14.6. Output Buffer Stage

It is evident from the gain result, (14.29), that the high gain depends on having a high resistance at the output; a relatively small load resistor can reduce the gain substantially. For example, suppose that the amplifier is to be used as a resistance feedback amplifier as shown in Fig. 14.6. The input is at the gate of M₂, and the feedback resistor is connected back to the gate of M₁. The load at the output is R_f + R₂ || R_O.

For a specific example, assume that R_O >> R_f >> R₂ such that the load is approximately R_f. With the additional load, the gain (14.30) becomes

Equation 14.31

which is the result of R_f being in parallel with the output resistance of the common-source stage. Suppose that R_f = 10 kΩ and I_D3 = 100 μA. In this case the gain would drop to about 400 (from about 4400), which would be unacceptable, as proper operation of opamps assumes a very high gain.

This loading problem is substantially eliminated with the addition of a buffer stage to isolate the load from the output node of the common-source stage, as shown in the circuit of Fig. 14.7. The source-follower buffer stage is made up of M₆ and M₁₃. The current-source bias transistor M₁₃ also uses the reference voltage provided by M₁₀. The load on the common-source stage of M₃ is now infinite. The feedback network is included in the circuit for comparing with the diagrammatic version of the circuit of Fig. 14.6.

The gain expression of the source follower requires that the body effect be taken into consideration, as the body of M₆ will be at signal ground. [The body of M₆ will be attached to the negative supply voltage (not shown).] From Unit 7 it was shown that the source-follower stage gain, with resistor bias [(7.12)] for this case is

For the present case of current-source bias for the source-follower stage, the gain expression becomes, with inclusion now of the feedback network and the external load resistance of Fig. 14.7,

Equation 14.32

To obtain a quantitative sense of the body effect on the gain of the source follower, suppose that η = 0.15. Thus, in the limit for g_m6 → ∞, a_vsf = 0.87. This is the best possible result, given the body effect.

For a more general assessment, we can substitute the relations g_m = 2I_D/V_eff, [(4.5)], and g_ds = I_Dλ_n [(14.15)] into (14.32) to obtain

Equation 14.33

We note that the terms associated with g_ds6 and g_ds13 are negligible. Minimization of the resistance term requires making g_m6[(R_f + R₂) || R_O] >> 1 (making the output resistance of the source follower much less than the load resistance). This involves a combination of a large I_D6 and k_n6.

Suppose that I_D6 = 400 μA, W₆ = 500 μm, KP_n = 100 μA/V², and that the gate length is L = 10 μm. Recalling that k_n = (KP_n/2)(W/L) (Table 3.1), we obtain k_n = 2500 μA/V² and 1/g_m6 = 500 Ω. Thus, for example, for R_O = 10 kΩ, R_f = 10 kΩ, and R_f >> R_G, a_vsf = 0.80, compared with the limiting value of a_vsf = 0.87. Thus the source follower functions very well to isolate the load from the high-gain common-source stage.

Finally, we write the overall gain expression, including a general load resistance, R_L, by combining (14.29) and (14.32) as

Equation 14.34

A good approximation for the case of a sufficiently large g_m6 is

Equation 14.35