T.R | Title | User | Personal Name | Date | Lines |
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1220.1 | One way... | ELESYS::JASNIEWSKI | just a revolutionary with a pseudonym | Tue Mar 21 1989 07:22 | 32 |
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First of all, you're not going to ruin your amp by putting
a line level into the input. Plugging the *AC* line into the input
of your amp might do the job, but the levels of the walkman will
not. It may sound distorted, however, as the level *is* considerably
higher than that from a guitar. You need about a 10:1 reduction,
which can be accomplished with a resistor divider network containing
2 resistors. If you want to "mix" the stereo outputs of the walkman
together for a single mono signal, you'll need three resistors.
Without going into the details of the physical construction,
I would get the appropriate parts plus two 47K ohm resistors and
one 4.7K ohm. This will load the line outs to about 50K ohms, which
is an easy load for the walkman to drive. If you use a metal box,
the "ground" or "shield" of the cabling will all be common - this
is what I'd recommend.
Left out <---------/\/\/\/\--------
|-----/\/\/\/\------->"ground".
Right out <--------/\/\/\/\-------- 4.7K
47K (Both) |
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|
^
Guitar amp.
The "schematic" of the line level reducer looks like this. The
output to the guitar amp is taken from the junction of all three
resistors. The 4.7K resistor *must* connect to ground for this to
work. Good Luck with your project!
Joe Jas
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1220.2 | | PNO::HEISER | Arizona: The Battle for Seattle | Tue Mar 21 1989 11:33 | 9 |
| Going the other way...
How about an adaptor to connect �" jack from the guitar to separate
Left and Right connectors to plug into a pre-amp on a home stereo?
I don't have a guitar amp but would like to play a guitar through
my home system.
Mike
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1220.3 | Simple! | VIDEO::TASSINARI | Bob | Tue Mar 21 1989 12:01 | 2 |
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Radio Shack lists a 1/4" to RCA phono adapter.
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1220.4 | this aint the answeer your lookin fer, but | DARTS::PELKEY | If my ancestors could see me now! | Tue Mar 21 1989 12:10 | 10 |
| Orrrr, better yet..
search the want adds and by urself a second hand Rockman, and use
*that* with your walkman, and into headphones....
Indispensible for learning songs in my book...
Probably get away cheap if you look hard enuff..
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1220.5 | Take It Easy | AQUA::ROST | DWI,favorite pastime of the average guy | Tue Mar 21 1989 12:51 | 10 |
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Re: .2
It's been said before, but I'll say it again....Use caution when
playing guitar through stereo systems. In particular be careful
to avoid boosting the bass. Home speakers are not as rugged as
guitar amp speakers and can be blown easily. When in doubt, keep
the volume down or use headphones.
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1220.6 | It Works!! (voice filled with wonderment...) | DNEAST::GREVE_STEVE | If all else fails, take a nap... | Mon Mar 27 1989 11:08 | 14 |
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RE: .1, thanks again... built this patch cable this weekend
and it works fine (BTW, there is plenty of room inside a 1/4" phone
jack for the 4.7k... I wired it across the leads inside the jack).
The only change I'd make (if I could) would be to create a little
more volume for the cassette. It used to distort no matter what
level it was at (without the divider)... now I wish it were just
a skosh higher.... Don't tell me the answer though, cos this little
item is one more (GOOD?) reason for me to buy that Princeton Chorus
that I truly need... <grin>..
Steve
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1220.8 | I Had To Say It | AQUA::ROST | DWI,favorite pastime of the average guy | Mon Mar 27 1989 11:47 | 7 |
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Re: .6
Use a smaller resistor value to get more volume. Replace the 47Ks
with a 22K or 33K, for instance.
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1220.9 | How about the other resistor?? | DNEAST::GREVE_STEVE | If all else fails, take a nap... | Mon Mar 27 1989 14:24 | 5 |
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And leave the 4.7k going to ground??
U gotta take us dummies by the hand <grin>
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1220.10 | Some Simple Math To Explain | AQUA::ROST | DWI,favorite pastime of the average guy | Mon Mar 27 1989 16:52 | 24 |
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Yes, you just replace the two 47K ohm resistors in the original circuit
with lower valued ones (like 22K). What we have here is a resistive
divider.
You have the signal going through a 47K ohm resistor then a 4.7K
ohm resistor to ground. By the principle of superposition, we can
consider each channel (left and right) as totally independent.
Ohm's law says V=IR, and in a series connection the current I is
constant. Since the 47K ohm resistor is ten times the 4.7K, you get
1/11 of the original signal passed into the amp:
V(walkman) = (47,000 X I1) + (4700 X I1)
where V(amp) equals 4700 X I1.
Now, if we switch to 22K ohms, we get about 1/5 of the original signal
passed to the amp (or roughly twice as much signal as before):
V(walkman) = (22,000 X I2) + (4700 X I2)
now V(amp) equals 4700 X I2. Since V(walkman) is constant, it's
easy to see that I2 will be larger, therefore, V(amp) will be greater.
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1220.11 | I get it! | DNEAST::GREVE_STEVE | If all else fails, take a nap... | Tue Mar 28 1989 15:23 | 8 |
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EXCELLENT EXPLANATION!!! Thanks, soooo.. if I were to replace
the two 47k with, ohhhhh, say a 100k ohm pot (although I don't know
if such a thing exists) then I could vary the output with the pot,
right??
Steve
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1220.12 | | AQUA::ROST | DWI,favorite pastime of the average guy | Tue Mar 28 1989 17:00 | 5 |
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Re: .11
Yes, but of course you either need *two* pots, or else a stereo
pot (two pots on a common shaft).
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