Conference napalm::commusic_v1

    Plugging in an unwired plug is probably OK.
    
    If you're really paranoid, get a pair of 10-ohm 1-watt resistors
    and wire them as follows to your stereo plug:
    
          ring     sleeve-----------------
    tip       \-----------------------   |
      \--------------------------    |   |
                               	|    \   |
                                /    /   |
                                \    \   |
                                /    /   |
                                \    |   |
                                |    |   |
                                ----------
    
    That'll make a decent dummy load.
    
    	-Bill

    dont bother with load.
    
    Just plug in dummy jack.
    
    seems the internal circuitry has it all covered........
    
    really.
    
    ron
    

    Thanks for the simple answers.  If only all electronics was this
    easy!

>>>   -< Can't you just turn the volume down? >-

    No.  I guess I should have mentioned that the only volume slider
    on the thing affects the line outputs, the internal speakers, and
    the headphones at the same time.  Thus if I want to use the internal
    sounds through my mixer, I can't turn the volume down.
    
    Nice idea, though.  Thanks for the suggestion.
    
    - John -

    Mr. Load Resistor speaks again...
    
    OK, I've solved the headphone defeat.  Now, we're going for the
    big time.  My guitar amp owner's manual says that I can use a dummy
    resistor of 50 watts, 8 ohms to let me use ONLY the direct out.
    Great.
    
    I've tried Radio Shack, Active/Future Electronics in Westborough,
    and Stark Electronics in Worcester.  None of these folks have a
    50 watt, 80 ohm load resistor.  The closest thing is a Radio Shack
    20 watt, 8 ohm resistor or a 25 watt, 10 ohm resistor at Stark
    Electronics.
    
    Is there a way these can be hooked up (e.g., series or parallel)
    to suit the purpose of an approx. 50 watt, 8 ohm load?
    
    Or, does anyone know another place to look for the "real" thing.
    
    Or, would an L-pad work to cut down the volume.  (I found a 30 watt,
    8 ohm L-pad in my travels.)
    
    Or, should I buy an old Scholz power soak and just be careful with
    the volume and the soak.  (I've heard that excessive use of these
    can burn out the tubes faster since you're running the amp hotter
    than it sounds.)
    
    By the way, the amp is a rather pricy Mesa/Boogie so I don't want
    to experiment on it too much.  The things we do to let small children
    sleep.
    
    Any and all help will be greatly appreciated.  (I can hear Bill
    Yerazunis's fingers starting to type already!)
    
    - John -

    How about an 80 W, 8 ohm configuration using the 20 W, 8 ohm resistors?
    
  	  	+---^v^v^v----+----^v^v^v----+
		|	      |		     |
	o-------+	      |		     +----o
		|	      |		     |
	    	+---^v^v^v----+----^v^v^v----+

    Steve

    Watts add in parallel or in series, it don'tmatter.
    You can use two 4-ohm 25W in sereies, or 4 * 2-ohm 12.5 W in series,
    or 2 * 16-ohm 25W in parallel, or 4 * 32-ohm 12.5 W in parallel.
    At least I think so.  Don't complain to me if it melts.
    I still don't understand the original problem.
    Tom

    You nmay wish to be very careful doing this...I built myself a power
    soak this way about 17 or 18 years ago (well before Tom S.) and
    one day when a wire accidentally got broken the amp tried to drive
    a 50 ohm load for at least 2 seconds before the output transformer
    melted into oblivion....my poor marshall....but yes the series/parallel
    resistor scheme will work just fine... 
    
    dbii

    Re: -.1, -.2, -.3
    
    It appears that the solution (presuming that I'm careful) is so simple
    that I didn't see it.  The problem was that, although I knew Ohm's law,
    I didn't know what to do with the wattages.  Knowing that they sum,
    regardless of series/parallel entanglements, makes it seem trivial. 
    
    I may end up building a ventilated box to hold the resistor network
    (to protect from heat) and then check it carefully before I use
    it.
    
    Thanks again for all the help.  I'll let you know how it turns out
    if/when I get the thing built.
    
    - John -

    Be a bit careful with wattages.  Although they do sum, if you don't
    do the series/parallel stuff right you can create a hot spot.  For
    example, if I solve the problem with a 4 ohm, 10W resistor and two
    16 ohm, 20W resistors (total of 50W):
    
					
    					20W, 16 ohm
				+--------^v^v^v---------+
		10W, 4 ohm	|			|
    	o-------^v^v^v----------+			+---------o
		I->		|	20W, 16 ohm	|
				+--------^v^v^v---------+

    I might think I can safely dissipate 50 W and my impedance is 8 ohms.  
    But, if I dissipate 50W with it:
    
    	50 = I�R = I�*8
        I� = 50/8 = 6.25
    
    Thus, the power dissipated by the 4 ohm resistor is:
    
	I�R = 6.25*4 = 25W
    
    And, you can kiss the 4 ohm, 10W resistor goodbye ...
    
    Steve

    re: -.1
    
    Note that it matters, but isn't the overall impedance of this:
    
>>>    ... if I solve the problem with a 4 ohm, 10W resistor and two
>>>    16 ohm, 20W resistors (total of 50W):
>>>    
>>>					
>>>    					20W, 16 ohm
>>>				+--------^v^v^v---------+
>>>		10W, 4 ohm	|			|
>>>    	o-------^v^v^v----------+			+---------o
>>>		I->		|	20W, 16 ohm	|
>>>				+--------^v^v^v---------+
>>>
    
    12 ohms?  I.e., the 4 ohms plus the 8 ohms created by the parallel
    16 ohm resistors.  If I follow the calculation, this would still
    cause problems since the 10 watt resistor would be trying to dissipate
    ~16.8 watts.
    
    I tried to mimic the calculation for the original suggestion of
    4, 20 watt 8 ohm resistors hooked up like this:
    
                +--R1--+    +--R3--+
                |      |    |      |
           o----+      +----+      +-------o
                |      |    |      |
                +--R2--+    +--R4--+

    
    and got the following:	80 watts = I�R = I�8  (for the circuit)
    				=> I� = 10
  
    				Therefore, for each resistor pair,
			    	I�R= 10 * 4 = 40
    
    This gets each parallel pair trying to dissipate 40 watts (which they
    should be capable of handling, especially if I don't crank the amp).
    
    Is this right?  Is this so dangerously prone to bad tolerances, shorts,
    cold solder joints, etc. that I should scrap the idea?
    
    - John - 

    John, this looks good on the surface, the only thing that may be a
    problem, and my more learned collegues can possibly help here, is
    reactance. By crude measurements I've come to the conclusion that 2
    ohms of the 8 ohms load impedance on every speaker I've ever metered
    out was reactive  (a result of the capcitance and inductance of the
    speaker internals) in other words they measured 6 ohms DC resistance
    and the other two ohms must have been AC resistance or reactance. Would
    this necessatate the use of non-inductive resistors if the new load was
    calculated for 8 ohms DC (that's what we're doing I think)?
    (non-inductive resistors are expensive in high wattage ratings and last
    I knew only one manufacturer in the States, I forget the name
    though...) 
    
    Since I blew my marshall away doing something similar to this I
    recommend caution.
    
    dbii 

    the inductance would u thee impedance and lower the current, therefore
    reducing stress on the resistor package.
    Tom

    ...and increeasing stress on the driver?
                               
    
    dbii

    
    (Sorry to be late, John :-) )
    
    Your 2x2 matrix of 10-ohm resistors ought to work fine.  It'll
    have an overall impedance of 10 ohms and 40 watts worth of power
    dissipation. 
    
    
    If you want to play with non-symmetric designs, remember what they
    taught you back in nursery school:
    
    	Twinkle, twinkle, little star,
    	Power equals I-squared R.
    
    -----
    
    My experimetation has led me to the conclusion that a good-quality
    driver is 2 ohms of resistance and 6 ohms of inductance.  Cheap
    drivers are the other way around (R<->L, not L<->C :-) )
                                                    
    -----
    
    Your biggest danger (IMHO) is that you'll overdrive the resistors
    in the heat (groan) of the moment, one of them will slag open, and
    the radical change in load will cause your entire system major
    problems.  Therefore, try for a design with at least a factor of 2 
    in terms of power safety margin (i.e. 100 watts capable.) and 
    possibly some redundancy if possible (i.e. one resistor fails, but
    it's failure doesn't destroy the world.)
                                  
    My Newark Electronics catalog lists mil-spec 8, 16, and 30 ohm 50 watt
    chassis-mount (i.e. extruded-aluminum metal-cased, the "good stuff")
    for $4.01 each.  Putting 4 of the 30 ohm resistors in parallel will
    give you 7.5 ohms at 200 watts... and if one fails, you still have a
    10-ohm 150-watt load, which will continue to soak the complete output
    of your M-B safely. 
    
    Newark also lists a 10-ohm 100 watt unit for $7.84 (ceramic, unarmored,
    certainly not as bounce-proof as the mil-spec units above.)  That
    would also work.  But I wouldn't trust it as much as I would the
    quadruple-redundant MIL-spec system.
    
    	-Bill
    
    

    Thanks, Bill.  I just knew that you'd come up with a suggestion
    that would point me in the safest direction.
    
    Thanks also to the other suggestions.  I may just order some of
    those MIL-spec reisitors and see what happens.
    
    - John -