- #71
michael1978
- 434
- 19
THIS is the same example without re2, but you add in serie re2Jony130 said:Yes for Vcc = 10V and hfe = 150
i mean this laste circuit, and the first example without R2 has the same value value of r1 and r2 is r1=22k, and r2=4.7K and power supply 10V ,
and this is correct R1 = (Vcc - Vb)/( 11*Ib) = 22KΩ
of this R1 = (Vcc - Vb)/( 10*Ib) = 22KΩ
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OK
Av = Rc/re = gm*Rc = Ic/26mV*Rc = (Ic*Rc)/26mV
50[V/V] = (Ic*Rc)/26mV
from this
(Ic*Rc) = 1.3
I choose Rc = 1K
So
Ic = 1.3/1K = 1.3mA
But as you can see this approach is not very good because give as low voltage swing.
We need to use Re1 or use a different type of a circuit. But lest as add Re1 resistor.
Now Av = Rc/(re+Re1)
And for Rc = 1K we can use this
Ic = 0.5Vcc/Rc = 5V/1K = 5mA
So re = 26mV/5mA = 5.2Ω
And
Re1 = Rc/Av - re = 1K/50 - 5.2Ω = 20 - 5.2Ω = 14Ω
This circuit is also not very good because low Re1 means that Hfe spread and temperature change will have significant impact on bias point stability. This is why I use Ve = 1V and add Re2 to reduce the voltage gain to 50[V/V]
thnx
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