# Shay High Voltage



## Treeman (Jan 6, 2008)

I know I have seen this before, but I can't find the old post. The new Bachman Shay with DCC and sound does not like high DCC voltage. What would be the best way to drop the voltage in the Shay.


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## Road Foreman (Jan 2, 2008)

Mike, 

A easy way to drop the voltage is with a bridge rectifier, install the bridges at the input of the DCC board in the Shay.. You will need 1 per track side.. This will drop about 2 volts.. If you need more voltage drop use more bridges rectifiers.. Hope this helps.. 

BulletBob


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## Treeman (Jan 6, 2008)

Thanks, Bob


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## Dennis Cherry (Feb 16, 2008)

Posted By Road Foreman on 18 Mar 2010 08:26 PM 
Mike, 

A easy way to drop the voltage is with a bridge rectifier, install the bridges at the input of the DCC board in the Shay.. You will need 1 per track side.. This will drop about 2 volts.. If you need more voltage drop use more bridges rectifiers.. Hope this helps.. 

BulletBob 

Actually one bridge rectifier drops 1.4 volts each, so if two are added in series like 1 one each track side. The drop will be 2.8 volts. Make sure the current rating of the bridge rectifier is higher than needed, would recommend using 3 amp bridge rectifiers. just in case you have a stall condition and the current increase significantly.


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## Greg Elmassian (Jan 3, 2008)

I think that what Bob is indicating is you take the FW bridge, connect the + and - outputs together. 

Now take the AC "inputs" and connect them together. 

What you now have is a single component that passes AC and has one diode drop. theat drop should be about .7 volts, maybe a bit higher. 

Use two of them in series with the DCC decoder, and you get 1.4 theoretically, or maybe a bit more. 

I'd use 5 amp at least. 

Bob, did I understand your wiring of the FW bridge to be a "bidirectional voltage drop" correctly? 

Regards, Greg


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## Dennis Cherry (Feb 16, 2008)

Posted By Greg Elmassian on 19 Mar 2010 01:25 PM 
I think that what Bob is indicating is you take the FW bridge, connect the + and - outputs together. 

Now take the AC "inputs" and connect them together. 

What you now have is a single component that passes AC and has one diode drop. theat drop should be about .7 volts, maybe a bit higher. 

Use two of them in series with the DCC decoder, and you get 1.4 theoretically, or maybe a bit more. 

I'd use 5 amp at least. 

Bob, did I understand your wiring of the FW bridge to be a "bidirectional voltage drop" correctly? 

Regards, Greg 

i have to disagree with you on the voltage drop. If you draw out a bridge rectifier and tie the + and - together you will have 2 diodes in series in both directions. since a diode has a normal drop of .7volts the two series diodes gives you 1.4 volt drop in both directions. 2 bridge rectifiers in series is still 2.8 volts. The 5 amp bridge is acceptable. The Shay normally uses less then 1 amp without a sound board. The 3 or 5 amp bridge will work, the 3 amp is smaller and might fit better in the shay. Just may opinion.


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## Greg Elmassian (Jan 3, 2008)

That's why I outlined how I wired it... and for that wiring I am right, one diode drop. As you can read in my post, I tied the AC inputs together as well, unnecessary, and it does reduce it to one diode drop. 

But there is a better way to wire it, as you pointed out, and in that case it is two diode drops. 

For some reason, I remembered the wrong wiring, and did not have the opportunity to draw it up. 

Thanks for that Dennis, I was making the "connections" too complex, ha ha. 

Regards, Greg 

p.s. I'd still go 5 amp, remember that you can get pretty high (albeit sort duration) current spikes.


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## Dennis Cherry (Feb 16, 2008)

Posted By Greg Elmassian on 20 Mar 2010 10:25 AM 
That's why I outlined how I wired it... and for that wiring I am right, one diode drop. As you can read in my post, I tied the AC inputs together as well, unnecessary, and it does reduce it to one diode drop. 

But there is a better way to wire it, as you pointed out, and in that case it is two diode drops. 

For some reason, I remembered the wrong wiring, and did not have the opportunity to draw it up. 

Thanks for that Dennis, I was making the "connections" too complex, ha ha. 

Regards, Greg 

p.s. I'd still go 5 amp, remember that you can get pretty high (albeit sort duration) current spikes. 


I guess i missed your wiring of the bridge, yes it would be .7 volts that way. Good way to use a bridge with a lower current ratings to increase the total ratings.

i hope we did not confuse anyone on the forum.


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## Greg Elmassian (Jan 3, 2008)

No, if anything, I helped create confusion. You have the better idea, and I need to look at the circuit before I open my mouth! 

Thanks for the correction... I need to put a picture on my site, it makes immediate sense when drawn out. 

Regards, Greg


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## Dan Pierce (Jan 2, 2008)

And a cheaper way is to use a diode in between the + and - to get the bridge 1.4 drop and then .7 for every diode added to the + and - in series.


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## Greg Elmassian (Jan 3, 2008)

Nope, TWO diodes in parallel at that location to keep it working in both polarities... 

and 2 diodes (in parallel) at a time wherever you add them... otherwise you no longer have a bidirectional voltage drop. 

The basic "unit" here is a pair of diodes in parallel, in opposite polarities. So, no matter what the polarity of the current, it flows through one or the other diode. 

what connecting the plus to the minus of the bridge does is make a circuit of 2 of these "Units" in series... so 2 voltage drops, bidirectional conduction. 

Putting a single diode between the plus and minus makes the whole thing polarity sensitive again, only one polarity will work. Adding 2 diodes in between, adds another, third "unit" in series. 

No magic here, just a slick way to turn a FW bridge into a bidirectional 1.4v voltage drop. 



Greg


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## Peter Osborne (Jan 5, 2008)

Try this: 

http://www.modelrectifier.com/search/product-view.asp?ID=1261 

Peter.


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## Greg Elmassian (Jan 3, 2008)

two series chains of 6 diodes... should be 4.2 volts... no mention of the amps. 3 full wave bridges would be more compact and have less bare wires, but might be more expensive... depends on the amps of this unit. 

Yep, another alternative, although I would not use it unless I knew the amp rating of the diodes. I'd guess 3 from the size of the diodes. 

Anyone have one and know the part numbers of the diodes? 

Regards, Greg


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## Dennis Cherry (Feb 16, 2008)

Posted By Greg Elmassian on 28 Mar 2010 09:16 PM 
two series chains of 6 diodes... should be 4.2 volts... no mention of the amps. 3 full wave bridges would be more compact and have less bare wires, but might be more expensive... depends on the amps of this unit. 

Yep, another alternative, although I would not use it unless I knew the amp rating of the diodes. I'd guess 3 from the size of the diodes. 

Anyone have one and know the part numbers of the diodes? 

Regards, Greg 
If you just want to use a standard diode the 1N5400 is rated at 3 amps.

If you want to use a bridge rectifier then the door is wide open to the amp ratings. 

I would suggest going to Mouser.com and using the keywords "bridge rectifier" Just remember the higher the current rting the bigger the bridge rectifier gets. So look over the specifications for the dimensions.

Mouser is fast on filling orders and shipping.


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## Greg Elmassian (Jan 3, 2008)

Dennis, I was more referring to Pete's suggestion for that MRC board, in that the amperage of the diodes is not specified. 

(I actually DO know the part numbers of the 1N series of diodes ;-) ) 

Regards, Greg


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