User:John R. Brews/Sample

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Block diagram for asymptotic gain model.

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Possible signal-flow graph for the asymptotic gain model.

The asymptotic gain model (also known as the Rosenstark method) is a representation of the gain of negative feedback amplifiers given by the asymptotic gain relation:

$G=G_{\infty }\left({\frac {T}{T+1}}\right)+G_{0}\left({\frac {1}{T+1}}\right)\ ,$

where T is the return ratio with the input source disabled (equal to the negative of the loop gain in the case of a single-loop system composed of unilateral blocks), G_∞ is the asymptotic gain and G₀ is the direct transmission term. This form for the gain can provide intuitive insight into the circuit and often is easier to derive than a direct attack on the gain.

A block diagram that leads to the asymptotic gain expression is shown in the upper figure at right. The asymptotic gain relation also can be expressed as a signal-flow graph. See lower of two figures. The asymptotic gain model is a special case of the extra element theorem.

Definition of terms

As follows directly from limiting cases of the gain expression, the asymptotic gain G_∞ is simply the gain of the system when the return ratio approaches infinity:

$G_{\infty }=G\ {\Big |}_{T\rightarrow \infty }\ ,$

while the direct transmission term G₀ is the gain of the system when the return ratio is zero:

$G_{0}=G\ {\Big |}_{T\rightarrow 0}\ .$

Advantages

• This model is useful because it completely characterizes feedback amplifiers, including loading effects and the bilateral properties of amplifiers and feedback networks.
• Often feedback amplifiers are designed such that the return ratio T is much greater than unity. In this case, and assuming the direct transmission term G₀ is small (as it often is), the gain G of the system is approximately equal to the asymptotic gain G_∞.
• The asymptotic gain is (usually) only a function of passive elements in a circuit, and can often be found by inspection.
• The feedback topology (series-series, series-shunt, etc.) need not be identified beforehand as the analysis is the same in all cases.

Implementation

Direct application of the model involves these steps:

1. Select a dependent source in the circuit.
2. Find the return ratio for that source.
3. Find the gain G_∞ directly from the circuit by replacing the circuit with one corresponding to T = ∞.
4. Find the gain G₀ directly from the circuit by replacing the circuit with one corresponding to T = 0.
5. Substitute the values for T, G_∞ and G₀ into the asymptotic gain formula.

These steps can be implemented directly in SPICE using the small-signal circuit of hand analysis. In this approach the dependent sources of the devices are readily accessed. In contrast, for experimental measurements using real devices or SPICE simulations using numerically generated device models with inaccessible dependent sources, evaluating the return ratio requires special methods.

Connection with classical feedback theory

Classical feedback theory neglects feedforward (G₀). If feedforward is dropped, the gain from the asymptotic gain model becomes

$G=G_{\infty }{\frac {T}{1+T}}$

$={\frac {G_{\infty }T}{1+{\frac {1}{G_{\infty }}}G_{\infty }T}}\ .$

while in classical feedback theory, in terms of the open loop gain A, the gain with feedback (closed loop gain) is:

$A_{FB}={\frac {A}{1+{\beta }_{FB}A}}\ ,$

Comparison of the two expressions indicates the feedback factor β_FB is:

$\beta _{FB}={\frac {1}{G_{\infty }}}\ ,$

while the open-loop gain is:

$A=G_{\infty }\ T\ .$

If the accuracy is adequate (usually it is), these formulas suggest an alternative evaluation of T: evaluate the open-loop gain and G_∞ and use these expressions to find T. Often these two evaluations are easier than evaluation of T directly.

Examples

The steps in deriving the gain using the asymptotic gain formula are outlined below for two negative feedback amplifiers. The single transistor example shows how the method works in principle for a transconductance amplifier, while the second two-transistor example shows the approach to more complex cases using a current amplifier.

Single-stage transistor amplifier

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MOSFET transresistance feedback amplifier.

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

Consider the simple MOSFET feedback amplifier at right. 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.

Under the circuit with the transistor is the small-signal equivalent circuit, where the transistor is replaced by its hybrid-pi model.

Return ratio

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Small-signal circuit with return path broken and test current i_t driving amplifier at the break.

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 the figure. In the figure, 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:

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

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 ∞.

Asymptotic gain

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. 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

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

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

Direct feedthrough

To find the direct feedthrough G₀ 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

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

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

Overall gain

The overall transresistance gain of this amplifier is therefore:

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

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.

Two-stage transistor amplifier

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

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Three schematics used to discuss the asymptotic gain model; parameter α = β / ( β+1 ); resistor R_C = R_C1.

A figure at left 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. 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.

Although the figure indicates the output node, it 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.

Return ratio

The circuit to determine the return ratio is shown in the top panel of the arrangement of three small-signal circuits. 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:

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

KVL provides the collector voltage at the top of R_C as

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

Finally, KCL at this collector provides

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

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

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

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

Gain G₀ with T = 0

The circuit to determine G₀ is shown in the center panel of the small-circuit array. In the center panel, 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:

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

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

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

or, rearranging terms,

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

Using KCL at the top of R₂:

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

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

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

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

Gain G_∞ with T → ∞

The circuit to determine G_∞ is shown in the bottom panel of the small-signal circuits. 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:

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

Comparison with classical feedback theory

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

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

and the open-loop gain A is:

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

Overall gain

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.

Gain using alternative output variables

In the figure showing the amplifier with transistors, 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:

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

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:

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

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