Driving Multiple Headphones
E. Hill
ouput options: 4 ohms, 8 ohms, 16 ohms, and 70v. This is one of those Radio
Shack MPA-25 three channel mixers. I found it laying around, and would like
to use it for a small overflow interpretation system.
Couple of notes:
- Sometimes only a couple of headphones are plugged in, so I assume I must
stick with parallel wiring.
- The wireless transmitter is one of those "listen to your TV without
disturbing others" units. It has a volume control on the transmitter, and
there are volume controls on the receivers. The input is a phono jack.
I'd like the technical answer, with the math. (I need to get out of the
newbie stage sometime. <grin>) Thanks much...
- Eric
E. Hill
(8.34 ohms, actually.) With 20 watts available (your amp uses a
transformer-coupled output stage, which is common for low-end commercial PA
stuff, and it can put the same power across all rated loads) you'll have
nearly 1.7 watts/earcup, which is more than adequate for a translation
system in a medium-noise environment. <<
LV:
Ok. I've got this math: 100 ohms / 12 = 8.34 ohm load. Where is the 1.7
watts per earcup derived from?
If I wanted to use normal "walk-man" style headphones: 8 ohms / 12 = .67 ohm
load. How would I derive how much power I would need?
I understand if a resistor is put in parallel with the load, the resistance
increases, but what happens if one is put in series? Would I put one
resistor in series with the "red," or both the "red and black?" How do I
determine the value to use?
Thanks for the help...
- Eric
Lord Valve
Lord Valve Speaketh:
The easy way to do this is obtain 6 *identical* pairs of
phones, preferrably in the 100-ohm/earcup range. I assume
that your Radio Shack amp is monaural; you'll need to wire
the output box with TRS jacks with the tips and rings tied
together, so the signal will show up in all the earcups.
This will give you 12 100-ohm earcups in parallel, for an
8-ohm load. (8.34 ohms, actually.) With 20 watts available
(your amp uses a transformer-coupled output stage, which is
common for low-end commercial PA stuff, and it can put the
same power across all rated loads) you'll have nearly 1.7
watts/earcup, which is more than adequate for a translation
system in a medium-noise environment. It's important to use
identical phones, because if you mix impedances, the phones
with lower impedances will be louder than the the others.
If you had more power, you could use low-impedance (8 ohms
per earcup) phones, and put a resistor in series with each
output jack to keep the system impedance up; with only 20
watts, you have no power to spare. You may want to consider
using a commercially-available headphone amp system, like those
made by Rolls or DOD; these have individual amps (and volume
controls) for each headset. Due to differences in hearing,
what one person considers thunderously loud may be perceived
as less than adequate by another; in order to use "volume"
pots on a passively-split distribution system, you need
more power available than you currently have.
Lord Valve
VISIT MY WEBSITE: http://www.freeyellow.com/members2/lord-valve/
Good tube FAQ for newbies. Click the e-mail link and join my
SPAM LIST; just put "SPAM ME" in the header and I'll sign you
up. (If you only want a set of e-mail catalogs, put "CATS ONLY"
in the header.) I specialize in top quality HAND-SELECTED NOS and
current-production vacuum tubes for guitar and bass amps. Good
prices, fast service. TONS of gear and parts in stock...let's DEAL!
NBS Electronics, 230 South Broadway, Denver, CO 80209-1510
Phone orders/tech support after 1:00 PM Denver time at 303-778-1156
NOW ACCEPTING VISA AND MASTERCARD
CHAT WITH LORD VALVE: Log onto any DALnet server and join
channel #CONELRAD. Look for me there most any night after
11:00 PM Denver (Mountain) time. Guitar-amp questions and
what-have-you are welcome.
"The only trouble with being blind is...you can't see a goddamn thing!"
-Ray Charles-
Lord Valve
<any...@anywhere.com> writes:
>
>>> This will give you 12 100-ohm earcups in parallel, for an 8-ohm
load.
>(8.34 ohms, actually.) With 20 watts available (your amp uses a
>transformer-coupled output stage, which is common for low-end
commercial PA
>stuff, and it can put the same power across all rated loads) you'll
have
>nearly 1.7 watts/earcup, which is more than adequate for a translation
>system in a medium-noise environment. <<
>
>
1.7
>watts per earcup derived from?
watts equally. 20/12 = 1.7 (almost).
>
>If I wanted to use normal "walk-man" style headphones: 8 ohms / 12 =
>load. How would I derive how much power I would need?
you'd need to put a resistor in series with each output
jack, to keep the overall resistance up.
>I understand if a resistor is put in parallel with the load, the
resistance
>increases,
LV: No, you have this backwards. Putting a resistor in parallel
with the load *decreases* the resistance; putting a resistor
in *series* with the load increases the resistance.
but what happens if one is put in series? Would I put one
>resistor in series with the "red," or both the "red and black?" How do
I
>determine the value to use?
LV: Assuming 8-ohm earcups, you'd wire the tip and ring contacts
at each output jack together (to put the mono signal
into both sides of the headsets); this would give a 4-ohm
load at each headset. You'd put a 43-ohm 2-watt resistor
in series with each output jack; this would give a 47-ohm
load at each station. 47/6 = 8 (close enough) so you'd
connect the box to the 8-ohm tap on the amp.
Lord Valve
(sig lite)
E. Hill
17.6 / 12 (ear buds) = 1.47 ohms
or
8.8 / 6 (head-sets) = 1.47 ohms
8 ohms (output on amp) * 6 (head-sets) = 48 ohms
x = 48 ohms - 8.8 ohms
x = 39.2 ohms
So I would need a 39-ohm, 2-watt resistor, in series with each jack? And,
the jacks are wired together in parallel?
But, you said that 20 watts wasn't enough. I don't understand, 20/12 =1.7,
at any constant impedance, right?
______
I also need to take into account the affect of connecting to the wireless
transmitter input. It's a phono plug designed to take input from a TV or
stereo. It has a "volume" control on it, which I believe attenuates the
input signal.
Thanks (I know this is Electronics 101)...
- Eric
Peter Larsen
Lord Valve kindly quoted:
>
> In <XIRJ3.2819$mN5.1...@typhoon01.swbell.net> "E. Hill"
> <any...@anywhere.com> writes:
> >
> >Can I attach 6 headphones and a wireless transmitter to a 20w amp?
Yes. There just may be a slight quality advantage in using the 16 Ohm
tap on the output transformer instead of the 8 Ohm tap.
> It
> has 4
> >ouput options: 4 ohms, 8 ohms, 16 ohms, and 70v. This is one of those
> Radio
> >Shack MPA-25 three channel mixers. I found it laying around, and would
> like
> >to use it for a small overflow interpretation system.
Don't know anything about it, all I know is that Radio Shack exists.
Thus I also do not know how important it could be to match the load
impedance. A standard headphone outlet has a couple of hundred Ohms in
series with the output.
The virtue of doing this is that the actual power delivered is
automatically adjusted to fit the requirements of headphones of varying
impedance. There is an additional virtue of doing it when you have
multiple paralleled outputs: a short circuit of any one output will be
irrelevant for the other outputs, they will still work.
This is intended to supplement the follow up, not replace it. It could
however - if the headphones are known - be a very good idea to test what
series resistor gives an adequate loudness, so as to have a crude,
simple limitation of how loud it can get in those headphones in case
something goes wrong somewhere.
Not wiring the headphones to mono and instead giving the amp a
monophonic signal may be a flexibility advantage.
--
******************************************************************
* This posting handcrafted by Peter Larsen, pla...@teliamail.dk *
* My homepage is at: http://w1.1358.telia.com/~u135801844/ *
******************************************************************
André Huisman
XIRJ3.2819$mN5.1...@typhoon01.swbell.net...
> Can I attach 6 headphones and a wireless transmitter to a 20w amp? It has
4
> ouput options: 4 ohms, 8 ohms, 16 ohms, and 70v. This is one of those
Radio
> Shack MPA-25 three channel mixers. I found it laying around, and would
like
> to use it for a small overflow interpretation system.
We need to consider a few things first (and I included some of the data you
give in follow ups).
You use walkman headphones (just hope you haven't gotten open mics closeby
(they pick the sound from these "open" units up pretty good). Their
impedance is 20 Ohms (around that) per cup.
First point: We want to protect the hearing of the people who put on the
headphones in the first place. In that context, the stuff LV wrote (1.3W per
headphone) is actually quite dangerous. That's more than enough to
completely ruin your ears FOR EVER if someone goofes.
SO: Use of 20 Ohms walkman headphones. Those often are run from 3V systems
(meaning a swing of around 3Vpp or 1Veff). That means power (P=U^2/Z)=
around 1/20W. For the sake of "headroom", let's say we're going to calculate
with 100mW (hey, now where have we seen that number before ;-).
You also stated that all headphones must be able to being plugged in and out
independant (thus NO series connection is allowed).
Your amp is a 20W @ 8 Ohms amp. That means a voltage swing of around 13V
eff. We need to bring that down to around 1.5V (devide by ~9 in other
words). We also have to consider that ANY use of series resistance is going
to interfere with the impedance of the headphone (for even a headphone is
NOT a resistor). SO, we should at least use a series resistor AND a parallel
resistor (to make the load more resistive and as such less sensitive to
changes in impedance on the part of the headphone(s)).
Now wouldn't it also be nice if we were able to make some form of volume
control? YES, it would be nice. SO, we also need a certain pot for that. We
can choose to make either the series resistor OR the parallel resistor
changeable. Since the bulk of the power is going to be dissipated by the
series resistor, choosing this one to be a pot is not the best way. SO:
We're going to use a parallel pot (also makes a nice regulatable output
right down to absolute silence (noise included!!!)).
Suggestion time? Yes, it is (and remember, these are mere suggestions). Use
47 Ohms series resistors and 22 Ohms parallel resistors. This means the
total of the parallel arrangement is 22 Ohms // 10 Ohms (the actual
headphone, L&R in parallel) -> 6.9 Ohms. Voltage divider now is:
6.9/(47+6.9)=0.127. Hey, that's pretty close now isn't it?
But, we also wanted some form of volume control. Easy. Replace the 22 Ohm
parallel resistor with a 22 Ohms pot. The parallel arrangement will actually
give quite a descent volume control (considering the wirewound pots are
always of the lineair type and not of the logarithmic type).
Now the wattage numbers: Worst case simplified calculation warning. If
headphone output is shorted, the full voltage is over the 47 Ohms series
resistor. With a voltage of 13Veff that means 13*13/47=3.6W. Just use a 5W
(or even 10W) resistor (those white ones used in speakers) and be sure to
give it some room (for heat radiation). NO problemo.
Now for the power rating of the pot meter. Worst case (no headphone
attached), the voltage over the pot is around 13*(22/(22+47)) (for a 22 Ohm
pot) -> 4.15V. Power -> 4.15^2/22=0.78W. Just get a 1W (or bigger) 22 Ohm
wirewound pot (they are quite easily obtainable) and you won't have any
problems. Examples: MEC MCU 2W cermet pot (manuf. code: MCU-22R) ($6 per
piece), OR abuse a speaker L-pad for it (maybe even cheaper). Don't need the
volume control (or not on every channel but just a few)? Replace the 2W 22R
pots with 2W (or bigger) 22R resistors.
Load on amp (worst case, shorted outputs or pots zet to 0 Ohms) = 47/6=7.8
Ohms. The first 8 Ohm amp to die from that is a WHIMP ;-) Highest load on
amp (pots full, no headphone): (47+22)/6=11.5. Won't the amplifier just
totally LOVE you for this load ;-)
SO: With all this work (actually 6 5W 47R resistors and 6 2W 22R wirewound
pots isn't that much to begin with) you have a pretty failsafe system with
volume control per headphone, with descent max power to headphones (test
with just one headphone and change resistor values if needed) AND pretty
much removing the NON resistive headphone out of the picture. In short, you
have a darn good suggestion that should work like a charm (if I may be as
frank (though I'm André) to say that ;-)
Side note: The cost of the 22R/2W pots are actually of such proportions that
designing a 6-way headphone distribution system becomes feasible. Yet, if
you're not that technically inclined, the above suggestion could be a valid
one.
> - Sometimes only a couple of headphones are plugged in, so I assume I must
> stick with parallel wiring.
> - The wireless transmitter is one of those "listen to your TV without
> disturbing others" units. It has a volume control on the transmitter, and
> there are volume controls on the receivers. The input is a phono jack.
For the wireless transmitter: Find out the actual sensitivity of it's input.
You KNOW the voltage swing of the amp (remember, you just set it to full
power before clipping and do the volume adjustments on the respective pots
of the respective headphones) which is 13V.
Say the transmitter wants 100mV input signal (not a too strange level). You
have 13V so you'll have to divide it by 130 (roughly). That _roughly_ means
a 10k series resistor and a 100 Ohm resistor in parallel with the input
(nice low impedance isn't it?). Where to place this? Best would be close to
the input, safer would be (and thus probably preferable) to do it close to
the amplifier (so you can actually short the output without shorting the
amp). Since the impedance (seen back from input of transmitter) is actually
quite low, you won't have any capacitance loading problems, noise problems
or such. IF cable has to be run over extensive lengths, then place the
divider network close to input (so swing on cable is big and thus
sensitivity to noise is low). Power rating of resistors: Well that's easy:
13*13/10k=17mW. Ha, next to every resistor you'll find will be able to
handle this whopping power ;-)
> I'd like the technical answer, with the math. (I need to get out of the
> newbie stage sometime. <grin>) Thanks much...
As you've seen, the math is actually quite simple.
If you need a picture of the (IMO pretty simple series/parallel
arrangement), just say so and I'll put one on my site with a link to it
here.
--
André Huisman
New Line licht & geluid
hui...@new-line.nl
http://www.new-line.nl
http://www.djfaq.cjb.net
--- pardon my French, I'm Dutch ---
André Huisman
E. Hill
to around 1.5V (devide by ~9 in other words). <<
Why 1.5v?
>> Voltage divider now is: 6.9/(47+6.9)=0.127. Hey, that's pretty close now
isn't it? <<
Sorry, I'm not following... Pretty close to?
Thanks...
- Eric
Dave Dingley
>
> >> That means a voltage swing of around 13V eff. We need to bring that down
> to around 1.5V (devide by ~9 in other words). <<
>
>
> >> Voltage divider now is: 6.9/(47+6.9)=0.127. Hey, that's pretty close now
> isn't it? <<
>
>
> Thanks...
>
> - Eric
Hi All
I have been Lurking for several months now , waiting to see if my
kind of live sound shows up, and here is my chance.
The amp has a 70 volt output - use this with suitable line
(70/100v) to 8 or 16 ohm transformers to each headset. 5 watt
transformers are available here for about NZ$15 (7.50 US) with
tappings for .5 , 1 , 2 , 5 watts . You can then have individual
volume controls (100 ohm pots) and up to 20 headphones (assuming
you use the 1 watt tap) with minimal variation in over all volume
if any user changes their setting.The transformers are wired in
parallel.
This is the basis of comical and industrial "distributed" sound
systems (paging , "Muzac" and the like).
See "Sound System Engineering " by Davis and Davis for an
explanation of how it works.
The wireless system can use one the transformers with say a 10k
pot as its volume control or , better still , check if the amp
has a Tape or Aux Out socket and use that.
I use 70 / 100 volt idstributed systems frequently for outdoor
events (sports meetings , horse races and occasional political
rallies etc.) with re-entrant horns and long cable runs (100
metres to over 1 km) and lowish power (100 - 300 watts ).
(flame-proof suit on mode)
Adds a new dimension to pro.live-sound!!! Yes ,simpler in some
respects , just as complex in others. Intelligibility is what I
strive for , the intelligence is (sometimes) supplied by the
commentators / announcers / politicians (NOT) .
We all have one thing in common - the a**holes we meet and deal
with !
Dave
E. Hill
need to see how much room is available inside.
- Eric
André Huisman
> >> That means a voltage swing of around 13V eff. We need to bring that
down
> to around 1.5V (devide by ~9 in other words). <<
> Why 1.5v?
Sorry. Thought the math was "self-explanatory".
P=U*I
U=I*R ->
P=U^2/R
100mW=1.5^2/20
(112.5mW to be precise)
> >> Voltage divider now is: 6.9/(47+6.9)=0.127. Hey, that's pretty close
now
> isn't it? <<
> Sorry, I'm not following... Pretty close to?
20W amp in 8 Ohms -> P=U^2/R -> 20=U^2/8 -> U^2=8*20 -> U=sqrt(8*20)=12.6V
12.6V * 0.127 (division factor of ladder network) = 1.6V
Hey, that's pretty close to 1.5V Power in headphone roughly: 1.6^2/20=129mW
(You asked for the "math" approach).
> Thanks...
When in doubt, always ask ;-)
E. Hill
Exactly. Some large mistakes have been made on assumptions. I'm sure you've
heard of NASA's blunder with the Mars probe.
As always, thank you...
- Eric