Expert of Black Magic Only Can Answer That !!!!!!!!

[ahmed osama]

Hi all

check the following circuit in the image :

I found strange thing , if you calc. the poles of this cir by ur hand u will find two poles
1: ~ 2 Mhz
2: ~2 Ghz

So the dominant pole (3 db point) is 2 Mhz but in simulation just one pole which is 2 GHZ ???

Note : i calc the pole by my hand but by the simulator values which in the circuit

any explain???

 

[safwatonline]

i didnt take a very good look but is Rc=16 or 16k , this may be ur 3 decade factor

 

[ahmed osama]

i didnt take a very good look but is Rc=16 or 16k , this may be ur 3 decade factor

No i am so sure it is 16 not 16K , i checked this circuit so so many before puting it here

thx anyway

 

[pisoiu]

I've duplicated the simulation and the results confirms. But be serious....120 V ?!? Something is very wrong.

/pisoiu

 

[safwatonline]

I've duplicated the simulation and the results confirms. But be serious....120 V ?!? Something is very wrong.

/pisoiu

well , this is AC analysis voltage can take any value.
btw what is the 16 ohm resistor , isnt this nothing , just short it.
also how did u get the values u r talking about 2Mhz and 2 Ghz.

 

[pixel]

Not a magic at all, maybe you were tyred?. It is quite normal that ac analysys gives you a rubbish if you give it to analyze a rubbish. ac analysys does not have any sense if dc operation points are not properly set, because non linear components are linearized aroud dc points, and simulation is valid ONLY at the linear range around them (small signals). From your schematic Vbe(dc)=Vdd, but it should be ~ 0.7V!!!
How to add one resistor from base to gnd?

 

[safwatonline]

Not a magic at all, maybe you were tyred?. It is quite normal that ac analysys gives you a rubbish if you give it to analyze a rubbish. ac analysys does not have any sense if dc operation points are not properly set, because non linear components are linearized aroud dc points, and simulation is valid ONLY at the linear range around them (small signals). From your schematic Vbe(dc)=Vdd, but it should be ~ 0.7V!!!
How to add one resistor from base to gnd?

well , what r talking about , we know that ac analysis is linearizing the dc point ,so the gain and non -linearity will not be true , but this BJT is in active mode re-check the file Vbe=o.79 and Vce≈9.
hint: note this is BJT not MOS so consider base current .

 

[tsb_nph]

Hi all

check the following circuit in the image :

I found strange thing , if you calc. the poles of this cir by ur hand u will find two poles
1: ~ 2 Mhz
2: ~2 Ghz

So the dominant pole (3 db point) is 2 Mhz but in simulation just one pole which is 2 GHZ ???

Note : i calc the pole by my hand but by the simulator values which in the circuit

any explain???

The value of collector current is huge - 288 mA. Are you sure the BJT model is valid at this value of collector current? The value of beta degrades at high collector current. The nominal value of beta in the model is 378.9, but the actual beta in the simulation is 176. Maybe the BJT model is not valid at this value of high current, please verify if model holds good for 288 mA of IC - maybe Gummel Poon model might not be accurate, you might have to use high current (hi-cum) model.

Coming to hand calculations, the two poles are at the input node (base) and at the output (collector) of the BJT. Can you please let us know the details of your hand calculations :

i) Input R=? C=?
You have a 5K Ohms in parallel with input resistance of BJT. What is the value of capacitance at the input node?
ii) Output R= 15.2 Ohms C=?
Ro for the transistor is Va/Ic, which is 315 Ohms, in parallel with 16 Ohms gives around 15.2 ohms. What is the value of capacitance at the output node?


Bharath

 

[pixel]

well , what r talking about , we know that ac analysis is linearizing the dc point ,so the gain and non -linearity will not be true , but this BJT is in active mode re-check the file Vbe=o.79 and Vce≈9.
hint: note this is BJT not MOS so consider base current .

OK you are wright Vbe=0.79V, not 9V. I work with much lower power and for me IB=1.6mA is something really wrong.

I think that tsb_nph has got a point. It seems like model is wrong, and it should be changed. Also you should add load capacitance, because you phyisically have it.
In data-sheet - Fig3. it could be seen that beta depends on current Ic and Vce. For currents greater than 200-300mA beta can have significant drop, which is greater for smaller Vce.
Maybe you should try to lower the current Ic (i.e x1/2), and increase Rc (x2) to get higher current gain and to maintain same Vce voltage (4.5V).

 

[VSMVDD]

you need ic=0 on the collector

 

[PaulHolland]

Hi, Simulation can sometimes be completely noncense: if you make a current source of let say 1 Amp and you put a 1 M Ohm resistor parallel to the current source and accross you put a volt meter you will see: 1 Megavolts... WOW, why doing difficulr in a TV Set. make your simulated currentsource and you can create 25kV for a few cents ....

Paul.

 

[aryajur]

Hello Ahmed,
It puzzled me at first also. I guess the following explanation will clear all points raised by others also:

1. The circuit has been biased in a self biasing scheme which is perfectly valid. The DC operating point is shown alongside the circuit and everything adds up there.
2. 120V is the AC answer given by the simulator for a input AC of 1V, ie. the small signal gain for the circuit is around 120, which also adds up if you calculate it using GM. RO
3. The pole positions as calculated by ahmed are also perfectly valid. Here is the calculation:
At the output R = 16||324 = 15.2
C ≈CBC = 5.76p
so f = 1/(2ΠRC) = 1.82GHz
At the base R = 5k||RPI = 5k||16.8 ≈16.8
C = (3.67n + 120*5.76p) = 4.36n (Miller effect)
so f = 1/(2ΠRC) = 2.17MHz

But the AC analysis shows only the output zero. Here it is not the issue of the model being valid or not since whatever the model is, whether representing reality or not, there is some model and according to that model we are not getting the answer we would expect. The plot should have showed a pole at 2MHz but it doesn't show it only shows one at 2GHz.
The reason is because at the input you are driving it directly from a source! So effectively the base capacitance is being charged through a zero resistance that is why the pole goes to f = ∞ and thus you won't see it in the simulation.
Hope this clears all the doubts

 

[moh2000]

By other words,
As long as the circuit is driven that way, the first pole does not exist. Because the capacitance at the base does see a very low impedance (voltage source).

 

[teteamigo]

To see a pole at f=2MHz choose a coupling capacitor with value lower than or equal to capacitance at Base node to ground.

 

[aryajur]

The input pole will be visible if the input driving source is a current source.