Common-emitter (CE) Amplifier

The CE amplifier with tuned output and input circuits is shown in Fig. 5.4.1(a). C₃and C₄are dc blocking capacitors that have negligible reactance at high frequencies. The bias resistor R_BIASsupplies bias current to the base, and this can also be assumed to have negligible effect on the high-frequency performance. The sig- nal source is shown as an equivalent current generator i_sand R_s. The equivalent circuit, using the hybrid- equivalent circuit for the transistor, is shown in Fig. 5.4.1(b), where r_bbhas been assumed negligible.

From the equivalent circuit of Fig. 5.4.1(b), it can be seen that the output resistance of the transistor and the load resistance are in parallel with the output tuned circuit. The output capacitance of the transistor, shown as C_c, is in parallel with the circuit tuning capacitance C₂and will form part of the resonant circuit.

Let the output inductor have a series resistance r₂and inductance L₂, then as shown in Fig. 5.4.1(c), the com- ponents on the output side can be grouped together in an admittance form as

(5.4.1)

What is not immediately apparent from the equivalent circuit is the effect that the feedback capacitance C_cbhas. To see the effect of this, the equivalent circuit in Fig. 5.4.1(c) can be analyzed. Admittance Y₁rep- resents the admittance of R_s,r_beand C_bein parallel with the input tuned circuit, and Y₂the output admit- tance as previously defined. The feedback admittance is Y_f j C_cb. The current equation for the output node is

(5.4.2) From this the voltage gain is

(5.4.3) Y_f g_m

Y_f Y₂ A_v v_o

v_i

v_i(g_m Y_f) v_o(Y₂Y_f) 0 g_mv_i (v_ov_i)Y_fv_oY₂ Y₂ 1

r_c 1

R_L 1

r₂j L₂j (C_cC₂) F 1

T

C_deplC_cb(1g_mr_cc) g_m

F 1

T

C_deplC_cb g_m

To compare Y_fand g_m, note that Y_fj C_cband g_m T(C_beC_cb). Since the transistor will be operated at a frequency Tto reduce the effects of feedback, and since C_beC_cb, then g_m|Y_f|, and the expression for gain becomes

(5.4.4) where Y_oY₂Y_f. The gain is maximum when Y_ois resonant, which means that C_cbmust be included in the output tuning. There is also a 180° phase shift in the gain under these conditions.

The output admittance can be written in the form from Eq.(1.4.2).

(5.4.5) Y_o1jyQ_{2 eff}

R_{D2 eff} g_m

Y_o A_v g_m

Y_fY₂

(a)

(b)

(c) RBIAS

C2

C1

R_s i_s

C4

C3

C_cb

Cbe g_mv_i

g_mv_i

r_c C_c C2 L2

R_L v_o +

–

vo

+

– vi

+

– v_i +

– C1 L1 rbe

Y_f

Y1 Y2

i_s Rs

is

L1

L2

VCC

Figure 5.4.1 Tuned CE amplifier (a) circuit and (b) equivalent circuit. (c) Equivalent circuit for nodal analysis.

where

(5.4.6) The last term on the right-hand side is the reciprocal of the dynamic resistance of the tuned circuit alone, but including the transistor capacitances. The damping effects of the transistor output resistance and load resistance are taken into account in the calculation of the effective dynamic resistance R_{D2 eff}. The effec- tive Q-factor of the output circuit is

(5.4.7) Hence the gain can be written as

(5.4.8) It should be noted that this is the voltage gain with reference to the input terminals, and for this reason the input admittance does not affect it.

Turning now to the input circuit, the equation for the input node is

(5.4.9) Hence the input admittance is

(5.4.10) Thus, to tune this to resonance, the effect of the feedback must be taken into account. In practice, this interaction with the output circuit can complicate the tuning procedure. The admittance term Y_f(1 A_v) is often referred to as the Miller input admittance,named after J. M. Miller, whose name is also given to a theorem dealing with the feedback case in general.

Substituting for gain expression gives

(5.4.11) At the resonant frequency of the output circuit, this becomes

(5.4.12) Thus, for this situation, the Miller admittance can be represented by a capacitor C_M:

(5.4.13) The Miller capacitance must be included in the tuning of the input circuit for this to be resonant at the same frequency as the output circuit. Note, however, that the frequency response of the input tuned circuit will not be that of a singly tuned circuit, because the Miller admittance is a function of frequency in general and, in fact, also introduces a conductance component at frequencies off resonance.

C_M (1g_mR_{D2 eff})C_cb Y_inY₁j C_cb(1g_mR_{D2 eff}) Y_inY₁j C_cb

^{11gmRjyQD2 eff2 eff}

Y₁Y_f(1A_v) Y_in i_s

v₁

v₁(Y₁Y_f (1 A_v)) v₁(Y₁Y_f A_vY_f) i_s v₁(Y₁Y_f) v₂Y_f

A_v g_mR_{D2 eff} 1jyQ_{2 eff}

Q_{2 eff} o(C_c C₂C_cb)R_{D2 eff} 1

R_{D2 eff} 1 r_c 1

R_L (C_cC₂C_cb)r₂ L₂

From this example, it is seen that the effect of the 0.6-pF capacitance translates to a Miller input capacitance of 47 pF, and this will be in addition to the already existing C_becapacitance.

Returning now to the input circuit, at resonance its dynamic resistance is obtained from

(5.4.14)

where R_D1is the dynamic resistance of the input tuned circuit by itself. Also, assuming the Miller admittance is constant over the frequency range of the 3-dB bandwidth, the effective Q-factor of the input circuit is

(5.4.15) Q_{1 eff} o(C₁C_beC_M)R_{1 eff}

1 R_{D1 eff} 1

R_s 1 R_D1 1

r_be EXAMPLE 5.4.1

A CE amplifier has a tuned circuit in the collector, which resonates at 5 MHz with a total tuning capacity of 100 pF. The undamped Q-factor of the tuned circuit is 150. The amplifier feeds a load resistance of 5 k, and the output resistance of the transistor is 40 k. Calculate the voltage gain referred to the input terminals and the Miller capacitance at the input. The transistor operates a collector current of 500 µA, and the collector-to-base capacitance is 0.6 pF.

SOLUTION

At resonance the output admittance is purely conductive and is

Therefore,

C_M (178)0.647 pF A_v g_m

Y_o 78 246 S 1

4010³ 1

47.7510³ 1 510³ Y_o 1

r_c 1 R_D 1

R_L R_D2 Q

oC 150

2 510⁶10¹⁰47.75 K g_m I_c

V_T 50010⁶

2610³ 0.019 S

In many situations the input signal source is represented by a voltage equivalent generator, and the voltage gain referred to the source emf is of importance. In terms of the equivalent current generator source, the emf v_si_sR_s, or i_sv_sG_s. Hence, from the input node equation,

(5.4.16)

The gain referred to the source emf is therefore

(5.4.17) G_s

Y_inA_v A_vs v₂

v_s v₂

A_vY_in

∴v_sG_s v₁Y_in i_sv₁Y_in

EXAMPLE 5.4.2

For the amplifier of Example 5.4.1, the input tuned circuit has a Q-factor of 100 at a frequency of 5 MHz, the inductance being 2 µH. The source resistance is 1000 . The transistor is 200, and C_be10 pF.

Calculate the effective Q-factor of the input circuit and the voltage gain referred to the source emf.

SOLUTION From Example 5.4.1,A_v 78 and C_M 47 pF. The dynamic resistance of the tuned circuit is

The effective dynamic conductance is

Hence,

The effective Q-factor is

The voltage gain referred to source is

60 78764

1000 R_{D1 eff}

R_s A_v A_vs G_s

Y_in A_v Q_effR_{D1 eff}

oL 24 R_{D1 eff}764 1

R_{D1 eff} 1 R_s 1

R_D1 1

r_be1.31 mS R_D1Q _oL4.71 K

This example shows how the Q-factor can be severely reduced. In practice, to avoid this, the source and the transistor are usually connected through capacitive or inductive taps on the input tuned circuit, as discussed in Chapter 1. Figure 5.4.2 shows the source connected through a capacitive tap. This reduces the damping effect of the source conductance on the tuned circuit and, in addition, provides matching. As described in Section 4.16, the effective source resistance may be optimized for minimum noise factor through input coupling.

In document Electronic Communications by Dennis Roddy, John Coolen .pdf (Page 154-159)