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Electronics · Ch 4 — Feedback in Amplifiers

Voltage series negative feedback

4.4

Voltage series negative feedback

In the voltage series negative feedback connection, a fraction of the output voltage is applied in series with the input signal through the feedback network. This is the most important of the four types, and here we analyse in detail how it changes the amplifier's voltage gain, gain stability, input impedance, output impedance and bandwidth.

Expression for voltage gain

Figure 4.4.1 shows the block diagram. Let VsV_s be the input to the whole system, ViV_i the input to the basic amplifier, VfV_f the feedback voltage and VoV_o the output voltage.

The open-loop gain gives A=VoViA = \dfrac{V_o}{V_i}, so

Vo=AVi...(1)V_o = A V_i \quad \text{...(1)}

The closed-loop gain is

Af=VoVs...(2)A_f = \frac{V_o}{V_s} \quad \text{...(2)}

The feedback factor gives β=VfVo\beta = \dfrac{V_f}{V_o}, so

Vf=βVo...(3)V_f = \beta V_o \quad \text{...(3)}

With negative feedback the input to the basic amplifier is

Vi=Vs−Vf=Vs−βVo=Vs−βAViV_i = V_s - V_f = V_s - \beta V_o = V_s - \beta A V_i

Rearranging,

Vs=Vi(1+Aβ)...(4)V_s = V_i(1 + A\beta) \quad \text{...(4)}

Substituting equation (4) into equation (2) and using equation (1),

Af=VoVi(1+Aβ)=A1+AβA_f = \frac{V_o}{V_i(1 + A\beta)} = \frac{A}{1 + A\beta}

Thus the voltage gain with negative feedback decreases by a factor of (1+Aβ)(1+A\beta).

Stability in Gain

The closed-loop gain is Af=A1+AβA_f = \dfrac{A}{1 + A\beta}. When Aβ≫1A\beta \gg 1, the '1' in the denominator is negligible, so

Af≈AAβ=1βA_f \approx \frac{A}{A\beta} = \frac{1}{\beta}

The gain then depends only on β\beta, and β\beta is set by stable passive elements (resistors, capacitors, inductors) in the feedback network. Hence the gain is not affected by changes in device parameters, supply voltage or ageing of components, and the feedback gain is far more stable.

We can quantify this stability. Differentiating Af=A1+AβA_f = \dfrac{A}{1+A\beta} with respect to AA:

dAfdA=(1+Aβ)×1−Aβ(1+Aβ)2=1(1+Aβ)2\frac{dA_f}{dA} = \frac{(1+A\beta)\times 1 - A\beta}{(1+A\beta)^2} = \frac{1}{(1+A\beta)^2}

Multiplying and dividing the right-hand side by AA and using Af=A1+AβA_f = \dfrac{A}{1+A\beta},

dAfAf=dAA×1(1+Aβ)\frac{dA_f}{A_f} = \frac{dA}{A} \times \frac{1}{(1+A\beta)}

Since (1+Aβ)>1(1+A\beta) > 1, the percentage change in the feedback gain AfA_f is much smaller than the percentage change in the internal gain AA. This is precisely why negative feedback stabilises the gain.

Increase in Input Impedance

It is desirable for an amplifier to have a high input impedance so that it does not load the preceding stage or the input voltage source. Voltage series negative feedback achieves exactly this (Figure 4.4.2).

Let IiI_i be the input current, ViV_i the input to the basic amplifier, VsV_s the source voltage, VfV_f the feedback voltage and VoV_o the output. The input impedance of the basic amplifier is

Zi=ViIi...(1)Z_i = \frac{V_i}{I_i} \quad \text{...(1)}

and the input impedance with feedback is

Zif=VsIi...(2)Z_{if} = \frac{V_s}{I_i} \quad \text{...(2)}

The input to the basic amplifier is

Vi=Vs−Vf=Vs−βVo=Vs−βAViV_i = V_s - V_f = V_s - \beta V_o = V_s - \beta A V_i

so

Vs=Vi(1+Aβ)...(3)V_s = V_i(1 + A\beta) \quad \text{...(3)}

Substituting equation (3) into equation (2) and using equation (1),

Zif=Vi(1+Aβ)Ii=Zi(1+Aβ)Z_{if} = \frac{V_i(1 + A\beta)}{I_i} = Z_i(1 + A\beta)

Thus the input impedance with negative feedback increases by a factor of (1+Aβ)(1+A\beta).

Decrease in output impedance

An amplifier with low output impedance can deliver voltage or power to the load without much loss. Voltage series negative feedback provides this too.

The output impedance of the feedback amplifier is the ratio of output voltage to output current with the input shorted. Let ZoZ_o be the output impedance of the basic amplifier and ZofZ_{of} that of the feedback amplifier. To find it, a hypothetical voltage source VoV_o is applied at the output (Figure 4.4.3), and the amplifier is modelled as a dependent source AViA V_i in series with its internal impedance ZoZ_o:

Zof=VoIo...(1)Z_{of} = \frac{V_o}{I_o} \quad \text{...(1)}

Applying KVL to the output loop,

Vo=IoZo+AVi...(2)V_o = I_o Z_o + A V_i \quad \text{...(2)}

The input with negative feedback is Vi=Vs−VfV_i = V_s - V_f, but the input is short-circuited so Vs=0V_s = 0, giving Vi=−VfV_i = -V_f. Substituting into equation (2),

Vo=IoZo−AVf=IoZo−AβVoV_o = I_o Z_o - A V_f = I_o Z_o - A\beta V_o

Vo(1+Aβ)=IoZoV_o(1 + A\beta) = I_o Z_o

Zof=Zo1+AβZ_{of} = \frac{Z_o}{1 + A\beta}

Thus the output impedance with negative feedback decreases by a factor of (1+Aβ)(1+A\beta).

Frequency response of negative feedback amplifier

The bandwidth (BW) of an amplifier without feedback is the separation between its 3 dB frequencies f1f_1 and f2f_2 (Figure 4.4.4):

BW=f2−f1BW = f_2 - f_1 …

Figure 1Block diagram of a voltage series negative feedback amplifier showing the source Vs in series with the feedback voltage, the basic amplifier, load resistor RL and the feedback network sampling the output voltage
Fig. 1 — Block diagram of a voltage series negative feedback amplifier showing the source Vs in series with the feedback voltage, the basic amplifier, load resistor RL and the feedback network sampling the output voltage

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Reproduces textbook Figure 4.4.1. The source VsV_s is in series with the input; input current IiI_i enters the basic amplifier of gain AA as ViV_i; the output appears as VoV_o across the load resistor RLR_L; the 'Feedback network β\beta' samples the output voltage across RLR_L and returns Vf=βVoV_f = \beta V_o in series with the input. This figure matters because it is the refe …

Formula 2Closed-loop gain of a voltage series negative feedback amplifier

Af=A1+AβA_f = \dfrac{A}{1 + A\beta} — the voltage gain with negative feedback, reduced from the open-loop gain AA by the factor (1+Aβ)(1+A\beta). Here AA is the open-loop gain and β\beta the feedback fraction; since (1+Aβ)>1(1+A\beta) > 1, the closed-loop gain AfA_f is always smaller than AA — the price paid for t …

Formula 3Stabilised gain when loop gain is large

Af≈1βA_f \approx \dfrac{1}{\beta} when Aβ≫1A\beta \gg 1 — the gain then depends only on the stable feedback fraction β\beta, making it independent of device parameters, supply voltage and ageing. Because β\beta is fixed by stable passive elements (resistors, capacitors, inductors) in the feedback network, the clos …

Formula 4Fractional change in feedback gain versus internal gain

dAfAf=dAA×11+Aβ\dfrac{dA_f}{A_f} = \dfrac{dA}{A} \times \dfrac{1}{1 + A\beta} — since (1+Aβ)>1(1+A\beta) > 1, a given percentage change in the internal gain AA produces a much smaller percentage cha …

Figure 5Block diagram used to derive the input impedance of a voltage series negative feedback amplifier, emphasising the input current Ii, source Vs and feedback voltage Vf
Fig. 5 — Block diagram used to derive the input impedance of a voltage series negative feedback amplifier, emphasising the input current Ii, source Vs and feedback voltage Vf

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Reproduces textbook Figure 4.4.2, the same voltage-series topology as Figure 4.4.1 but drawn to analyse input impedance. The input current IiI_i flows from the source VsV_s into ViV_i of the basic amplifier; the output VoV_o across RLR_L is sampled by the 'Feedback network β\beta', which returns Vf=βVoV_f = \beta V_o in series with the input. This figure matters because it defines the loop used to compute $Z_i …

Formula 6Input impedance with negative feedback

Zif=Zi(1+Aβ)Z_{if} = Z_i(1 + A\beta) — voltage series negative feedback raises the input impedance by the factor (1+Aβ)(1+A\beta), where ZiZ_i is the input impedance of the basic amplifier. A high input impedance is desirable so the amplifier does not load the preceding stage or the input source; this result fo …

Figure 7Equivalent circuit for deriving the output impedance of a voltage series negative feedback amplifier, with the input shorted and a hypothetical voltage source applied at the output
Fig. 7 — Equivalent circuit for deriving the output impedance of a voltage series negative feedback amplifier, with the input shorted and a hypothetical voltage source applied at the output

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Reproduces textbook Figure 4.4.3, an equivalent circuit for the output-impedance derivation. The input is short-circuited (Vs=0V_s = 0); the basic amplifier is modelled as a dependent voltage source AViA V_i in series with its internal output impedance ZoZ_o; a hypothetical external source VoV_o is applied at the output, driving current IoI_o into the loop; the feedback network returns Vf=βVoV_f = \beta V_o to the shorted input. This figure matters because applying KVL around the o …

Formula 8Output impedance with negative feedback

Zof=Zo1+AβZ_{of} = \dfrac{Z_o}{1 + A\beta} — voltage series negative feedback lowers the output impedance by the factor (1+Aβ)(1+A\beta), where ZoZ_o is the output impedance of the basic amplifier. A low output impedance lets the amplifier deliver voltage or power to the load with little loss; it is derived with the …

Figure 9Voltage gain versus frequency response curves of an amplifier before and after negative feedback, showing reduced gain but wider bandwidth
Fig. 9 — Voltage gain versus frequency response curves of an amplifier before and after negative feedback, showing reduced gain but wider bandwidth

Drawn by us to help you understand the concept clearly, and verified to make sure it's accurate. For exams, practice from your textbook's own diagram.

Reproduces textbook Figure 4.4.4, a gain-versus-frequency graph. The 'Before FB' curve peaks at AmaxA_{max} with 3 dB points at f1f_1 and f2f_2 (the 0.707 AmaxA_{max} level) and bandwidth BWBW. The 'After FB' curve is lower (peak Af maxA_{f\,max}) but wider, with 3 dB points f1′f_1' (below f1f_1) and f2′f_2' (above f2f_2) and bandwidth BWfBW_f. This figure matters because it shows visually that negative feedback reduces gain but increases …

Formula 10Bandwidth and gain-bandwidth product with negative feedback

BWf=BW(1+Aβ)BW_f = BW(1 + A\beta) and A×BW=Af×BWfA \times BW = A_f \times BW_f — the bandwidth increases by the same factor (1+Aβ)(1+A\beta) by which the gain decreases, so the gain-bandwidth product stays constant. The lower cut-off falls to f1′=f1/(1+Aβ)f_1' = f_1/(1+A\beta) and the upper rises to f2′=f2(1+Aβ)f_2' = f_2(1+A\beta), so th …