Headphones' varied impedances and USB limited power => class G adsl drivers ?

The recent discussion on headphones' largely different impedances and their varied voltage requirements got me thinking. So, a bit of brainstorming...

USB ports and chargers are widely available but the current they can deliver is often quite limited. When I designed a small DAC+amp made to be entirely usb powered from a laptop port, I had the choice between using a dc-dc converter at either +/-5V or +/-9 or 12V. The first had enough current for low impedance headphones, the second enough voltage for high impedance headphones. But the reverse wasn't true. Since I mostly use hd650, no big deal at the time.

But for the sake of universality, a chipset like the ths6032 looks perfect. It's an adsl line driver so it's designed to drive low impedance loads at low distortion. It incorporates a class G system so we can use a low current/high voltage supply for high impedance headphones (or even small bursts of high power in lower impedance ones) and a beefier low voltage supply for low impedance headphones while staying in a smaller power budget. And I see no reason at first sight why it couldn't be used in multiloop to provide ease of integration and guarantee proper audio performances. As it is a current feedback amp, it offers a further degree of play for compensation.

I'm honestly a bit surprised to see how little attention the ths6032 (a quite old device) has received compared to the tpa6120a2 (only 2 hits in the whole forum).

Your thoughts ?
 
Interesting thought, I think I came across that IC when browsing TIs Line Drivers a while ago. I think the reason it has received little attention compared to TPA6120A2 is the much higher price and that it falls slightly short of the TPA6120A2/THS6012 in basically all other aspects. But I might be wrong on that one.
 
Large amounts of current noise though(*), you'd need to drive this from an opamp buffer to swallow that up - the voltage noise isn't too bad though (rather surprising actually). Such an input buffer can double as LPF to prevent RF getting into the device and boosted.

Note that these high-speed current-feedback amps are very fussy as to feedback networks and gains, so follow the datasheet recommendations, but it does look interesting - the distortion for low frequencies isn't given as anything under 100kHz isn't of interest for its intended use, but ought to be at least as good as the 100kHz values which are OK.

(*) RF circuitry has to operate at low impedance, so bias currents are large and current noise is big - just drive it at low impedance (50 to 100 ohms is a safe range) so the current noise doesn't bite you.

The Johnson current noise of a 50 ohm resistor is 18pA/√Hz and is somewhat higher then this device's non-inv input current noise - 50 ohms translates this to a very quiet 0.9nV/√Hz. The feedback network impedance at the inverting input also must handle its current noise without turning it into a large voltage.
 
Thanks for the comments.

After further consideration, I'm going to rain on this parade a bit, all by myself. 😱

To elaborate a bit on the possible power output, p.13 of the datasheet is key. Here's the main quote (fig 37 is attached below).

This is where the THS6032 class-G architecture is useful.The class-G output stage utilizes both a high supply voltage [VCC(H) typically ± 15V] and a low supply voltage [(VCC(L) typically ± 6V].

As long as the output voltage is less than [VCC(L)–2.5V], then part of the output current will be drawn from the VCC(L) supplies. If the output signal goes above this cutoff point [forexample,VO>VC(L)–2.5V], then all of the output current will be supplied by VC(H).

To ensure that the cutoff point does not introduce distortion into the system, the entire output stage is always biased on. This constant biasing scheme will cause a decrease in the efficiency over hard switching class-G circuits, but the very low distortion results tend to outweigh the efficiency loss. The biasing scheme used in the THS6032 can be shown by the currents being supplied by the VCC(L) power supplies in Figure37. This graph shows there is no discrete current transfer point between the VCC(L) supplies and the VCC(H) supplies. This was done to ensure low distortion throughout the entire output range. By changing the VCC(L) supply voltage, the system efficiency can be tailored to suit almost any system with high crest factor requirements.

In practical terms, let's consider we have a 5V/500mA output from a USB port. 2.5W continuous. But we're drawing sinewaves not DC across the whole output, so even considering peak values and efficiency losses, we can use about 4W worth of dual DC-DC converters.

Let's split that between a 3W +/-12V converter (giving us 60ma per channel to play with) and a 1W +/-5V converter (giving us 50ma per channel). So, from ballpark quick calculations:

- with a 16R load, we get about 100ma peak, about 1.1Vrms.
- with a 32r load, we get about 70ma peak, about 1.6Vrms
- with a 64r load, VCCL becomes pretty much useless in terms of boosting max output power.
- with a 300R or 600R load, no problems hitting 7vrms on VCCH alone of course.

But if we use a 1W +/-5V converter and a more common 2W +/-12V converter (giving us 40ma per channel), not going above the converter ratings for the peak current values :

- with a 16R load, we could get a touch above 80ma peak, just under 1Vrms.
- with a 32R load, we can reach about 1.4V rms or just above 60ma peak.

Not really worth going for the 3W converter in terms of extra dB gained.

Also worth noting is that a 3W +/-9V converter would give us the same max current of 80ma and actually more power into 32 or 64r loads (as this could sustain higher voltage outputs). Are the efficiency gain and the 3 extra volts of headroom really worth it ?
 
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