Ee3101 Experiment 6

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High-frequency Behavior of Bi-Polar Junction Transistors Matt xxxxx Student ID : xxxxxxx 10/2/06 – 10/16/06

Abstract In this experiment we will measure the input and output of various BJT-based amplifiers to determine the value of the BJT’s internal components, such as

Cπ , rπ , rx , and C µ which are included in the hybrid-π

small signal model. We will also investigate the effects of different amplifier designs on the midband gain and bandwidth of the circuit. We will show that the product of gain and bandwidth of amplifier designs is a constant value. This experiment is important because we will learn about the internal components of the BJT and the different designs of amplifiers that utilize the BJT.

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Introduction In previous experiments, we looked at how the BJT can be used in amplification and current mirroring applications; these were for medium to low frequency situations where the high frequency effects of the BJT were not eminent. The focus of this experiment is the analysis of BJT-based amplifiers, and how the hybrid –π model describes the BJT in these situations. This model is well suited for small signal analysis and accounts for the internal capacitances of the BJT at high frequencies. During transistor operation, electric fields build up within the BJT and create intrinsic capacitances; we can see these capacitances represented as capacitors in our hybrid- π model In this experiment we measure and calculate the components of the hybrid-π model (Figure 6.0.1) of the BJT which is commonly used to analyze the response of small signal inputs of a circuit at various frequencies, especially at high frequencies. The capacitance effect in the BJT does not show much effect at low frequencies, because they work as an open circuit. However, as the frequency may increase these capacitances work as a low-pass filter, which results in a decrease of the current or voltage gain. The break frequency, same as cutoff frequency, is calculated at the -3dB value of the max gain, which is the value of

1 Amax . The relationship between the cutoff frequency and the components of the Hybrid-π Model is 2 given in Figure 6.0.2 Base

Collector

R x

R pi

C pi

C u

1 / f H = 2πrπ [Cπ + (1 + g m RL )Cµ ]

R o

Emitter

Fig 6.0.1

Fig 6.0.2

In the first part of the experiment we gather information of the CE Amplifier and calculate the break frequency of the circuit. Using that data, with the equation in Figure 6.0.2, we are able to calculate the values of the unknown components in the Hybrid-π Model. Further along, as we get to understand the Hybrid-π Model, we will then focus on other amplifiers, such as the Cascode, and the CC-CB, to investigate the relationship between midband gain and bandwidth. We will find that the greater the gain is, the smaller the bandwidth will be and vice versa.

Experiment

Part 1 – Hybrid- π Model Experiment 6.1 For the first part of our experiment, we will use a simple common-emitter transistor design as shown in Fig 6.1.1.

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BJT Schematic of A Common-Emitter Amplifier Vcc

R b

R s

R l

C s

Q 1 C A 3096AE

Vs R e

C e

Given Values VCC = +15V VB=7.5V RL=100Ω RS=10K IS=1mA Ce=CS=10uF Solved For Values RB = 2MΩ RE = 6.77KΩ

0

Fig 6.1.1

We will use the Q1 BJT contained within the CA3096AE IC chip for our transistor as seen in Fig 6.1.2. This BJT has a theoretical β value of around 250.

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Fig 6.1.2

Notes on this (Fig 6.1.1) design. 1. At low frequencies, the internal BJT capacitances will have negligible effects 2. CE and CS are intended to be short circuits at the measurement frequency and open circuits during DC bias. 3. We will attempt to bias the transistor so that the collector current is about 1mA 4. RL=100Ω so that we may increase it by a factor of 10 later on and still have it be < 10K Ω 5. CE=CS=10uF so that we may have shorts at lower frequencies, but not at DC. We still need to solve for some component values. To do this, we will look at the circuit in its DC bias mode as seen in Fig 6.1.3. DC Bias Mode of Common Emitter Amplifier Vcc

R b

R l

Q 1 C A 3096AE

R e

Given Values VCC = +15V VB=7.5V RL=100Ω RS=10K IS=1mA Ce=CS=10uF Solved For Values RB = 2MΩ RE = 6.77KΩ

0

Fig 6.1.3 We need to solve for RB and RE, to calculate RB, we will solve for IB first using Ic.

IC = (1 + β ) I B

1mA = (1 + 250) I B I B = 3.98uA Since we know we want VB to be 7.5V and VCC = 15, we know we will have a voltage drop across RB of 7.5V and the current through the resistor is

RB =

I b = 3.98uA .

VCC − V B 7.5V = = 1.88MΩ ≈ 2 MΩ IB 3.98uA

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ We can also solve for Re in the following way,

I E = I B + I C = 1mA + 3.98uA ≈ 1mA

0 = V B − V BE − I E R E = 7.5V − 0.7V − (1mA) RE R E = 6.77 KΩ ≈ 4.7 KΩ + 2 KΩ We can verify our design by making sure that VB = 7.5V. When measured, it turns out to be 7.4V. This is close enough for our purposes as this is only a bias voltage, we just want it near 7.5V so that we have enough room for voltage swing when we begin amplifying. Our error percentage is 100 x

7.5V − 7.4V = 13% 7.5V

We will now measure Rin which is the input resistance into the base node of the BJT. To do this, we will measure

Vin = Rin . Vin will be VB and Iin will be the current though the RS resistor. I in 169mV − 75mV = 9.4mA 10 KΩ V 75mV Rin = in = = 8 KΩ I in 9.4mA I in =

We can verify our circuit operation by measuring our β Value using the following equation.

I C = (1 + β ) I B

We need to know IB and IC. To measure these values, we will just take the voltage drop across their respective resistors and divide by the resistor value

V  I =  . R 

15V − 7.4V = 3.81uA 2 MΩ 15.07V − 14.96V IC = = 1.1mA 100Ω IB =

We can now solve for β.

I C = (1 + β ) I B ⇒ 1.1mA = (1 + β ) 3.81uA

β = 287 Our error percentage is 100 x

287 − 250 = 14.8% 250

We also need to solve for gm which we will need later on. gm is defined by the following equation.

gm =

IC . VT is always 26mV for a BJT so we have the following equation. VT

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gm =

1.1mA = 4.4mS 26mV

(Fig 6.1.4)

These results are reasonable considering the equipment we are using. Experiment 6.2 In the next three experiments, we will derive three equations using the break frequency of the circuit to obtain the unknown values of

Cπ , rπ , rx , and C µ as represented in the hybrid-π model of the small signal

representation of a BJT. We can see the small signal representation in Fig 6.2.1 below. Small Signal Representation of a BJT Base

Collector

R x

R pi

C u

C pi

R o

Emitter

Fig 6.2.1 We will use the following equation (Fig 6.2.2) to do this.

1 = 2πrπ [Cπ + (1 + g m R L ) C µ ] fH Fig 6.2.2 We will modify the circuit slightly in each portion of the experiment to obtain different measurements for f H while knowing corresponding variables in the equation. This way, we can develop three equations with three unknowns in a system of linear equations.

f H as out -3dB point. Where our current gain, AI , has dropped 3dB. Also note point will occur when AI = .707 xAmidband

Note that we will be using that the

fH

We will need a method of measuring current gain. We know that current gain is the output current divided by input current. Here, our output current is the collector current, and our input current is the base current. We can measure these currents by measuring our voltage drops over respective resistors and diving by the resistance value. See Fig 6.2.3 Derivation of AI Measurement

AI =

iout iin

− vC i RL = C = iB vS − v B RS Fig 6.2.3

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ In the first part of this experiment, we will take our data when RL=100Ω We will now take out current gain measurements at various frequencies to find our -3dB point. See Fig 6.2.4(a) for out numerical data and Fig 6.2.4(b) for a graphical representation of our data.

Results of AI Measurement at -3dB Point and Various Frequencies When RL=100 Ω Frequency (kHz) 1 10 100 350

ib (µA pk->pk) 4.5 5.4 5.41 6.5

ic (mA pk->pk) Gain (IC/IB) 1.57 322.2 1.66 302 1.63 302 1.48 227.8 Fig 6.2.4(a)

<- Midband Gain

<- -3dB point

Results of A I Measurement at -3dB Point and Various Frequencies When RL = 100Ω 350 RL = 100Ω

300

Gain (I C /I B )

250 200 150 100 50 0 0

100

200 Frequency (kHz)

Fig 6.2.4(b)

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300

400

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Using this data, we can determine that our -3dB point lies at

f H = 350kHz , and therefore we know that

1 1 = 2πrπ [Cπ + (1 + g m R L ) C µ ] = = 2πrπ [Cπ + (1 + g m 100Ω ) C µ ] fH 350kHz

(Fig 6.2.5)

by plugging measured values into our equation in Fig 6.2.2. We will use Fig 6.2.5 later on when we solve for our unknowns. Experiment 6.3 For the second portion of the experiment we will add capacitance to Fig 6.1.1 between the base and the collector of the BJT (Cx=1nF). Our new circuit will look the schematic see in Fig 6.3.1 New Schematic with CX Added

Vcc

R b

R l C x

R s

C s

Q 1 C A 3096AE

Vs R e

Given Values VCC = +15V VB=7.5V RL=100Ω RS=10K IS=1mA Ce=CS=10uF Cx=1nF Solved For Values RB = 2MΩ RE = 6.77KΩ

C e

0

Fig 6.3.1 Our new small signal model will change from what is seen in Fig 6.3.2(a) to Fig 6.3.2(b). Note that after the addition of CX, our Cµ value is negligible since it is much smaller than CX. We can therefore neglect Cµ in our f H equation (Fig 6.2.2). Hybrid-π representation of BJT before and after the addition of CX Capacitor

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ C x Base

Collector

R x

R pi

C pi

C u

Base

Collector

R x

R o

R pi

C u

C pi

Emitter

R o

Emitter

Fig 6.3.2(a)

Fig 6.3.2(b)

We will use the exact same process as described in experiment 6.2 to take the measurements of the current gain over frequency and then discover our new f H . See Figures 6.3.4(a)&(b) for our results. Results of AI Measurement at -3dB Point and Various Frequencies When RL=100 Ω & Cx=1nF Frequency (kHz) 1 3.8 10

ib (µA pk->pk) 3.6 5.2 6.55

ic (mA pk->pk) 1.56 1.59 1.05 Fig 6.3.3(a)

Gain (IC/IB) 433 306.1 160

<- Midband Gain <- -3dB point

Results of AI Measurement at -3dB Point and Various Frequencies When RL=100 Ω & Cx=1nF 500 450

RL=100 Ω & Cx=1nF

400

Gain (I C /I B )

350 300 250 200 150 100 50 0 1

10 Frequency (kHz)

Fig 6.3.3(b) Using this data, we can determine that our -3dB point lies at

f H = 3.8kHz , and therefore we know that

1 1 = 2πrπ [Cπ + (1 + g m R L )C µ ] = = 2πrπ [Cπ + (1 + g m 100Ω)( C µ + C x ) ] by plugging fH 3.8kHz

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ measured values into our equation in Fig 6.2.2. Since we know that

C µ + C x ≈ C x = 1nF , we can write

1 1 = = 2πrπ [Cπ + (1 + ( 4.4mS )100Ω )(1nF ) ] (Fig 6.3.4) f H 3.8kHz We will use this equation (Fig 6.3.4) later on when we solve for our unknowns. We can also conclude that inserting the 1nF capacitor, Cx decreased our f H from 350kHz to 3.8kHz, and that Iin increases with frequency because our circuit impedance decreases with frequency. Experiment 6.4 For the third portion of our experiment, we will remove the capacitor, C x that we added in the previous analysis and we will be increasing our RL by a factor of 10. Our circuit should look like the schematic shown in Figure 6.4.1.

Schematic of our circuit after we have removed Cx and have increased RL by a Factor of 10.

Vcc

R b

R s

R l

C s

Q 1 C A 3096AE

Given Values VCC = +15V VB=7.5V RL=100Ω RS=10K IS=1mA Ce=CS=10uF

Vs R e

C e

Solved For Values RB = 2MΩ RE = 6.77KΩ

0

Fig 6.4.1 We will use the exact same process as described in experiment 6.2 to take the measurements of the current gain over frequency and then discover our new f H . See Figures 6.4.1(a)&(b) for our results. Results of AI Measurement at -3dB Point and Various Frequencies When RL=1KΩ Frequency (kHz)

ib (µA pk->pk)

ic (mA pk->pk)

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Gain (IC/IB)

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4.5 5.22 5.23 5.5 6.61

1.57 1.57 1.56 1.35 1.26

348 301 299 246 190

<- Midband Gain

<- -3dB point

Fig 6.4.1(a) Results of A I Measurement at -3dB Point and Various Frequencies When RL=1KΩ 400 350

RL=1KΩ

Gain (I C /I B )

300 250 200 150 100 50 0 1

10

100

1000

Frequency (kHz)

Fig 6.4.1(b) Using this data, we can determine that our -3dB point lies at

f H = 200kHz , and therefore we know that

1 1 = 2πrπ [Cπ + (1 + g m R L )C µ ] = = 2πrπ [Cπ + (1 + ( 4.4mS )(1KΩ ) ) C µ ] fH 200kHz (Fig 6.4.2) by plugging measured values into our equation in Fig 6.2.2. We can conclude that increasing the load resistor, RL decreased our f H from 350kHz to 200kHz. This is most likely because increasing the load resistor will increase the input impedance and therefore slow the charging time of the capacitors. This reduces the gain because the capacitors cannot charge as quickly as before at higher frequencies.

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ Experiment 6.5 We can use the three test case equations, Figures 6.2.5, 6.3.4, and 6.4.2, to solve for our unknown values of

Cπ , Rπ , andC µ . To calculate the values, we use a matrix of the three unknown variables with the three equations.

From the first equation, (Figure 6.2.5), we have:

A.

1 = 2πrπ [Cπ + (1 + (4.4mS )(100Ω))C µ ] 350 KHz 1 1 = Cπ + 1.44C µ 350 KHz ( 2π ) rπ 0 = Cπ + 1.44C µ −

1 1 350 KHz (2π ) rπ

From the second equation, (Figure 6.3.4), we have:

B.

1 = 2πrπ [Cπ + (1 + ( 4)(100Ω))(C µ + 1nF )] 3.8 KHz 1 1 = Cπ + 1.44(C µ + 1nF ) 3.8 KHz ( 2π ) rπ 1 1 = Cπ + 1.44C µ + 1.44 ×10 −9 3.8 KHz ( 2π ) rπ − 1.44 ×10 −9 = Cπ + 1.44C µ −

1 1 3.8 KHz ( 2π ) rπ

From the third equation, (Figure 6.4.2), we have:

C.

1 = 2πrπ [Cπ + (1 + (4.4mS )(1KΩ))C µ ] 200 KHz 1 1 = Cπ + 4.4C µ 200 KHz (2π ) rπ 0 = Cπ + 4.4C µ −

1 1 200 KHz (2π ) rπ

If we solve these equations using a system of equations solver, we will get:

Cπ = 11.5 pF Cµ = 3 pF rπ = 28.0 KΩ calculate rx the following occurs:

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By looking at these results, we know there is something wrong. rπ is much too high a value. We know this because when we

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rx = Rin − rπ = 8 K − 28.0 K = −20 KΩ We can’t have a negative resistance so this result isn’t plausible. However, all of our other data checks out so there is nothing we can do here. Given the condition of our equipment we can’t expect to have perfect answers on high frequency measurements that require very precise results.

Part 2 – Other Cases with Different Amplifiers We will now inspect the relationship, or tradeoff between the midband gain and its bandwidth. Experiment 6.6 We will now construct a cascade amplifier using the schematic shown in Figure 6.6.1. The objective of this experiment is to measure the gain and bandwidth of the circuit to compare it with the data gathered earlier in this lab. Cascode Amplifier Schematic

Vcc

R5

R1 C1

Vout Q1 2N3933

Va R2

0 C2

Q2 2N3933

Vb Vs

R3

R4

C3

Given Values VCC=20V IR5=1mA VA=10V VB=5V R3=1KΩ C1=C2=10µF C3=µ47 Solved for Values R1=2KΩ R2=1KΩ R4=4.3KΩ

0

Fig 6.6.1 Since we were having many problems with the BJTs contained within the CA3096 IC chip, we will switch to the 2N3933 discrete BJT at this point. This BJT has a theoretical β value of around 150. We know that in DC bias mode, our circuit will behave like the circuit seen in Figure 6.6.2

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Cascode Amplifier in DC Bias Mode Vcc

R5

R1

Vout Q1 2N3933

Va R2

Q2 2N3933

Vb R3

R4

Given Values VCC=20V IR5=1mA VA=10V VB=5V R3=1KΩ C1=C2=10µF C3=µ47 Solved for Values R1=2KΩ R2=1KΩ R4=4.3KΩ

0

Fig 6.6.2 Now we need to solve for some unknown values in our schematic using the DC bias mode schematic (Figure 6.6.2). Since we want VODC to be biased 15V, VCC at 20V, and we want IR5=1mA, we can solve for R5:

R5 =

VCC − VODC 20V − 15V = = 5 KΩ I5 1mA

Since we know VB=5V We also know that the equation for the current though R3 is

I R3 =

VB 5V = = 5mA R3 1KΩ

The current through R3 is the same current though R1 and R2 in DC operation if we neglect base current. So IR1 and IR2 are both 5mA as well. We can therefore solve for our values of R 2 and R1 by using our known values for VCC, VA, and VB.

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R2 =

V A − V B 10V − 5V = = 1KΩ I R2 5mA

R1 =

VCC − V A 20V − 10V = = 2 KΩ I R1 5mA

Since the voltage at the emitter of Q2 is 5V-0.7V=4.3V, the current though R4 is 1mA during DC operation if we neglect base currents. This is because we are therefore assuming the entire right side of the amplifier has a current of 1mA.

R4 =

V EQ 2 I R4

=

4.3V = 4.3KΩ ≈ 2.2 KΩ + 2.2 KΩ 1mA

Now that we know all of our component values, we will measure the peak to peak voltage gain over a range of frequencies. Keep in mind that Vin is the positive terminal of the voltage source VS, and Vout is marked on Figure 6.6.1. Our results are shown in Figures 6.6.3(a)&(b). Results of Av Measurement at -3dB Points and Various Frequencies Frequency (kHz) 0.02 0.1 0.21 1 10 100 1000 5000 10000 12200 15000 20000

vin (mV pk->pk) 94 88 88 88 82 75 75 75 82 88 88 63

vout (V pk->pk) 0.44 1.13 1.45 1.75 1.75 1.75 1.69 1.69 1.63 1.45 1.25 0.57 Fig 6.6.3(a)

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Gain (Vout/Vin) 4.7 12.8 16.5 19.9 21.3 23.3 22.5 22.5 19.9 16.5 14.2 9.05

<- -3dB point

<- Midband Gain

<- -3dB point

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Results of A v Measurement at -3dB Points and Various Frequencies

25

Gain (V out/V in)

20

15

10

5 Cascode Amplifier 0 0.01

0.1

1

10

100

1000

10000

100000

Frequency (kHz)

Fig 6.6.3(b)

f L = 210 Hz and f H = 12.2 MHz , and therefore we know that our bandwidth is 12.2 MHz − 210 Hz ≈ 12.2 MHz we also know that our Using this data, we can determine that our -3dB points lie at midband gain is 16.5 V/V. Experiment 6.7 We will now construct a CC-CB amplifier to compare with the previous cascade amplifier we analyzed in experiment 6.6. The schematic for the CC-CB amplifier is shown below in Figure 6.7.1. Note that we will be using the 2N3393 discrete transistors that were used in experiment 6.6. CC-CB Amplifier Schematic Vcc

R 5

R 3

R 1

Q 1 2N 3933

C 1 Q 2 2N 3933 Vs

C 2

0 Va

Vb R 2

-16-

C 3

0

R 4

Vout

Given Values VCC = 15V VA=5V VB=5V VODC=10V ICQ1=1mA IR5=1mA C1=C2=10µF C3=47µF Solved For Values

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___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ ___ __ R4=5KΩ R2=5KΩ R1=1MΩ R3=1MΩ IBQ1=IBQ2=6.6µA Fig 6.7.1 We know that in DC bias mode, our circuit will behave like the circuit seen in Figure 6.7.2 CC-CB Amplifier in DC Bias Mode Vcc

R 5

R 3

R 1

Q 1 2N 3933

Q 2 2N 3933

Va

Vout

Vb R 2

R 4

0

Given Values VCC = 15V VA=5V VB=5V VODC=10V ICQ1=1mA IR5=1mA C1=C2=10µF C3=47µF Solved For Values R4=5KΩ R2=5KΩ R1=1MΩ R3=1MΩ IBQ1=IBQ2=6.6µA

Fig 6.7.2 Now we need to solve for some unknown values in our schematic using the DC bias mode schematic (Figure 6.7.2). We know that VB=5V and IR5=1mA and therefore IR4=1mA if we neglect base current, so we can solve for the resistance of R4

R4 =

VB 5V = = 5 KΩ I R 4 1mA

We can also solve for R2 in the same fashion.

R2 =

VA 5V = = 5 KΩ I R 2 1mA

We also know that VBQ1 and VBQ2 are both 5V+.7V=5.7V because of VBE=.7V so we can solve for both R1 and R3 by first solving for IBQ1 and IBQ2.

I BQ1 = I BQ 2 =

I C 1mA = = 6.6 µA β 150

We can now solve for R1 and R3.

R1 = R3 =

VCC − VBQ I BQ

=

15V − 5.7V = 1.4 MΩ ≈ 1MΩ 6.6 µA

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We will now measure our peak to peak voltage gain over various frequencies just as we did in experiment 6.6. Once again, our input voltage is the positive terminal of V S and our Vout terminal is shown on Figure 6.7.1. Our results are shown in Figures 6.7.3(a)&(b).

Results of Av Measurement at -3dB Points and Various Frequencies Frequency (kHz) 0.02 0.095 0.1 1 10 100 1000 1310 2000

vin (mV pk->pk) 108 113 108 109 113 117 128 113 135

vout (V pk->pk) 2.69 9.02 9.33 12.31 12.31 12.3 10.78 9.02 8.65

Gain (Vout/Vin) 24.9 79.8 86.4 112.9 108.4 105.1 84.22 79.8 64.1

<- -3dB point <- Midband Gain

<- -3dB point

Fig 6.7.3(a) Results of A v Measurement at -3dB Points and Various Frequencies

120 100 Gain (V out/Vin )

80 60 40 20

CC-CB Amplifier

0 0.01

0.1

1

10

100

1000

10000

Frequency (kHz)

Fig 6.7.3(b) Using this data, we can determine that our -3dB points lie at

f L = 95 Hz and f H = 1.31MHz , and therefore we know that our bandwidth is 1.31MHz − 95 Hz ≈ 1.31MHz we also know that our midband gain is 79.8 V/V.

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Experiment 6.8 From the former experiments we can notice that the larger the gain will be, the smaller the bandwidth turns out to be. If we multiply our midband gain by out bandwidth, we can obtain a measurement called the gainbandwidth product. This measurement is the relative effectiveness of the amplifier. Since we already know we can sacrifice one quality for the other, the gain-bandwidth product will tell the amount of both qualities we can have. Figure 6.8.1 is a set of the results of the previous experiments. Previous Results and Calculated Gain-Bandwidth Products. Cascode CC-CB Midband Gain 23.3 112.9 Bandwidth 12.2M 1.31M Gain-Bandwidth Product 284.2M 237.1M Fig 6.8.1 Our percent deviation is 100 x

284.2M − 237.1M = 16% 284.2 M

We can see that the gain-bandwidth product of the cascode amplifier was larger, this means that the cascade amplifier was slightly more efficient than the CC-CB amplifier. The main idea, however, is that even though the basic schematics of the circuit were different, we can see a similar trade off in gain and bandwidth.

Conclusion First we calculated β as 287 of the common-emitter amplifier. We then obtained three equations relating the unknown values of the Hybrid-π Model and the cutoff frequencies that were gathered at the 3dB (70.7%) point through modifying the CE amplifier with three different composition. We added a capacitor parallel with

C µ to change the effect of the capacitance, and then changed the value of RL to see

the effect of the load resistance and the break frequency. From a calculation using a matrix of the three equations, with the three unknown values, we got result. The values of

Cπ = 11.5 pF , C µ = 3 pF , rπ = 28.0 KΩ as a

Cπ and C µ seem to reasonable values, however, rπ was bigger than expected. We

know this because when we calculate rx it turns out to be a negative value (-20KΩ) which isn’t possible. Next we measured the voltage gain of the cascode amplifier design and the CC-CB amplifier design. With the voltage gain we were able to calculate the break frequency of the amplifiers and the bandwidth of them. By comparing the data of the two, we were able to check the inverse proportional relationship between the midband gain and bandwidth. See Figure 6.8.1 for comparison of the two different amplifiers. With an amplifier that has a greater gain; it is hard to expect a large bandwidth. When designing an amplifier we must consider the trade off of the two core aspects within the amplifier. We have successfully gathered the information on how to interpret the high frequency response of the Hybrid-π Model and characteristic of the model. This information should be used on further amplifiers’ analysis and designing projects to gather more accurate results.

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