I have searched and there is no information about the Jung SR output impedance vs load current.
The only information available is Jack's measurements based on 100mA current.
So what is the "minimum" amount of current necessary to lower the output impedance sufficient enough for analogue stage of audio use?
We want large enough current (artificially induced if necessary, like the 51mA current in my implementation) to have acceptable low impedance, but we don't want too large the current as it is not practical - dissipating too much heat.
My 51mA is basically arbitrary.
What is the "optimal" current?
The only information available is Jack's measurements based on 100mA current.
So what is the "minimum" amount of current necessary to lower the output impedance sufficient enough for analogue stage of audio use?
We want large enough current (artificially induced if necessary, like the 51mA current in my implementation) to have acceptable low impedance, but we don't want too large the current as it is not practical - dissipating too much heat.
My 51mA is basically arbitrary.
What is the "optimal" current?
Not necessarily. A mismatch only will cause a very small deviation in Vout from the value with matching. For example, at 10nA Iin and a mismatch of 5k, the Vout deviation is in the area of 50uV. Hard to imagine a case where this is a problem.
Jan
Thank you Andrew. Let me be more explicit as to what I'm thinking about.
The project I am in the very early stages of thinking about is just a headphone amp, namely "The Wire" balanced headphone amp from the vendor area here. (The headphones have an impedance of 18 Ohms.) If I take the project on I would like to do the +/-12V power supply myself rather than purchase the 1A one offered by the designers of The Wire.
I was thinking of a full wave bridge rectifier** to RC input filter feeding an LM317/337 pair and on to one (or possibly a pair?) +/- Jung/Didden DIYaudio store regulator boards. 1A is quite likely way more than enough.
I'm just trying to understand the need for any capacitance between the two regulators. The LM317 data sheet says that a cap of 0.1uF improves transient response but is not needed for stability. But in the project I have in mind the two regulators are 'back to back' and so I ask this question.
BTW I also note that HiFiNut has just 2.4V difference between Vin and Vout. I understood the preferred differential was still 5V for best performance.
** I note that Jan's states in his article that it is better to have separate rectifier for +ve and -ve.
The project I am in the very early stages of thinking about is just a headphone amp, namely "The Wire" balanced headphone amp from the vendor area here. (The headphones have an impedance of 18 Ohms.) If I take the project on I would like to do the +/-12V power supply myself rather than purchase the 1A one offered by the designers of The Wire.
I was thinking of a full wave bridge rectifier** to RC input filter feeding an LM317/337 pair and on to one (or possibly a pair?) +/- Jung/Didden DIYaudio store regulator boards. 1A is quite likely way more than enough.
I'm just trying to understand the need for any capacitance between the two regulators. The LM317 data sheet says that a cap of 0.1uF improves transient response but is not needed for stability. But in the project I have in mind the two regulators are 'back to back' and so I ask this question.
BTW I also note that HiFiNut has just 2.4V difference between Vin and Vout. I understood the preferred differential was still 5V for best performance.
** I note that Jan's states in his article that it is better to have separate rectifier for +ve and -ve.
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It's a good idea to have plenty of capacitance at the DC_input of every voltage regulator. When the regulator wants a big slug of transient current, the input cap delivers.
If that frightens you in a cascade connection, you could always install an explicit series resistor between OUT #(n-1) and IN #
, whose value is bigger than the ESR of even the crummiest capacitor you've ever seen (e.g. 0.33R). Now regulator #(n-1) is loaded by something that appears to be a crummy, extra-high-ESR capacitor.
Also don't forget that you can choose the next higher secondary voltage (5V more?) and not pay any extra money for your transformer, since you're playing around in the sub-100VA-transformer world of headphone amps. Now you've got all the voltage room you need, to install higher values of series resistance. These are much easier for the upstream regulator to drive, but they do have bigger voltage drop and higher power dissipation.
If that frightens you in a cascade connection, you could always install an explicit series resistor between OUT #(n-1) and IN #
, whose value is bigger than the ESR of even the crummiest capacitor you've ever seen (e.g. 0.33R). Now regulator #(n-1) is loaded by something that appears to be a crummy, extra-high-ESR capacitor.Also don't forget that you can choose the next higher secondary voltage (5V more?) and not pay any extra money for your transformer, since you're playing around in the sub-100VA-transformer world of headphone amps. Now you've got all the voltage room you need, to install higher values of series resistance. These are much easier for the upstream regulator to drive, but they do have bigger voltage drop and higher power dissipation.
The SR will be hard pressed to deliver the current needed to drive 1W into one 18 ohm load, let alone two (cans)! (My ears can't take that many SPL's anyway.) A decently designed output stage should have sufficient PSRR to not require regulation.
Be careful about adding anything around the error amplifier in the SR --
Be careful about adding anything around the error amplifier in the SR --
Thanks Mark.
Re current load, as you can perhaps tell I struggle to compute fully through from SPL-with-headroom-at-the-ear, back through efficiency (in my case 118dB) to voltage and finally to current need. I guess the short cut is to simply note that a +/-12V supply is specified and 12/18 = 0.67A plus headroom for quiescent current draw. (I guess that is each can and hence my "or possibly a pair?" in post 667.) Of course, it will never be asked to output anything close to 12V but let's design for that (at least for now).
Jack, I thought the SR was capable of around an amp (provide the 5V headroom x 1A = 5W is dissipated with adequate heat sinking)? The Wire balanced headphone amp is a rather simple device: the output stage is a pair of LME49610 (within the feedback loop of a OPA1632) for each channel.
If I am heading in the wrong direction do please let me know.
(I have looked at other designs but I understand many would be challenged by the 18R load.)
EDIT: PS the heat sink is not listed on Jan's BoM. I have some of these but it's not clear they will fit.
Re current load, as you can perhaps tell I struggle to compute fully through from SPL-with-headroom-at-the-ear, back through efficiency (in my case 118dB) to voltage and finally to current need. I guess the short cut is to simply note that a +/-12V supply is specified and 12/18 = 0.67A plus headroom for quiescent current draw. (I guess that is each can and hence my "or possibly a pair?" in post 667.) Of course, it will never be asked to output anything close to 12V but let's design for that (at least for now).
Jack, I thought the SR was capable of around an amp (provide the 5V headroom x 1A = 5W is dissipated with adequate heat sinking)? The Wire balanced headphone amp is a rather simple device: the output stage is a pair of LME49610 (within the feedback loop of a OPA1632) for each channel.
If I am heading in the wrong direction do please let me know.
(I have looked at other designs but I understand many would be challenged by the 18R load.)
EDIT: PS the heat sink is not listed on Jan's BoM. I have some of these but it's not clear they will fit.
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I understand that even if I desired an SPL of 130dB, the threshold of pain - which I definitely would not - we're talking about 0.5V RMS of just less than 30mA of required current.
I'm not sure how the 118dB @ 1mW efficiency compares to other high quality headphones. I will have a google.
118dB/mW for your heaphone is very high sensitivity.
We normally see a range of 90dB to 110dB/mW
If you want 120dB for threshold of pain, then you are looking for +2dB ref 1mW, or about 1.6mW
P=I²R
Therefore I= sqrt(P/R) = sqrt(0.0016/18) = 9.4mAac.
That's equivalent to 13.3mApk.
If one assumes that the worst case current demand is 150% of that figure for a reactive load, then you can design for 20mApk, not 500mApk.
If you want to add on 10dB for headroom, then multiply your worst case current demand by sqrt(10) i.e. 63mApk.
Let's check my arithmetic.
0.063mApk divided by 1.5 to get back to a resistive equivalent gives 0.042mApk into 18r
Pmax = 0.042² * 18 / 2 = 0.015876 W = 15.9mW
10*Log (15.876mW) = 10*1.2 = +12dB
add on 118dB sensitivity and you get your required 130dB = +10dB ref pain!
We normally see a range of 90dB to 110dB/mW
If you want 120dB for threshold of pain, then you are looking for +2dB ref 1mW, or about 1.6mW
P=I²R
Therefore I= sqrt(P/R) = sqrt(0.0016/18) = 9.4mAac.
That's equivalent to 13.3mApk.
If one assumes that the worst case current demand is 150% of that figure for a reactive load, then you can design for 20mApk, not 500mApk.
If you want to add on 10dB for headroom, then multiply your worst case current demand by sqrt(10) i.e. 63mApk.
Let's check my arithmetic.
0.063mApk divided by 1.5 to get back to a resistive equivalent gives 0.042mApk into 18r
Pmax = 0.042² * 18 / 2 = 0.015876 W = 15.9mW
10*Log (15.876mW) = 10*1.2 = +12dB
add on 118dB sensitivity and you get your required 130dB = +10dB ref pain!
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yes I think I have the maths now. 118dB/mW does seem to be very efficient. These are in-ear.
Of course, I wouldn't want to swap headphones and find the amp and its supply was wildly deficient.
Pick some lower efficiency cans like, say, some HiFiMan HE6 with 83.5db/mW efficiency and 50R impedance and a 12V supply couldn't deliver the voltage for (a painful) 120dB.
At any rate, I thought the SR was good for about an amp and obviously so is the (correctly selected) LM317/337 and hence I thought I had a big field to play in...
Of course, I wouldn't want to swap headphones and find the amp and its supply was wildly deficient.
Pick some lower efficiency cans like, say, some HiFiMan HE6 with 83.5db/mW efficiency and 50R impedance and a 12V supply couldn't deliver the voltage for (a painful) 120dB.
At any rate, I thought the SR was good for about an amp and obviously so is the (correctly selected) LM317/337 and hence I thought I had a big field to play in...
The problem with the SR and high load current is that the AD797/AD825 run out of steam with the D44/D45H11 with hfe of 40-50. I haven't tested the SR with a Darlington transistor like TIP142/147, but you could give it a try and report back on your results. I will defer to Mark Johnson who has far more expertise than I in these matters!
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I agree with Jack but I suggest that if you're laying out a brand new PCB, please consider making your Darlington out of two discretes instead of buying an all-in-one-package Darlington. Walt Jung's latest 2015 vintage shunt super duper, does this with two PNPs: D45H11 and PN2907A. Here in this DIYA thread, to make a positive regulator using series pass device, you could consider the D44H11 (80 MHz, 10A) for the NPN power output half of the Darlington and the venerable PN2222A (300 MHz, 600mA) for the weakling first half. Go ahead and install bootstrapped base-to-emitter bias resistors like Walt does (1K for the 2222A, 49.9R for the D44H11) and be happy.
Those wishing to be more adventuresome can consider using the much newer, much fancier, Sanyo/ON Semiconductor 2SC6144 in place of the D44H11, and to hell with Darlingtons! The 2SC6144 has an fT of 330 MHz(!!), Beta of 200, Icmax of 10 amperes, and VCEmax of 50 volts. Octopart.com knows about these parts, DigiKey and Verical have stock on the shelf, and the price is USD 0.62 per piece. I myself bought fifty, just in case these get discontinued. (Here is the datasheet).
Another possibility is to use the Belleson trick of building a Darlington-like pair, using a D44H11 or 2SC6144 for the big device, and using a very high IDSS JFET (like the J105) as a source follower to drive the output xitor's base. Since the JFET is a depletion mode device, it operates with VGS < 0 volts, i.e., the gate is below the source. So you get some amount of level shifting for free. Since no current flows into the JFET's gate, you get what amounts to a Darlington with infinite Beta. Fool around with it in SPICE, but beware, it's difficult to find a .MODEL of a J105 FET.
Yet another possibility is to replace the NPN emitter follower, by an NMOS source follower. This is an advanced project for advanced builders.
_
Ok. I was hoping for something a bit easier this time around. The NMOS source follower project I am (hopefully) finishing up was arduous to say the least (and not just for me ). And I need the negative supply. I was hoping I could use these boards given the much lower current need than my existing project.
With the headphones I have in mind for purchase the current need is very small but if I want some flexibility I need to work towards a significantly greater current spec. Oh joy.
PS Based on a browse of Mouser it looks like the closest PNP to the 2SC6144 might be the 2SA2222...
). And I need the negative supply. I was hoping I could use these boards given the much lower current need than my existing project.With the headphones I have in mind for purchase the current need is very small but if I want some flexibility I need to work towards a significantly greater current spec. Oh joy.
PS Based on a browse of Mouser it looks like the closest PNP to the 2SC6144 might be the 2SA2222...
The best looking single-package Darlington transistors seem to be the 2SD2014(N) / 2SB1257(P) . fT = 75 MHz, Beta = 2000, 4 amps, 60 volts, 25 watts, Safe Operating Area includes 20V@1A. Plastic TO-220 so you'll need a somewhat better heatsink.
Same pinout as the D45H11. A drop-in replacement.
Same pinout as the D45H11. A drop-in replacement.
That exceptional headphone would require ~4.5W to generate peaks of 120dB.
4.5W into 8ohms is equivalent 21.2Vpk & 424mApk.
Is the headphone specified for transient levels that high?
If it is, then build a power amplifier !
BTW a normal headphone amplifier operating from 12Vdc could give a maximum output signal of around 4Vac. That would give ~108dB in those exceptionally inefficient headphones
I would like to make sure I understand this properly. I understand that Q2 and Q4 deliver the principal required base currents to Q1 and Q3. The op amps "trim" (i.e. fine tune) the current to provide the regulation. The op amps have to be able to sink the entire base current only when the load is disconnected. I understand that as beta falls more base current is required to achieve incremental Ic. When I look at the data sheet for the D44/D45H11 the DC current gain doesn't begin to droop until beyond 1A of Ic albeit the inflexion point isn't far beyond 1A and shifts lower as temp rises (Fig 4 and 5.) Both the AD825 and AD817 have a minimum output current of 50mA. Just what runs out of steam?
If I look at the Darlington replacements suggested by Mark, I presume I would need to test the circuit for stability etc. Having only ever looked at +ve regs I'm not yet sure how I test the -ve.
I still don't know what the footprint of the board is for the heat sink. Can it accommodate the ones I linked to in post 671?
Yes, Andrew, those particular headphones are exceptional in their lack of efficiency. Were one to purchase them one would need to think very differently about how to drive them - even if I doubt anyone would want 120dB peaks. I don't know what their peak SPL rating is as it doesn't appear to be listed in their promotional material. I just threw that out as an extreme example. As noted previously, I would have thought that a 1A +/-12V supplied amp would accommodate a lot of options (albeit not necessarily all).
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