Saturday, February 13, 2021

Audio Amplifier feedback - LTP with Frequency Dependent Load

In a number of previous posts, I looked at shaping the gain of a simple common emitter stage by applying frequency dependent local feedback. Local feedback, however, is not the only available gain-shaping tool. Another commonly used way is to add frequency dependent load.

Consider a simple long tail pair (LTP):


The analysis of the LTP can be found in many places online; my favorite is the ECE 3050 class notes by Marshall Leach. For small signals, the LTP is essentially a voltage dependent current source with transconductance $g_m={1 \over r_e} = {I_E \over V_T} = {I_{LTP} \over {2 V_T}}$ loaded with $R_1$ and $R_2$:


Clearly, by making $R_1+R_2$ frequency dependent, the gain $${v_o \over v_{diff}} = {{v_{C1}-v_{C2}} \over v_{diff}} = {{v_{diff} \times g_m(R_1+R_2)} \over v_{diff}} = g_m(R_1+R_2)$$ can also be made frequency dependent. For example, one can connect an RC network across $R_1+R_2$:

This modifies the load impedance from $R_1+R_2$ to $${(R_1+R_2)||({R_c+{1 \over {s C_c}}})}={(R_1+R_2){{(s R_c C_c+1)} \over {s(R_1+R_2+R_c)C_c+1}}}$$ with a zero at $\omega_z={1 \over {R_c C_c}}$ and a pole at a zero at $\omega_p={1 \over {(R_1+R_2+R_c) C_c}}$.

With the values shown, the zero is at 1.6MHz and the pole is at 2.65kHz (blue trace); for comparison, the green trace shows the response of the same LTP without the RC network:

In principle, the input impedance of the next stage (e.g. a common emitter amplifier) should be included into the calculation as connected in parallel to $R_1$ and/or $R_2$. That does not affect the zero, but the position of the pole can be more difficult to find. Note that the input impedance of a typical second stage of an audio amplifier with Miller compensation falls with frequency and can be quite low, pushing the pole to a higher frequency. This usually is a desirable outcome as it limits the phase lag introduced by the pole within the amplifier's bandwidth.

If the LTP is loaded not with resistors but with a current mirror, the formulas above still hold if you use the actual load impedance instead of $R_1+R_2$. The load would include the input impedance of the next stage in parallel with the output impedances of the differential pair itself and of the current mirror. For higher load impedance, the pole moves to a lower frequency. At the limit, if the differential pair is a pure current source with infinite output impedance, and the RC network is its only load, the pole appears at $w_z=0$.

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