13.5. Shunt – Series Cascade Amplifier

The cascading of two common-source amplifier stages will now be explored. This will expand the discussion on feedback amplifiers to include the shunt – series (current) amplifier. As shown in Fig. 13.5, the current I_o is sensed, with the feedback network consisting of R_S2 and R_f, and a fraction of I_o is summed at the gate input node along with the input source current. Gate resistor R_G1 provides for an addition bias design variable but otherwise is simply combined with the source resistance, R_s, in the amplifier performance analysis or design.

13.5.1. DC Design

For the design, assume that a choice has been made for I_D1 and I_D2 and V_G2 = V_D1. The latter must be such that V_G2 > V_GS1 + V_GS2. Then, for R_D1, use

Equation 13.10

As will be indicated below, the ideal current gain is

Equation 13.11

Use this, based on the feedback-amplifier design current gain goal, to obtain a relation between R_f and R_S2. This gives

Equation 13.12

Now determine the value of R_S2 which statisfies the design value of I_D2. This is obtaned from

Equation 13.13

where 13.12 has been used for R_f and where V_S2 = V_G1 – V_GS2 and

Equation 13.14

and

Equation 13.15

Finally, select R_G1 to satisfy the V_GS1 requirement from

Equation 13.16

13.5.2. DC Stability

The feedback network provides significant bias stabilization and the effect can be assessed by the same formulation as for the opamp. In Unit 11.4, it was shown that the output voltage for a given offset voltage is [(11.16)]

As applied to our shunt – series circuit (by equivalency), A_VNI = 1 + R_f/R_G1, and a_vo is the voltage-gain magnitude (minus terminal equivalent) from the gate, V_g1, to the source V_s2, that is, the gain of the case of a common-source stage (M₁) and a source-follower stage (M₂). Also, for this case, ΔV_o ≡ ΔV_S2. Note that for this case, ΔV_o is a bias-voltage change, whereas for the opamp, it is the output voltage (different from zero volts).

Suppose that the dc (bias) is at a set state. Then the bias is altered such as by a change of a transitor parameter (either by temperature change or substituting one transistor for a new one). For example, a change in V_tno of M₁ is equivalent to installing an offset voltage, and [(11.16)] can be applied directly to this case. Make R_G1 = R_f for A_VNI = 2. Also assume a typical a_vo = 40 (common-source stage). For a change of δV_tno ≡ –V_off = –0.1 V, ΔV_O ≈ 2 · 0.1 V = 0.2 V. Note that based on a_vo = 40, with no feedback stabilzation, the change in output would be approximately a_voδV_tno = 4 V.

13.5.3. Signal Current Gain

We first compute the open-loop transconductance and then the open-loop current gain. This is done, for a good approximation, by disconnecting the right side of R_f from the souce of M₂ and connecting it to ground. (This disables the shunt – series feedback and otherwise alters the circuit only slightly.) The result is

Equation 13.17

where R_i = R_s||R_G1||R_f. (This assumes that R_f >> R_S2 which is consistent with A_iI >> 1.) Transconductance G_m is for the complete cascade circuit (open loop), that is,

Equation 13.18

Note that the series – series feedback effect is retained in the open-loop computation and that R_S2 has been used as an approximation for R_S2||R_f. This once again assumes that R_f >> R_S2. The loop gain is

Equation 13.19

Following the discussion in Unit 11, which led to (11.3), the current gain with feedback is then

Equation 13.20

where (ideal current gain of the feedback amplifier)

Equation 13.21

As a design example, suppose the design goal is A_iI = 10. Assume that the transistors have V_tno = 1 V, k_n1 = 1000 μA/V², and k_n2 = 2000 μA/V². We select I_D1 = 100 μA, I_D2 = 200 μA, V_DD = 10 V, and V_G2 = 3.5 V. From (13.12) and (13.13) we obtain R_S2 = 11.4 kΩ and R_f = 103 kΩ. Based on (13.16), the value of R_G1 is R_G1 = 156 kΩ. To satisfy the drain voltage requirements, R_D1 = 65 kΩ and R_D2 = 25 kΩ. The latter is based on V_D2 = V_DD/2. Also, we assume that R_s = 100 kΩ.

The open-loop transconductance is G_m = 3372 μA/V and R_i = 38 kΩ for a_i = 129. Loop gain T = 12.9 such that A_i = 9.28. We note that since A_i ≈ A_iI = 10, the current gain is very predictable, despite bias and parameter variations.