Calculating output impedance of Morgan-Jones headphone amp? - diyAudio
 Calculating output impedance of Morgan-Jones headphone amp?
 User Name Stay logged in? Password
 Home Forums Rules Articles diyAudio Store Gallery Wiki Blogs Register Donations FAQ Calendar Search Today's Posts Mark Forums Read Search

 Please consider donating to help us continue to serve you. Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving
 23rd October 2012, 08:25 PM #1 diyAudio Member   Join Date: Oct 2012 Calculating output impedance of Morgan-Jones headphone amp? When I read this page: HeadWize - Project: The Morgan Jones Mini Tube Headphone Amplifier by Chu Moy it says that the calculated output impedance of the first amp is 6 ohms. I'm trying to teach myself how to calculate output impedances but I can't arrive at that 6 ohm figure no matter what I try. Let me explain what I'm doing and hopefully someone can enlighten me? Looking at that schematic, it looks like the output is seeing the Ra of one triode (I'm using Ra=2600 ohms) in parallel with (Ra of the other triode - 2600 ohms - in series with R4 which is 3300 ohms). So I'm taking: 1/(1/2600 + 1/(2600+3300)) = 1804 ohms Please forgive my low level of knowledge, I'm trying to learn. Thanks in advance for anyone willing to help me figure this out!

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is OffTrackbacks are Off Pingbacks are Off Refbacks are Off Forum Rules

 Similar Threads Thread Thread Starter Forum Replies Last Post pingfloid Tubes / Valves 0 11th May 2012 07:50 AM chokesrule Tubes / Valves 8 18th December 2010 07:02 PM trevorjn Headphone Systems 7 6th December 2009 10:52 PM NotInTheMafia Tubes / Valves 13 21st January 2004 12:33 AM Prune Tubes / Valves 4 13th May 2003 07:00 PM

 New To Site? Need Help?

All times are GMT. The time now is 04:27 PM.