Could someone here please help me gain some insight regarding amplifier input sensitivity, CDP output sensitivity, and their effect on volume control position? I will be very grateful.🙂
I'm running my Oppo BDP95 with 2v output using se directly into my LSR&D mono amps with 2.4v input for rated power. Is there a compromise here, ie: loss of signal/information? Would I gain any performance by using the 4v balanced out on the Oppo with an adapter to se in to the amps? Or is what I have actually ideal since I have to turn the volume higher than otherwise for equivalent spl? Thank you in advance.
I'm running my Oppo BDP95 with 2v output using se directly into my LSR&D mono amps with 2.4v input for rated power. Is there a compromise here, ie: loss of signal/information? Would I gain any performance by using the 4v balanced out on the Oppo with an adapter to se in to the amps? Or is what I have actually ideal since I have to turn the volume higher than otherwise for equivalent spl? Thank you in advance.
You can't drive your amplifier to full power now, but do you ever want to? Have you ever felt the need to turn the volume knob all the way to the right?
My question refers to resolution. Is it being compromised with this set up? Which is better, to attenuate the Oppo or turn it up?
True, you are losing 1.6dB, but only if you have the volume control set at maximum (which is normally the minimum attenuation, 0dB).
Once you set the control just a little below maximum volume, (-1.6 dB or more), the theoretical 1.6dB loss is of no consequence.
Thanks guys. I eliminated the pre. It sounds better, a bit more resolution. But the s/n question hadn't occurred to me. I'll check.
The manual does not differentiate s/n ratio between se and XLR. However, it states that.."by transmitting a pair of differential signals, the balanced output provides better common-mode noise rejection and improves sound quality." If there's an improvement to be realized here, would it still be the case without balanced inputs on the amp? I mean by using an XLR to rca adapter to the amps.
I don't think there will be much difference in noise between single ended and differential.
You are using very low impedance connections.
Unless they are run next to mains cables then there shouldnt be a problem.
Even then the shielding should stop most noise.
You are using very low impedance connections.
Unless they are run next to mains cables then there shouldnt be a problem.
Even then the shielding should stop most noise.
The CMRR part is true, the sound quality part is marketing. However, with an XLR to rca adapter, you will loose the CMRR advantage. Therefore, it all boils down to the s/n ratio of both outputs. But, as Nigel remarks, the difference will not be to loose sleep over.
I think your existing source voltage of ~2.2Vac and your existing sensitivity for full power of 2.4Vac are an almost perfect match. The difference of <0.8dB will not be noticed.
They prevent you overdriving the power amplifier and that in turn means you can never be responsible for clipping any signal sent to your speakers.
This existing arrangement allows you to "use" the full range of the vol pot. That gives you the best noise performance and results in the best resolution of the available signal.
They prevent you overdriving the power amplifier and that in turn means you can never be responsible for clipping any signal sent to your speakers.
This existing arrangement allows you to "use" the full range of the vol pot. That gives you the best noise performance and results in the best resolution of the available signal.
A balanced output only provides a significant advantage when used with a balanced cable to a balanced input. The advantage is less noise and interference, not more 'resolution'. If you can't hear any noise and interference then it is already good enough.
You mean solving the clipping issue/problem is as simple as matching output/input sensitivity? Matching sensitivity yields true 'power on demand'?, ie: as long as headroom is sufficient?
Clipping occurs when you put in too much signal. If you can't do this, then you can't have clipping.
if the maximum output just before clipping is 28.28Vac = 40Vpk = 100W into 8r0
and
The gain is 26times (+28.3dB) then the maximum input signal is 28.28/26 =1088mVac.
There is often an attenuator in front due to the two input filters. Assuming 1k in series and 47k in parallel we have the input sensitivity = 1088*{47k+1k}/47k = 1111mVac
1111Vac at the input terminals will take the unloaded amplifier to it's maximum output voltage of 28.28Vac.
No "power on demand" whatever that means.
Apply more than 1111mVac and the output clips.
Attach a load and the maximum output voltage will usually be slightly less. This results in slightly less input voltage. Suppose the maximum voltage into a 6r0 load is 27.5Vac, then the maximum input signal will be 27.5/26*{47+1}/47 = 1080mVac.
Simple arithmetic, no magic numbers required.
and
The gain is 26times (+28.3dB) then the maximum input signal is 28.28/26 =1088mVac.
There is often an attenuator in front due to the two input filters. Assuming 1k in series and 47k in parallel we have the input sensitivity = 1088*{47k+1k}/47k = 1111mVac
1111Vac at the input terminals will take the unloaded amplifier to it's maximum output voltage of 28.28Vac.
No "power on demand" whatever that means.
Apply more than 1111mVac and the output clips.
Attach a load and the maximum output voltage will usually be slightly less. This results in slightly less input voltage. Suppose the maximum voltage into a 6r0 load is 27.5Vac, then the maximum input signal will be 27.5/26*{47+1}/47 = 1080mVac.
Simple arithmetic, no magic numbers required.
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