Dispositivos amplificadores
Primer amplificador
Análisis a c-d
Como por Re circula una
le
l
=
1.55 mA, entonces:
V R
=
Re .
le
=
6 8
kQ .
1 55 mA
=
1054 V
el
I
l'
.
..
.
1.55 mA
120
=
12.9 IlA
lE,
=
lB,
+
le,
=
(0.0129 + 1.55) mA
=
1.56 mA
(20 - 8.65 - 10.54) V
=
5.19
Q
1.56 mA
12
=
10
lB
I
=
10· 12.91lA
=
0.129 mA
VB
I
(VE + 0.7) (0.81 +0.7) V
Rz
=
¡;-
=
0.129 mA
=
0.129 mA
11.7
kQ
(comercial 12
kQ)
Segundo amplificador
(20 - 1.5) V
11 . 12.9IlA
130.3
kQ
(valor comercial 120
kQ)
RB,
=
R¡II R2
=
1211120
=
10.9
kQ
120· 26mV
hiel
=
1.56
mA
2
kQ
V
RC2
=
Re, . le,
=
3.3
kQ .
3.2
mA
=
10.56 V
le,
lB
=-"
, P2
3.2 mA
=
26 66 A
120
. Il
lE,
=
lB,
+
le,
=
(0.0266 + 3.2) mA
=
3.226 mA
(20 - 7.3 - 10.56) V 664
Q
3.22mA
183
1...,174,175,176,177,178,179,180,181,182,183 185,186,187,188,189,190,191,192,193,194,...259