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Amp capable of powering a Modhouse Tungsten

No clue if I'm really missing out by not getting something higher end.
Absolutely not.

More volume is all that a new Amp could bring. Everything else stays the same.
 
Absolutely not.

More volume is all that a new Amp could bring. Everything else stays the same.
Yeah that's what I figure. I have no clue how people could say it doesn't get loud enough for them, I feel like how hard these are to drive is really over stated, I can get these loud enough just barely even on low gain.
 
Looks like the newly launched Topping A900 is the new top choice to drive the Tungsten. Appears like they would make a wonderful pairing.

The A900 is insane, I can’t imagine why Topping would make another headphones amplifier, but they will.
 
Another 3rd party verification of Valencia measurements:
 
I’m using the Flux Mentor with my DS Tungsten. An absolutely perfect pairing and powers them just fine.


I have a Flux FA-10 Pro that does a great job with the Tungsten and HE6SEv2.

However, I recently purchased an inexpensive banana plug to balanced connector cable from Alibaba to try speaker amp outputs for harder to drive headphones. I am testing an old Krell 300i integrated as a headphone amp via this cable with good results. I am using Roon with a -6db digital preamp output and am quite happy with the results. It would seem that some of these older high powered integrated amps could be a good headphone amp for harder to drive headphones.
Krell.jpg


Per ChatGPT, the Krell should be able to deliver more than enough power for any headphone:

Given the Krell KAV-300i is rated 150 W into 8 Ω and 300 W into 4 Ω, those two specs imply it’s basically voltage-limited (same max voltage swing), because:
  • Vrms=P⋅R=150⋅8=1200≈34.64 VV_{\mathrm{rms}}=\sqrt{P\cdot R}=\sqrt{150\cdot 8}=\sqrt{1200}\approx 34.64\text{ V}Vrms=P⋅R=150⋅8=1200≈34.64 V
  • Check with 4 Ω: 300⋅4=1200≈34.64 V\sqrt{300\cdot 4}=\sqrt{1200}\approx 34.64\text{ V}300⋅4=1200≈34.64 V ✅
So into 155 Ω, assuming it can still swing ~34.64 Vrms:

P=V2R=1200155≈7.74 WP=\frac{V^2}{R}=\frac{1200}{155}\approx 7.74\text{ W}P=RV2=1551200≈7.74 W
Answer: ~7.7 watts into 155 Ω (idealized, based on its 8/4-ohm specs).
 
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