Originally Posted by Navyblue
Does amplifier fix it output voltage, output current or none of the above? Since they are normally rated with their power rating.
Typically amplifiers don't fix output voltage or current (so to answer: none of the above). There are certain cases where that's not true -- for example, the amplifier section of an FM radio -- but that's a very unique situation. They do that since the signal should be only changing in terms of frequency, the amplitude would need to remain the same, otherwise you'd get wonky output -- like if you drew a sine wave of two different amplitudes (but same frequency) and found to slope at the same point, it would be different... and FM is all about the slope, so they design it to keep the amplitude constant.
If it's a power amplifier (with no volume control), it's just got a constant gain, some maximum output voltage, and maximum output current. So if you had a 50W power amplifier, it would list a maximum RMS or peak-to-peak input voltage (so that the amplified output wouldn't be clipping).
Originally Posted by Navyblue
Why does R1 power matters? Is it equivalent to what the headphone is getting? Or it has some indirect relationship with what the headphone is getting?
R1 power matters for one reason. In the schematic, it accounts for most of the power dissipation. So you have to make sure it doesn't explode, and you want to keep the power "loss" as low as possible. I came up with my values through guess-and-check, but there might be a relationship. But really, if R1 and R3 increase in value, then less power will be delivered to the headphone impedance... oppositely, if R2 increases in value, more power will be delivered to the headphone. Just think of R1 and R3 as series loads (getting in the way), and R2 as a shunt load (diverting power away).
Originally Posted by Navyblue
Why does the author fix the Zout at 120 ohms and V out at 5V? Is it a standard requirement for headphone?
I think it was just an effort to make a matched-impedance case (the author is designing the thing for 120 ohm headphones). Where Zout = Zload. For the 120 Ohm case, if you had a Vrms of 5V at the headphones, you'd be getting about 200mW -- which would blow your ears off I think. The author chose this value so that even if you cranked the amplifier up really loud, it wouldn't clip on the output. So 5V is relative for this design.
In RMS values:
V(desired) = sqrt(Power(max)*Zload) -- assuming V(desired) isn't larger than the amplifier can source
Originally Posted by Navyblue
Why does the author did not take into account of the impedance of the headphone? Does it matter at all?
The author does in a way... that's the reason for choosing Zout to be 120. The impedance does matter, if you had the same output circuit and you switched from a 120 Ohm headphone to a 600 Ohm headphone, the max power output would be seriously diminished.
Originally Posted by Navyblue
Since amplifiers are designed to push 8 ohms (or around there) speakers, would it be best if the circuit is designed so that the amplifier "sees" 8 ohms? Is it too high or too low impedance that would damage the amplifier?
Normally I would say yes... but that would be if you wanted to deliver maximum power. What that would result in would be a matching network, so you'd have the input impedance of the circuit set to the output impedance of the amplifier, and the output impedance of the circuit set to the load impedance of your headphones. However, you don't want 20W or more going into your headphones
! So instead we ignore the input impedance of the circuit so that it gets pretty mis-matched and you don't get the full power transmitted into the headphone load. If you have too low of an impedance you can damage the amplifier -- it would basically be shorting the output, so all of the power of the amplifier would get dissipated in its own output transistors, and they would burn up. Usually you can't blow up an amplifier by having too high of an impedance, otherwise your amplifier would explode every time it didn't have a load or speaker attached.
Originally Posted by Navyblue
Assuming the above are true, should the circuit be designed so that the headphone "sees" 120 ohms while the amplifier "sees" 8 ohms, while at the same time providing a maximum of 50 mW to the headphone? Is it possible at all?
The headphone should see its own impedance from the output of the circuit. So if it's a 300 Ohm can, design the circuit for 300 ohm output impedance. If you plan on using a bunch of different impedance headphones, design for the worst case -- the lowest impedance -- that way you don't blow them up. If you made it for say 200mW for a high impedance headphone (say 600 Ohms), and then threw some 32 Ohm headphones on there, they would get a LOT more power than 200mW.
I'm sure it's possible to design a matched circuit, but in this case, you don't want to. We're trying to cut down the power from >20W to less than 100mW or so. I used 50mW earlier just because -- the specifications on the beyerdynamic website said 100mW max, so I just went conservative (better not to blow something up, plus I doubt you'd be using them full power).
Originally Posted by Navyblue
Why isn't it simply Req = 1 / (1/R2 + 1/R3) ?
Well you want to know the voltage when the headphone is connected (it doesn't really matter if the circuit is just sitting there doing nothing). If the headphone isn't connected, then the Req would just be "R2". If the headphone is connected, it's a pretty large resistance/impedance (and I'll just go ahead and mention that for this stuff, you can use the two terms interchangeably -- if you add in capacitors and inductors, then you wouldn't be able to use them interchangeably) that you need to account for. It's in series with R3, so you just add them together. Then find the parallel combination of that blob and R2.
Originally Posted by Navyblue
Btw, when resistor fries, do they just stop letting current through? I guess they won't let everything pass when they malfuction?
Yep, usually resistors fail "open," meaning they burn up and nothing is left connected. Keep in mind the "usually" part, I guess they could potentially burn up and get shorted together. My experience in how parts fail is kind of limited
Originally Posted by Navyblue
50mW is the rating for 250 ohms DT880, can it be assumed that it is the same for the 600 ohms version?
I don't know. I wouldn't assume anything when expensive parts are involved
So you could either try and find a specification sheet for the headphones, or just call Beyerdynamic and ask.
Originally Posted by Navyblue
Please pardon me for these basic questions. I'm just a non electronic guy happen to be brave enough to tinker with my equipment. I just hope I won't blow up anything.
I'm just impressed whenever someone tries to figure this stuff out on their own. I recently got my degree in electrical engineering, and I remember how utterly confused I was when I first started learning. Heck, I'm still confused most of the time
Originally Posted by Navyblue
I'm thinking of:
R1 = 3 ohms
R2 = 5 ohms
R3 = 120 ohms
Would this work? This way the amplifier is seeing about 8 ohms and the headphone will be seeing about 120 ohms. And a 300 ohms headphone will see about a maximum of 400 mW rms. Is my calculation correct?
The problem will be the 3 and 5 ohms resistor would need a high power rating, which are huge and expensive. And do they introduce a lot of interference?
I wouldn't use it
. In theory it should work, but it would be so inefficient that it's not even funny. (I'm assuming you're using a 50W amplifier which has a 20V rms output voltage)
You are right though, the amplifier sees about 8 ohms, and the headphone would see about 120 ohms. I think you're off by a factor of sqrt(2) somewhere on the power. I used pspice to simulate and got about 263mW rms -- which when you multiply by sqrt(2) you get reasonably close to 400.
You would be hard-pressed to find any reasonably priced resistors with those power ratings (you'd have to either put a bunch in parallel or series to get the right resistances and combined power rating). I don't think they would cause any interference since they're just passive, although you never really know until you test them.
I calculated R1 = 300 Ohm, R2 = 750 Ohm, and R3 = 250 Ohm. That won't have a matched output impedance, but it will give pretty consistent results (within 10% of 100mW) for anything higher than about 200 Ohm impedance.
Originally Posted by violante
I'll apologize in advance for the partial thread jack, but do any of you know of any particularly useful resources for an introduction to power from an audio perspective? Electronic or print is fine, it's just a bit overwhelming to know where to start!
I don't know of anything off the top of my head. I'd start with a book on circuit analysis, one that would cover (at a minimum) these topics:
-Series/Parallel combinations (resistors)
-Thevenin/Norton equivalent circuits
-Mesh current analysis (need to be able to do math with matrices [linear algebra] or solve simultaneous equations)
-Node voltage analysis (just algebra)
-R/C circuits (Resistor-capacitor filters)
-R/L circuits (resistor-inductor)
-RLC circuits (resistor, inductor, capacitor)
Once you've got that down, you could pick up a book on audio or power. It's really really helpful to know the other stuff first though.
Also google "Pspice student" and try and find a copy of the software for free. It's useful if you don't feel like doing the math, or if it's a complicated circuit where doing the math would be almost impossible. It's got something of a learning curve to it, but it will save you time later.
Need help with calculation and speaker output
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