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Not a bassheadbut I do prefer clean sub-bass(100hz and below) that gives you the impact,weight of a song and a punchy,chest-thumping for the bass itself but fairly neutral mids and highs...reason for me liking the M-100 so much. Had the BH and E11 and now the Apex Glacier,an old Xenos X0HA-REP and lastly the C5.
Not really sure what would fit you best but will definitely steer you to the C5. But I do miss the BH(don't ask me why it's no longer with me).Heck it!! why don't you just get both as the BH is really a fun amp to play with.
I think the ZO2 has the best rumble for the sub-bass, but I like the wide-band "theater" bass of the C5's bass boost. The ZO2 definitely sounds great with the M-100 and is a better bass boost in the sense that it can be adjusted, but the C5's bass boost is actually quite handy for the M-100 as well for noisier environments.
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Nope.
Lets first model the HD800 with a circuit having a similar impedance curve:
Now, We will connect an ideal signal source to it.
Ideal meaning no output impedance and voltage limits.
Headphone sensitivity is usually given in dB/mW (or loudness / power).
This implies loudness is proportional to power, so we'd want to find out the power consumed by the headphone.
This can be done using ohm's law: W=V^2/R
We know that V is constant for all frequencies because an ideal amplifier is used.
For this explanation, let V=1.
We also know R or in this case Z for impedance instead of resistance.
Since we modeled the headphone using a circuit.
Where f is frequency.
Now we can determine the relationship between power and frequency:
As we can see, the power consumed is actually least at the highest impedance.
This is the opposite of what you suggested.
Now let's talk about amplifier output power.
If the amp is unable to provide enough power (either because max current or voltage is reached) then is causes clipping.
On the other hand, the ability to provide more than enough power has no effects. It won't change the FR.
Alight. Now lets talk about non-ideal amps with output impedance.
Consider a signal source with an output impedance of 100Ω:
Now this is connected to our model of the HD800 with a probe connected to the amp's output:
To determine the power, we will use the same method as above: Ohm's law.
But this time Voltage is NOT constant due to the amp's output impedance.
Voltage can be determined by modeling the amp and headphone as a potential divider.
Giving us a voltage curve that looks like this:
Let's try more different values of output impedances:
So it seems that when driving non-linear loads, it's best to have a low output impedance.
Originally Posted by Joe Bloggs /img/forum/go_quote.gif
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1. You may really be running out of volume knob. Of course it would sound bland when it's not loud enough.
2. The amp may be running out of voltage swing. A high impedance load is easy to drive if the amp can swing enough voltage to drive it. Note the if.
An amp may have high gain allowing you to crank the volume really high, but not enough voltage swing to back it, thus causing distortion.
Power and voltage are NOT equivalent. Here are a few equations for you to chew on:
P = IV (Power = Current x Voltage)
P = I^2R (Power = Current^2 x Resistance)
P = V^2/R (Power = Voltage^2 / Resistance)
Power is, ahem, the power put through the drivers, ie energy pumped in per unit time. Voltage is, well, http://en.wikipedia.org/wiki/Voltage
A power source holding constant voltage pumps MORE into a circuit the lower the resistance (P = V^2 / R note that power is divided rather than multiplied by R). Which is why short circuits (which is when resistance goes to near zero) blows things up.
3. As I said "easy to drive" should most properly be determined by the efficiency measurement, which has dB/mW as the unit. High or low impedance phones can all possibly have high or low efficiency. Note that "efficiency" is often stated in dB/V instead; this is not a true power efficiency measurement and favours low impedance cans.
4. Your brain may be messing with you from all the reports you heard about high-Z cans and the requirement for an amp that can do arc welding
#!/usr/bin/python import math vUnld = 14.0 * math.sqrt(0.125) # unloaded voltage v150 = 3.337 # voltage into 150 ohms v600 = 4.146 # voltage into 600 ohms iSC = 0.045 * math.sqrt(0.5) # short circuit current of OPA2227 in Arms rOut = ((vUnld / v150) - 1.0) * 150.0 # total open loop output impedance rOut = (v600 - v150) / ((v150 / 150.0) - (v600 / 600.0)) vUnld = v600 * ((600.0 + rOut) / 600.0) def calcPower(r): global vUnld, iSC, rOut v = vUnld * r / (r + rOut) i = v / r if i > iSC: i = iSC v = i * r p = v * v / r print "%3.0f ohms: V = %.3f Vrms, Ipeak = %.1f mA, P = %.1f mW" \ % (r, v, i * math.sqrt(2.0) * 1000.0, p * 1000.0) for r in [16.0, 32.0, 50.0, 62.0, 150.0, 300.0, 600.0]: calcPower(r) 16 ohms: V = 0.509 Vrms, Ipeak = 45.0 mA, P = 16.2 mW 32 ohms: V = 1.018 Vrms, Ipeak = 45.0 mA, P = 32.4 mW 50 ohms: V = 1.591 Vrms, Ipeak = 45.0 mA, P = 50.6 mW 62 ohms: V = 1.973 Vrms, Ipeak = 45.0 mA, P = 62.8 mW 150 ohms: V = 3.337 Vrms, Ipeak = 31.5 mA, P = 74.2 mW 300 ohms: V = 3.836 Vrms, Ipeak = 18.1 mA, P = 49.0 mW 600 ohms: V = 4.146 Vrms, Ipeak = 9.8 mA, P = 28.6 mWQuote:
Those calculations are not quite right. First, 14.0 Vp-p is actually 4.95 Vrms, because 1 Vrms = 2 * sqrt(2) = 2.8284 Vp-p. Second, the method of linearly extrapolating power is wrong, because power is inversely proportional to the impedance with constant voltage. With your formula, the power output into loads above ~900 Ω would be negative, and into a near-infinite impedance open circuit, an extreme negative value.Now the difficult part is modeling how the maximum output voltage drops with decreasing load impedance. A simple model is shown below that assumes a linear open loop output impedance, and a hard peak current limit of 45 mA taken from the datasheet of the OPA2227. Using the following Python script to perform all the calculations:
I get these results:Code:#!/usr/bin/python import math vUnld = 14.0 * math.sqrt(0.125) # unloaded voltage v150 = 3.337 # voltage into 150 ohms v600 = 4.146 # voltage into 600 ohms iSC = 0.045 * math.sqrt(0.5) # short circuit current of OPA2227 in Arms rOut = ((vUnld / v150) - 1.0) * 150.0 # total open loop output impedance rOut = (v600 - v150) / ((v150 / 150.0) - ((v600 / 600.0))) vUnld = v600 * ((600.0 + rOut) / 600.0) def calcPower(r): global vUnld, iSC, rOut v = vUnld * r / (r + rOut) i = v / r if i > iSC: i = iSC v = i * r p = v * v / r print "%3.0f ohms: V = %.3f Vrms, Ipeak = %.1f mA, P = %.1f mW" \ % (r, v, i * math.sqrt(2.0) * 1000.0, p * 1000.0) for r in [16.0, 32.0, 50.0, 62.0, 150.0, 300.0, 600.0]: calcPower(r)
Of course, due to the simplicity of the model, the numbers above are still not very reliable. Also, the maximum power into low impedances depends largely on the actual (rather than specified) short circuit current.Code:16 ohms: V = 0.509 Vrms, Ipeak = 45.0 mA, P = 16.2 mW 32 ohms: V = 1.018 Vrms, Ipeak = 45.0 mA, P = 32.4 mW 50 ohms: V = 1.591 Vrms, Ipeak = 45.0 mA, P = 50.6 mW 62 ohms: V = 1.973 Vrms, Ipeak = 45.0 mA, P = 62.8 mW 150 ohms: V = 3.337 Vrms, Ipeak = 31.5 mA, P = 74.2 mW 300 ohms: V = 3.836 Vrms, Ipeak = 18.1 mA, P = 49.0 mW 600 ohms: V = 4.146 Vrms, Ipeak = 9.8 mA, P = 28.6 mW
