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Jim Hoff
 
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Default "Fixed" 8 volt output o-loads eq

My deck (Eclipse 8053) has biamp outputs - provided the "rear" channels are
switched to become "mid" and the front channels become the "high" outputs.

So, in biamp mode I have no rear channels---but the deck also has a "Fixed
Out" that runs at full volume only.

I ran this fixed out to an old alpine 10-band that I am using to control the
rear speakers. Fuzztone. On the EQ, there is an led array that shows total
overload unless I lower all the bands to minus 12 db (or -15 db, maybe--I
forget) and it's still slightly distorted at those settings.

My plan is to try a 10k resistor (guessing on the value here) inline with
the tip of the interconnect rca before the EQ.

Is this plan valid???

Another idea, better, would be to able to sum the mids & highs for the back
speakers *without having them be summed to the front speakers. That way,
fronts and rear would be controlled by the head unit's volume control. I
don't know how to build this circuit, other than buying a little mixer.


  #2   Report Post  
MZ
 
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Default "Fixed" 8 volt output o-loads eq

My plan is to try a 10k resistor (guessing on the value here) inline with
the tip of the interconnect rca before the EQ.


Check your amp's manual. A lot of times there's a value for input
impedance. It's generally in the 10-30k range. Then just treat it as a
voltage divider.

PS Are you saying that this output isn't controlled by the volume knob?


Is this plan valid???

Another idea, better, would be to able to sum the mids & highs for the

back
speakers *without having them be summed to the front speakers. That way,
fronts and rear would be controlled by the head unit's volume control. I
don't know how to build this circuit, other than buying a little mixer.


It's a pain. Leave it alone.

Why not run just the low output to the rears? Where's the cutoff?


  #3   Report Post  
Jim Hoff
 
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Default "Fixed" 8 volt output o-loads eq


Yes the "fixed" output I am trying to tone down is not controlled by
any voulme control on the head unit, other than the mute control.

I need a full range signal and not just mid-bass at the request of my
daughter passengers in the back. The xover is set for 1.6k hz, nice
and low

The spec in the manual says 55 ohm output impedance. (not 55k)

So does that mean if I build an interconnect with a 55 ohm resistor in
the tip, the signal will be reduced 3 db??




On Wed, 18 Feb 2004 08:11:31 -0500, "MZ"
wrote:

My plan is to try a 10k resistor (guessing on the value here) inline with
the tip of the interconnect rca before the EQ.


Check your amp's manual. A lot of times there's a value for input
impedance. It's generally in the 10-30k range. Then just treat it as a
voltage divider.

PS Are you saying that this output isn't controlled by the volume knob?


Is this plan valid???

Another idea, better, would be to able to sum the mids & highs for the

back
speakers *without having them be summed to the front speakers. That way,
fronts and rear would be controlled by the head unit's volume control. I
don't know how to build this circuit, other than buying a little mixer.


It's a pain. Leave it alone.

Why not run just the low output to the rears? Where's the cutoff?


  #4   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

The spec in the manual says 55 ohm output impedance. (not 55k)

So does that mean if I build an interconnect with a 55 ohm resistor in
the tip, the signal will be reduced 3 db??


No, that's the output impedance of the head unit. What you need is the
input impedance of the amplifier. Oh yeah, and you may want to reduce by
more than 3dB.


  #5   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

The spec in the manual says 55 ohm output impedance. (not 55k)

So does that mean if I build an interconnect with a 55 ohm resistor in
the tip, the signal will be reduced 3 db??


No, that's the output impedance of the head unit. What you need is the
input impedance of the amplifier. Oh yeah, and you may want to reduce by
more than 3dB.


Oh yeah, the other way to do it is to use 2 resistors and create a voltage
divider.

Really, you don't need to know the input impedance. All it does for you is
give you a good starting point. If you can't find it out, just assume 20k
ohms.




  #6   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

PS-

If you use a potentiometer (say, 200k ohms or more range), you can dial in
the ideal value of resistance, then use a resistor of that value as the
permanent solution.

--
Mark
remove "remove" and "spam" to reply
"MZ" wrote in message
...
The spec in the manual says 55 ohm output impedance. (not 55k)

So does that mean if I build an interconnect with a 55 ohm resistor in
the tip, the signal will be reduced 3 db??


No, that's the output impedance of the head unit. What you need is the
input impedance of the amplifier. Oh yeah, and you may want to reduce

by
more than 3dB.


Oh yeah, the other way to do it is to use 2 resistors and create a voltage
divider.

Really, you don't need to know the input impedance. All it does for you

is
give you a good starting point. If you can't find it out, just assume 20k
ohms.




  #7   Report Post  
Kevin McMurtrie
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it.



In article ,
"MZ" wrote:

PS-

If you use a potentiometer (say, 200k ohms or more range), you can dial in
the ideal value of resistance, then use a resistor of that value as the
permanent solution.

--
Mark
remove "remove" and "spam" to reply
"MZ" wrote in message
...
The spec in the manual says 55 ohm output impedance. (not 55k)

So does that mean if I build an interconnect with a 55 ohm resistor in
the tip, the signal will be reduced 3 db??

No, that's the output impedance of the head unit. What you need is the
input impedance of the amplifier. Oh yeah, and you may want to reduce

by
more than 3dB.


Oh yeah, the other way to do it is to use 2 resistors and create a voltage
divider.

Really, you don't need to know the input impedance. All it does for you

is
give you a good starting point. If you can't find it out, just assume 20k
ohms.




  #8   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it.


No, 1kohm would be worthless. It would provide somewhere on the order of
0.2dB attenuation, which would be inaudible.

I suggested 200k ohm as a POT value, so that he has sufficient range to make
the determination. I suspect the resistance value that will provide the
results he's after won't exceed ~50kohm.


  #9   Report Post  
Jim Hoff
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq


I'll try the pot.

How do I wire it? Like this:

_____~~~~~~~~_
to head __ ^
| |________________
|_______________________to eq


Or do I want the wiper to sweep to ground?



On Thu, 19 Feb 2004 14:18:40 -0500, "MZ"
wrote:

200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it.


No, 1kohm would be worthless. It would provide somewhere on the order of
0.2dB attenuation, which would be inaudible.

I suggested 200k ohm as a POT value, so that he has sufficient range to make
the determination. I suspect the resistance value that will provide the
results he's after won't exceed ~50kohm.


  #10   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

No, don't run it to ground. You don't want to short things. Just run the
wiper out so that it forms a series circuit with the amplifier input.
Essentially, you're just treating it as a variable series resistor.

--
Mark
remove "remove" and "spam" to reply
"Jim Hoff" wrote in message
...

I'll try the pot.

How do I wire it? Like this:

_____~~~~~~~~_
to head __ ^
| |________________
|_______________________to eq


Or do I want the wiper to sweep to ground?



On Thu, 19 Feb 2004 14:18:40 -0500, "MZ"
wrote:

200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it.


No, 1kohm would be worthless. It would provide somewhere on the order of
0.2dB attenuation, which would be inaudible.

I suggested 200k ohm as a POT value, so that he has sufficient range to

make
the determination. I suspect the resistance value that will provide the
results he's after won't exceed ~50kohm.






  #11   Report Post  
Jim Hoff
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

Ok I'll try it like my diagram above and post if it works.

The only stereo potentiometer in town (Juneau, AK) was a 100k ohm. Thanks.




"MZ" wrote in message
...
No, don't run it to ground. You don't want to short things. Just run the
wiper out so that it forms a series circuit with the amplifier input.
Essentially, you're just treating it as a variable series resistor.

--
Mark
remove "remove" and "spam" to reply
"Jim Hoff" wrote in message
...

I'll try the pot.

How do I wire it? Like this:

_____~~~~~~~~_
to head __ ^
| |________________
|_______________________to eq


Or do I want the wiper to sweep to ground?



On Thu, 19 Feb 2004 14:18:40 -0500, "MZ"
wrote:

200K is way too high. The impedance will be too high to overcome

cable
capacitance and electric noise. 1K would be more like it.

No, 1kohm would be worthless. It would provide somewhere on the order

of
0.2dB attenuation, which would be inaudible.

I suggested 200k ohm as a POT value, so that he has sufficient range to

make
the determination. I suspect the resistance value that will provide

the
results he's after won't exceed ~50kohm.






  #12   Report Post  
Kevin McMurtrie
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

One side to the 8V audio output, one side to the signal ground, and the
wiper to the low voltage amp input. Use a pot in the area of 1K to 4K.

The 200K series-only pot idea doesn't work because the impedance will
become too high. The impedance to the amp should be no higher than a
few K Ohms to avoid loss of trebble, electrical noise pickup, and
distortion due to non-linearities of the amp input's impedance.


In article ,
Jim Hoff wrote:

I'll try the pot.

How do I wire it? Like this:

_____~~~~~~~~_
to head __ ^
| |________________
|_______________________to eq


Or do I want the wiper to sweep to ground?



On Thu, 19 Feb 2004 14:18:40 -0500, "MZ"
wrote:

200K is way too high. The impedance will be too high to overcome cable
capacitance and electric noise. 1K would be more like it.


No, 1kohm would be worthless. It would provide somewhere on the order of
0.2dB attenuation, which would be inaudible.

I suggested 200k ohm as a POT value, so that he has sufficient range to make
the determination. I suspect the resistance value that will provide the
results he's after won't exceed ~50kohm.


  #13   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

One side to the 8V audio output, one side to the signal ground, and the
wiper to the low voltage amp input. Use a pot in the area of 1K to 4K.


dB = 10ln(A/B) = 10ln(20/21) = -0.4dB

Sorry. Won't be effective.

The 200K series-only pot idea doesn't work because the impedance will
become too high. The impedance to the amp should be no higher than a
few K Ohms to avoid loss of trebble, electrical noise pickup, and
distortion due to non-linearities of the amp input's impedance.


What nonlinearities?

Um...and he can't set the amp's input impedance. It's predetermined by the
10kohm or greater resistor inside the chassis. He'll modify the input
impedance some by providing a series resistance, but it still shouldn't be
too high unless he has to introduce some major attenuation.


  #14   Report Post  
Kevin McMurtrie
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

In article ,
"MZ" wrote:

One side to the 8V audio output, one side to the signal ground, and the
wiper to the low voltage amp input. Use a pot in the area of 1K to 4K.


dB = 10ln(A/B) = 10ln(20/21) = -0.4dB

Sorry. Won't be effective.


It will be perfectly effective when wired correctly. This isn't a
series resistor. It's a voltage divider where the wiper varies between
the signal and the signal's ground.


The 200K series-only pot idea doesn't work because the impedance will
become too high. The impedance to the amp should be no higher than a
few K Ohms to avoid loss of trebble, electrical noise pickup, and
distortion due to non-linearities of the amp input's impedance.


What nonlinearities?


Amplifiers with the classic differential bipolar transistor inputs can
present a varying impedance if the circuit is not designed well. Since
car amps are often cheaply designed and a low impedance input is
expected, there could be significant distortion if a high value series
resistor is used.


Um...and he can't set the amp's input impedance. It's predetermined by the
10kohm or greater resistor inside the chassis. He'll modify the input
impedance some by providing a series resistance, but it still shouldn't be
too high unless he has to introduce some major attenuation.


  #15   Report Post  
MZ
 
Posts: n/a
Default "Fixed" 8 volt output o-loads eq

One side to the 8V audio output, one side to the signal ground, and
the
wiper to the low voltage amp input. Use a pot in the area of 1K to

4K.

dB = 10ln(A/B) = 10ln(20/21) = -0.4dB

Sorry. Won't be effective.


It will be perfectly effective when wired correctly. This isn't a
series resistor. It's a voltage divider where the wiper varies between
the signal and the signal's ground.


He'd be better off using a resistor for the final solution rather than a
pot, considering his application. Again, I only recommended the pot so that
he could pinpoint the correct resistance value. As I mentioned in a
previous post, a v divider would work fine. But it's not necessary.

The 200K series-only pot idea doesn't work because the impedance will
become too high. The impedance to the amp should be no higher than a
few K Ohms to avoid loss of trebble, electrical noise pickup, and
distortion due to non-linearities of the amp input's impedance.


What nonlinearities?


Amplifiers with the classic differential bipolar transistor inputs can
present a varying impedance if the circuit is not designed well. Since
car amps are often cheaply designed and a low impedance input is
expected, there could be significant distortion if a high value series
resistor is used.


Low impedance input isn't expected. The vast majority of amplifiers have
input impedances greater than 10kohms. Most of the time, the inputs look
like a load resistor and an op amp.


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