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G8MNY > TECH 07.04.26 15:51l 85 Lines 3175 Bytes #10 (0) @ WW
BID : 56657_GB7CIP
Subj: An AF amplifier stage
Path: JH4XSY<N3HYM<GB7YEW<GB7CIP
Sent: 260407/0641Z @:GB7CIP.#32.GBR.EURO #:56657 [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 ﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾂﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄ _
Rc / \
ﾚﾄﾄﾄﾄﾄﾄｴ Cout Output ﾞ ﾝ
Rb ﾃﾄﾄﾄﾄｴﾃﾄﾄﾄﾄ ｳ ｳ ｳ
/'\,/ ｳ ｳ/ ﾞ ﾝ ﾞ
Inputﾄﾄﾄｴﾃﾄﾄﾄﾁﾄﾄﾄﾄｴ NPN \_/
Cin ｳ\e
ｳ /'\,/
Re
0Vﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾁﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄ
BASE BIAS R = Hfe x (Rc+Re) Approx
For ｫ the DC swing on the output. This is because we want the same voltage
CﾄE (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 // ((Gﾄ1) 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 ﾄ3dB and
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, and
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 ﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾂﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄ
1K
ﾚﾄﾄﾄﾄﾄﾄｴ + Cout
100K ﾃﾄﾄﾄﾄｴﾃﾄﾄﾄﾄ Output
+ ｳ ｳ/ 0.5uF ｳ
Inputﾄﾄｴﾃﾄﾄﾄﾁﾄﾄﾄﾄｴ Hfe=100 10k Load
Cin NPNｳ\e ｳ
1uF ｳ ｳ
100R ｳ
0V ﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾁﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄﾄ
So in the above example Collector should be around +6V
Gain about 9 times
Output Z about 900R + XCout
Input Z about 5K + XCin
LF response with input source Z of zero, and output load of 10K...
Input ﾄ3dB LF roll off, @ 31Hz where Xc = 5K
Output ﾄ3dB LF roll off, @ 29Hz where Xc = 10.9K
Giving ﾄ6dB @ 30Hz & 12dB/Octave LF cut.
Why don't U send an interesting bul?
73 De John, G8MNY @ GB7CIP
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