Distortion in Feedback Amplifiers

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Consider an amplifier, shown in Fig.1, whose input-output relationship under small signal conditions given by,

(1)   \begin{equation*}v_{\tiny o} = a_{\tiny 1}v_{\tiny e} + a_{\tiny 2}v_{\tiny e}^{\tiny 2} + a_{\tiny 3}v_{\tiny e}^{\tiny 3} + \ldots \end{equation*}

Substituting v_{\tiny e}=v_{\tiny i}-\beta v_{\tiny o}in Eq.(1.1),

(2)   \begin{equation*}v_{\tiny o} = a_{\tiny 1}(v_{\tiny i}-\beta v_{\tiny o}) + a_{\tiny 2}(v_{\tiny i}-\beta v_{\tiny o})^{\tiny 2} + a_{\tiny 3}(v_{\tiny i}-\beta v_{\tiny o})^{\tiny 3} + \ldots \end{equation*}

To solve for v_{\tiny o}, represent the transfer function of the feedback amplifier by power series as,

(3)   \begin{equation*}v_{\tiny o} = b_{\tiny 1}v_{\tiny i} + b_{\tiny 2}v_{\tiny i}^{\tiny 2} + b_{\tiny 3}v_{\tiny i}^{\tiny 3} + \ldots \end{equation*}

and evaluate for b_{\tiny 1}, b_{\tiny 2}, b_{\tiny 3},\ldots

From Eq.(1.3),

(4)   \begin{eqnarray*} b_{\tiny 1} &= \frac{\partial v_{\tiny o}}{\partial v_{\tiny i}}{\Huge|}_{v_{\tiny i}{\tiny =0}}\\ 2b_{\tiny 2} &= \frac{\partial^{\tiny 2} v_{\tiny o}}{\partial v_{\tiny i}^{\tiny 2}}{\Huge|}_{v_{\tiny i}{\tiny =0}}\\ 6b_{\tiny 3} &= \frac{\partial^{\tiny 3} v_{\tiny o}}{\partial v_{\tiny i}^{\tiny 3}}{\Huge|}_{v_{\tiny i}{\tiny =0}}\end{eqnarray*}

Partially differentiation of Eq.(1.2) w.r.t v_{\tiny i} gives,

(5)   \begin{equation*}\frac{\partial{v_{\tiny o}}}{\partial{v_{\tiny i}}} = a_{\tiny 1} \left( 1-\beta\frac{\partial{v_{\tiny o}}}{\partial{v_{\tiny i}}}\right) + 2a_{\tiny 2}\left(v_{\tiny i}-\beta v_{\tiny o}\right) \left( 1-\beta\frac{\partial{v_{\tiny o}}}{\partial{v_{\tiny i}}}\right) + 3a_{\tiny 3}\left(v_{\tiny i}-\beta v_{\tiny o}\right)^{\tiny 2}\left( 1-\beta\frac{\partial{v_{\tiny o}}}{\partial{v_{\tiny i}}}\right) \end{equation*}

From Eq.(1.3), if v_{\tiny i} = 0 \rightarrow v_{\tiny o} = 0.

(6)   \begin{equation*}\Rightarrow b_{\tiny 1} = \frac{\partial v_{\tiny o}}{\partial v_{\tiny i}}{\Huge|}_{v_{\tiny i}{\tiny =0}} = \frac{a_{\tiny 1}}{1+\beta a_{\tiny 1}}\end{equation*}

Differentiate Eq.(1.4) w.r.t v_{\tiny i} and solving for b_{\tiny 2} yields,

(7)   \begin{equation*}\Rightarrow b_{\tiny 2} = \frac{1}{2}\frac{\partial^{\tiny 2} v_{\tiny o}}{\partial v_{\tiny i}^{\tiny 2}}{\Huge|}_{v_{\tiny i}{\tiny =0}} = \frac{a_{\tiny 2}}{(1+\beta a_{\tiny 1})^{\tiny 3}}\end{equation*}

Similarly solving for b_{\tiny 3},

(8)   \begin{equation*}\Rightarrow b_{\tiny 3} = \frac{1}{6}\frac{\partial^{\tiny 3} v_{\tiny o}}{\partial v_{\tiny i}^{\tiny 3}}{\Huge|}_{v_{\tiny i}{\tiny =0}} = \frac{a_{\tiny 3}(1+\beta a_{\tiny 1})-2a_{\tiny 2}^{\tiny 2}\beta}{(1+\beta a_{\tiny 1})^{\tiny 5}}\end{equation*}

Amplifier without Feedback

For an amplifier without feedback defined by Eq.[1], the output due to input v_{\tiny e} = A \cos(\omega_{\tiny 1}t) is given by,

(9)   \begin{equation*}v_{\tiny o} = \(a_{\tiny 1} + \frac{3}{4}a_{\tiny 3}A^{\tiny 2}\)A\cos(\omega_{\tiny 1}t) + \frac{1}{2}a_{\tiny 2}A^{\tiny 2}\cos(2\omega_{\tiny 1}t) + \frac{1}{4}a_{\tiny 3}A^{\tiny 3}\cos(3\omega_{\tiny 1}t) + \dots \end{equation*}

So the second and third order harmonic distortions for small signals for an amplifier without feedback are given by

(10)   \begin{equation*} HD_{\tiny 2}= \frac{1}{2}\frac{a_{\tiny 2}}{a_{\tiny 1}}A \end{equation*}

(11)   \begin{equation*} HD_{\tiny 3}= \frac{1}{4}\frac{a_{\tiny 3}}{a_{\tiny 1}}A^{\tiny 2} \end{equation*}

Amplifier with Feedback

    \begin{equation*}HD_{\tiny 2fb}= \frac{1}{2}\frac{b_{\tiny 2}}{b_{\tiny 1}}A = \frac{1}{2}\frac{a_{\tiny 2}}{a_{\tiny 1}}\frac{1}{(1+\beta a_{\tiny 1})^{\tiny 2}}A \end{equation}

    \begin{equation*}\Rightarrow HD_{\tiny 2fb} = \frac{1}{{a_{\tiny 1}(1+\beta a_{\tiny 1}})^{\tiny 2}}HD_{\tiny 2}\end{equation*}

(12)   \begin{equation*}HD_{\tiny 3fb}= \frac{1}{4}\frac{b_{\tiny 3}}{b_{\tiny 1}}A^{\tiny 2}\end{equation*}

(13)   \begin{equation*}\Rightarrow HD_{\tiny 3fb} = \frac{1-\frac{2a_{\tiny 2}^{\tiny 2}\beta}{a_{\tiny 3}\(1+\beta a_{\tiny 1}\)}}{\(1+\beta a_{\tiny 1}\)^{\tiny 3}} HD_{\tiny 3} \end{equation*}

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