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Increasing headphone impedance to reduce background noise (using iPAQ)

Posted on 2004-09-10
Last Modified: 2012-06-27
Hi - Our engineer is unavailable and I thought someone might be able to answer this quickly.  My electrical engineering skills are rusty!
To reduce background noise, we have a need to increase the impedance of a set of professional headphones used by our hearing testing software that runs on an iPAQ.  We purchase the transducers and build a custom cable with a jack capable of fitting into the iPAQ, and we want to insert some circuitry to increase the impedance so we reduce the background noise.  We're using 50 ohm transducers, and we want to bump up the impedance to something over 100 ohm --- the exact value to be determined as we test different impedances and output levels.  Here's a rough sketch of the circuit we are using in each side of the headphone cable.  We get a total impedance of 100 ohms (Z) when R1=120 ohm, R2= 390 ohm, R3=100 ohm, and C1=100,000pf.  Question: What would the R and C values need to be for Z=110,120,130,140,150,200,250 ohms?  The frequencies we use range from 250Hz to 8kHz.  Thanks in advance!
PS Any recommendations on companies or individuals who could create the cables and/or printed circuitry for us in small make to order quantities? (under 50 units at a time)
\_____/ speaker 50 ohm
 +| -|
   |  |
   |  +--------------+
   |                      |
   |      R1             |
   |                       |
   +----------+        |
   |             |         |
   |             |         |
   |            ---        |
   >           /-\ C1   |    
   >            |         |
   > R2       |         |
   >            |         |
   |             |        |
   +----------+       |
   |                      |
   |                      |
   |      R3            |        
   |                      |
   |  +--------------+
 +| -|
   |  | Z=total impedance
   \  /
    || 3.5mm jack
Question by:Brett8

Expert Comment

ID: 12032294
So essentially you have to work out 1/(1/(R1+R3) + 1/R2 + 1/Rc) + 50. Where Rc is the resistance of the capacitor (which is frequency dependent) and the 50 is the impedance of the speaker itself. In your case you must have measured the capacitor at 78 ohms.

To bump the overall resistance up you can just bump up a single R value at a time.

I'm curious as to how your solution will lead to reduced noise. An impedance mismatch will cause reflections on the cable and will lead to increased noise. You should measure the impedance of the source and match your speaker and cable to that.

Expert Comment

ID: 12032299
As a silly suggestion, why not just use a variable resistor/potentiometer to adjust the impedance as you listen? When you find the best value you can just measure the setting and use a fixed resistor of the same value.

Author Comment

ID: 12033436
Thanks Neil.  That's a big help.  We're getting rid of the background noise --- noise which is there when the headphones are plugged in but no signal is provided --- higher impedance seems to work for this.  Since the capacitor resistance seems to quickly become the variable that most influences the circuit impedance (e.g. the impedance is roughly the same for R1,R2,R3 either 10k each or 100k each), I'm curious as to what happens at the various frequencies.  How widely does the capacitor resistance vary?  What does the capacitor do for us?  Reduce those reflections?  What would happen if we use a resistor in series (just R2, and eliminate R1, R3, and C1)?
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LVL 31

Assisted Solution

rid earned 50 total points
ID: 12033530
The capacitor reactance is: capacitance / 2 x [pi] x frequency.
I was wondering about that C... it would make higher frequencies come through with greater loudness.

Accepted Solution

Neil_Simpson earned 400 total points
ID: 12033735

If you eliminanted all but one reistor you would have a total impedance of 50ohms (for the speaker) + R (the single resistor value). It's a whole lot simpler. You may find that it sounds better with the capacitor as it will be acting as a high pass filter (http://www.st-andrews.ac.uk/~www_pa/Scots_Guide/experiment/highpass/hpf.html). My electronics is a little rusty (and I've just woken up) so I can't work this out exactly at the moment.

The reason the capacitor dominates the resistance at the moment is that in the calculation you'll find it impossible to get a value higher than the smallest parallel resistor, it's just the nature of things.

The reflections will be reduced when the impdance of your cable is matched to (i.e. the same as) the imput impedance and the impedance of the load.
LVL 14

Assisted Solution

tmj883 earned 50 total points
ID: 12035135
Hope the link works, as this is a very nice free utility to construct and simulate circuits prior to building the same.

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