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Analog Electronic Circuits Lab-module1

Module by: Gowra P S. E-mail the author

Summary: This module includes amplifier and oscillator circuits

Experiment No.1

RC Coupled Amplifier

AIM: Wiring of RC coupled single stage (a) BJT and (b) FET amplifier and determination of gain, frequency response, Input and Output impedances.

a. BJT Amplifier :-

Components :- Resistors, Capacitors, Audio Signal Generator(ASG), 0-30V DC Regulated Power Supply (RPS), Cathode Ray Oscilloscope (CRO) with probes, SL100 Transistor, Decade Resistance Box (DRB), Multimeter, Connecting Wires and Board.

Circuit Diagram:-

**Figure 1**

R_E = 470Ω

R_c= 2.2.KΩ

R₁ = 12KΩ

R₂ = 2.2KΩ

C_E = 47µF

C_c = 1µF OR 0.1µF

Design:-

a. To find R E :-

Let V_CC= 10V, I_E = I_C= 2mA

β = 100 (for SL100)

V_CC

Choose V_CE = ------- = 5 V

V_RE= ------ V_CC= 1V

But V_RE= I_ER_E

V_RE1

R_E = ------- = ------------ = 500Ω (Select 470Ω)

I_E 2 X 10^-3

b. To find R C :-

Choose V_CE = ------- V_CC = 5 V

Applying KVL to the output loop

V_CC = V_RC+V_RE+V_CE

V_RC = V_CC - V_RE- V_CE = 10 - 1 - 5 = 4V

R_C = V_RC/I_C = 2KΩ (Select 2.2KΩ)

c. To find R 1 & R 2 :-

Stability factor ‘S’ is given by,

1+ β

S = -------------

β. R_E

1+ --------

R_th+R_E

Assuming S=5 and substituting for R_E, β and S in the above equation we get

R_th=1.9KΩ --------- (1)

By applying KVL to the input loop,

V_R2 = V_BE+V_RE= 0.6+1 = 1.6V

V_CC

Also V_th = R₂ -------------

R₁ + R₂

R₂

= ----------- = 0.16 ------------ (2)

R₁ + R₂

From equations (1) and (2), we get R₁ and R₂ as,

R₁ = 12.KΩ (Select 12KΩ)

R₂ = 2.3KΩ (Select 2.2KΩ)

d. To find by pass capacitor C E :-

X_CE = 0.1R_E = 47Ω

C_E = --------- = 33.87µF (Select 47 µF)

2πX_CE

PROCEDURE :-

a. To get frequency response:-

The connections are made as shown in the circuit diagram.
Before applying the input signal, the DC conditions are checked by setting V_CC = 10V. V_CE must be close to ½ V_CC ≈ 5V.
Input sinusoidal signal (around 20mV peak to peak ) should be applied using audio signal generator.
Keeping input signal V_inconstant, the frequency of the input sine wave is varied from 100hz to 1MHz in suitable steps while measuring the output voltage for different frequencies. The gain of the amplifier is calculated from these values.
Also the gain in db is calculated and tabulated. The graph of frequency versus gain in db is plotted on a semi log sheet.
Bandwidth is calculated from the frequency response. Also gain bandwidth product is computed.

Tabular Column:-

V_i = ---------- mV

Table 1
Frequency (hz)	V o	A v = V o /V i	Gain in db= 20log10 Vo/Vi
100Hz1MHz

Frequency Response Curve:-

Mid freq. region

3dbBandwidthf₂Frequency in HzGain in dbf₁Low freq. regionHigh freq. Region

b. To measure input impedance Z i :-

V_iVo

**Figure 2**

DRB is connected as shown in the circuit and let the resistance of DRB be zero.
Input sine wave (20 to 50mV peak –peak) is applied keeping the frequency in the mid band region.
Output voltage V_o is measured.
Resistance on DRB is increased till V_o reduces to half of its value. The DRB value gives the input impedance Z_i of an amplifier.

c. To measure output impedance Z o :-

ViVo

**Figure 3**

DRB is connected as shown in the circuit and let the resistance of DRB be high in terms of several kilo ohms or mega ohms.
Input sine wave (20 to 50mV peak –peak) is applied keeping the frequency in the mid band region.
Output voltage V_o is measured.
Resistance on DRB is decreased till V_o reduces to half of its value. The DRB value gives the output impedance Z_o of an amplifier.

Results:-

a. Band width (BW) = --------- Hz

b. Midband gain = A_mid= ---------------

c. Gain Bandwidth Product = BW X A_mid = ------------------

d. Input Impedance = ---------Ω

e. Output impedance = -------- Ω

b. FET Amplifier:-

Circuit Diagram:-

**Figure 4**

Design :-

From data sheet of BFW10/11

I_DSS = 10mA

V_P = -3V or -4V

Let I_D = 2mA

a. To find R S :-

From equation for current

I_D = I_DSS(1 – V_gs/V_p)²

I_D/I_DSS = (1 - V_gs/V_p)²

Substituting for I_D, I_DSSandV_P, we get

V_gs = - 1.65V

V_gs = I_DR_S

R_S = V_gs/I_D = 0.820KΩ = 820Ω

Choose R_S = 1KΩ

b. To find R G :-

I_gs R_{G =}V_gs

R_G= V_gs/I_gs = 1.65/1µA = 1.65MΩ

c. To find R D :-

Applying KVL

V_DD= I_D R_D + I_DR_S+V_DS

V_DD– V_DS – I_DR_S

R_D = ----------------------------

I_D

10 – 5 – 2

= --------------------- = 1.5KΩ

Let R_D= 1.5KΩ

d. To find C S and C C :-

Let f = 100Khz

X_cs = 1/100 R_S = 1000/100 = 10Ω

X_cs = ---------------

2πfC_s

C_s = ----------------------------- = 0.15µF (Select 0.22µF)

2πX100X10³ X 10Ω

PROCEDURE:-

Same as BJT amplifier. V_DS should be ½ V_DD≈ 5V as DC Condition.

Tabular Column:-

V_i = ----------------- mV

Table 2
Frequency	V o	A V = V o /V i	Gain in db = 20 log 10 V o /V i
100hz 1Mhz

Results:-

a. Band width (BW) = --------- Hz

b. Midband gain = A_mid= ---------------

c. Gain Bandwidth Product = BW X A_mid = ------------------

d. Input Impedance = ---------Ω

e. Output impedance = -------- Ω

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This work is licensed by Gowra P S under a Creative Commons Attribution License (CC-BY 3.0), and is an Open Educational Resource.

Last edited by Gowra P S on Nov 17, 2009 12:19 am US/Central.

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