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#11 | |
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diyAudio Member
Join Date: Aug 2012
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Quote:
P RMS = [ (Ipeak x 2)^2 * 8ohms ] / sqrt(2) P RMS = Sqrt ( 3^2 * 8 ) /sqrt(2) P RMS = 72/1.414 P RMS = 50.91 yeah it takes time to stabilise the bias. I read the link that you gave to me... I am looking forward for working on two pole compensation improving the HF distortion. so I have a doubt that the way you recommended to measure the Bias current is continuous not peak is it? the above calculation is nelsons method of calculating class A.. he first calculates everything in peak and the divides it by 1.414 to get the final power... Last edited by rhythmsandy; 30th January 2013 at 03:49 PM. |
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#12 |
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diyAudio Member
Join Date: Jan 2008
Location: Carp
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When you measure bias current you measure DC current flowing through the output transistors with the input shorted. There is no peak current, just DC.
The current into your speakers will vary with frequency as nearly all speakers are a reactive load with varying phase angles between voltage and current and varying impeadeance with frequency. A good capacitor will nearly dissapate 0 watts of heat while having tens of amps of current flowing through it and having hundreds of volts across it. Same with a good inductor. Think about where the currents flow in a class A output stage, with no load, with no load and signal, with a load and no signal, with a load and signal. |
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#13 |
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diyAudio Member
Join Date: Aug 2012
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according to your way it seems correct as its all about the I^2 x R which is right but how come nelson is measuring peak but why is he going that way rather than direct I^2 x R...
strange... does anybody has an explanation? why is nelson considering peak first? but how is he calculating peak when one can measure the DC directly.. |
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#14 |
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diyAudio Member
Join Date: Aug 2012
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what will be the Vce for the above amp is it some value less than the psu like if my psu is 78V so Vce could be 70V?
what about Ic? Im just calculating the max transient current that can be drawn for a duration of 100ms and 10ms so that the psu can be optimised... so how to calculate the Vce? |
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#15 |
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diyAudio Member
Join Date: Aug 2012
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this is the equation but we dont have the values of Rb and Re values are not given in the datasheet.
Ib = Ic * beta. Vcc - IbRb - Vbe - (beta + 1)Re = 0 |
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#16 |
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diyAudio Member
Join Date: Aug 2012
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Vcc - IcRc - Vce - IcRe = 0 how to calculate the Rc and Re values? its not in datasheet
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#17 |
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diyAudio Member
Join Date: Jan 2008
Location: Carp
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I don't understand your last few posts. Are you making a new design? I thought you are you using a prebuilt board.
Doug Self has discussed the value of Re in his book. Look for his articles on "Distortion in amplifiers" maybee you shouldn't as you will find other problems to loose sleep over The power transformer is usually chosen depending on desired power output. If you will have a lot of bias current add that to the transformer ratting and then a bit more for piece of mind. example stereo 100 watt/channel power amplifier @8ohms transformer = 200-300 VA same amplifier but with 500 ma bias. to get 100 watt @8ohms the rails will be about +/- 50 volts DC 500ma bias x 100VDC = 50 watts per channel in standby ie no signal so add another 100 va to get 300-400 va transformer. Some people here use KVA transformers for 50 watt amplifiers. Most music has a 10-1 dynamic range. You will almost never run your amplifier constantly to clipping, maybe sometimes when you have had too many drinks and there is a loud party. just about every transformer can run 200% rating for short times. Ie you can pull double the VA out of it with no real problems. For capacitors in the power supply 2000uf per amp of output current is enough. There are tens of thousands of amplifiers that have less and they sound good. There is a long thread here somewhere that discusses this to death. for 100 watts per channel you need about 3.5 amps @ 8 ohm, no speaker is exactly 8 ohms so lets say 5 amps. so we get 5 x 2 x 2000, = 20,000uf per rail. I just took apart an old kenwood amplifier and I reused the chasis for a project. The kenwood amplifier was rated at 130 watts /channel. The capacitors were 7,500 uf in the old Kenwood. I have 2 other similar kenwood amplifiers that I use and they sound fine. see post 2024 in the folowing thread "The Wire AMP" Class A/AB Power Amplifier based on the LME49830 with Lateral Mosfets the transient current will depend on the load and the resistance of all the wiring and all the series parts. Could be hundreds of amps esp if the output is shorted, I have done it and parts dissapear only to leave a black smudge and smoke and melted wires. Just build your amp and have fun. Add some changes later, does it sound better or different or worse if you have 1,000 or 100,000 uf capacitors. some people don't like to have relay contacs as the say the relay contacts change the sound. Some people use solid copper wire some use braided wire some silver wire. Some use gold connectors some brass etc etc. Some people like to align the amplifier so the earth's magenetic field only crosses signal wires at 90 degrees. Hey maybe I should start a new thred on that idea. |
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#18 |
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diyAudio Member
Join Date: Aug 2012
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i had a strange HF rolloff after 14khz so i suspected that its becoz of the RL network on the output so I just connected the output straight before the RL netwoked and all worked super good and even the transparency increased alot.
what RL values are recommended at the output. what problems one would face if its completely removed? |
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#19 |
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diyAudio Member
Join Date: Jan 2008
Location: Carp
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the output R and L should not cause any HF roll off
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#20 | |
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diyAudio Member
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Quote:
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