First impression, cleaner than C421, more sparkle in the highs. Will be updating as I listen more
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JDS Labs C5/C5D (pg96) portable amp/amp+DAC  Page 31
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 Apo0th3karY
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Not a basshead but I do prefer clean subbass(100hz and below) that gives you the impact,weight of a song and a punchy,chestthumping for the bass itself but fairly neutral mids and highs...reason for me liking the M100 so much. Had the BH and E11 and now the Apex Glacier,an old Xenos X0HAREP 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 subbass, but I like the wideband "theater" bass of the C5's bass boost. The ZO2 definitely sounds great with the M100 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 M100 as well for noisier environments.
+1
I think all this just sold me on the C5 sooner.
 miceblue
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Thanks for posting your initial impressions. I'm kind of interested to see how the C5 improves over the c421's sound quality.
How is the power on the C5 compared to others like the E12 and BH?
 miceblue
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I can't say much about the E12 other than they provide 880 milliwatts at 32 Ω, and the BH just has a single power measurement, 300 milliwatts, at an unknown impedance.
From JDS Lab’s official C5 blog post:
3.337 VRMS @ 150 Ω = 74.24 milliWatts
4.146 VRMS @ 600 Ω = 28.65 milliWatts
14.0 Vpp (peaktopeak voltage) = 4.214 Volts (maximum output voltage)
Extrapolated values (assuming linear behavior):
86.19 milliWatts @ 32 Ω (typical portable headphones)
87.81 milliWatts @ 16 Ω (typical earphones)
59.05 milliWatts @ 300 Ω (Sennheiser HD 650, HD 600, and HD 800 [from Sennheiser's website])
84.37 milliWatts @ 50 Ω (approximate impedance for planar magnetic HiFiMAN and Fostex T50RP Mad Dog headphones)
82.95 milliWatts @ 62 Ω (AKG K 701/ K 702/ Q 701)
Math (Click to show)
Power Equation:
Power = Voltage * Current
P = V * I (1)
Units: Watts = Volts * Amperes
Ohm’s Law:
Voltage = Current * Resistance
V = I * R
Rearranged: I = V / R (2)
Units: Amperes = Volts / Ohms
Rearranging Equation (1) to include Equation (2):
Power = Voltage * (Voltage / Resistance)
P = V * (V / R)
P = (V^2) / R (3)
Units: Watts = Volts^2 / Ohms
Voltage_RMS = peaktopeak Voltage * log(2) (4)
Units: Volts = Volts
From JDS Lab’s official C5 blog post:
3.337 VRMS @ 150 Ω
4.146 VRMS @ 600 Ω
14.0 Vpp (peaktopeak voltage)
Therefore, from Equation (4):
Voltage_RMS = 14.0 Volts peaktopeak * log(2) = 4.214 Volts [this is basically the maximum voltage the C5 can output without clipping the source’s signal]
RMS (root mean square) is basically an average value
From Equation (3):
P = ((3.337 Volts_RMS)^2) / 150 Ω = 0.07424 Watts, or 74.24 milliWatts @ 150 Ω
P = ((4.146 Volts_RMS)^2) / 600 Ω = 0.02865 Watts, or 28.65 milliWatts @ 600 Ω
Extrapolating values assuming the behavior is linear (straightline approximation):
Slope = Rise / Run
Slope = ΔPower / ΔImpedance
Slope = (P2  P1) / (600 Ω  150 Ω) = (28.65 milliWatts  74.24 milliWatts) / (600 Ω  150 Ω)
Slope = 0.1013 milliWatts/Ω
Using the slope value, the power supplied to headphones of other impedances can be extrapolated (again, assuming linear behavior):
Slope = Slope
0.1013 milliWatts/Ω = (74.24 milliWatts  X) / (150 Ω  32 Ω)
X = 0.1013 milliWatts/Ω * (150 Ω  32 Ω) + 74.24 milliWatts = 86.19 milliWatts @ 32 Ω
0.1013 milliWatts/Ω = (74.24 milliWatts  X) / (150 Ω  16 Ω)
X = 0.1013 milliWatts/Ω * (150 Ω  16 Ω) + 74.24 milliWatts = 87.81 milliWatts @ 16 Ω
0.1013 milliWatts/Ω = (74.24 milliWatts  X) / (150 Ω  300 Ω)
X = 0.1013 milliWatts/Ω * (150 Ω  300 Ω) + 74.24 milliWatts = 59.05 milliWatts @ 300 Ω
0.1013 milliWatts/Ω = (74.24 milliWatts  X) / (150 Ω  50 Ω)
X = 0.1013 milliWatts/Ω * (150 Ω  50 Ω) + 74.24 milliWatts = 84.37 milliWatts @ 50 Ω
0.1013 milliWatts/Ω = (74.24 milliWatts  X) / (150 Ω  62 Ω)
X = 0.1013 milliWatts/Ω * (150 Ω  62 Ω) + 74.24 milliWatts = 83.15 milliWatts @ 62 Ω
A linear plot of all of these values:
If any of my calculations are wrong, please let me know. :)
Off topic about power (Click to show)
I did some more digging behind the whole power "requirements" for headphones:
The clearest picture of how easy or hard a phone would be to drive would be a plot of dB/mW efficiency against frequency but I have never seen such a graph presented. You would have to infer it from the dB/mW at 1kHz, frequency response and impedance response charts and calculate the result yourself. But if you're not EQing, the lowest impedance frequencies will always be the hardest to drive and the highest impedan frequencies the easiest.
For phones with overall high impedance, there will be few amps that drive them to volume but almost all such amps will be able to drive them to their "full potential". OTOH for low impedance but low efficiency phones, almost all amps will drive them loud enough but many may distort.
Since the HD800 seems to have a weird bump in the bass region for its impedance measurement, does that mean an amp would need to supply more power for that specific load at that specific frequency? I'm guessing the single impedance measurement for headphones is the average value.
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 nonideal 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 nonlinear loads, it's best to have a low output impedance.
Originally Posted by Joe Bloggs
LOLOLOL......
@b1o2r3i4s5: If the highest peak of an impedence point is easy to drive then why is it sometimes when we drive such a headphone with an amp capable of driving 600ohm loads fairly easily sound underpowered or very bland when compared to when the headphone is driven by something capable of outputting more power? Can you give a thorough explanation as to why the claims of the HE6 sounding best when powered via the speaker taps of an integrated amplifier compared to a conventional head amp?
I'm getting a bit confused but when you put things such as power and voltage, aren't these technically both the same? Or are you referring to power as the total amount of capable output from the amp overall and voltage just one thing applied to the components/circuitry etc? A bit of clarification.
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 highZ cans and the requirement for an amp that can do arc welding
Edited by miceblue  3/5/13 at 7:57pm
 customNuts
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The C5 works well for the K 701 for the most part.
I was doing some testing between the C5 and O2 again and the C5 seems to be a tad bit brighter than the O2. So in terms of synergy, the O2 might be a better choice for the K 701, but it doesn't sound all that bad. I'm not too familiar with synergy with the K 701, but they do sound good with the C5.
Test tracks:
Battles  Futura
Dj CUTMAN  8Bit Brawl (ft. S&CO)
Thx for the impressions. I wouldn't doubt that the 02 would be the better performing amp for the k701 but the bass or lack there of worries me. So with the bass boost on the c5 surely that would end up being the more 'full/musical' presentation??
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Micro details are good and heard easily, and the sound is dynamic, I would actually say it sounds a bit fuller than O2, but I dont have my O2 next to me to compare.
 putente
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Been using the amp for a few hours now, it seems very good so far, but even on the lowest volume with my CIEM's it is a tad too loud, only a little though. Yes the mids do sound a little forward, but it has full bodied bass and detailed highs, very close to the sound of the O2. The bass boost is perfect for when you want some extra lows, and doesn't muddy up the sound.
Micro details are good and heard easily, and the sound is dynamic, I would actually say it sounds a bit fuller than O2, but I dont have my O2 next to me to compare.
Which opamp version is your C5?
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 putente
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Strange... From what I've read since the c421 was released I was under the impression the AD8620 was the more detailed and neutral opamp option. Maybe that has something to do with the C5 circuit being different than the c421...
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I will be doing a comparison, at the moment I'm on a school trip for 3 days in the Algarve with only the C5
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anyone with a audio quality comparison of c5 and e12?????
From JDS Lab’s official C5 blog post:
3.337 VRMS @ 150 Ω = 74.24 milliWatts
4.146 VRMS @ 600 Ω = 28.65 milliWatts
14.0 Vpp (peaktopeak voltage) = 4.214 Volts (maximum output voltage)
Extrapolated values (assuming linear behavior):
86.19 milliWatts @ 32 Ω (typical portable headphones)
87.81 milliWatts @ 16 Ω (typical earphones)
59.05 milliWatts @ 300 Ω (Sennheiser HD 650, HD 600, and HD 800 [from Sennheiser's website])
84.37 milliWatts @ 50 Ω (approximate impedance for planar magnetic HiFiMAN and Fostex T50RP Mad Dog headphones)
82.95 milliWatts @ 62 Ω (AKG K 701/ K 702/ Q 701)
Those calculations are not quite right. First, 14.0 Vpp is actually 4.95 Vrms, because 1 Vrms = 2 * sqrt(2) = 2.8284 Vpp. 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 nearinfinite 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:
#!/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)
I get these results:
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
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.
Edited by stv014  3/6/13 at 3:43pm
 miceblue
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From JDS Lab’s official C5 blog post:
3.337 VRMS @ 150 Ω = 74.24 milliWatts
4.146 VRMS @ 600 Ω = 28.65 milliWatts
14.0 Vpp (peaktopeak voltage) = 4.214 Volts (maximum output voltage)
Extrapolated values (assuming linear behavior):
86.19 milliWatts @ 32 Ω (typical portable headphones)
87.81 milliWatts @ 16 Ω (typical earphones)
59.05 milliWatts @ 300 Ω (Sennheiser HD 650, HD 600, and HD 800 [from Sennheiser's website])
84.37 milliWatts @ 50 Ω (approximate impedance for planar magnetic HiFiMAN and Fostex T50RP Mad Dog headphones)
82.95 milliWatts @ 62 Ω (AKG K 701/ K 702/ Q 701)
Those calculations are not quite right. First, 14.0 Vpp is actually 4.95 Vrms, because 1 Vrms = 2 * sqrt(2) = 2.8284 Vpp. 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 nearinfinite 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:
#!/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)
I get these results:
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
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.
And this is why I leave the math to the people who actually know what they're doing.
Thank you for explaining that, but I'm not quite sure what the rOut equation is doing. Also, why did you define rOut and vUnld, and then redefined them later in the scrypt?
 JDS Labs C5/C5D (pg96) portable amp/amp+DAC
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