# Frequency Response of Amplifiers (BJT and FET transistors)

by cyeokpeng
Tags: amplifiers, frequency, transistors
 P: 69 Hi, I have a question pertianing to the effect of capacitance of BJT amplifier circuits on the low and high frequency response of the amplifier. Why is it that the coupling and the decoupling capacitors, in effect, only affects the low frequency response of the amplifier, and why is it that the junction capacitances between any two terminals affect only the high frequency response? Because in the analysis treatment and design techniques of BJT amplifiers on the frequency response, we only consider the short-circuit time constant method of the coupling and decoupling capacitors for low frequency cutoff, ignoring its effect on the the high frequency response, and vice versa for the treatment of junction capacitances (open-circuit time constant method) on the high frequency response. But my question is why? Is there any phyiscal explanation to such estimation?
 P: 19 Well, since English is not my mother's tongue, I'll try to be brief and not get tangled in terminology. As you probably know, complex impedance of capacitor is 1/(jCw), w being circular frequency of current flowing through the capacitor. On higher frequencies, impedance of a capacitor is relatively small and on low frequencies, it grows large. When you analyse circuit on low frequencies, capacitors connected in series with the signal flow start making difference, as your input voltage is distributed between the capacitor impedance and input impedance (resistance) of the amplifier (or some stage of amp). (e.g. that capacitor will completely block DC signal). On high frequencies, capacitors connected in parallel to resistances come into play, as their impedances decrease. As parallel connection of impedances is dictated by the smaller one, on HF those capacitors determine input impedance of amplifier stage (this is one of the reasons for having trouble with HF amplifiers). This works for any parallel/serial connection but I explained it on the input stage of an amplifier. I hope this was understandable, as it has been some time since my Electronics course, and, again, I apologise for my English. Please respond, either to correct me or to agree :))
P: 69
Attached documents are diagrams of a CE amplifier.
I will use the CE amplifier to illustrate my problem stated in this thread.

Your explanations seems to be quite agreeable, one first thought.

Refer to diagram of CE amplifier in the attacnment
At low frequency of the input signal (same frequency of signal output due to linear system), the coupling capacitors C1 and C2 are somewhat connected in series at the input stage and output terminal respectively. (I do not use two-or multiple stages amplifier to complicate analysis) Hence C1 and C2 in effect present relatively high impedances (1/jwC increases as w->0) at the input and output stages, taking the largest "cake" or portion of the voltage of the signal. For example, looking at the small-signal equivalent circuit for low frequency analysis, the base current signal ib is reduced due to the finite impedance of C1 in series with rpi, thereby decreasing the voltage across the output stage (B*ib*Rc//RL). At the output stage, the finite impedance of C2 takes away a significant portion of the output voltage (B*ib*Rc//RL) due to the potential divider. Both the effect of C1 and C2 reduces the output signal voltage, thereby reducing the gain.
Therefore, at low frequency of the signal, C1 and C2 cannot technically be considered as short-circuits in our usual analysis of the CE amplifier, because the impedances across C1 and C2 become significantly large enough to not to neglect it. The magnitude gain of the CE amplifier is decreasing as the frequency of the signal -> 0 Hz and that is the reason why the lower 3-dB frequency uses the coupling and decoupling capacitors' corresponding short-circuit time constants to determine.
OK, now comes the question! Why for low frequency analysis, we use short-circuit time constant method instead of open-circuit time constant?