16th April 2012 07:12 PM
BJT Cascode Amplifier
During my project in uni I have come across some things I am having difficulty understanding.
I have to design an amplifier to work up to atleast 2MHz, with a voltage gain of atleast 100.
I started my design with a common emitter amplifier using a BC107 BJT transistor. Unfortunately it only operated at 200Khz with gain of about 30dB.
I have now designed a cascode amplifier and Im having difficulty calculating the bypass capacitors.
My DC design is for 2SC2166:
Vcc = 10V
Ic = 5mA
gm = 70
RL = (10-7.5)/5mA = 500
Re = 1/5mA = 200R
R3 = (1+0.7)/714uA = 2380R
R2 = (7,5-5)+0.7/786uA = 4071R
R1 = (10-3.2)/865uA = 7861R
The rules I have used are:
Vre = 10% Vcc
Ir1 = 10 X Ib
I would really appreciate if someone could explain how the capacitors are calculated, I understand that I might have to make a small signal model, but still having difficulty with it.
16th April 2012 08:41 PM
Cascode is usually used with high voltage. Vcc 50 volts.
Cascode is usually used with low voltage gain. 2 to 10 (some times 20)
If you need 2mhz at a voltage gain of 100 then the Ft of the transistors needs to be >>200mhz.
RE X C3 = RL X Capacitance at output Capacitance at output= load cap + collector capacitance of top transistor
RL and the output capacitance reduces the gain at high frequencies.
RE and C3 causes the gain to go up at high frequencies.
Lowest frequency used = 1/2*3.24*C2*(R2//R3) C2 and the resistors make a high pass filter.
Lowest frequency used = 1/2*3.24*C1*(R1//R2)*10 C1 needs to keep the voltage constant at low frequency.
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