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Frequency Response of Amplifiers (BJT and FET transistors)

  1. Feb 15, 2005 #1

    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?
  2. jcsd
  3. Feb 15, 2005 #2
    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 :))
  4. Feb 17, 2005 #3
    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?

    Thanks for your help!

    Attached Files:

  5. Feb 17, 2005 #4
    I'm not sure what do you refer to as open/short circuit methods for determining time constants, probably because we may use some different names for the same methods in my language, or maybe because I forgot all about them.

    Still, I'll try to help (these are my assumptions of your actual questions :))

    You think: "Why shouldn't we just say C-s are open circuit and make our life easier?"
    Answer: As we usually concern with mid-ranged frequencies, we approach low frequencies from "the right" and high freqs from "the left" (if you would plot a Bode diagram of transfer function). Because you gradually decrease your frequency, you are concerned what is the lower bound of "mid-freq" range. On that point you can't disregard coupling C-s. You can concern them as open circuits only on even lower frequencies (around 0) where attenuation is large and signal really does appear cut.

    You think: "OK, I have three or four possible constants - which to use?"
    As I see it, two constants are of interest here - T_in = (Rp+Rs)*C_1, T_out = (R_L+R_C)*C. The constant given by C_E is not of an interest as R_E || 1/sC_E is fed by a constant current and does not contribute to output transfer function. Rp = R_B || r_pi
    As both transfer functions (input and output) damp the signal on before their characteristic frequencies, and pass it after, your lower-bound frequency is the higher one (as, coming from mid-freq range, you first encounter amplification drop at that frequency). So you compute T_in and T_out, and take the smaller constant as that one will give the higher frequency.

    There is quite a possibility that I didn't understand your question and that it wasn't any of the above. If that is so, I'll try again... :)

    Hope it helped
  6. Feb 18, 2005 #5
    The joys of Low, mid and high freq response

    Well, at "low frequencies" the response is dominated by the high vaue capacitors because the low value capacitors have an impedance that remains high compared to other impedances over the "low frequency response" range. At the mid frequency response, the high vaue capacitors have a low impedance compared to the other impedances, so the response is dominated by the changing impedance of the BJTs capacitance which is small.

    Simple as that.
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