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G8MNY  > TECH     16.04.24 10:01l 88 Lines 3325 Bytes #999 (0) @ WW
BID : 5568_GB7CIP
Read: GUEST
Subj: An AF amplifier stage
Path: IW8PGT<IZ3LSV<DB0ERF<OK0NAG<OK0NBR<SR1BSZ<EA2RCF<EA5URD
Sent: 240416/0710Z @:EA5URD.EAMU.ESP.EU #:2559 XFBB7.07 Bid:5568_GB7CIP

R:240416/0710Z @:ED1ZAC.EAC.ESP.EU #:35769 [EANET1G.06] $:5568_GB7CIP
R:240416/0705Z @:GB7CIP.#32.GBR.EURO #:5568 [Caterham Surrey GBR] $:5568_GB7CIP

From: G8MNY@GB7CIP.#32.GBR.EURO
To  : TECH@WW

By G8MNY                                (Updated Dec 04)
(8 Bit ASCII graphics use code page 437 or 850, Terminal Font)

This simple amplifier circuit is easy for calculations.

 +9V ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄ        _
                    Rc                 / \
             ÚÄÄÄÄÄÄ´   Cout   Output Þ   Ý
            Rb      ÃÄÄÄÄ´ÃÄÄÄÄ   ³   ³   ³
  /'\,/      ³    ³/              Þ   Ý   Þ 
InputÄÄÄ´ÃÄÄÄÁÄÄÄÄ´ NPN            \_/
        Cin       ³\e
                    ³  /'\,/
                    Re
  0VÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄ

BASE BIAS R = Hfe x (Rc+Re) Approx
     For « the DC swing on the output. This is because we want the same voltage
     CÄE (almost the same as across Rb) as across the total load R of Rc+Re.

GAIN = Rc/Re approx (Rc may be lower due to external load).
     With high transisitor current gain Hfe, then Ie approx = Ic, so the
     emitter NFB Re controls the collector current making the voltage gain just
     the voltage drop ratio of Rc/Re. Assuming no external loads. For high gain
     applications Re includes the internal emitter R of the transistor
     (typically a few ohms).

Output Z = XCout + (Rc // ((GÄ1) x Rb))
     This is the added components, including the apparent fraction of the bias
     Rb with load current in it.
     "//" means in parallel, many of the paralleled terms are insignificant.
     Technically the amount that (G-1)x Rb component that affects the output Z
     it will also depend the input source Z.
 
Input Z = XCin + ((Hfe x Re) // (Rb/(G+1)))
     This is the added components, including the apparent fraction of the bias
     Rb with input current in it.
     "//" means in parallel, many of the paralleled terms are insignificant.

LF Roll off
     Cin & Cout affect the LF response. Basically each one will give Ä3dB &
     6dB/Octave roll off when Xc equals the source + load Zs.

HF Response
     Intrinsically limited by the transistor's FT when the Hfe becomes 1, &
     component layout (inter capacitance) causing Miller HF N.F.B. effects
     between output & input.

HF Compensation
     HF loss can be compensated for by putting a suitable C across Re to give
     +3dB boost were Xc=Re, e.g. where the measure drop is -3dB. The 6dB/Octave
     lift after that should flatten the amp losses out. The input Z will be
     reduced at HF though. Not often used!

EXAMPLE

+12V ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄ
                  1kê
            ÚÄÄÄÄÄÄ´   + Cout
          100kê    ÃÄÄÄÄ´ÃÄÄÄÄ Output
         +  ³    ³/    0.5uF       ³
InputÄÄ´ÃÄÄÄÁÄÄÄÄ´ Hfe=100       10k Load
      Cin     NPN³\e               ³
      1uF          ³               ³
                  100ê             ³
 0V ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ

So in the above example Collector should be around +6V
Gain about 9 times
Output Z about 900ê + XCout
Input Z about 5kê  + XCin

LF response with input source Z of zero, & output load of 10k...
    Input Ä3dB LF roll off, @ 31Hz where Xc = 5kê
    Output Ä3dB LF roll off, @ 29Hz where Xc = 10.9kê
    Giving Ä6dB @ 30Hz & 12dB/Octave LF cut.


Why don't U send an interesting bul?

73 De John, G8MNY @ GB7CIP


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