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G8MNY > TECHNI 04.12.21 11:05l 85 Lines 3174 Bytes #999 (0) @ WW
BID : 54735_GB7CIP
Read: GAST
Subj: An AF amplifier stage
Path: DB0FFL<OE5XBL<DB0RBS<DB0ERF<DK0WUE<F1OYP<VE3CGR<OK2PEN<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 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄĀÄÄÄÄÄÄÄÄÄÄÄ _
Rc / \
ŚÄÄÄÄÄÄ“ Cout Output Ž Ż
Rb ĆÄÄÄÄ“ĆÄÄÄÄ ³ ³ ³
/'\,/ ³ ³/ Ž Ż Ž
InputÄÄÄ“ĆÄÄÄĮÄÄÄÄ“ NPN \_/
Cin ³\e
³ /'\,/
Re
0VÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄĮÄÄÄÄÄÄÄÄÄÄÄ
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 ÄÄÄÄÄÄÄÄÄÄÄÄÄÄĀÄÄÄÄÄÄÄÄÄÄ
1kź
ŚÄÄÄÄÄÄ“ + Cout
100kź ĆÄÄÄÄ“ĆÄÄÄÄ Output
+ ³ ³/ 0.5uF ³
InputÄÄ“ĆÄÄÄĮÄÄÄÄ“ Hfe=100 10k Load
Cin NPN³\e ³
1uF ³ ³
100ź ³
0V ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄĮÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄ
So in the above example Collector should be around +6V
Gain about 9 times
Output Z about 900ź + XCout
Input Z about 5kź + XCin
LF response with input source Z of zero, & output load of 10k...
Input Ä3dB LF roll off, @ 31Hz where Xc = 5kź
Output Ä3dB LF roll off, @ 29Hz where Xc = 10.9kź
Giving Ä6dB @ 30Hz & 12dB/Octave LF cut.
Why don't U send an interesting bul?
73 De John, G8MNY @ GB7CIP
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