CAVEMaN 0 Report post Posted August 6, 2003 need help N@Z? I built my own amp, maybe I can give you a hand... Share this post Link to post Share on other sites
N@Z 0 Report post Posted August 6, 2003 I certainly don't mind the assistance. I PM you when I get round to it. Share this post Link to post Share on other sites
Wortel 0 Report post Posted August 6, 2003 Im gonna start off cheap. My dad is not audiophile just very skilled with electronics. I wanna build that Jan Maier Corda Maybe someday.. I need some skills in reading circuit diagrams... Share this post Link to post Share on other sites
Firefox 0 Report post Posted August 6, 2003 however it's quite a nice amp for portable use .i would suggest you build a Gilmore instead. Not everyone can build a gilmore, dude.. I add a filter to the power in, to reduce the noise, and it does its job quite well. and for the input cap, 0.1uF is too small in the DC filter, changing to 5uF would be better. Think of this, 0.1uF, the cut off frequency is 15Hz, which is quite high. When change it to 5uF, the cutt off- freq drop to 1.5Hz. It's not a filter to reduce noise actually. The capacitor's purpose is to reduce DC offset from the inputs. Given a 100K ohms resistor and a 0.1uF cap forming a high pass filter, there is significant phase distortion & attenuation up to 100Hz. That explains why the change to a 5uF cap improved the sound. It's not because the cut-off at 15.9Hz is too high. BTW, a 5uF cap. with 100K ohms resistor forms a cut off freq. @ 0.318Hz, not 1.5Hz. At the output, i suggest you put a 120ohm resistor there for AKG K501 for maximum power transfer.( basically the sound remains the same). To obtain nominal power transfer to the headphones, you need to measure the impedance of the output and add a resistor in series with the output to obtain a TOTAL resistance/impedance equal to the impedance of the headphones. Secondly, obtaining maximal power transfer to the headphones doesn't mean better sound. Most headphone amps have lower output impedance because that allows the amp to better control the drivers in the headphones. Share this post Link to post Share on other sites
Wortel 0 Report post Posted August 6, 2003 I'll take it onboard Firefox. I read your post carefully but don't know enough about electronics to really follow. Share this post Link to post Share on other sites
huangyong 0 Report post Posted August 7, 2003 0.1uF is too small in the DC filter, what i mean is DC filter which block any DC at the signal input. and forgot to put in the calculation which is 1/(2*pie*R*C) pie=3.1416 R=the reisistor value in the filter C=the capacitor value in the filter. and very sorry for miscalculation. what i mean for the "reduce noise" is include with a power filtering which including caps and buffer. and after experimenting, the maximum power transfer design is only good for AKG K501...i think... Share this post Link to post Share on other sites
jasonhanjk 0 Report post Posted August 7, 2003 PI=3.1415927 It is highly recommended to use the opamp DC offset rather than using the capacitors. It is introducing multiple phase-shift to audio signal as the opamp is drawing micro amps of current from the source. Share this post Link to post Share on other sites
Firefox 0 Report post Posted August 7, 2003 It is highly recommended to use the opamp DC offset rather than using the capacitors. Opamp DC offset? Care to explain? Share this post Link to post Share on other sites
jasonhanjk 0 Report post Posted August 7, 2003 A bit hard... Share this post Link to post Share on other sites
Firefox 0 Report post Posted August 7, 2003 A bit hard... Frankly, if opamps could get rid of DC voltage on the input, you're going to have major problems using them in amps. The capacitors are for removing DC offset from the source, not from the output of the opamp as you might have thought. Most Fet opamps are pretty okay but bi-polar opamps do need some special configurations to prevent offset on the output. Share this post Link to post Share on other sites
jasonhanjk 0 Report post Posted August 7, 2003 I see.... Share this post Link to post Share on other sites
Firefox 0 Report post Posted August 7, 2003 I see.... To conclude, the input capacitors are still essential when it comes to removing DC offset from your sources. However, for personal amps where you know the source(s) do not have DC offset on the output, you may safely omit the caps. Thus, removing the problem(s) of degraded sound quality (if any) & the high-pass filter. Share this post Link to post Share on other sites
jasonhanjk 0 Report post Posted August 7, 2003 Is the best way to 0 dc offset by matching resistor in the source? Share this post Link to post Share on other sites
Firefox 0 Report post Posted August 7, 2003 Is the best way to 0 dc offset by matching resistor in the source? No, it depends on how the output stage is in the source. Usually, there are already capacitors in the CDP for opamps based outputs. Hence, any DC offset would have been removed. Some outputs are tube outputs and those have their own compensation for removing DC. Some use transistors in push-pull configurations and the DC is removed using other methods. Share this post Link to post Share on other sites
Wortel 0 Report post Posted August 7, 2003 Now Im even more confused. I gotta read more headwize. :| Share this post Link to post Share on other sites
