ANALYSIS OF A TRANSISTOR AMPLIFIER USING H-PARAMETERS:
To form a transistor amplifier it is only necessary to connect an external load and signal source as indicated in fig. 1 and to bias the transistor properly.
Consider the two-port network of CE amplifier. RS is the source resistance and ZL is the load impedance. The h-parameters are assumed to be constant over the operating range. The ac equivalent circuit is shown in fig. 2. (Phasor notations are used assuming sinusoidal voltage input). The quantities of interest are the current gain, input impedance, voltage gain, and output impedance.
For the transistor amplifier stage, Ai is defined as the ratio of output to input currents.
1.3.2 Input impedance:
The impedance looking into the amplifier input terminals ( 1,1' ) is the input impedance Zi
The ratio of output voltage to input voltage gives the gain of the transistors.
1.3.4 Output Admittance:
It is defined as
Av is the voltage gain for an ideal voltage source (Rv = 0).
Consider input source to be a current source IS in parallel with a resistance RS as shown in fig. 3.
In this case, overall current gain AIS is defined as
To analyze multistage amplifier the h-parameters of the transistor used are obtained from manufacturer data sheet. The manufacturer data sheet usually provides h-parameter in CE configuration. These parameters may be converted into CC and CB values. For example fig. 4 hrc in terms of CE parameter can be obtained as follows.
For CE transistor configuaration
Vbe = hie Ib + hre Vce
Ic = hfe Ib + hoe Vce
The circuit can be redrawn like CC transistor configuration as shown in fig. 5.
Vbc = hie Ib + hrc Vec
Ic = hfe Ib + hoe Vec
Hybrid model for transistor in three different configurations
Typical h-parameter values for a transistor:
Analysis of a Transistor amplifier circuit using h-parameters
A transistor amplifier can be constructed by connecting an external load and signal source and biasing the transistor properly.
The two port network of Fig. 1.4 represents a transistor in any one of its configuration. It is assumed that h-parameters remain constant over the operating range.The input is sinusoidal and I1,V1,I2 and V2 are phase quantities.
Fig. 1.5 Transistor replaced by its Hybrid Model
Current Gain or Current Amplification (Ai)
For transistor amplifier, the current gain Ai is defined as the ratio of output current to input current,i.e,
Ai =IL /I1 = -I2 / I1
From the circuit of Fig 1.5
I2= hf I1 + hoV2
Substituting V2 = ILZL = -I2ZL
I2= hf I1- I2ZL ho
I2 + I2ZL ho = hf I1
I2( 1+ ZL ho) = hf I1
Ai = -I2 / I1 = - hf / ( 1+ ZL ho)
Ai = - hf / ( 1+ ZL ho)
Input Impedance (Zi)
In the circuit of Fig 1.5, RS is the signal source resistance .The impedance seen when looking into the amplifier terminals (1,1’) is the amplifier input impedance Zi,
Zi = V1/I1
From the input circuit of Fig V1 = hi I1 + hrV2
Zi = ( hi I1 + hrV2) / I1
= hi + hr V2 / I1
V2 = -I2 ZL = A1I1ZL
Zi = hi + hr A1I1ZL / I1
= hi + hr A1ZL
Substituting for Ai
Zi = hi - hf hr ZL / (1+ hoZL)
= hi - hf hr ZL / ZL(1/ZL+ ho)
Taking the Load admittance as YL =1/ ZL
Zi = hi - hf hr / (YL + ho)
Voltage Gain or Voltage Gain Amplification Factor(Av)
The ratio of output voltage V2 to input voltage V1 give the voltage gain of the transistor i.e,
Av = V2 / V1
V2 = -I2 ZL = A1I1ZL
Av = A1I1ZL / V1 = AiZL / Zi
Output Admittance (Yo)
Yo is obtained by setting VS to zero, ZL to infinity and by driving the output terminals from a generator V2. If the current is I2 then Yo= I2/V2 with VS=0 and RL= ∞.
From the circuit of fig 1.5
I2= hf I1 + hoV2
Dividing by V2,
I2 / V2 = hf I1/V2 + ho
With V2= 0, by KVL in input circuit,
RSI1 + hi I1 + hrV2 = 0
(RS + hi) I1 + hrV2 = 0
Hence, I2 / V2 = -hr / (RS + hi)
= hf (-hr/( RS + hi)+ho
Yo= ho- hf hr/( RS + hi)
The output admittance is a function of source resistance. If the source impedance is resistive then Yo is real.
Voltage Amplification Factor(Avs) is taking into account the resistance (Rs) of the source.
Fig. 5.6 Thevenin’s Equivalent Input Circuit
This overall voltage gain Avs is given by
Avs = V2 / VS = V2V1 / V1VS = Av V1/ VS
From the equivalent input circuit using Thevenin’s equivalent for the source shown in Fig. 5.6,
V1 = VS Zi / (Zi + RS)
V1 / VS = Zi / ( Zi + RS)
Then, Avs = Av Zi / ( Zi + RS)
Substituting Av = AiZL / Zi
Avs = AiZL / ( Zi + RS)
Avs = AiZL RS / ( Zi + RS) RS
Avs = AisZL / RS
Current Amplification (Ais) taking into account the source Resistance(RS)
Fig. 1.7 Norton’s Equivalent Input Circuit
The modified input circuit using Norton’s equivalent circuit for the calculation of Ais is shown in Fig. 1.7
Overall Current Gain, Ais = -I2 / IS = - I2I1 /I1 IS = Ai I1/IS
From Fig. 1.7 I1= IS RS / (RS + Zi)
I1 / IS = RS / (RS + Zi)
and hence, Ais = Ai RS / (RS + Zi)
Operating Power Gain (AP)
The operating power gain AP of the transistor is defined as
AP = P2 / P1 = -V2 I2 / V1 I1 = AvAi = Ai AiZL/ Zi
AP = Ai2(ZL/ Zi)
Small Signal analysis of a transistor amplifier