BJT high frequency analysis

Question

I have a little bit of confusion on how the textbook author is determining the high frequency cutoffs for the parasitic capacitance parameters for a common emitter amplifier. This is Boylestad Electronic Devices and Circuit Theory 11th edition.

For the analysis, he treats \$ C_s\$, \$ C_c\$, and \$C_E\$ as shorts since, at these higher frequencies, these high pass-filter capacitors will be low impedance. This makes sense to me.

He ends up with two high frequency cutoff frequencies (both a low-pass filter effect) as follows:

\$ f_{Hi} = \frac{1}{2 \pi R_{THi}C_i}\$

Where

\$ R_{THi} = R_{sig}//R_1//R_2//\beta r_e\$

with

\$ r_e \approx \frac{26mV}{I_{EQ}} \$

and

\$ C_i = C_{Wi} + C_{be} + C_{Mi} = C_{Wi} + C_{be} + (1-A_{VL})C_{bc} \$

So \$ C_{Wi} \$ is input wiring capacitance, \$C_{be}\$ is the base to emitter junction capacitance, and \$ C_{Mi}\$ is the Miller effect capacitance due to \$ C_{bc}\$ feedback from input to output.

\$ A_{VL} \$ is the loaded pass-band gain (though not accounting for the AC voltage source series resistance)

The other cutoff frequency is determined by:

\$ f_{Ho} = \frac{1}{2\pi R_{THo}C_o} \$

where

\$ R_{THo} = R_L//R_C \$

and

\$ C_i = C_{Wo} + C_{ce} + C_{Mo} = C_{Wo} + C_{ce} + (1-\frac{1}{A_{VL}})C_{bc}\$

Again the miller effect is seen on the output side.

The author goes through a numerical example, and using these formulas I get the same results.

However, when I simulate this circuit in LTSpice (being sure to convert the NPN model to an "ideal" NPN, and manually insert the shown capacitors), I'm seeing that only one of the break frequencies is close (the break frequency closest to the pass-band, aka the -3dB frequency). However, the other break frequency seems to consistently off by almost exactly +1 decade (higher frequency than calculated).

I'm curious if someone can explain why this is, or indicate the correct way to calculate both cutoff frequencies.

Here is my LTSpice simulation:

And here is the plot I'm seeing:

So from the math, I get a cutoff at 689kHz and 18MHz, but checking where the asymptotic lines cross, I am seeing more like 700kHz (good) and 120 MHz (bad).

On the other hand, if I compute the values of Ci and Co using the equations, and reform the simulation to use these values directly:

I am getting a much closer match to the textbook example (700kHz and 18MHz) :

So it's either the textbook isn't calculating the equivalent miller capacitors correctly, or the simulation isn't handling the feedback capacitor (Cbc) quite correctly.

Answer 1

From what I remember, the Cbe depends on the emitter current.

Higher current causes lower reac (1/gm); at 1mA, its 26 ohms. at 2mA its 13 ohms 2mA is 4 volts / 2Kohm (ignoring the emitter bypass cap).

Thus your circuit has an incremental emitter-base small-signal resistance of about 13 ohms.

If the 2N2222 has 160MHz Ftau, the emitter-base needs 1nanosecond time-constant, or 1,000 picoSeconds TAU.

Given 1,000 pS TAU, and R of 13 ohms, the needed capacitance is 80 pisoFards.

However the large emitter bypass capacitor may influence this.

Answer 2

I decided to simulate this amplifier in LTspice as well.

The circuit:

And the frequency response matches the OP's original frequency response.

Since in AC analysis LTspice is using linearize BJT's mode. I decided to draw (in LTspice) my own small-signal model and compare the results.

And I've got exactly the same result as previously.

So the frequency responses are perfectly matched. Very good.

Now let us see what we get if we apply the Miller approximation to the Cbc.

And the frequency response is

And the close-up

As we can see the frequency response above the first pole frequency does not match anymore with the original circuit after we applied the Miller approximation. And this is what OP observed in his simulation. So, we can conclude that the Miller approximations give us the correct placement of a first pole only (dominant pole).

Answer 3

Cbc + Cbe = 3.6 pF is in parallel with Cco=1 pF , net 4.6 pF with load reduces to ;

Req=1419, Ceq =5.6 pF f(-3dB)=20.02 MHz might be closer to textbook

Answer 4

EE&O

My simulator gives this result, a "little" different behavior.
So I searched different functions ( Vo versus {Vb, Vg, V2}).

For reference, here are two pictures with the use of a CCCS (linear).
Note the influence of the Ccb and Ceb capacitor.

Answer 5

It's almost as though the simulation is only taking into account Cwo (1pF) and ignoring Cbc and Cce in its determination of a value for Co..

Try running a simulation with Cwo removed and see if that highest frequency pole disappears altogether.