i heven't finish it,I am considering the overall arrangement and structure now.The type of my opa548 package is DDPK,so It will be difficult to install.
on the another hand,K1000 is a very special headphone,need more large power.
these is some specifications of K1000:
Impedance: 120 Ohms
Sensitivity: 74dB for 1mW free field
Frequency Response: 30Hz to 25kHz
Power Requirement : 100mW for 1Pa
Distortion at 400mW: around 100dB
- 2nd harmonic <0.5% 200Hz to 2kHz
<1.0% 100Hz to 200Hz
Power Rating: 1000mW (test tone DIN 45582) corresponds to approx 104dB
So,I think the range of K1000 suitable powers is between 100 and 1000mws.Pmax=10lg(1000/1)=30dBm.
If the source has a 0dB output such as the portable cd player,Then the amp needs 30-10lg(600/120)=23dB to get the MAX 104dB SPL.
I am using BB's opa548 making this amp now.The OPA548 is a high-voltage/high-current op amp£¬and it can driving a wide variety of loads. Because sensitivity of K1000 is relatively low,and Its power range is relatively wide(1mw~1000mw).So a single gain stage will be not good.i have to do some more thinking after system matching tests...
now,I set my k1000 amp in this way,a +10dB pre-amp and a +26dB output driver amp.The pre-amp can be bypass.I wanted to know if anyone had any thoughts on the design
thanks a lot.
on the another hand,K1000 is a very special headphone,need more large power.
these is some specifications of K1000:
Impedance: 120 Ohms
Sensitivity: 74dB for 1mW free field
Frequency Response: 30Hz to 25kHz
Power Requirement : 100mW for 1Pa
Distortion at 400mW: around 100dB
- 2nd harmonic <0.5% 200Hz to 2kHz
<1.0% 100Hz to 200Hz
Power Rating: 1000mW (test tone DIN 45582) corresponds to approx 104dB
So,I think the range of K1000 suitable powers is between 100 and 1000mws.Pmax=10lg(1000/1)=30dBm.
If the source has a 0dB output such as the portable cd player,Then the amp needs 30-10lg(600/120)=23dB to get the MAX 104dB SPL.
I am using BB's opa548 making this amp now.The OPA548 is a high-voltage/high-current op amp£¬and it can driving a wide variety of loads. Because sensitivity of K1000 is relatively low,and Its power range is relatively wide(1mw~1000mw).So a single gain stage will be not good.i have to do some more thinking after system matching tests...
now,I set my k1000 amp in this way,a +10dB pre-amp and a +26dB output driver amp.The pre-amp can be bypass.I wanted to know if anyone had any thoughts on the design
thanks a lot.
Attachments
I was asking the above question because I've got OPA541 and OPA549 laying around, I wanted to use them for an active speaker system.
I pretty much like your amp. Although I usually tend to dislike P2P wiring, I find your's actually beautiful. I guess it's because of the copper.
Let me just calculate that through. The K1000 has a nominal impedance of 120R. Do you have (or can you measure) an accurate impedance curve, btw.?
Given a maximum desired output of 1000mW, the good old equation P=U*U/R lets me think of a voltage between 10V and 11V to achive 1000mW into 120R.
You're planning to implement an overall gain of +36dB, including pre- and power stage. Man, what kind of sources do you use?
The power level of 0dBm is defined to be a power of 1mW (into a 600R load, according to definition). A voltage level of 0dBV is defined to be 1Vrms. That means, calculating a resulting power level out of a given voltage level involves your (K1000's) input impedance.
Calculating is done with the equations (in the voltage domain) dB = 20 log (V1/V2) and (in the power domain) dB = 10 log (P1/P2).
Also, have a look at a handy tool here: http://www.sengpielaudio.com/calculator-db-volt.htm
Well, 0dB in respect to what? As a Bel scale always has to be referenced, it's important to be sure of the correct calculation here. The K1000 is just too expensive to expose it to an amp with a wrong gain structure...
Considering a rough 0dBV as the output of your cd player, your 36dB of gain (voltage gain of 63, according to the above equation) would result in an opa's output of 63V. Assuming a flat 120R impedance curve, this would result in an output power of P = 63V*63V/120R = 33W.
Your cd player will probably easily deliver you an output voltage of 1Vrms (and more). Who's making the mistake? Me, I hope.
As you probably want to use your head-amp independently of your portable cd player (not glued together, that is), you should stick with an input sensitivity that fits some standards. Or one that fits your source(s), if you want to be individual.
An output power of 1000mW with a standard source of -10dBV (0.316V, older HiFi-standard, according to DIN and ISO) would require an overall gain of a rough 30dB (voltage amplification factor of nearly 32).
OTOH, an output power of 1000mW with a standard source of +4dBu (1.224V, professional standard, according to IEEE, ITU and whatever) would require an overall gain of a rough 18dB (voltage amplification factor of 8).
What? More? 
Well, just one: Have you ever wondered why next to nobody's actually using a power amp for headphone amplification?
BTW: 1000mW is not extraordinarily much power for a headphone. The BUF634, for example, can deliver a rough 250mA, works with a +/-15V supply and integrates perfectly with most opamps as a (voltage) gain stage. That makes the BUF634 deliver (say, at 12V output voltage capability) 1.200mW into 120R. It's too low for the K1000 to have headroom left (because of the relatively high 120R), but a 50W power chip-amp as the next bigger step? Think about it...
I pretty much like your amp. Although I usually tend to dislike P2P wiring, I find your's actually beautiful. I guess it's because of the copper.
Let me just calculate that through. The K1000 has a nominal impedance of 120R. Do you have (or can you measure) an accurate impedance curve, btw.?
Given a maximum desired output of 1000mW, the good old equation P=U*U/R lets me think of a voltage between 10V and 11V to achive 1000mW into 120R.
You're planning to implement an overall gain of +36dB, including pre- and power stage. Man, what kind of sources do you use?
The power level of 0dBm is defined to be a power of 1mW (into a 600R load, according to definition). A voltage level of 0dBV is defined to be 1Vrms. That means, calculating a resulting power level out of a given voltage level involves your (K1000's) input impedance.
Calculating is done with the equations (in the voltage domain) dB = 20 log (V1/V2) and (in the power domain) dB = 10 log (P1/P2).
Also, have a look at a handy tool here: http://www.sengpielaudio.com/calculator-db-volt.htm
Well, 0dB in respect to what? As a Bel scale always has to be referenced, it's important to be sure of the correct calculation here. The K1000 is just too expensive to expose it to an amp with a wrong gain structure...
Considering a rough 0dBV as the output of your cd player, your 36dB of gain (voltage gain of 63, according to the above equation) would result in an opa's output of 63V. Assuming a flat 120R impedance curve, this would result in an output power of P = 63V*63V/120R = 33W.
Your cd player will probably easily deliver you an output voltage of 1Vrms (and more). Who's making the mistake? Me, I hope.
As you probably want to use your head-amp independently of your portable cd player (not glued together, that is
), you should stick with an input sensitivity that fits some standards. Or one that fits your source(s), if you want to be individual. An output power of 1000mW with a standard source of -10dBV (0.316V, older HiFi-standard, according to DIN and ISO) would require an overall gain of a rough 30dB (voltage amplification factor of nearly 32).
OTOH, an output power of 1000mW with a standard source of +4dBu (1.224V, professional standard, according to IEEE, ITU and whatever) would require an overall gain of a rough 18dB (voltage amplification factor of 8).
What? More?
Well, just one: Have you ever wondered why next to nobody's actually using a power amp for headphone amplification?
BTW: 1000mW is not extraordinarily much power for a headphone. The BUF634, for example, can deliver a rough 250mA, works with a +/-15V supply and integrates perfectly with most opamps as a (voltage) gain stage. That makes the BUF634 deliver (say, at 12V output voltage capability) 1.200mW into 120R. It's too low for the K1000 to have headroom left (because of the relatively high 120R), but a 50W power chip-amp as the next bigger step? Think about it...
I agree with Sek's assessment: way too much gain overall. The standard line level is 2 V p-p or 0.7 Vrms. Caculate the RMS voltage that you need for "full" power with your headphones (keeping in mind that it will probably be ridiculously loud) and determine gain from that. P=(V^2)/R
I would strongly recommend using a much lower gain than 26 dB for the amplifier (OPA548) stage. This amp has a much lower gain bandwidth product than any of National's chip amps. You said you'd use a gain of 26 dB, which is a voltage gain of 20. The gain-bandwidth product is 1 MHz, so you have a bandwidth (-3 dB) of 50 kHz. That is close enough to the audible region to affect it. Your response will be down 1 or 2 dB at 20 kHz. I very strongly recommend running a much lower voltage gain (like 5) and getting the rest of the required gain from a pre-amp stage. Also, try the chip in inverting mode. It seems to perform well that way (see the overshoot vs. load capacitance chart).
I would strongly recommend using a much lower gain than 26 dB for the amplifier (OPA548) stage. This amp has a much lower gain bandwidth product than any of National's chip amps. You said you'd use a gain of 26 dB, which is a voltage gain of 20. The gain-bandwidth product is 1 MHz, so you have a bandwidth (-3 dB) of 50 kHz. That is close enough to the audible region to affect it. Your response will be down 1 or 2 dB at 20 kHz. I very strongly recommend running a much lower voltage gain (like 5) and getting the rest of the required gain from a pre-amp stage. Also, try the chip in inverting mode. It seems to perform well that way (see the overshoot vs. load capacitance chart).
thanks sek,you are right.
Let me just show my calculation again.See where to make mistakes.
The K1000 has 74dB SPL for 1mW,104dB SPL for 1000mW.Calculating the required gain from 1mw to 1000mw is done with the equations(in the power domain) dB = 10 log (P1/P2).
that is 10log(1000mw/1mw)=30dBm.
It is incorrect that I want to calculate in this way..that my gain set up too high.
I think me mixing up voltage and power levels.
thanks macboy for you point out the characteristic of opa548's gain bandwidth product, let me think deeply again.
this is my revised sch version,the opa548's relatively high input offset is cancelled by use of an opa2604 as an input stage. The gain is set at 10(Vout/Vin) overall by R3 and R4. the PSU uses a series of batteries.
Let me just show my calculation again.See where to make mistakes.
The K1000 has 74dB SPL for 1mW,104dB SPL for 1000mW.Calculating the required gain from 1mw to 1000mw is done with the equations(in the power domain) dB = 10 log (P1/P2).
that is 10log(1000mw/1mw)=30dBm.
It is incorrect that I want to calculate in this way..that my gain set up too high.
I think me mixing up voltage and power levels.
thanks macboy for you point out the characteristic of opa548's gain bandwidth product, let me think deeply again.
this is my revised sch version,the opa548's relatively high input offset is cancelled by use of an opa2604 as an input stage. The gain is set at 10(Vout/Vin) overall by R3 and R4. the PSU uses a series of batteries.
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://www-k.ext.ti.com/srvs/cgi-bin/webcgi.exe?Company={c7698c3f-af4f-4acc-bbc3-e615ef792d18},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.
Okay, the spec sheet says that, we believe it for now.
Let's call that our desired output power.
Yup. But while the calculation is "valid", it's not the one you want to use here...
Okay, I get your point.
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 dimension-less, 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).
Everybody mixes things up with dB's once in a while
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 headphone-amplifiers 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).
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://www-k.ext.ti.com/srvs/cgi-bin/webcgi.exe?Company={c7698c3f-af4f-4acc-bbc3-e615ef792d18},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.
Okay, the spec sheet says that, we believe it for now.
Let's call that our desired output power.
Yup. But while the calculation is "valid", it's not the one you want to use here...
Okay, I get your point.
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 dimension-less, 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).
Everybody mixes things up with dB's once in a while
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 headphone-amplifiers 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).
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?
I find these formulae:
£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.
Here have one queries, According to input voltage and amp's gain set,we can calculate the output voltage value.but You have not consider the voltage range of the power supply.
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
I can't get your on line calculation tool.Are there any other ways that visit this url?
I find these formulae:
£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.
Here have one queries, According to input voltage and amp's gain set,we can calculate the output voltage value.but You have not consider the voltage range of the power supply.
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
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