I am analyzing common emitter amplifier where operating point is set by feedback resistor between b-c.
Voltage gain showed by symulator is much lower than (-)R2/R1 and input resistance is lower than B*R1 (B is beta, hfe).
What is R3 impact? I am looking for exact formulas (for "by hand" calculating) for k and R_inp. Knowledge about R_out would be also educating.
My goal is understanding voltage and input impedance drop in circuit like below:
Predicted: k=U_Rc/25mV=6V/25mV=240 V/V, R_input=(25mV/Ib || 1k)*B || 82k=109k || 82k=47k
Simulated: k=113 V/V, R_input=13k
Answer
For this circuit
simulate this circuit – Schematic created using CircuitLab
The input impedance is equal to Rin=(RB+RC)⋅(rπ(β+1)RE)rπ+(β+1)(RC+RE)+RB
Or google the Miller effect How does a Miller cap physically create a pole in circuits? or this Simple op amp question, finding gain and input resistance
AS for your second circuit
$Q_2$ emitter current will be around $700µA$ (If I ignore the base current)
And $I_{C1} \approx \frac{(V1+V2)- 2V_{BE}(1+\frac{R_5}{R_6})}{R_2} \approx 52mA $
Which means that the in a real circuit the $Q_2$ emitter current will be around $800µA$
and $Q_2$ base current will be around $I_{B2} = 2µA$ hence $I_{C1} \approx\frac{(V1+V2)- (2V_{BE}(1+\frac{R_5}{R_6})+I_{B2}R_5) }{R_2}\approx\ 32mA $
So the AC small-signal parameters are:
$r_{e2} = \frac{26mV}{I_E2} = 32\Omega$ and $r_{e1} = \frac{26mV}{I_E1} = 0.8\Omega$
The voltage gain will be around
AV≈R7||(β+1)re1R7||(β+1)re1+re2⋅R2re1≈221V/V
We get such big difference because of the BC547C model you used in the simulation.
In your model we see $R_E = 0.6\Omega$
Which means, that $Q_1$ voltage gain stage is
R2re1+RE=200Ω0.8Ω+0.6Ω=143
Therefore the overall voltage gain is around $AV = 143*0.9 = 129 V/V$
And the input resistance is equal around:
RIN≈R6||R5AV+1||[(β+1)∗(re2+R7||(β+1)∗re1)]≈12kΩ
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