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#1 |
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diyAudio Member
Join Date: Jul 2008
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Hi,
I realize this question may be suitable for another forum, but this is the only place I've found where people have the technical knowledge to answer it. I would appreciate some insight into the questions I have. On one particular headphone forum I visit there has been some discussion about using an integrated speaker amplifier to power headphones instead of using a dedicated headphone amplifier. A lot of integrated amps power the headphone jack by using resistors between the headphone jack and the main speaker out. One particular issue with doing this is that the impedance at the headphone jack is can be fairly high (>60 ohms). So, in theory, if we use a headphone with a 25 ohm impedance, then Z output (60 ohms) is greater than Z load (25 ohms). But here is what I don't understand. Why is Z output measured at the headphone jack? Since the headphone jack is connected to the main speaker out via resistors, why isn't Z output measured at the speaker terminals, at the point before the resistors connected to the headphone jack? So, if we have a 60 ohm headphone jack and a 25 ohm headphone, isn't the Z load 85 ohms? Isn't not like the speaker terminals know they are going to a headphone jack and the headphones are a separate entity. Don't the speaker terminals just see one load? If I have a 600 ohm headphone (assuming the impedance is flat at all frequencies) isn't that creating the same load at the speaker out as a 300 ohm headphone jack and a 300 ohm headphone? If you haven't guessed already I'm not an engineer. I know some basic electronic theory and electronics DIY. So please spell things out a little |
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#2 |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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As always, I am bound to say, Rod Elliott has got some advices.
Seems like whatever issue somebody brings up here, the man has been there - has written something about it Headphone Adaptor for Power Amplifiers I am quite sure he deals with those issues of impedance matching you enquire about. http://sound.westhost.com/project100.htm I could write some story of my own, if you gave me all the facts of your output arrangement and your setup. But I leave you for some explorations of your own. And later, if you have any further questions after reading that project by our Australian Diy Audio friend I will try to answer as best I can. Have some good reading at ESP http://sound.westhost.com/ Here is the best #1 page of them all: http://sound.westhost.com/site-map.htm Lineup
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lineup |
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#3 |
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diyAudio Member
Join Date: Apr 2008
Location: Carlisle, England
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I have never seen headphones connected to the louspeaker o/p before. But i tend to use amps of many hundreds of watts so that could be why.
I have driven headphones on a disco from an op-amp with reasonable results. I could certainly hear Ok with the disco going full bore.
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http://www.murtonpikesystems.co.uk PCBCAD40 pcb design software. |
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#4 | |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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Quote:
Power amplifiers for loudspeakers and headphones amplifiers/line amplifiers for headphones. This is the sound thing. Dedicated amplifiers, like for 600 Ohms headphones, would outperform any power amplifier. When somebody has bought a good 600 Ohm headphones pair (probably Sennheiser) it would be sort of a shame not to give them a good chance. After all, hi-fi headphones performs almost 20 dB lower in distortion than hi-fi loudspeakers. This level is certainly something humans can possible hear a difference of. A dedicated high quality headphone amplifier, will walk miles around even the best 100 Watt integrated power amplifiers. Because i think your 600 Ohm's Headphones might deserve it ![]() May I suggest some High Fidelity project from either: http://www.headwize.com/projects/index.htm http://www.head-fi.org/forums/f6/ Lineup
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#5 | |
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Banned
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Quote:
Your basic understanding of the situation is correct. Engineers sometimes describe things loosely inter se, however, and the exact interpretation of the phraseology used to describe an arrangement may depend on context. If the writer is primarily concerned with the analysis from the headphone POV the output impedances may be lumped together at the headphone output. w |
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#6 |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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which actually means what?
practically?
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#7 |
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diyAudio Member
Join Date: Jul 2008
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Hi lineup,
Unfortunately that page doesn't answer my question ![]() I'm not looking to make a integrated -> headphone amp adapter. But in the case of the headphone adapter on that page, I'm wondering why the Z output is measured at the end of the headphone adapter and not where it is connected to the speaker terminals. In the case of that page, isn't Z load 120 ohm (or near that) and and Z output close to 0? |
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#9 | ||
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Banned
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Quote:
Quote:
w |
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#10 |
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diyAudio Member
Join Date: Jul 2008
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Hi Everybody.
Thanks for the answers. Unfortunately I'm still not getting it. Let me put out an example. Let's say I have a 50W (at 8 ohms) amp. The specifications of this amp state that the impedance of the headphone jack is 68 ohms. The engineers who built the amp did this by connecting the headphone jack to the main speaker out via two 68 ohm resistors (one for each channel). Now, I have two headphones. Headphone A has an impedance of 25 ohms. Headphone B has an impedance of 300 ohms. Some basic rules of thumb present a problem here. If I plug in headphone A (25 ohms) into the 68 ohm headphone jack, the Z load (headphone impedance) is lower than the Z output (headphone jack impedance). This causes changes in the frequency response of the audio and the damping factor is close to zero. The 25 ohm headphone will probably sound bad. However, when we use the headphone B (300 ohm) the Z load is higher than the Z output. So the problems seen with headphone A are not as great. But...why is Z output taken at the headphone jack? I understand why the amp maker would want to specify it, but isn't the impedance in series? Wouldn't a 68 ohm headphone jack and a 25 ohm headphone make a 93 ohm load on the amp? Thus, by what I've said, even the Z load now (93 ohms) is greater than Z output (close to 0) at the speaker out. My theory is that regardless of the headphone, the Z load will always be greater than the Z output. I'm sorry if I'm being so detailed about this. It's a reflection of how much I need it explained to me. If somebody wants a basic schematic I suppose I could draw one up. |
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