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Chip Amps Amplifiers based on integrated circuits 

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20th February 2004, 03:52 AM  #11 
diyAudio Member
Join Date: Aug 2002
Location: Washington

hi digi01
could you verify the value of the R in the attenuator?
1) 100 Ohm 2) 330 Ohm 3) 1K 4) 5.1 K 5) 4.7 K is that correct? JP 
20th February 2004, 04:29 AM  #12  
diyAudio Member
Join Date: May 2003
Location: Berlin

Okay, let me split the discussion in the two departments of (1) finding (and understanding) the right gain structure and (2) finding a (correct) implementation of it with the opa2604 and the opa548.
While I can't confirm your circuit topology or components choice (not enough experience with the 548), I'll try to solve the first problem first. In the meantime i've found a more complete calculation tool over here: http://wwwk.ext.ti.com/srvs/cgibin...i.exe?Company={c7698c3faf4f4accbbc3e615ef792d18},kb=analog,case=obj(32624),new The DIYAudio software can't handle the syntax of the url, please copy the whole string into a new browser window. It features all neccessary calculation rules (in the bottom area of the site) an does a conversion of "Voltage <> Power". As one can clearly try out, the relation between dBm (Power) and Vrms (Voltage) of course depends on the impedance. Quote:
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The scale of dBm references a desired power level to a given power level, assuming equal impedances (or 600Ohm according to definition). The point is that you don't want to implement an actual amplification between two powers but between a voltage source and an impedance load! "Amplification" is defined as a relation (division) between an input unit and an output unit in the same domain. Consequently, the resulting amplification factor is dimensionless, thus handled in dB (without an "m" or "V"). This makes your 10*log(1000mW/1mW) = 30dB. In your application, You've got a given source voltage (or a nominal source level, referenced to a voltage) and a given load impedance (of a rough 120Ohm). And now you're trying to assume an amplification between an input and an output power level, and that's what doesn't fit!!! You could either do a correct calculation into the dBm domain by referencing the 0dBm level to the given impedance (difficult and uncool...) or calculate it the easy way: find the correct amplification factor to achieve your desired output level in the domain of the input signal, because that's the way it is intended. In order to make this possible, opamps have a high input impedance (so that the impedance of the source becomes negligible and ideally no current flow is needed, hence abstracting from the input power to the input voltage) and a low output impedance (so that the output voltage doesn't depend on the load impedance and all neccessary current can be supplied and becomes irrelevant again, hence abstracting again). In normal opamps, like the opa2604, the output impedance is low in relation to the input impedance of other opamps. But as the K1000 has an input impedance that is significantly lower than that (and too low for an opamp like the opa2604), the above working principle would be violated due to insufficient current capability: the output current would become important and the above abstraction would no longer apply. That's where the opa548 comes into play: it is capable of delivering the neccessary current, thus fulfilling the above abstraction scheme. It is basically "just" an opamp with a fat ouput stage. Consequently the output voltage is (ideally) assumed to be independend of the flowing current, the limit being when the conducted current is overloading the internal circuit and the opamp doesn't work in a "linear" way any longer. That's the exact explanation why people correctly want an amplifier that could supply more current than it will have to, because the amplifying circuit shall be kept in it's linear range and prevented from overheating or power compression (note: this relevant ratio between load impedance and amp output impedance is called the amp's damping factor, which people tend to like high ). To resume the above said:  Amplification is always done from an input signal domain into the same domain as output, e.g. input voltage to ouput voltage.  Output power is regulated by the output voltage, not the output impedance (ideally assuming a constantly low output impedance). Thus the effort of calculating the output power can easily be reduced to calculating the output voltage using Ohm's law (U=R*I > I=U/R; P=U*I; U*I=U*U/R; > P=U*U/R). Quote:
Now lets get some things calculated: depending on your given input voltage (which I assumed to be around 1V for the ease of calculation) and your given load impedance of 120Ohm, we can at first calculate the desired output voltage to U=sqrt(P*R)=10.95V (as 10.95V*10.95V/120Ohm=1W) and then calculate the desired amplification factor (in the voltage domain) as 20*log(10.95V/1V)=20.78dB. Please take this rough 20dB as a starting point only. The exact value you want will depend on your input voltage (as e.g. 2V would be amplified to 20.93V, blowing the beloved K1000 away with 20.93V*20.93V/120Ohm=4W. Start with a lower gain and try to listen to some music (after testing with a cheap speaker, of course). You will probably recognize that you don't want it louder (as the maximum output voltage does not depend on the gain setting, but on the maximum input voltage below clipping). I'd suggest you have a look at other well respected headphoneamplifiers and investigate their gain structure. As we have just learned, the desired gain does only depend on used input signals and desired output voltage (and thus power into a certain impedance) and not on the used devices (assuming they are good enough). Correct me if I'm wrong Sebastian. PS: The mentioned 1mW for an SPL of 74dB is just a reference to the phone's efficiency. This value is not used (or needed) to determine the desired gain structure of an amp. It basically just helps you to compare the K1000's efficiency (sound pressure level per power input) with other headphones (e.g. louder/more quiet when connecting to the same amplifier at the same volume setting). 

24th February 2004, 06:46 AM  #13  
diyAudio Member

sek,good article,it is so long
I can't get your on line calculation tool.Are there any other ways that visit this url? Quote:
£Ðm=10 lg(£Ð/1)dBm £Ðv=20Lg(£Õ/0.775) (dB) £Ðm=£Ðv£«10Lg£¨600/£Ú£©(dBm) I think that these can supplemented you refer to the calculation based on dBm (Power),and the relation between dBm and Vrms. Quote:
Even if I spend the voltages of +  18V , I think the op's voltage output swing can not export all PSU voltage value. Like usual GC, the voltage of the power is in +18 to +22 volts. voltage Gain set to 22. We get the voltage it is low too in fact. There is a efficiency question of the power chip stage inside. In fact, the voltage of the audio is a dynamic signal, can only use Ohm's law to be estimated cheers 

24th February 2004, 06:49 AM  #14  
diyAudio Member

Quote:


24th February 2004, 07:19 AM  #15 
diyAudio Member
Join Date: Aug 2002
Location: Washington

ok
thanks for the info
JP 
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