Asymptotic gain model

Figure 3: FET feedback amplifier

Consider the simple FET feedback amplifier in Figure 3. The aim is to find the low-frequency, open-circuit, transresistance gain of this circuit G = v_out / i_in using the asymptotic gain model.

Figure 4: Small-signal circuit for transresistance amplifier; the feedback resistor R_f is placed below the amplifier to resemble the standard topology

Figure 5: Small-signal circuit with return path broken and test voltage driving amplifier at the break

The small-signal equivalent circuit is shown in Figure 4, where the transistor is replaced by its hybrid-pi model.

It is most straightforward to begin by finding the return ratio T, because G₀ and G_∞ are defined as limiting forms of the gain as T tends to either zero or infinity. To take these limits, it is necessary to know what parameters T depends upon. There is only one dependent source in this circuit, so as a starting point the return ratio related to this source is determined as outlined in the article on return ratio.

The return ratio is found using Figure 5. In Figure 5, the input current source is set to zero, By cutting the dependent source out of the output side of the circuit, and short-circuiting its terminals, the output side of the circuit is isolated from the input and the feedback loop is broken. A test current i_t replaces the dependent source. Then the return current generated in the dependent source by the test current is found. The return ratio is then T = −i_r / i_t. Using this method, and noticing that R_D is in parallel with r_O, T is determined as:

${\displaystyle T=g_{\mathrm {m} }\left(R_{\mathrm {D} }\ ||r_{\mathrm {O} }\right)\approx g_{\mathrm {m} }R_{\mathrm {D} }\ ,}$

where the approximation is accurate in the common case where r_O >> R_D. With this relationship it is clear that the limits T → 0, or ∞ are realized if we let transconductance g_m → 0, or ∞.[5]

Finding the asymptotic gain G_∞ provides insight, and usually can be done by inspection. To find G_∞ we let g_m → ∞ and find the resulting gain. The drain current, i_D = g_m v_GS, must be finite. Hence, as g_m approaches infinity, v_GS also must approach zero. As the source is grounded, v_GS = 0 implies v_G = 0 as well.[6] With v_G = 0 and the fact that all the input current flows through R_f (as the FET has an infinite input impedance), the output voltage is simply −i_in R_f. Hence

${\displaystyle G_{\infty }={\frac {v_{\mathrm {out} }}{i_{\mathrm {in} }}}=-R_{\mathrm {f} }\ .}$

Alternatively G_∞ is the gain found by replacing the transistor by an ideal amplifier with infinite gain - a nullor.[7]

To find the direct feedthrough ${\displaystyle G_{0}}$ we simply let g_m → 0 and compute the resulting gain. The currents through R_f and the parallel combination of R_D || r_O must therefore be the same and equal to i_in. The output voltage is therefore i_in (R_D || r_O).

Hence

${\displaystyle G_{0}={\frac {v_{out}}{i_{in}}}=R_{D}\|r_{O}\approx R_{D}\ ,}$

where the approximation is accurate in the common case where r_O >> R_D.

The overall transresistance gain of this amplifier is therefore:

${\displaystyle G={\frac {v_{out}}{i_{in}}}=-R_{f}{\frac {g_{m}R_{D}}{1+g_{m}R_{D}}}+R_{D}{\frac {1}{1+g_{m}R_{D}}}={\frac {R_{D}\left(1-g_{m}R_{f}\right)}{1+g_{m}R_{D}}}\ .}$

Examining this equation, it appears to be advantageous to make R_D large in order make the overall gain approach the asymptotic gain, which makes the gain insensitive to amplifier parameters (g_m and R_D). In addition, a large first term reduces the importance of the direct feedthrough factor, which degrades the amplifier. One way to increase R_D is to replace this resistor by an active load, for example, a current mirror.

Figure 6: Two-transistor feedback amplifier; any source impedance R_S is lumped in with the base resistor R_B.

Figure 7: Schematics for using asymptotic gain model; parameter α = β / ( β+1 ); resistor R_C = R_C1.

Figure 6 shows a two-transistor amplifier with a feedback resistor R_f. This amplifier is often referred to as a shunt-series feedback amplifier, and analyzed on the basis that resistor R₂ is in series with the output and samples output current, while R_f is in shunt (parallel) with the input and subtracts from the input current. See the article on negative feedback amplifier and references by Meyer or Sedra.[8][9] That is, the amplifier uses current feedback. It frequently is ambiguous just what type of feedback is involved in an amplifier, and the asymptotic gain approach has the advantage/disadvantage that it works whether or not you understand the circuit.

Figure 6 indicates the output node, but does not indicate the choice of output variable. In what follows, the output variable is selected as the short-circuit current of the amplifier, that is, the collector current of the output transistor. Other choices for output are discussed later.

To implement the asymptotic gain model, the dependent source associated with either transistor can be used. Here the first transistor is chosen.

The circuit to determine the return ratio is shown in the top panel of Figure 7. Labels show the currents in the various branches as found using a combination of Ohm's law and Kirchhoff's laws. Resistor R₁ = R_B // r_π1 and R₃ = R_C2 // R_L. KVL from the ground of R₁ to the ground of R₂ provides:

${\displaystyle i_{\mathrm {B} }=-v_{\pi }{\frac {1+R_{2}/R_{1}+R_{\mathrm {f} }/R_{1}}{(\beta +1)R_{2}}}\ .}$

KVL provides the collector voltage at the top of R_C as

${\displaystyle v_{\mathrm {C} }=v_{\pi }\left(1+{\frac {R_{\mathrm {f} }}{R_{1}}}\right)-i_{\mathrm {B} }r_{\pi 2}\ .}$

Finally, KCL at this collector provides

${\displaystyle i_{\mathrm {T} }=i_{\mathrm {B} }-{\frac {v_{\mathrm {C} }}{R_{\mathrm {C} }}}\ .}$

Substituting the first equation into the second and the second into the third, the return ratio is found as

${\displaystyle T=-{\frac {i_{\mathrm {R} }}{i_{\mathrm {T} }}}=-g_{\mathrm {m} }{\frac {v_{\pi }}{i_{\mathrm {T} }}}}$

${\displaystyle ={\frac {g_{\mathrm {m} }R_{\mathrm {C} }}{\left(1+{\frac {R_{\mathrm {f} }}{R_{1}}}\right)\left(1+{\frac {R_{\mathrm {C} }+r_{\pi 2}}{(\beta +1)R_{2}}}\right)+{\frac {R_{\mathrm {C} }+r_{\pi 2}}{(\beta +1)R_{1}}}}}\ .}$

The circuit to determine G₀ is shown in the center panel of Figure 7. In Figure 7, the output variable is the output current βi_B (the short-circuit load current), which leads to the short-circuit current gain of the amplifier, namely βi_B / i_S:

${\displaystyle G_{0}={\frac {\beta i_{B}}{i_{S}}}\ .}$

Using Ohm's law, the voltage at the top of R₁ is found as

${\displaystyle (i_{S}-i_{R})R_{1}=i_{R}R_{f}+v_{E}\ \ ,}$

or, rearranging terms,

${\displaystyle i_{S}=i_{R}\left(1+{\frac {R_{f}}{R_{1}}}\right)+{\frac {v_{E}}{R_{1}}}\ .}$

Using KCL at the top of R₂:

${\displaystyle i_{R}={\frac {v_{E}}{R_{2}}}+(\beta +1)i_{B}\ .}$

Emitter voltage v_E already is known in terms of i_B from the diagram of Figure 7. Substituting the second equation in the first, i_B is determined in terms of i_S alone, and G₀ becomes:

${\displaystyle G_{0}={\frac {\beta }{(\beta +1)\left(1+{\frac {R_{f}}{R_{1}}}\right)+(r_{\pi 2}+R_{C})\left[{\frac {1}{R_{1}}}+{\frac {1}{R_{2}}}\left(1+{\frac {R_{f}}{R_{1}}}\right)\right]}}}$

Gain G₀ represents feedforward through the feedback network, and commonly is negligible.

The circuit to determine G_∞ is shown in the bottom panel of Figure 7. The introduction of the ideal op amp (a nullor) in this circuit is explained as follows. When T → ∞, the gain of the amplifier goes to infinity as well, and in such a case the differential voltage driving the amplifier (the voltage across the input transistor r_π1) is driven to zero and (according to Ohm's law when there is no voltage) it draws no input current. On the other hand, the output current and output voltage are whatever the circuit demands. This behavior is like a nullor, so a nullor can be introduced to represent the infinite gain transistor.

The current gain is read directly off the schematic:

${\displaystyle G_{\infty }={\frac {\beta i_{B}}{i_{S}}}=\left({\frac {\beta }{\beta +1}}\right)\left(1+{\frac {R_{f}}{R_{2}}}\right)\ .}$

Using the classical model, the feed-forward is neglected and the feedback factor β_FB is (assuming transistor β >> 1):

${\displaystyle \beta _{FB}={\frac {1}{G_{\infty }}}\approx {\frac {1}{(1+{\frac {R_{f}}{R_{2}}})}}={\frac {R_{2}}{(R_{f}+R_{2})}}\ ,}$

and the open-loop gain A is:

${\displaystyle A=G_{\infty }T\approx {\frac {\left(1+{\frac {R_{f}}{R_{2}}}\right)g_{m}R_{C}}{\left(1+{\frac {R_{f}}{R_{1}}}\right)\left(1+{\frac {R_{C}+r_{\pi 2}}{(\beta +1)R_{2}}}\right)+{\frac {R_{C}+r_{\pi 2}}{(\beta +1)R_{1}}}}}\ .}$

The above expressions can be substituted into the asymptotic gain model equation to find the overall gain G. The resulting gain is the current gain of the amplifier with a short-circuit load.

In the amplifier of Figure 6, R_L and R_C2 are in parallel. To obtain the transresistance gain, say A_ρ, that is, the gain using voltage as output variable, the short-circuit current gain G is multiplied by R_C2 // R_L in accordance with Ohm's law:

${\displaystyle A_{\rho }=G\left(R_{\mathrm {C2} }//R_{\mathrm {L} }\right)\ .}$

The open-circuit voltage gain is found from A_ρ by setting R_L → ∞.

To obtain the current gain when load current i_L in load resistor R_L is the output variable, say A_i, the formula for current division is used: i_L = i_out × R_C2 / ( R_C2 + R_L ) and the short-circuit current gain G is multiplied by this loading factor:

${\displaystyle A_{i}=G\left({\frac {R_{C2}}{R_{C2}+R_{L}}}\right)\ .}$

Of course, the short-circuit current gain is recovered by setting R_L = 0 Ω.