transformer question

Started by c-rok holmes, September 11, 2009, 05:27:34 PM

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c-rok holmes

I live in toronto and am having a real hard time finding a 12.6v 300ma filament transformer.  I'm building a  leslie simulator and am hoping that I can use an alternate transformer if it allows.  Can anyone give me any pointers? I was told that this kind of transformer shouldn't be hard to find- if you haven't guessed already, i am new to pedal building.  Here's the link to the schematic and parts list.
Thanks in advance,

http://www.generalguitargadgets.com/diagrams/les_sim2.jpg

.Mike

What an interesting bypass method. I would probably bypass that bypass method and use a true bypass, eliminating the LDR/LED-or-bulb.

Anyway, I'm no expert, but it looks like this circuit runs on DC. The AC voltage coming from the transformer is immediately rectified into DC before being fed to the rest of the circuit. I see no reason why you can't use a standard wall-wart DC adapter to power the circuit. The question is, how many volts and how much current must the wall-wart put out? That is where I'm not sure.

There is a formula that I've seen (many times) on the forum here that deals with voltage and rectifying AC. A 12.6V AC transformer, when rectified, will give you 12.6V multiplied by a factor (that I forget), minus the diode drops caused by the rectifying. I'm sure someone can chime in with the formula, or you can search the forum here to find it. Once you know that value, you can figure out how much actual DC voltage/current you need your wall-wart to put out.

Hopefully that will get you started down the right path, and hopefully someone else can chime in to confirm if I am right or wrong. :)

Mike

If you're not doing it for yourself, it's not DIY. ;)

My effects site: Just one more build... | My website: America's Debate.

maarten

You could use just a 12 volt transformer (and rectify this, as is done in the schematic), or you could use 2 9 volt batteries connected in series (for 18 volts) and connect this to R1 and to the lamp (omitting C1). Instead of the 2 batteries you might try whether it runs well enough for you at 9 volts. If not, you might try a wall wart that gives you 15 or 18 volts DC, if it works good enough for you at 9 volts you might use about any wall wart that gives you between 9 and 18 volts. The 12,6 filament transformer will give you approximately 17 volts after rectification, but probably you will have a hard time finding a wall wart providing that voltage as well as a 12,6 transformer.
Maarten

George Giblet

#3
You would be better off just using a 12V DC unregulated plug pack rated at 300 to 500mA.
The supply at the left side of R1 is going to be in the order of 13 to 15V, say 14V, depending on the
specifics of the transformer.

Most of the current is required to drive the lamp!   It is possible you could adapt the circuit to work with LED with lower current

From what I can work out the #46 lamp is 6.3V 0.25A.  The lamp is driven from the unregulated supply
to the left of R1.  The 47ohm resistor to the lamp limits the current so the 14V rail doesn't pop the lamp.
On full brightness the lamp is going to be 25ohm, add that to 47ohm and you get 72ohm.
Then 14V driving the total 72ohm limits the current to 14/72 = 190mA, which is in the right order.

The rest of the circuit, to the right of R1, is only going to pull around 5mA which will cause a drop of about 5V
across the 1k R1, so that part of the circuit would be running at 14-5 = 9V.  (Most of the current for tyhis part of the circuit is going down the zener!)

It's therefore feasibly to convert the circuit to 9V.  You would drop R1 to a small value like 47ohm.
You would also reduce R24 to limit the current  to 190mA by changing R24 to say 22ohm.


[Ed:
       I just found the rest of the article.  The bottom left side of page 3 (page 53) confirms the LDR is only used for bypass.
      The whole transformer thing is only used to drive the lamp, and the lamp is used for the bypass circuit.
      As it stands this is an extremely wasteful design.   The annoying parts like the transformer, LDR and lamp
      can all be disposed with with no side effects to the basic workings of the circuit.

       These mods will let you run it from 9V (even a 9V battery):
       - remove the LDR and the lamp and replace them with the usual foot bypass footswitch
       - replace R1 with 47R, or even 0R
       - run the circuit from 9VDC

       It will pull about 5mA from the 9V rail.
]

George Giblet

I wouldn't be running this circuit of 9V batteries.  Drawing 100mA 2x9V batteries is going to last about 5hrs a costly circuit to run.

The most energy efficient and convenient is to move towards and LED version running from 9V.

I'd be looking at tweaking the circuit around the zener too.

davidallancole

#5
Hello,

Hammond Manufacturing makes a transformer with those specs.  They are one if not the most popular transformer manufacturers in Canada.  Their power transformers and audio transformers are used in many diy projects for amplifiers and what not.

The part number for a model that will work is 166F12.  You would not be using the center tap wire in this project, so just tape the exposed wire and your good to go.  You should be able to get this at local electronics stores or online from digikey.ca