| | G8MNY > TECH 22.04.04 17:23l 81 Lines 2915 Bytes #999 (0) @ WW
BID : 53941_GB7CIP
Read: GUEST OE7FMI
Subj: An AF amplifier stage
Path: DB0FHN<DB0FOR<DB0MRW<DB0ERF<DB0HDF<DB0HOT<OK0PBX<OK0PAD<OK0PPL<DB0RES<
ON0AR<GB7CIP
Sent: 040421/2236Z @:GB7CIP.#32.GBR.EU #:53941 [Caterham] $:53941_GB7CIP
From: G8MNY@GB7CIP.#32.GBR.EU
To : TECH@WW
By G8MNY (new graphics Dec 03)
This simple amplifier circuit is easy for calculations.
+9V ÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÂÄÄÄÄÄÄÄÄÄÄÄ
Rc
ÚÄÄÄÄÄÄ´ Cout
Rb ÃÄÄÄÄ´ÃÄÄÄÄ
³ ³/
I/P ÄÄÄ´ÃÄÄÄÁÄÄÄÄ´ NPN
Cin ³\e
³
Re
0VÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÄÁÄÄÄÄÄÄÄÄÄÄÄ
To get « the DC swing on the O/P then biasing Rb = Hfe x (Rc+Re)
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 G approx = Rc/Re (Rc may be lower due to external load).
With high 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).
O/P 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 O/P Z
it will also depend the I/P source Z.
I/P 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.
HF Compensation
HF loss can be compensated for by putting a suitable C across Re to give
+3dB boost were Xc=Re eg 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
I/P ÄÄÄ´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
O/P Z about 900ê +XCout
I/P Z about 5Kê +XCin
LF response with Input source Z of zero, & O/P load of 10K...
I/P Ä3dB LF roll off, @ 31Hz where Xc = 5Kê
O/P Ä3dB LF roll off, @ 29Hz where Xc = 10.9Kê
Giving Ä6dB @ 30Hz & 12dB/Octave LF cut.
Why don't U send an interesting bul?
/QSL
73 De John, G8MNY @ GB7CIP
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