AM receiver

Thread Starter

sima7

Joined May 20, 2016
7
hi
i simulated a AM receiver circuit (shown bellow). transistor 1&2 are off but should be on . i don't know what to do?
please hellp me.
 

Thread Starter

sima7

Joined May 20, 2016
7
Of course, the main circuit is


just the name of elements are different and i changed the r4 10k ohm to 1kohm just for testing.
someones set R2 , 2.2k and someones set 330 ohm.
Q1&2 are off in this photo.
 

Thread Starter

sima7

Joined May 20, 2016
7
i also simulate the circuit with HSPICE .
DC analysis is:
**** bipolar junction transistors


subckt
element 0:q2 0:q1 0:q3
model 0:bc109c-x 0:bc109c-x 0:bc109c-x
ib 7.7651p -250.3103p 1.7210u
ic 16.6853n 250.3032p 738.5131u
vbe 369.9050m -369.9050m 651.0822m
vce 8.8818 8.5413 1.5977
vbc -8.5119 -8.9112 -946.5764m
vs -8.8818 -8.9112 -1.5966
power 148.1988n 2.2305n 1.1810m
betad 2.1488k -999.9718m 429.1066
gm 639.2468n 24.0493f 28.7194m
rpi 133.7118x 5.239e+13 16.7054k
rx 0. 0. 0.
ro 2.2301g 9.601e+14 39.3839k
cpi 13.8119p 9.8936p 28.1532p
cmu 2.2225p 2.1914p 3.9691p
cbx 0. 0. 0.
ccs 0. 0. 0.
betaac 85.4749 1.2602 479.7693
ft 6.3448k 369.6009m 142.2952x
 
Last edited:

Thread Starter

sima7

Joined May 20, 2016
7
netlist:


c2 m 0 100n ic=10m
c3 vc2 b3 100n ic=10m
c4 vc3 n 100n ic=10m
c5 vcc 0 10u ic=10m
c6 vc1 0 100n ic=10m
c7 vi m 500p ic=10m
l1 vi m 200u ic=10m
r1 m vc2 100k
r2 e1 b2 1k
r3 vc1 vc2 330
r4 vc3 b3 550k
r5 vcc vc3 10k
r6 n 0 25
r7 vcc vc1 1k
vcc vcc 0 9
q2 vc2 b2 0 BC109C-X
q1 vc1 vi e1 BC109C-X

q3 vc3 b3 0 BC109C-X
.model BC109C-X NPN
+Is=7.049f
+Xti=3
+Eg=1.11
+Vaf=28.14
+Bf=677
+Ise=7.049f
+Ne=1.38
+Ikf=96.23
+Nk=0.5
+Xtb=1.5
+Br=2.209
+Isc=250.3p
+Nc=2.002
+Ikr=10.73
+Rc=1.433
+Cjc=5.38p
+Mjc=.329
+Vjc=.6218
+Fc=.5
+Cje=11.5p
+Mje=.2717
+Vje=.5
+Tr=10n
+Tf=437.8p
+Itf=3.097
+Xtf=12.85
+Vtf=10
* PHILIPS pid=bc108c case=TO18
* 91-08-02 dsq
*$


vi vi 0 sin(0,1u ,500k)

.op
.tran 1u 50m
.end
 
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