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Adding impedance makes the Denon's D5000 bass a little bit less prominent?

post #1 of 16
6/17/09 at 7:04pm
Thread Starter 
Hi, i've tried to add an impedance adapter (33 ohms) to my Denon D5000, and it seems to my ears that the bass tends to be a little bit less prominent. It's a very subtle difference, but it seems enought to be noticed.

Could this be possible?
Anyone tried something like this?

Just a curiosity.
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post #2 of 16
6/17/09 at 7:14pm
Are you adding 33 ohm resistors in series between the amplifier and headphone?

Certainly increasing the source impedance driving the headphone will alter their frequency response and tonal balance some.
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post #3 of 16
6/17/09 at 7:24pm
Thread Starter 
Quote:
Originally Posted by Donald North View Post
Are you adding 33 ohm resistors in series between the amplifier and headphone?
I've used the Etymotic resistor/adaptor, which can be easily bought on ebay.
I've putted this between the amplifier and the headphone.
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post #4 of 16
6/17/09 at 7:29pm
capacitor, inductors will alter the resultant frequency by lagging or leading, putting it out of phase. For a straight up resistor however, i doubt it will alter the freq at all, at least there won't be an audible difference except maybe lowered volume. Also resistors have tolerances, unless they are super high precision ones, and that can never be good. There should be better ways to make the bass less prominent such as EQ or markl mod (although I haven't tried markl mod myself, I've read what it can do).

Hope this helps.

JY
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post #5 of 16
6/17/09 at 7:40pm
No, an added resistor forms a voltage divider with the impedance of the headphone, which usually varies with frequency, thereby modifying the frequency response some. Headroom's impedance measurement of the D5000 is shown with such a large scale that it looks flat. However if zoomed into the range of 0 - 50 ohm, I'm sure we'll see some variation.
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post #6 of 16
6/17/09 at 7:53pm
Donald North,

Are you sure it forms a voltage divider?
To form a voltage divider, you much have parrellel connection, and from what I've read the resistor is in series in OP's attempt.

Also I am not sure how having a voltage divider would effect the frequency response... Say, if you added x ft. length cable between the amp and headphones that has 33ohm resistance, would that really alter the frequency response of the headphone?

EDIT:
ok, above example with the cable is a bad one. How about a audio splitter? if you split the source signal into 2 and use 2 headphones to listen it the source, would they sound different? I would think the volume would be much lower but I don't think the sound itself will change at all.

JY
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post #7 of 16
6/17/09 at 8:26pm
A signal splitter is more complicated to explain and different that the OP, so let's stick with discussing the addition of a series resistor.

Let's say a headphone has an impedance of 50 ohm with max at bass resonance of 100 ohm. Power this with an amplifier with has zero output impedance. The voltage delivered across the headphone will be the same at all frequencies. Now let's add a series resistor between the amplifier and headphone. Calculating with Ohm's law, the added series resistor now forms a voltage divider with the headphones, and the voltage delivered across the headphone will be maximum at the 100 ohm resonant frequency and lower elsewhere. This greater voltage at one frequency versus the others will alter the frequency response delivered to and produced by the headphone.

Here's a recent thread which measures the frequency response on the headphone output of several different amplifiers with several different headphones. For a given headphone, some amplifiers will have more contoured frequency response than others due to their higher output impedance.
http://www.head-fi.org/forums/f133/h...ne-abx-429619/
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post #8 of 16
6/17/09 at 10:26pm
I'm having a little difficulty understanding what you are trying to explain

I've included the picture I had drawn in my had, (and excuse me for my poor paint drawing)

I don't really see a voltage divider happening in the circuit.
And when you've mentioned Ohm's law, were you referring to the voltage divider equation?

Vout = Vin * (R2 / (R1 + R2))

I fail to see how a frequency can be altered in anyway with the equation above.

To my understanding,
Original signal -> voltage divider -> 2 signals with different peak voltages, but frequency still in phase -> combined -> Original signal

Please correct me if I am wrong! I am curious and willing to learn
LL
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post #9 of 16
6/17/09 at 10:42pm
No problem, I'll try to explain further:

R1 is the added resistor; R2 is the headphone.

Let's say R1 is 33 ohm and R2 is 50 ohm nominal headphone with impedance peak of 100 ohm and 100 Hz (bass resonant frequency). Remember: Nearly every dynamic headphone is reactive impedance; not purely resistive.

Vin = 5Vrms

Vout (voltage across headphone) = 5V * (50/33+50) = 3V at all frequencies BUT 100Hz

Vout = 5V * (100/33+100) = 3.76V at 100 Hz bass resonant frequency

The result is a contouring effect of the frequency response based on the headphone's impedance curve and added series resistor value. The larger the resistor, the greater the contouring.
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post #10 of 16
6/17/09 at 11:11pm
Donald North,

Of course!

Thank you! It makes very good sense now!

Quote:
Originally Posted by Donald North View Post
Remember: Nearly every dynamic headphone is reactive impedance; not purely resistive.
I totally forgot about that

Thanks again for bearing with me

Cheers,
JY
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post #11 of 16
6/18/09 at 5:52am
Thread Starter 
It should be interesting if somebody who owns the Denons and the adapter could try to do this.
To my ears, the bass decreases a little bit.
I'm curious to know if somebody else can experience this.
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post #12 of 16
6/18/09 at 11:19am
Quote:
Originally Posted by archigius View Post
It should be interesting if somebody who owns the Denons and the adapter could try to do this.
To my ears, the bass decreases a little bit.
I'm curious to know if somebody else can experience this.
I've tried it with an amplifier with a ~50ohm output impedance. The bass was audibly increased with a measured bass hump (I think at about 50 or 60 hz) of about 3db.

Are you sure you are listening at matched volumes? Could it be possible you are listening at a slightly lower volume because adding a resistor (which is what you are doing) will decrease the volume a bit if you keep the volume setting on the amp the same. That would result in a perceived lowering of bass level.
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post #13 of 16
6/19/09 at 4:43am
Thread Starter 
Quote:
Originally Posted by odigg View Post
Are you sure you are listening at matched volumes? Could it be possible you are listening at a slightly lower volume because adding a resistor (which is what you are doing) will decrease the volume a bit if you keep the volume setting on the amp the same. That would result in a perceived lowering of bass level.
Yes, when i put the adapter, i raise the volume. I'm pretty sure the perceived bass is less, this is very strange. It seems, to my ears, that the also upper hights become less energetic.
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post #14 of 16
6/19/09 at 10:30am
Quote:
Originally Posted by archigius View Post
Yes, when i put the adapter, i raise the volume. I'm pretty sure the perceived bass is less, this is very strange. It seems, to my ears, that the also upper hights become less energetic.
Adding a resistor, especially with a resistance higher than the impedance of the Denons, will change the frequency response. It's also dependent on an amp.

Is it possible to try it with another amp to see if you get the same result?
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post #15 of 16
6/19/09 at 12:59pm
Thread Starter 
Quote:
Originally Posted by odigg View Post
Adding a resistor, especially with a resistance higher than the impedance of the Denons, will change the frequency response. It's also dependent on an amp.

Is it possible to try it with another amp to see if you get the same result?
No, it isn't possible, i have only one amp...
But the headphone output of my cd player gives the same results.
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