# BJT high frequency analysis

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.

• The problem could be with your simulation, if you add pictures of it that makes it easier to help you. – Vinzent Jan 19 '20 at 19:10
• Good idea. Updated to add some simulation schematics and results. – Sittin Hawk Jan 19 '20 at 19:23
• @SittinHawk Why do you have 406.29 pF on one schematic and 6 pF on the other? Is that because of your calculation that you believe it should be? – jonk Jan 19 '20 at 20:19
• The first schematic has the original configuration of the capacitors. The 2nd schematic has the "equivalent" capacitors using the Ci and Co equations. If the textbook theory holds true, I believe I should get identical results. But I'm not. – Sittin Hawk Jan 19 '20 at 20:26

## 1 Answer

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.