hmm...I missed it? \[ -16.2/5 \] is AvOK Try this circuit:
View attachment 194092
Assume β = 100 and the circuit is "driven" from a ideal voltage source.
Av = ?
hmm...I missed it? \[ -16.2/5 \] is AvOK Try this circuit:
View attachment 194092
Assume β = 100 and the circuit is "driven" from a ideal voltage source.
Av = ?
yeah : was just thinking it..will be Rb || 0 = 0If the circuit is "driven" from an ideal voltage source the gain will be equal to (20k||5k)/1k = 4V/V if we ignore re.
The RB will not be "reflected" at the emitter side due to the ideal voltage source driving the amplifier input.
first is -19.8....or you can say -20
Oki. but why 19.8? I would say 20.first is -19.8....or you can say -20
No, wrong, This one is a tricky.2nd is about -200?
19.8 because of β/ (β+1)...20 if ratio is 1Oki. but why 19.8? I would say 20.
No, wrong, This one is a tricky.
Assume ideal current source, hence ro = ∞The resistance in the current source is not important?
no gain then? is the answer a number?Assume ideal current source, hence ro = ∞
lol aii...Need study my first circuit then read the links and seeGood job, no gain.
Exactly this is what you need to do now. And I have to go to sleep.lol aii...Need study my first circuit then read the links and see
lol...Sleep well... hopefully you have a dreamExactly this is what you need to do now. And I have to go to sleep.
I think there is an error(s) in the book.
First: No gain?? I think 330k should be 330 ohms. Now we have gain.
Second: Bias is not right, never was. 6.1k should be 61k
In SPICE I changes the 1V signal on input to 0.5V.
Now the circuit is just strange.
Now Ve2 is at 6.8V or about 1/2 of the 12V supply. That is good.
I think "Vo" is the output not Ve2. Also strange.
With a 0.5V signal in Vin we get Ve2=about 4V and VO=3V peak.
If I was making this amplifier: (note 0.33 ohms could be 0)
View attachment 194076
hi Z,in post #12. Do you know how he did the model ".n NPN(BF = 99)? "
Hmm E,hi Z,
In the LTS cmp folder you should find 'standard.bjt'
I have taken a clip of the 2N3904 transistor, BF is the 'Gain' of the transistor
E
C:\Users\ericg\Documents\LTspiceXVII\lib\cmp\standard.bjt
.model 2N3904_100 NPN(IS=1E-14 VAF=100 Bf=100 IKF=0.4 XTB=1.5 BR=4 CJC=4E-12 CJE=8E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12 ITF=1 VTF=2 XTF=3 Vceo=40 Icrating=200m mfg=Philips)
.model 2N3904_200 NPN(IS=1E-14 VAF=100 Bf=200 IKF=0.4 XTB=1.5 BR=4 CJC=4E-12 CJE=8E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12 ITF=1 VTF=2 XTF=3 Vceo=40 Icrating=200m mfg=Philips)
Yeah the quick fix.. Unfortunately, still do not follow...Can do something like in the image below?Another quick fix way is to paste this into your circuit, it works but it looks messy IMO.
.model My_2N3904_100 NPN(IS=1E-14 VAF=100 Bf=100 IKF=0.4 XTB=1.5 BR=4 CJC=4E-12 CJE=8E-12 RB=20 RC=0.1 RE=0.1 TR=250E-9 TF=350E-12 ITF=1 VTF=2 XTF=3 Vceo=40 Icrating=200m mfg=Philips)

Thanks but how did you find it? The My_2n3904 ?hi Z,
I have modified your original circuit to demo the BF option.
E
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