IZ3LSV

G8MNY  > TECH     03.05.16 10:52l 65 Lines 2550 Bytes #999 (0) @ WW
BID : 3123_GB7CIP
Read: GUEST
Subj: Dual phase AF AMP explained
Path: IZ3LSV<ED1ZAC<GB7CIP
Sent: 160503/0914Z @:GB7CIP.#32.GBR.EURO #:3123 [Caterham Surrey GBR] $:3123_GB
From: G8MNY@GB7CIP.#32.GBR.EURO
To  : TECH@WW

By G8MNY                                    (Updated May 06)
(8 Bit ASCII graphics use code page 437 or 850, Terminal Font)

               /'\./ ³\   /'\./     \./'\   /³
                 ÚÄÄij '>ÄÄÄÄÄÄ L S ÄÄÄÄÄÄ<' ÃÄÄÄ¿
                 ³   ³/'                   `\³   ³
/'\./ ÚÄÄÄÄÄÄÄÄ¿ ³                               ³
______³  PHASE ÃÄÙ                               ³
      ³INVERTERÃÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ
      ÀÄÄÄÄÄÄÄÄÙ       \./'\
   (phase splitter)

The LS sees 2x the voltage of a normal single amplifier arrangement. That is 4
times the power! e.g. a 4ê LS with 0V & 13V DC on the amplifier may give ñ 12V
peaks to the LS. Peak crest power = 12 x 12 / 4 = 36W peak, = 18W RMS.

Or for typical commercial disco amplifier may use ñ90V @ 22A PSU to power a
pair of amplifiers in bridged to 8ê LS load, to give a peak power of
179 x 179 / 8 = 4kW peak crest power (USA rating), or 2kW RMS. Although this
sounds big they are made this size! But the power rails usually dip a lot under
heavy load, but 4kW peak pulse is what the loud speaker system has to handle!

The phase inverter/splitter could be just this....

                   ÚÄÄÄÄÄÄÄÄÄÄÄsmoothed +ve
                  1kê
              ÚÄÄÄÄ´    1uF
            220kê  ÃÄÄÄÄ´ÃÄÄÄÄÄ phase inverted by 180ø
/'\./         ³  ³/    +         \./'\
Input ÄÄÄÄÄ´ÃÄÁÄÄ´ NPN
         0.1uF   ³\e    1uF      /'\./
                   ÃÄÄÄÄ´ÃÄÄÄÄÄNON inverted (sames as I/P)
                  1kê  +
      ÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄ

The identical 1kê in emitter & collector give the same output if there are no
significant following different circuit loads. The bias 220kê should be
selected to give about 1/2 the voltage across the transistor. e.g. on a 12V
supply emitter = 3V, collector = 9V, so there is 6V across the transistor, then
about 2V peak to peak can be handled.

But it is more usual to attenuate the output of the 1st amplifier by the gain
of the 2nd amplifier & feed it into the inverting input of the 2nd amplifier.

           AMP             AMP
/'\./     ³\ 1  /'\./     \./'\  2 /³                R1+R2
  I/P ÄÄÄÄ´ `>ÄÄÂÄÄÄÄÄ L S ÄÄÄÄÄÄ<'-ÃÄ¿  AMP2 GAIN = ÄÄÄÄÄ
          ³/'   ³                 `\³ ³               R2
               R1                     ³
                ÃÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÙ
               R2
               _³_
              ////

But with this simple method you do get 2x the distortion out of the 2nd
amplifier!


Why don't U send an interesting bul?

73 De John, G8MNY @ GB7CIP


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