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Advance Electronics. Prof . Rajput Sandeep Assist. Prof., EC Dept. HCET , Siddhpur. Chapter 1: Transistor at High Frequency Lecture : 1. The hybrid- pi CE Transistor Model. The hybrid π model at low frequencies. Chapter 1: Transistor at High Frequency Lecture : 2. - PowerPoint PPT Presentation

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Advance ElectronicsProf. Rajput SandeepAssist. Prof., EC Dept.HCET ,SiddhpurChapter 1: Transistor at High FrequencyLecture : 1The hybrid- pi CE Transistor Model

The hybrid model at low frequencies Chapter 1: Transistor at High FrequencyLecture : 2The Input Conductance - gbeTo obtain the expression for the input conductance gbe , refer to the low frequency hybrid model shown in figure 1.

Figure 1:The hybrid model at low frequencies Note that all the capacitor from the high frequency hybrid- model have been removed because the capacitances at low frequencies are of negligible values. The Input Conductance - gbeFigure 2 shows the h parameter model at low frequency, for the same transistor.

Now, rbc > rbe therefore almost all current Ib will flow through rbe . so, V be = Ib x rbe (1)Figure 2:The hybrid model at low frequencies

The Input Conductance - gbeSo that, the short circuit collector current in figure 1 is given by , Ic = gm Vbe = gm ( Ib x rbe ) (2)The short circuit current gain is defined as, hfe = Ic / Ib = gm rbe or, rbe = hfe / gm = (hfe VT) / | Ic | (3)Now, input conductance ,gbe = Ib / Vbe = 1 / rbe By substituting the value of rbe we can get, gbe = gm / hfe (4) This is the required expression for the input conductance.

7The Input Conductance - gbeConclusion : By referring the below equation , rbe = hfe / gm = (hfe VT ) / | Ic | (5) It can be concluded that if hfe is constant over a certain range of current, then rbe is directly proportional to temperature (due to V T ) and it is inversely proportional to current Ic .

Therefore input conductance gbe will be inversely proportional to temperature and directly proportional to current if hfe is constant. The feedback Conductance gbc Let us define the reverse voltage ratio hre using the low frequency hybrid pi circuit of figure 1 as follows:

hre = (Vbe / V ce ) | for Ib = 0(6)

Lets see the modified low frequency hybrid pi model in figure 3.

Figure 3: Modified hybrid pi model at low frequencies The feedback Conductance gbc From the figure 3 it is evident that rbe and rb c form a potential divider across Vce .

So that , Vbe = { rbe / (rbe + r bc ) } x Vce (7) so, hre = Vbe / V ce = { rbe / (rbe + r bc ) } (8)

Rearrange this expression ,

h re rbe + hre rbc = rbe so, h re rbc = rbe - hre rbe

= rbe ( 1 hre )(9) The feedback Conductance gbc But, hre < < 1. Hence (1- hre ) = 1.

so that , hre rbc = rbe

The feedback conductance gbc is defined as,

gbc = Ib / Vce

But , Ib = Vbe / r be so that gbc = Vbe / ( Vce * rbe )

But, Vbe / Vce = hre and 1 / rbe = gbe

So that, The feedback conductance = hre * gbe . (11) The Base Spreading Resistance rbb

Let us define hie from the low frequency hybrid- pi model.From the figure 3 it is evident that rbe and rb c from a potential divider across Vce and rbc > > r be so that,

(rbe || rbc ) = rbe (12)Figure 4: Modified hybrid pi model at low frequencies The Base Spreading Resistance rbb Hence hie = (Vbe / Ib ) | Vce =0

= {Ib rbb + Ib * (rbe || rbc )}/ Ib

So that , hie = rbb + (rbe || rbc ) = rbb + rbe (13)

This is the expression for hie in terms of hybrid pi parameter.

The base spreading resistance is given by, rbb = hie rbe In the expression for hie, Substitute the expression of

rbe = (hfe VT ) /|Ic | to get,

hie = rbb + rbe = rbb +{ hfe VT / |Ic |}(14)13Chapter 1: Transistor at High FrequencyLecture : 3The output conductance gce

The output conductance hoe can be obtained from the low frequency hybrid pi equivalent circuit by open circuiting the input terminal, i.e. Ib = 0. Figure 5 : Modified hybrid pi model at low frequencies From figure 5 , hoe = ( Ic /Vce ) | Ib = 0 ,(16)

But Ic = I1 + I2 + I3

Now, I1 = Vce / r ce , I2 = gm V be = gm hre Vce and I3 = Vce / (rbe + rbc) Now Substituting (1 / r ce ) = gce and (1/rbc ) = gbc to get,

hoe = gce +gm hre + gbc (17)

Hence output conductance, gce = hoe gm hre gbc

Substitute gm = hfe gbe and hre = gbc / gbe to get,

gce = hoe [ hfe gbe * (gbc / gbe ) ] gbc (18)

so that , gce = hoe hfe gbc gbc

gce = hoe ( 1 + hfe ) gbc (19)

This is the required expression for the output conductance.The output conductance gce The Hybrid pi Capacitance The Basic high frequency hybrid pi model includes two capacitance. Cbc or Cc and Ce.

Capacitance Cbc or Cc :

Cbc or Cc is the collector junction capacitance. It is measured CB capacitance between collector ( C ) and base ( B ) with the input (E) open.Cbc or Cc is usually specified by the manufacturer as Cob. In the active region of the transistor operation , the CB junction is reverse biased. Hence Cc represents the transition capacitance CT.Cc is inversely propositional to VCE, because as VCE is increased, the width of the depletion region also increases and Cc decreases. The Hybrid pi Capacitance Capacitance Ce : Ce capacitance appears between base and emitter, in the hybrid pi model.Ce represents the sum of emitter diffusion capacitance (CDe ) and the emitter junction capacitance CTe.But the emitter junction is forward biased and the diffusion capacitance of a forward biased junction is much higher than its transition capacitance. i.e. CDe > > CTe Therefore, Ce = (CDe + CTe ) = CDe (20) Ce or CDe is propositional to the emitter bias current Ic and it is almost independent of temperature.The Hybrid pi Capacitance The diffusion capacitance CDe is mathematically expressed as, CDe = gm (W2 / 2DB ) (21)Where DB = Diffusion constant for minority carriers in the base.Experimentally Ce or CDe is determined from the measurement of fT , a frequency at which the CE short circuit current gain reduces to 1. The value of Ce in terms of fT is given by, Ce = gm / 2 fT (22) But, Ce = CDe + Cte (23) Therefore , (CDe + CTe ) = gm / 2 fT So that fT = gm / 2 (CDe + CTe ) .(24)Simplified Hybrid pi Model Figure 6 : Simplified hybrid pi model If we neglect the resistance rbb because it is too small and rbc because it is too high, Then we get simplified hybrid model as shown in figure 6.

Validity of Hybrid pi Model Consider the CDe and Ce equations, CDe = gm (W2 / 2DB )(25) = (W2 / 2DB re)

= (W2 IE/ 2DB VT )

And g m = IE / VE , (26)

In both the equations we have assumed that VBE varies slowly. This is done to maintain the minority carrier distribution in the base region triangular as shown in figure 7. Validity of Hybrid pi Model If the distribution remains triangular under the condition of varying VBE then the slope at x =0 and x = W are equal. Hence the emitter and collectors currents will also be equal.

Figure 7 : Modified- carrier charge distribution in the base region If the rate of change of VBE is small such that the base incremental current IB is small in comparison to the collector incremental current IC then under such dynamic condition the hybrid pi model is valid. Validity of Hybrid pi Model Scientist Giacollecto has shown that the network elements of hybrid pi model are frequency independent provided that,

2f W2 / 6DB < < 1 (27) Now we know that, CDe = gm (W2 / 2DB )(28) = (W2 IE / 2DB VT)

And Ce = gm / 2 fT

Therefore, W2 /6DB = Ce / 3gm = 1/ 6fT

So that f < < 6fT / 2 = 3fT (29)

Thus the hybrid model is valid for frequencies approximately up to fT / 3. Chapter 1: Transistor at High FrequencyLecture : 4The CE Short Circuit Current Gain Consider the single stage CE transistor amplifier or the last stage of cascade configuration with load RL on this stage.

The hybrid pi equivalent circuit for a single transistor with a resistive load RL is show in figure 8.

Figure 8 : The hybrid pi equivalent circuit for a single transistor with resistive load RL The CE Short Circuit Current Gain In this as we want to obtain the short circuit current gain. So we will short the output terminals, C and E. Therefore RL = 0.

The approximate equivalent circuit with RL = 0 is as shown in figure 9. Figure 9 : Equivalent circuit with RL = 0

Note that the resistance rbc has been neglected as it is of very high value, and rce is short circuited.

The capacitance Ce and Cc will appear in parallel with each other. The CE Short Circuit Current Gain The short circuit load current IL is given by,

IL = - gmVbe (30)

Now let us calculate Vbe from figure 9. Let the parallel combination of rbe and Cc+Ce be represented by Z.

Z = { rbe * 1/