required, which most of you seem to be saying they are. Again, how important is the diode?


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## Greg Elmassian (Jan 3, 2008)

I would not use the diode... I cannot find an example circuit with it anywhere. I've been using the lm78xx series since it came out (must be 30 years or more)... 

Greg


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## Madman (Jan 5, 2008)

Thanks Greg


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## Greg Elmassian (Jan 3, 2008)

I usually start with the data sheets from the manufacturers, they all recommend in various situations the caps on the input and output... never seen a diode on the "ground" pin. 

look on page 22-on of the Fairchild (manufacturer) data sheet: *http://www.fairchildsemi...trong>**

non of them have a diode... maybe someone wanted a bit over 5 volts, I seem to remember some LGB units wanting 5.3 volts but maybe I am wrong. 


Regards, Greg*


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## Tom Bray (Jan 20, 2009)

The diode doesn't do much except raise the output voltage by 0.7V (assuming a standard silicon diode). It won't hurt the circuit since the diode is in series with with another pair of diodes in the bridge. 

The way it is wired you are picking off between the two diodes, sort of like a voltage divider. 

The person creating the circuit in the first place probably really needed 6V and that got to within 0.3V of that value. It was what was in the junk drawer.

The capacitors are to keep the circuit happy with various loads on it. Most circuits assume that for signal levels (or noise for that matter) the power supply is "0" volts - the capacitors do that in the input. Since most of the devices that the voltage regulator feeds has a similar requirement, the outputs are designed to drive heavy capacitive loads. 


Tom


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## Madman (Jan 5, 2008)

I received one of the DC converters that I posted links to earlier. It works very well in my Stainz. I will post a video of it soon. Now I want to buy more 5 volt or close to that voltage, smoke generators. I see LGB smoke units anywhere from $27.00 to $50.00. Being the cheapskate that I am, I found a Firma-Seuthe generator at Hobbylinc.com. However, I can find no specifications for it anywhere on the web. It's priced at $10.07 at Hobbylinc. But this can be an HO generator. Is there any more information on these units, or does anyone know anything about them?


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## Madman (Jan 5, 2008)

Here, as promised, is the first video of the Stainz with the 2.5-27 volt DC-DC converter powering the LGB 5 volt smoke unit.


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## Greg Elmassian (Jan 3, 2008)

Look great Dan. Well, it's great to see that all the discussion netted a great solution. I'm getting some of those inexpensive converters too, just to keep on hand for various unusual situations. 

Regards, Greg


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## Madman (Jan 5, 2008)

Thanks for your help greg, and everyone that participated in this topic. I have three more of those converters comming soon. You can't beat the price.


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