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G8MNY  > TECH     04.12.21 10:00z 85 Lines 3174 Bytes #29 (0) @ WW
BID : 54735_GB7CIP
Subj: An AF amplifier stage
Path: SR8BBS<SR1BSZ<IW0QNL<IW2OHX<UA6ADV<LU4ECL<GB7CIP
Sent: 211204/0952Z @:GB7CIP.#32.GBR.EURO #:54735 [Caterham Surrey GBR]
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 컴컴컴컴컴컴컴컫컴컴컴컴컴        _
                    Rc                 / \
             旼컴컴캑   Cout   Output    
            Rb      쳐컴캑쳐컴         
  /'\,/          /                     
Input컴캑쳐컴좔컴캑 NPN            \_/
        Cin       \e
                      /'\,/
                    Re
  0V컴컴컴컴컴컴컴컴좔컴컴컴컴컴

BASE BIAS R = Hfe x (Rc+Re) Approx
     For  the DC swing on the output. This is because we want the same voltage
     C횱 (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 // ((G1) 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 3dB &
     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 컴컴컴컴컴컴컴쩡컴컴컴컴
                  1k
            旼컴컴캑   + Cout
          100k    쳐컴캑쳐컴 Output
         +      /    0.5uF       
Input컴늘컴컨컴컴 Hfe=100       10k Load
      Cin     NPN\e               
      1uF                         
                  100             
 0V 컴컴컴컴컴컴컴컨컴컴컴컴컴컴컴컴

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

LF response with input source Z of zero, & output load of 10k...
    Input 3dB LF roll off, @ 31Hz where Xc = 5k
    Output 3dB LF roll off, @ 29Hz where Xc = 10.9k
    Giving 6dB @ 30Hz & 12dB/Octave LF cut.


Why don't U send an interesting bul?

73 De John, G8MNY @ GB7CIP


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