+ Post New Thread
Results 1 to 3 of 3

17th February 2014, 19:18 #1
 Join Date
 Apr 2012
 Posts
 11
 Helped
 2 / 2
 Points
 469
 Level
 4
Parasitic Capacitances in BJT
I want to understand the frequency response of a Common Emitter Amplifier. I am using the transistor 2N2222. How can I use the capacitances given in the datasheet namely, C_{obo} and C_{ibo} to estimate the bandwidth of my designed common emitter amplifier? I would be very thankful if somebody could explain how the Current Gain Bandwidth Product can be used to estimate the bandwidth for a given voltage gain also.

Advertisment

17th February 2014, 20:49 #2
 Join Date
 Apr 2010
 Location
 Netherlands
 Posts
 1,591
 Helped
 593 / 593
 Points
 10,891
 Level
 25
Re: Parasitic Capacitances in BJT
Cobo (almost) equals the capacitance between collector and base (also called miller capacitance). This is a voltage dependent capacitance and increaes with decreasing collector base voltage.
Cibo is the emitter base capacitance for the reversed biased BE junction (mostly Veb = 0.5V, or Vbe=0.5V). This is also a voltage dependent capacitor (compare with a varicap diode for oscillator tuning).
There is a hidden capacitance that is much larger the emitter base capacitance. It is related to the transition frequency of the transistor and is a diffusion capacitance.
As a rough estimate: Cbe.dif = 40*Ic/(2*pi*ft), Ic = DC collector current [A], ft = transition frequency [Hz]. So a transistor with ft=300 MHz (at 10 mA) running at 10 mA has a diffusion capacitance between B and E of around 210 pF. If you run the transistor at very low current, the diffusion capacitance may not be large with respect to the BE space charge capacitance and Ccb (Cobo). This is one of the reasons that ft drops at low collector current.
For small signal analysis (for example for oscillators) I just use total Base Emitter capacitance = 40*Ic/(2*pi*ft).
Though Cbe is generally large with respect to Ccb, Ccb is mostly the limiting factor in BW. The collector voltage swing is large w.r.t. the BE voltage swing. So 50 mV Vbe may result in 1V across Ccb, hence the current through this capacitor is relatively large, despite its small value with respect to total Cbe.
Ccb (miller capacitor) appears as a capacitor parallel to the input of the amplifier as: C = (1Av)*Ccb. Av = Voltage gain of amplifier with C as output and B as input.
For a common emitter (or degraded emitter), the voltage gain between B and E is negative, henc (1Av) is a positive number.
Effect of Ccb can be reducesd with a socalled cascode stage (two transistors required), and this increases the 3 dB bandwidth of an amplifier.
At RF, most of the AC base current is not because of hfe*IC, but because of Ccb that has to be charged and discharged. This relative high AC base current causes heat loss in the base resistance. This is not the differential resistance based on 1/(40*Ib), but because of a series ohmic resistance due to the thin base area (so called rbb'). So the actual power gain depends on ft, Ccb, Rbb' (and Ree', Rcc').

Advertisment

17th February 2014, 22:06 #3
 Join Date
 Jan 2008
 Location
 Toronto area of Canada
 Posts
 8,560
 Helped
 1995 / 1995
 Points
 49,170
 Level
 54
Re: Parasitic Capacitances in BJT
Isn't the base pin beside the collector pin so wiring capacitance adds to the transistor's capacitance.
Did you consider other stray wiring capacitances at the input and output?
+ Post New Thread
Please login