Advance Electronics Prof. Rajput Sandeep Assist. Prof., EC Dept. HCET,Siddhpur

Chapter 1: Transistor at High Frequency Lecture : 1 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 Chapter 1: Transistor at High Frequency Lecture : 2

The Input Conductance - g b’e  To obtain the expression for the input conductance g b’e, refer to the low frequency hybrid – π model shown in figure 1. Figure 1:The hybrid π model at low frequencies  Note that all the capacitor from the high frequency hybrid- π model have been removed because the capacitances at low frequencies are of negligible values.

The Input Conductance - g b’e  Figure 2 shows the h parameter model at low frequency, for the same transistor.  Now, r b’c > r b’e therefore almost all current I b will flow through r b’e. so, V b’e = I b x r b’e (1) Figure 2:The hybrid model at low frequencies

The Input Conductance - g b’e  So that, the short circuit collector current in figure 1 is given by, I c = g m V b’e = g m ( I b x r b’e ) (2)  The short circuit current gain is defined as, h fe = I c / I b = g m r b’e or, r b’e = h fe / g m = (h fe V T ) / | I c | (3)  Now, input conductance,g b’e = I b / V b’e = 1 / r b’e  By substituting the value of r b’e we can get, g b’e = g m / h fe (4)  This is the required expression for the input conductance.

The Input Conductance - g b’e  Conclusion :  By referring the below equation, r b’e = h fe / g m = (h fe V T ) / | I c | (5)  It can be concluded that if h fe is constant over a certain range of current, then r b’e is directly proportional to temperature (due to V T ) and it is inversely proportional to current I c.  Therefore input conductance g b’e will be inversely proportional to temperature and directly proportional to current if h fe is constant.

The feedback Conductance – g b’c  Let us define the reverse voltage ratio h re using the low frequency hybrid – pi circuit of figure 1 as follows: h re = (V b’e / V ce ) | for I b = 0(6)  Let’s see the modified low frequency hybrid – pi model in figure 3. Figure 3: Modified hybrid – pi model at low frequencies

The feedback Conductance – g b’c  From the figure 3 it is evident that r b’e and r b’ c form a potential divider across V ce. So that, V b’e = { r b’e / (r b’e + r b’c ) } x V ce (7) so, h re = V b’e / V ce = { r b’e / (r b’e + r b’c ) } (8) Rearrange this expression, h re r b’e + h re r b’c = r b’e so, h re r b’c = r b’e - h re r b’e = r b’e ( 1 – h re )(9)

The feedback Conductance – g b’c  But, h re < < 1. Hence (1- h re ) = 1. so that, h re r b’c = r b’e  The feedback conductance g b’c is defined as, g b’c = I b / V ce But, I b = V b’e / r b’e so that g b’c = V b’e / ( V ce * r b’e ) But, V b’e / V ce = h re and 1 / r b’e = g b’e  So that, The feedback conductance = h re * g b’e. (11)

The Base Spreading Resistance – r b’b  Let us define h ie from the low frequency hybrid- pi model.  From the figure 3 it is evident that r b’e and r b’ c from a potential divider across V ce and r b’c > > r b’e so that, (r b’e || r b’c ) = r b’e (12) Figure 4: Modified hybrid – pi model at low frequencies

The Base Spreading Resistance – r b’b  Hence h ie = (V be / I b ) | V ce =0 = {I b r b’b + I b * (r b’e || r b’c )}/ I b So that, h ie = r b’b + (r b’e || r b’c ) = r b’b + r b’e (13)  This is the expression for h ie in terms of hybrid – pi parameter.  The base spreading resistance is given by, r b’b = h ie – r b’e  In the expression for h ie, Substitute the expression of r b’e = (h fe V T ) /|I c | to get, h ie = r b’b + r b’e = r b’b +{ h fe V T / |I c |}(14)

Chapter 1: Transistor at High Frequency Lecture : 3 Chapter 1: Transistor at High Frequency Lecture : 3

The output conductance – g ce  The output conductance h oe can be obtained from the low frequency hybrid – pi equivalent circuit by open circuiting the input terminal, i.e. I b = 0. Figure 5 : Modified hybrid – pi model at low frequencies From figure 5, h oe = ( I c /V ce ) | I b = 0,(16) But I c = I 1 + I 2 + I 3 Now, I 1 = V ce / r ce, I 2 = g m V b’e = g m h re V ce and I 3 = V ce / (r b’e + r b’c )

 Now Substituting (1 / r ce ) = g ce and (1/r b’c ) = g b’c to get, h oe = g ce +g m h re + g b’c (17) Hence output conductance, g ce = h oe – g m h re – g b’c Substitute g m = h fe g b’e and h re = g b’c / g b’e to get, g ce = h oe – [ h fe g b’e * (g b’c / g b’e ) ] – g b’c (18) so that, g ce = h oe – h fe g b’c – g b’c g ce = h oe – ( 1 + h fe ) g b’c (19) This is the required expression for the output conductance. The output conductance – g ce

The Hybrid – pi Capacitance  The Basic high frequency hybrid – pi model includes two capacitance.  C b’c or C c and C e.  Capacitance C b’c or C c :  C b’c or C c is the collector junction capacitance. It is measured CB capacitance between collector ( C ) and base ( B ) with the input (E) open.  C b’c or C c is usually specified by the manufacturer as C ob.  In the active region of the transistor operation, the CB junction is reverse biased. Hence C c represents the transition capacitance CT.  C c is inversely propositional to V CE, because as V CE is increased, the width of the depletion region also increases and C c decreases.

The Hybrid – pi Capacitance  Capacitance C e : C e capacitance appears between base and emitter, in the hybrid – pi model.  C e represents the sum of emitter diffusion capacitance (C De ) and the emitter junction capacitance C Te.  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. C De > > C Te Therefore, C e = (C De + C Te ) = C De (20)  C e or C De is propositional to the emitter bias current I c and it is almost independent of temperature.

The Hybrid – pi Capacitance  The diffusion capacitance C De is mathematically expressed as, C De = g m (W 2 / 2D B ) (21)  Where D B = Diffusion constant for minority carriers in the base.  Experimentally C e or C De is determined from the measurement of f T, a frequency at which the CE short circuit current gain reduces to 1.  The value of C e in terms of f T is given by, C e = g m / 2 π f T (22) But, C e = C De + C te (23) Therefore, (C De + C Te ) = g m / 2 πf T So that f T = g m / 2 π (C De + C Te ).(24)

Simplified Hybrid – pi Model Figure 6 : Simplified hybrid – pi model  If we neglect the resistance r b’b because it is too small and r b’c because it is too high, Then we get simplified hybrid model as shown in figure 6.

Validity of Hybrid – pi Model  Consider the C De and C e equations, C De = g m (W 2 / 2D B )(25) = (W 2 / 2D B r ‘ e ) = (W 2 I E / 2D B V T )  And g m = α ∂I E / ∂ V E, (26)  In both the equations we have assumed that V BE 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 V BE 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 V BE is small such that the base incremental current I B is small in comparison to the collector incremental current I C 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, 2πf W 2 / 6D B < < 1 (27) Now we know that, C De = g m (W 2 / 2D B )(28) = (W 2 I E / 2D B V T ) And C e = g m / 2 π f T Therefore, W 2 /6D B = C e / 3g m = 1/ 6πf T So that f < < 6πf T / 2π = 3f T (29)  Thus the hybrid π model is valid for frequencies approximately up to f T / 3.

Chapter 1: Transistor at High Frequency Lecture : 4 Chapter 1: Transistor at High Frequency Lecture : 4

The CE Short Circuit Current Gain  Consider the single stage CE transistor amplifier or the last stage of cascade configuration with load R L on this stage.  The hybrid – pi equivalent circuit for a single transistor with a resistive load R L is show in figure 8. Figure 8 : The hybrid – pi equivalent circuit for a single transistor with resistive load R L

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 R L = 0.  The approximate equivalent circuit with R L = 0 is as shown in figure 9. Figure 9 : Equivalent circuit with R L = 0  Note that the resistance r b’c has been neglected as it is of very high value, and r ce is short circuited.  The capacitance C e and C c will appear in parallel with each other.

The CE Short Circuit Current Gain  The short circuit load current I L is given by, I L = - g m V b’e (30)  Now let us calculate V b’e from figure 9. Let the parallel combination of r b’e and C c +C e be represented by Z. Z = { r b’e * 1/(jw ( C e + C c ) )} / { r b’e + 1/(jw ( C e + C c )) } (31) = r b’e / {r b’e * (jw ( C e + C c ) + 1)} = r b’e / {1 + r b’e * (jw ( C e + C c ))}(32)  Including this term into the equivalent circuit we et the simplified hybrid – pi equivalent circuit as shown in figure 10

The CE Short Circuit Current Gain Figure 10 : Equivalent circuit with Z included  From figure 10 we can write that, V b’e = I i x Z = (I i * r b’e ) / (1+ jw r b’e (C c + C e ))(33) Substituting this value into equation (31) we get, I L = - g m r b’e I i /{1 + r b’e * (jw ( C e + C c ))}

The CE Short Circuit Current Gain Figure 11: The short circuit CE current gain Vs frequency (plotted on the Log- Log scale  Therefore, the current gain is given by, A I = I L / I i = - g m r b’e /{1 + r b’e * (jw ( C e + C c ))}  But h fe = g m r b’e So that, A I = - h fe /{1 + r b’e * (jw ( C e + C c ))} where,w = 2πf

The CE Short Circuit Current Gain  That means the current gain is inversely propositional to frequency.  At very low frequencies, the term jw r b’e (C c + C e ) is very small as compared to 1.Thereforecurrent gain A I ~ h fe.  But as the frequency increases the, current gain goes on decreasing.  Figure 11 shows the variation in the current gain A I expressed in decibels (i.e. 20log|A I |) against frequency on a logarithmic frequency scale.  Note that A I (dB) = 0 dB at frequency f = f T and for f > > f B the graph is a straight line having a slope of 6 dB / octave or 20 dB/decade.  Let us now define various frequencies as f α, f β and f T.

Chapter 5: Operational Amplifiers Lecture : 5 Chapter 5: Operational Amplifiers Lecture : 5

 “operational” was used as a descriptor early-on because this form of amplifier can perform operations of adding signals subtracting signals integrating signals,  The applications of operational amplifiers ( shortened to op amp ) have grown beyond those listed above. Operational Amplifiers  At this level of study we will be concerned with how to use the op amp as a device.  The internal configuration (design) is beyond basic circuit theory and will be studied in later electronic courses. The complexity is illustrated in the following circuit.

Operational Amplifiers  The op amp is built using VLSI techniques. The circuit diagram of an LM 741 from National Semiconductor is shown below. Internal circuitry of LM741

Fortunately, we do not have to sweat a circuit with 22 transistors and twelve resistors in order to use the op amp. The circuit in the previous slide is usually encapsulated into a dual in- line pack (DIP). For a single LM741, the pin connections for the chip are shown below. Pin connection, LM741. Operational Amplifiers

The basic op amp with supply voltage included is shown in the diagram below. Basic op am diagram with supply voltage. Operational Amplifiers

In most cases only the two inputs and the output are shown for the op amp. However, one should keep in mind that supply voltage is required, and a ground. The basic op am without a ground is shown below. Outer op am diagram. Operational Amplifiers

A model of the op amp, with respect to the symbol, is shown below. Op Amp Model Operational Amplifiers

The previous model is usually shown as follows: Working circuit diagram of op amp. Operational Amplifiers

Application: As an application of the previous model, consider the following configuration. Find V o as a function of V in and the resistors R 1 and R 2. Op amp functional circuit. Operational Amplifiers

In terms of the circuit model we have the following: Total op amp schematic for voltage gain configuration. Operational Amplifiers

Circuit values are: R 1 = 10 k  R 2 = 40 k  A = 100,000 R i = 1 meg  Operational Amplifiers

We can write the following equations for nodes a and b. Operational Amplifiers

Equation simplifies to; simplifies to; Operational Amplifiers

From Equations we find; This is an expected answer. Fortunately, we are not required to do elaborate circuit analysis, as above, to find the relationship between the output and input of an op amp. Simplifying the analysis is our next consideration. Eq 8.5 Operational Amplifiers

For most all operational amplifiers, R i is 1 meg  or larger and R o is around 50  or less. The open-loop gain, A, is greater than 100,000. Ideal Op Amp: The following assumptions are made for the ideal op amp. Operational Amplifiers

Ideal Op Amp: (a)i 1 = i 2 = 0: Due to infinite input resistance. (b) V i is negligibly small; V 1 = V 2. Figure 8.9: Ideal op amp. Operational Amplifiers

Ideal Op Amp: Find V o in terms of V in for the following configuration. Figure 8.10: Gain amplifier op amp set-up. Operational Amplifiers

Ideal Op Amp: Writing a nodal equation at (a) gives; Eq 8.6 Operational Amplifiers

Ideal Op Amp: With V i = 0 we have; With R 2 = 4 k  and R 1 = 1 k , we have Earlier we got Eq 8.7 Operational Amplifiers

Ideal Op Amp: When V i = 0 in Eq 8.7 and we apply the Laplace Transform; Eq 8.8 In fact, we can replace R 2 with Z fb (s) and R 1 with Z 1 (s) and we have the important expression; Eq 8.9 Operational Amplifiers

Ideal Op Amp: At this point in circuits we are not able to appreciate the utility of Eq 8.9. We will revisit this at a later point in circuits but for now we point out that judicious selections of Z fb (s) and Z in (s) leads to important applications in Analog Filters Analog Compensators in Control Systems Application in Communications Operational Amplifiers

Ideal Op Amp: Example 8.1: Consider the op amp configuration below. Figure 8.11: Circuit for Example 8.1. Assume V in = 5 V Operational Amplifiers

At node “a” we can write; From which; V 0 = -51 V Eq 8.10 Example 8.1 cont. Operational Amplifiers

Example 8.2: Summing Amplifier. Given the following: Figure 8.12: Circuit for Example 8.2. Eq 8.11 Operational Amplifiers

Example 8.2: Summing Amplifier. continued Equation 8.11 can be expressed as; Eq 8.12 If R 1 = R 2 = R fb then, Eq Therefore, we can add signals with an op amp. Operational Amplifiers

Example 8.3: Isolation or Voltage Follower. Applications arise in which we wish to connect one circuit to another without the first circuit loading the second. This requires that we connect to a “block” that has infinite input impedance and zero output impedance. An operational amplifier does a good job of approximating this. Consider the following: Figure 8.13: Illustrating Isolation. Operational Amplifiers

Example 8.3: Isolation or Voltage Follower. continued Figure 8.14: Circuit isolation with an op amp. It is easy to see that: V 0 = V in Operational Amplifiers

Example 8.4: Isolation with gain. Figure 8.15: Circuit for Example 8.4: Writing a nodal equation at point “a” and simplifying gives; Operational Amplifiers

Example 8.5: The noninverting op amp. Consider the following: Figure 8.16: Noninverting op am configuration. Operational Amplifiers

Example 8.5: The noninverting op amp. Continued Writing a node equation at “a” gives; Remember this Operational Amplifiers

Example 8.6: Noninverting Input. Find V 0 for the following op amp configuration. Figure 8.17: Op amp circuit for example 8.6. Operational Amplifiers

Example 8.6: Noninverting Input. The voltage at V x is found to be 3 V. Writing a node equation at “a” gives; or Operational Amplifiers