3 aspect model RR signalling help plse

Raiway Pete said:

I'd like to know what kind of error. Pls make sure that the diodes are back to back. If not then all of the DCC current will be diverted through a signal transistors PN junction during one half AC cycle of the DCC booster's output.

Click to expand...

It doesn't work, unless I'm missing something.

Raiway Pete said:

Also I sent a diagramatic explanation to MrDEb. You can check that out too.

Click to expand...

I checked out your pdf. And I did some searching on the web on rectifier circuits. I can't find one that uses only two diodes, they all use 4 unless it's connected to a trasformer.

Raiway Pete said:

If you just took out two diodes from the PC board diagram and have not replaced them with a jumper All the DCC power will try and go through the little teeny Twin-T transistors.

Click to expand...

Not sure where you want a jumper when there's only two diodes, maybe this is what I'm missing?

Attachments: Working.jpg (4 diodes) and NotWorking.jpg (2 diodes)

mneary said:

The signal transistors will receive 1.4V on their bases, which will force a lot more than a token current into the base. With one diode drop, the problem is less serious, but it depends heavily on the Vbe and base resistance of the transistor. A base resistor of a few hundred ohms would do no harm and would improve transistor life dramatically.

Click to expand...

Mr MNeary,

This is why I asked the question about using a shorted bridge rectifier. I have tried it and it actually works. I have used a scope and found that the drop across the silicon transistor PN junction remains pretty close to 0.7 volts. A meter showed that the base current, at about 1.5 amps through the diodes, somehow remained low. (I seem to remeber about 8 Ma.) With germanium transistors it was somewhat higher.

There is a small point to make here. It seems, I'm told, that the PN junctions in parallel don't exactly obey the DC parallel resistors rules and some form of compromise occurs. It's physics, I guess. However, sages have suggested that the proof of the pudding is in the eating. I have used silicon (0.7 volt drop)and Germanium (0.3 volt drop) transistors without any base resistors across two back to back 3 Amp diodes. Under those conditions I have not blown a transistor yet.

Today, in Slovakia, I have some Twin-T circuit boards that use power transistors (BD 442s without shunting diodes). This is the most sensitive version of the Twin-T. Also, passing 2 to 4 amps through their PN base emiiter junctions causes no problems. I also use 2N130X germanium signal (switching) transistors ( I have a lot of them) withtwo back to back diodes across them. I have never blown a transistor. Indeed my germanium signal transistor protptype in Minnesota is still working after 25 years of applied power. Proof of reliability is indicated bythe layer of dust on the mother board which contains ten Twin-Ts. On my test rig I tried 2N3906s for the Twin-T. Sensitivity was down a bit but it also worked quite well.

This is how the circuit should work. Using two back to back diodes the current devides itself in an indeterminate ratio between the a power diode and transistor PN junction. The diode actually drops between 0.6 and 0.8 volts. A silicon signal transistor usually drops a little less than 1 volt at about 5 Ma. So 0.7volts at the base should cause a current of less than 5Ma. That is the theory at least.

Although I know that the rectifier bridge drops 1.4 volts. I dont know why this actually causes no problem. At least I can say that I have not constructed any permanent versions of this circuit and cannot give any definitive "proof of the pudding" so to speak.

This is why I asked Angie and MrDeb to try the two diode version using silicon signal transistors. I'd hate to have Angie build the circuit only to have it blow six months from now. Besides two diodes are cheaper than a rectifier. I am also concerned that their simulators are not working with a full deck either. It seems that a two diode version won't work and a four diode version does. Somehow the theoretical abuse I'm imposing on my poor transistors has no practical negative results. Linn Wescot invented this circuit in 1958 and it has had tremendous success. The theorists claim that it doesn't work. But then the bumble bee can't fly either.

If you want more techy communcations I am at plowry10@hotmail.com

Regards

angie1199 said:

It doesn't work, unless I'm missing something.



I checked out your pdf. And I did some searching on the web on rectifier circuits. I can't find one that uses only two diodes, they all use 4 unless it's connected to a trasformer.



Not sure where you want a jumper when there's only two diodes, maybe this is what I'm missing?

Attachments: Working.jpg (4 diodes) and NotWorking.jpg (2 diodes)

Click to expand...

Your theoretical non-working circuit works fine in practice. Read my reply to Mr Mneary. I think your simulator has not enough data to work on. Is this circuit not working because there is insufficient current for the LED? Is it not working because the Twin-T voltage drop and the Diode pair drop are considered to be exactly the same. Take out the motor and put an 16 ohm resistor in its place. Better still save some cash by not buying a bridge rectifier or a pair of 3 amp diodes and just use two BD442s for the twin-Ts.
It will be much more forgiving and sensitive.

angi, your circuit looks almost like the oneI posted (see post 215)
really tired right now, kept up all night w/ cub scout pack overnighter.

Angie, MrDeb, Mneary,

have built a Twin-T using power transistors. Total components is 5.

There is no such thing as a two diode rectifier. There are high current diodes. Take two of them and put them back to back across the Twin-T where the bridge (4 diodes in one capsule) used to be. No jumpers needed.

Check out pictures I've sent

Expect more pictures in next message.

Secon two messages.

Pictures. The twin T LED lights when there is either 27 ohms or 10K across the track. Even wet fingers will trigger it.

Raiway Pete said:

have built a Twin-T using power transistors. Total components is 5.

Click to expand...

I see, I see

Raiway Pete said:

There is no such thing as a two diode rectifier. There are high current diodes. Take two of them and put them back to back across the Twin-T where the bridge (4 diodes in one capsule) used to be. No jumpers needed.

Click to expand...

Ok, so this works, yes, On DCC, yes, and this is with one tansistor, yes? (2 diodes, 1 resistor, 1 led and 1 transistor) ??

Angie,

No, the circuit has two power transistors, one capacitor (which you may not need), one resistor and an LED. I used a big one so it would be photogenic.

Build the circuit (attached) as is and start playing with it. Replace the resistor and LED with your optical isolator and get that working. Once that's working build your signal controller.

NOW!!! Do you have your signals yet? Did you buy them? Are they commercial, common anode Types. If you don't have them yet try and build your own. Shop bought ones are EXPENSIVE (or do you allready know about that?). My signal controller uses 4 signal diodes and hooks to the TWIN-T power supply. See if you can get a kit. We'll figure sommat out.

Regards

pete

Raiway Pete said:

No, the circuit has two power transistors, one capacitor (which you may not need), one resistor and an LED. I used a big one so it would be photogenic.

Build the circuit (attached) as is and start playing with it. Replace the resistor and LED with your optical isolator and get that working. Once that's working build your signal controller.

Click to expand...

Ok, so this isn't gonna blow my dcc command station up is it? It wasn't cheap you know

Will build a three section test track to try this out before I start building them into the model proper.

Raiway Pete said:

NOW!!! Do you have your signals yet? Did you buy them? Are they commercial, common anode Types. If you don't have them yet try and build your own. Shop bought ones are EXPENSIVE (or do you allready know about that?). My signal controller uses 4 signal diodes and hooks to the TWIN-T power supply. See if you can get a kit. We'll figure sommat out.

Click to expand...

I got 8 of the searchlight.jpg, not US but they'll do for now.

I changed the grain of wheat bulbs to LED's. They're definately bright enough. Not sure how to make my own but have been searching google for ideas. Found one for yard two colour ground signalling, but not the disc type.

The PDMarch.jpg ones look promising if I can work out how to fit 2mm leds to them.

Dear Angie,

First things first. Even if there is a short on the TWIN-T it is in 'series' with the DCC to the track. That means it is just as if it isn't there any more and your engine limits the current as per normal. The Twin-T power unit supplies only 15 milli amps to the LED through a 3.9 K resistor. Your DCC unit won't even notice that and as a matter of fact we can use that situation to detct a train on a track that has no power on it. (Later, get this working first).

Two, have you tried an electronics house that has subminiature LEDs. These are less than 2mm in dia. Also can you research 'surface mount' LEDs. They have no leads. They have little tabs that you can solder fine wire to and look just like the prototypical electrical housings on the real thing. They need about 7 miliamps and are still very bright. If you had to replace the lamps with LEDs (good idea). BUT did you have to feed wires down a mast. If so then your controller problems could be over. I have a control system that uses 4 diodes for a three light signal and one for a two light one.

Later

Pete

PS Does the P & D kit come complete or is it only metal parts? I can't see from the picture.

angie1199 said:

Ok, so this isn't gonna blow my dcc command station up is it? It wasn't cheap you know

Will build a three section test track to try this out before I start building them into the model proper.



I got 8 of the searchlight.jpg, not US but they'll do for now.

I changed the grain of wheat bulbs to LED's. They're definately bright enough. Not sure how to make my own but have been searching google for ideas. Found one for yard two colour ground signalling, but not the disc type.

The PDMarch.jpg ones look promising if I can work out how to fit 2mm leds to them.

Click to expand...

Hi Angie,

I hope the searchlites were not too expensive. I take it that your layout is 'N' gauge. I also understand that the sigals are "not US and that they will do for now". Does that mean you are modelling an American railroad?

There's quite a few companies that make 'N' signals. ATLAS is one. Nj international is another. Have you checked any of these out.

The searchlight signal looks as if it has a sub base housing that contains the lamps and uses some kind of fiber optics to 'chanel' the light to the signal target. Am I right? For three lamps (LEDs) how many wires are there? (4 or six)

As for the PDMarch kit. I enhanced the photo to get more details. Not much luck. There is a mast, I think, and a ladder. The rest I couldn't make out. I'm surethe UK must have some kind of electronics mail order house. 2mm LEDs are fairly common. I will check with my electronics store here in Slovakia.

Pete

This is the pn I'm looking at but it's 1A base current and emmiter-base voltage is 5V and the circuit is minimum of 12, if not 15. Not sure what I'm looking at to know the voltage drop.

https://www.electro-tech-online.com/custompdfs/2009/12/TIP31_A_B_C20TIP32_A_B_C.pdf

The 1N5400-1N5408 dodes seem right if dioes are used but, again, I have no idea how I should know what volts are lost across each diode.

As for the signals, yes, they are lit through a fibre optic, about 1" long with a piece of angled plastic to change the light's direction at the top. I had three grain of wheat bulbs in each aluminium tube which I replaced with 3 led's which will be either common anode, and cathodes going to the signalling circuit.

The PDMarsh one is base, mast, ladder and disc where the light should be placed. Not sure if I could modify/change the mast to house a fibre optic. Housing an ledin the disc would me it would need to be multicolored.

More on signals if we can get this circuit working tho'.

Raiway Pete said:

I take it that your layout is 'N' gauge. I also understand that the sigals are "not US and that they will do for now". Does that mean you are modelling an American railroad?

Click to expand...

Yes and yes.

angie1199 said:

This is the pn I'm looking at but it's 1A base current and emmiter-base voltage is 5V and the circuit is minimum of 12, if not 15. Not sure what I'm looking at to know the voltage drop.

https://www.electro-tech-online.com/custompdfs/2009/12/TIP31_A_B_C20TIP32_A_B_C-1.pdf

The 1N5400-1N5408 dodes seem right if dioes are used but, again, I have no idea how I should know what volts are lost across each diode.

As for the signals, yes, they are lit through a fibre optic, about 1" long with a piece of angled plastic to change the light's direction at the top. I had three grain of wheat bulbs in each aluminium tube which I replaced with 3 led's which will be either common anode, and cathodes going to the signalling circuit.

The PDMarsh one is base, mast, ladder and disc where the light should be placed. Not sure if I could modify/change the mast to house a fibre optic. Housing an ledin the disc would me it would need to be multicolored.

More on signals if we can get this circuit working tho'.

Click to expand...

Dear Angie,

Please check the pictures of the Twin-T circuit I sent you. There are NO power diodes on the board. There are only Power transistors. Those two square things are power transistors. The little white square is a small capacitor which is only there in case your DCC is a bit noisy. (Nice cornered squarewaves.) You can see the resistor and the LED.

As for the diodes. It does not matter what kind of diodes you "WOULD" use if you were using signal transistors. But you ar not. You are using power transistors which can handle the high current flow and don't need to shunt it through 2 or 3 amp diodes. Believe me. Try and build the circuit. Play with it. You will find it quite forgiving. I have build a number of them using unknown transistors, both power and signal. The circuit just works. When you have something working we will take it from there.

If you have not bought the transistors yet get BD442s and stay with the PNP option. If you want NPN (More understandable I guess) get the BD442s NPN equivalent or ones that have the highest base/emitter current carrying capability. There are only five components in the circuit I have sent you. You can just wire them together, hang them on a piece of string on a temporary basis (so to speak) and just try it.

Signals,

Ah I see your problem. 'N' gauge stuff at best is clunky looking. You are facing the same problem I did 20 or so years ago. That's when I made my own. They are US type and look great. I made them to scale according to a Model Railrouder Mag editor called Odie Odegard. (RIP). I used photoetching for the parts. We can discuss that later though.

Question.

How easy would it be to change the common anode mode (4 wires) to two individual wires to each LED? (6 wires). ?

Raiway Pete said:

Please check the pictures of the Twin-T circuit I sent you. There are NO power diodes on the board. There are only Power transistors. Those two square things are power transistors. The little white square is a small capacitor which is only there in case your DCC is a bit noisy. (Nice cornered squarewaves.) You can see the resistor and the LED.

Click to expand...

Power transistors are on the datasheet I put a link to. The emitter base voltage says 5V, this is a 12v/16v circuit. How does that work?

Raiway Pete said:

As for the diodes. It does not matter what kind of diodes you "WOULD" use if you were using signal transistors. But you ar not. You are using power transistors which can handle the high current flow and don't need to shunt it through 2 or 3 amp diodes.

Click to expand...

All well and good saying 'you're not' but I like to know how things work, not just copy a circuit like a hail mary. How do I know, from a datasheet, what volts I lose, with any component, transistor or diode? I think I might need to know that to be able to buy the right components.

Raiway Pete said:

If you want NPN (More understandable I guess) get the BD442s NPN equivalent or ones that have the highest base/emitter current carrying capability.

Click to expand...

The 441 is the NPN equivalent, but again, both the 441 and the 442 state Emitter-Base voltage (Vebo) of 5V. The circuit is interacting with a 12VDC and 15VAC power supply. Did I misunderstand the datasheet? Don't want smoke here!

Raiway Pete said:

Question.
How easy would it be to change the common anode mode (4 wires) to two individual wires to each LED? (6 wires). ?

Click to expand...

Why change it? I have a signalling circuit with a 12VDC out to the signals, and each cathode of the LED's goes to the green/yellow/red connections on the signalling circuit board.

QUESTION?
why are we changing the circuit again, using different transistors, only 2 diodes instead of 4 etc.
Angi is a novice and I sense a bit of frustration when a circuit is modified
keep it simple. Go with the 4 diodes. the output is cleaner IMO

more confussion or more understandable?

here are three different rectifier circuits using AC input of 5v
I included a dc circuit as well.
the A and B outputs are seperate for clarification (offset by 10ms) but exactly the same.
voltage loss almost 1 volt on the C and D circuits.
hope this helps describe what the different rectifier circuits do.

amperage wave forms

using 4 diodes the components draw less (no surprise)

angie1199 said:

Power transistors are on the datasheet I put a link to. The emitter base voltage says 5V, this is a 12v/16v circuit. How does that work?



All well and good saying 'you're not' but I like to know how things work, not just copy a circuit like a hail mary. How do I know, from a datasheet, what volts I lose, with any component, transistor or diode? I think I might need to know that to be able to buy the right components.



The 441 is the NPN equivalent, but again, both the 441 and the 442 state Emitter-Base voltage (Vebo) of 5V. The circuit is interacting with a 12VDC and 15VAC power supply. Did I misunderstand the datasheet? Don't want smoke here!



Why change it? I have a signalling circuit with a 12VDC out to the signals, and each cathode of the LED's goes to the green/yellow/red connections on the signalling circuit board.

Click to expand...

Hi Angie,

Transistors. Nice sine waves but the way the circuit is put together you will not see 5v across the emitter/base. Most of your voltage will be borne by the load across the track. Second the power dissipation is calculated as I²R. The square wave goes from one extreme to the other very quickly. When the transistor is off I (current) is zero (or close to it) makiing I²R close to zero. When the transistor is on I is high (2 amps or abouts max) but resistance R is 0 (or close to it). All in all the transistor is not dissipating any power (heat). Now when using AC or DC that is another matter. That Chopped half sine wave can be a killer because as it goes up and down it places the PN junction into transition mode and during transition I²R has a "resistive" value. Now power (ie heat) is generated. P.S The circuit I put together is still working on my DCC layout. The 15 ohm resistor gets quite hot. The transistors dont even get warm. If I was using DC (half sine waves) I would probably need heat sinks.

As far as the common collectors are concerned at 18v DC they are drawing about 5 milli amps. That is nothing when compared to the Emmiter/base current with a draw of an amp or so. Your photo-isolator LED current lmiting resistance is probably about 680 ohms (?????) the total resistance should be not less than 1K ohm. (assuming that the Opto Isolators LED needs about 10 milli amps to operate(illuminate).

The signals. It was just a question. I am assuming that like us you don't have deep pockets. The controler you put on this post is quite component rich because you are using a common connection to Vcc (12 volts). I have used the "current limiting resistor" requirement for LEDs to good advantage. If each LED uses one resistor to limit the current to about 7 Ma (my LEDs), then all that is needed to control a two aspect signal is one (yes 1) signal diode. To control a three aspect signal 4 are required. To control a three aspect permissive 5. (permissive is used in shunting -- yellow and red).

The signal diodes 1N914s (?) cost pennies.

Will write again tonite. I have to go teach

Regards

pete

MrDEB said:

QUESTION?
why are we changing the circuit again, using different transistors, only 2 diodes instead of 4 etc.
Angi is a novice and I sense a bit of frustration when a circuit is modified
keep it simple. Go with the 4 diodes. the output is cleaner IMO

Click to expand...

Hi Mr Deb.

I also understand Angie. Hell I went through it too. But there seems to be too much simlation. So the best way is to actually go build a circuit. Then using a meter find out what is going on. I have built a working circuit. Took me all of half an hour to solder it together. It is a very basic Twin-T, with no debouncing on the output. The circuit uses three components, not including an opto-isolator or relay or lamp. It is connected in series with the track so shorting is is not a problem. Besides the DCC booster has a fast cct breaker.

We are down to two, maybe three components. try it. I think you will like it.


ircuit