IZ3LSV

G8MNY  > TECH     10.03.08 09:00l 65 Lines 2530 Bytes #999 (0) @ WW
BID : 49599_GB7CIP
Read: GUEST
Subj: Dual phase AF AMP explained
Path: IZ3LSV<IW2OHX<OE6XPE<DB0RES<F5GOV<ON4HU<ON0BEL<GB7CIP
Sent: 080310/0020Z @:GB7CIP.#32.GBR.EU #:49599 [Caterham] $:49599_GB7CIP
From: G8MNY@GB7CIP.#32.GBR.EU
To  : TECH@WW

By G8MNY                                    (Updated May 06)
(8 Bit ASCII Graphics use code page 437 or 850)

               /'\./ 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 4V peak to peak can be handled.

But it is more usual to attenuate the O/P 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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