__Equivalent Circuits__

The voltage gain of a transistor amplifier can be calculated using an equivalent circuit, or model, of the transistor. An equivalent circuit is one that behaves in exactly the same way as the device it represents. Two equivalent circuits are often employed for audio-frequency amplifier calculations; these are the *h* parameter circuit and the mutual conductance circuit.

__h Parameter Circuit __

The *h* parameter equivalent circuit of a bipolar transistor is shown by Fig. 27. The three *h* parameters have been met previously. *h _{ie}* is the input resistance with

*V*constant,

_{CE}*h*is the current gain with

_{fe}*V*constant, and

_{CE}*h*is the output admittance with 18 constant.

_{oe}When a collector load resistance *R _{L}* is connected across the output of the equivalent circuit it will appear in parallel with the output resistance

*1/h*of the transistor 10 give an effective load resistance

_{oe}*R*

_{L}_{(eff)}of(

*R*x 1/

_{L}*h*)/(

_{oe}*R*+ 1/

_{L}*h*). The collector current

_{oe}*I*

*=*

_{c}*h*flows in the total resistance to produce the output voltage

_{fe}I_{b}*V*=

_{out}*V*. Therefore

_{ce}*V _{out}* =

*h*X

_{fe}l_{b}*R*

_{L(eff) }Very often *I/h _{oe}* is much larger than

*R*so that

_{L}*R*≈

_{L(eff) }*RL*which means that

*h*can be neglected. The input voltage is

_{oe}*V*=

_{IN}*V*=

_{be}*I*

_{b}h_{ie}_{ }so that the voltage gain

*A*is

_{v}*Av = V _{out}/V_{in} = (h_{fe}I_{b}R_{L})/I_{b}h_{ie}) = (h_{fe}R_{L})/h_{ie}* (6.11)

Fig. 27 h parameter equivalent circuit of a bipolar transistor

Alternatively, the presence of the collector load resistance reduces the current gain of the transistor to a value less than *h _{fe} . *Applying Kirchhoff's law to the collector circuit:

*I _{C} = h_{fe}I_{b} + h_{oe}V_{ce}* (6.12)

An increase in *Ic* produces an increased voltage drop across the collector resistance and so *V _{ce}* falls. Therefore,

*V*

_{ce}= - I_{c}R_{L}_{ }. Substituting into equation (6.12) gives

*I _{c} = h_{fe}I_{b} – h_{oe}I_{c}R_{L}*

*I _{c}(1 + h_{oe}R_{L}) = h_{fe}I_{b}*

*I _{c} = h_{fe}I_{b}/(1 + h_{oe}R_{L})*

Therefore, current gain *A _{i}*

_{ }=

*I*) (6.13)

_{c}/I_{b}= h_{fe}/(1 + h_{oe}R_{L}__Example 12 __

__Example 12__A bipolar transistor has the following *h* parameters: *h _{ie}* = 1200 Ω ,

*h*= 150. and

_{fe}*h*= 60 X 10

_{oe}^{-6}S. The transistor is used as an amplifier with a collector load resistance of2000 O. Calculate the voltage gain of the circuit (a) taking s; into account. and (b) neglecting

*h*.

_{oe}__Solution __

__Solution__(*a*) Output resistance = 1/*h _{oe}* = 1/(60 x 10

^{-6}) = 16.67 kΩ Effective load resistance = 16.67 kΩ in parallel with 2 kΩ, i.e. 17670 Voltage gain = (150 x 1767)11200 = 221 (

*Ans*.)

(*b*) Alternatively. AI = 150/(1 + 60 x 10^{-6} x 2(00) = 134 Voltage gain = (134 x 2000)11200 = 223 (*Ans*.)

Voltage gain = (150 x 2(00)/1200 = 250 (*Ans*.)

The *h* parameters of a bipolar transistor are not constant quantities but instead they vary with both the collector current and the temperature. Figs. 28(*a*) and (*b*) show typical variations of the three *h* parameters.

__Mutual Conductance Circuit __

__Mutual Conductance Circuit__

The ratio of *h _{ff}* to

*h*. which is the mutual conductance 8m. is however, much more constant since

_{ie}*h*and hif vary in different direclions . The value of 8m is primarily determined by the d.c. collector current (gm = 38 mS per mA of collector current) and it is easily predictable. Calculations of amplifier gain are therefore often based on the mutual conductance model of the transistor shown by Fig. 29. The output resistance

_{fe}*r*

_{out}_{ }is often omitted since its effect on the voltage gain is generally small.

Consider the circuit given in Fig. 30. At signal frequencies the reactances of the three capacitors are negligibly small. The collector supply line is effectively at earth potential as far as a.c. signals are concerned and so the bias resistors *R _{1}* and

*R*are in parallel both with one another and with the base-emitter terminals of the transistor. The

_{2}Fig. 28 Variation of *h* parameters with (*a*) collector current and (*b*) temperature

Fig. 29 Mutual conductance equivalent circuit of a bipolar transistor

emitter resistor *R _{4}* is effectively short circuited by capacitor

*C*and the collector resistor appears between the collector and the emitter. Thus, the

_{3}*h*parameter and mutual conductance equivalent circuits of the amplifier are as given in Fig. 31(

*a*) and (

*b*).

The bias resistors shunt the signal path and so they reduce the input resistance of the circuit. To limit this shunting effect the bias resistors should have as high *a* resistance as possible.

If the parameters of the transistor are *h _{ie}* = 2000 Ω and

*h*= 120 and gm = 120/2000 = 60 mS , then the voltage gain of the circuit is

_{fe}*A _{v}*

_{ }=

*V*= 120 x 4700/2000 = 282

_{out}/V_{in}The presence of the bias resistors does not affect the voltage gain of the circuit but it does reduce the input voltage that appears across the input terminals Suppose that the source voltage has an e.m.f. of 10 mV and an internal resistance of 1000 Ω. Then, Fig. 32(*a*), the input voltage is *V _{in}* = 6.67 mV,

*V*= 1.88 V, if the bias resistors are neglected. If, Fig. 32(b), the bias resistors are taken into account. the input resistance of the amplifier rill is found from 1/

_{out}*r*= 1/2000 + 1/(56 x 10

_{in}^{3}) + 1/(10 X 10

^{3}) and

*r*= 1620 Ω. Now the input voltage is 18 mV and

_{in}*V*= 1.74 V.

_{out}The difference between the two output voltages obtained may seem to be fairly significant but it must be remembered that the quoted resistor values are all nominal values and are subject to tolerance

variations. Because of this, for practical purposes 8m is often taken as being equal to le/25 mS, or 40 mS per mA of collector current.

__Voltage Gain of a FET Amplifier __

At audio-frequencies the performance of a FET can be described by its mutual conductance *g _{m}* =

*I*and the equivalent circuit shown in Fig. 33. The resistance

_{d}/V_{gs}*r*is the output resistance of the FET and is equal to δ

_{ds}*V*/δ

_{DS}*I*=

_{D}*V*with

_{ds}/I_{d}*V*constant. The output voltage

_{GS}*V*of the circuit is given by

_{ds}*V _{ds} = (g_{m}V_{gs}R_{L}r_{ds}/(R_{L} + r_{ds}) *

Therefore the voltage gain Ay is given by

*A _{v} = V_{ds}/V_{gs} = g_{m}R_{L}r_{ds}/(r_{ds} + R_{L}) *(6.14)

If, as is usual, *r _{ds}* >>

*R*

_{L}*A _{v}* =

*g*

_{m}

*R*

_{L}(6.15)

The mutual conductance of a FET is not a constant quantity but instead is a function of the drain current as shown in Fig 34. (loss is the drain current for *V _{GS}* = 0.)

__Example 13 __

Calculate the drain load resistance required to give the circuit of Fig. 35 a voltage gain of 20. The FET used has *g*_{m}= 4 X 10^{-3} S and *r _{ds}*

_{ }= 100 kΩ .

__Solution __

From equation (6.14), *A _{v}* = 20 = (4 X 10

^{-3}X l0

^{5}

*R*

_{2})/(l0

^{5}+

*R*

_{2})

Therefore *R _{2}* = (20 x 10

^{5})/380 = 5.26 kΩ (

*Ans*.)

Alternatively, using the approximate expression given by equation (6.15) *A _{v}* = 20 = 4 X l0

^{-3}

*R*or

_{2}*R*= 20/(4 x 10

_{2}^{-3}) = 5000 Ω (

*Ans*.)