ATTENUATOR AND IMPEDANCE TRANSFORMER

Attenuator

OBJECTIVES:

The purpose of an Attenuator is to cause loss of a known dB. An attenuator is a 2-port network consisting of resistors only. Here, you shall study

a) T-network Attenuatorb) Variable t-network Attenuatorc) Variable -network Attenuator

THEORY

T-network Attenuator and Variable T-network Attenuator

The Ra , Rb T-network is to be used as an attenuator. The design equations may be shown to be follows:

Ra = N – 1 R 0 N + 1

Rb = 2N R 0

N2 – 1

Where R0 is the characteristic or image impedance of the T-network and N is the insertion loss ratio measured by V1 / V2 . Or in dB, take 20 log10 N dB.

The same design equation also applies to the variable T-network attenuator on different levels of decibel.

Variable -Network Attenuator

The R1, R2 -network is to be used as an attenuator. The design equations may be shown to be as follows:

R1 = N + 1 R0

N – 1

R2 = N 2 – 1 R0

2N

EXPERIMENT

T-network Attenuator

a) Calculation : The T-network is designed to be a 5 dB attenuator.If R0 = 600 , find Ra and Rb.

b) Measure Ra and Rb as in the T-network. The measurements are compared to the calculation.

c) To verify the internal resistance of the Signal Generator Rg is 600 . Signal generator is set to 10 kHz (say) and V0 = 6 volts (use oscilloscope). A load resistor RL (decade box) is connected and the resistance is varied so that V1 = 3 volts. The RL value is recorded.

d) Attenuator Insertion Loss Measurement

i) For RL = 100, 200, … , 600, … , 1000 and V1 = 6 volts, V2 is measured.

ii) The attenuator insertion loss is calculated in dB or 20 log10 V1/V2 dB. A value of 5 dB is to be procured for all cases of load resistance RL.

Results:

a) Calculation of Ra and Rb :

20 log10 N = 5 N = 1.778

With given value R0 = 600

Ra = N – 1 R 0 = 168.0 N + 1

Rb = 2N R 0 = 987.2 N2 – 1

a) Measurements of Ra and Rb :

Ra = 163.0 and Rb = 960.0

b) With V1 = 3 V, the value of RL = 600

c) With V1 = 6 volts for RL = 0 , the attenuator insertion loss is 5 dB.

RL () V1 (V) V2 (V) dB100 2.97 1.67 5.00200 5.16 2.88 5.07400 8.25 4.59 5.09600 10.25 5.72 5.07800 11.69 6.53 5.061000 12.81 7.25 4.94

For all cases of load resistance RL , the attenuator insertion loss are approximately 5 dB. This is because the T structure resistance are very small.

Variable T-network Attenuator

A 0 to 15 dB variable T-network Attenuator in steps of 5 dB is provided.

a) Calculation : Table is to be filled.

b) The variable t-network attenuator is given having SSDT (single switch double throw) switch

c) The values of Ra1, Ra2, Ra3, Rb1, Rb2 and Rb3 is calculated.All resistors of the variable T-network Attenuator are measured.

d) The insertion loss is measured in dB and verified for 0, 5, 10 and 15 dB.

Results:

dB N Ra Rb

0 1 0 5 1.778 168.0 987.110 3.162 311.6 421.715 5.623 418.8 220.4

d) Calculation:

Ra1 = 168.0 Ra2 = 311.6 Ra3 = 418.8 = 168.0 = 311.6 = 418.8 Rb1 = 565.4 Rb2 = 201.3 Rb3 = 220.4

Measurement:

Ra1 = 140 Ra2 = 300 Ra3 = 415 = 140 = 300 = 415 Rb1 = 530 Rb2 = 200 Rb3 = 220

The values measured and calculated are quite similar to each other.

d) Measurements of insertion loss in dB whereby V1 = 6 V :

dB V1 V2 20 log10 V1

V2

0 6.0 6.0 05 6.0 3.3 5.1910 6.0 1.9 9.9915 6.0 1.1 14.74

Variable -network Attenuator

a) Calculation : Table is to be filled.

b) The variable -network Attenuator having 0 to 15 dB in step of 1 dB. It consists of four -networks connected in cascade. All the resistor values are entered. Connection of the four SSDT switches is shown clearly.

c) Measurement of the insertion loss in dB.A 600 load is connected. The variable -network attenuator insertion loss is measured and verified in dB, namely 0, 1, 2, 3, … ,15 dB.

Results:

a) Calculations :

dB N R1 R2

1 1.12 10600.0 68.22 1.26 5215.4 140.04 1.58 2669.0 284.18 2.51 1394.7 633.5

b) Measurements of the insertion loss whereby V1 = 6 V.

dB V2 20 log V1 V2

0 6.00 01 5.23 1.192 4.70 2.123 4.18 3.144 3.75 4.085 3.31 5.146 3.00 6.027 2.68 7.008 2.38 8.059 2.13 9.0210 1.89 10.0311 1.69 11.0212 1.50 12.0413 1.38 12.7914 1.19 14.0515 1.06 15.06

c) Sketching of the variable -network Attenuator from 0 to 15 dB in step of 1 dB.

Impedance Transformer

THEORY

Consider the following circuit

In the above case maximum power transfer does not occur. The purpose of an impedance transformer is to have impedance matching so that maximum power transfer can take place.

EXPERIMENT

a) Calculation

i) If V0 = 6 volts, Rg = 600 and RL = 600, show that Pmax = 15mW.If V0 = 6 volts, Rg = 600 and RL = 1800, show that PL = 11.25 mW.

ii) Show that the power transfer ratio

T ( f ) = PL = 4 RgRL = 0.75 Pmax (Rg + RL)2

b) An Impedance Transformer

i) An expression for Z (f) is derived, hence its real part R (f) and imaginary part X (f) in terms of L, C and RL.

ii) Then, show that

L = 1 Rg ( RL – Rg ) 2f0

C = 1 RL –Rg

2f0RL Rg

c) Calculation

If the operating frequency f0 is 30 kHz, calculate the values for L and C.The L and C values given in experiment are measured (using an LCR meter )

d) Measurement and calculation of the Power Transfer Ratio.

For frequency f = 5, 10, 15, …. , 50 kHz and V0 = 6 volts. V2 , which is function of frequency, V2 (f) is measured. Clearly it is known that

PL = V22

RL

Pmax = V02

4Rg

t( f ) = PL

Pmax

e) t(f) against f is plotted.

Results :

a) Calculations :

i) If Vo = 6 volts, Rg = 600 and RL = 600 ,

Pmax = Vo2 / 4 Rg

= 36 / 4 (600)= 15 mW

If Vo = 6 volts, Rg = 600 and RL = 1800 ,

Voltage divider, VL = [ Vo / (Rg + RL) ] RL = [ 6 / (600 + 1800) ] 1800

= 4. 5

PL = VL / RL

= 20.25 / 1800 = 11.25 mW

ii) The power transfer ratio

t( f ) = =

=

= 11.25 / 15

= 0.75

b) An impedance transformer

i) Deriving an expression for Z(f),

Z (f) = jL + (1 / jC) // RL = j L - 2 LR LC + RL

1 + jRLC = j L + R L ( 1 - 2 LC ) 1 + jRLC

PL

Pmax

Vo . RL 2 1 (Rg + RL) RL

Vo2

4 Rg

4 Rg RL

( Rg + RL )

= [ RL(1 - 2 LC) + 2 LR LC ] + j[ L - R L2 C (1 - 2 LC)]

1 + 2RL2C2

= RL + j ( L - RL2C )

1 + 2RL2C2

ii) For maximum power transfer,

Z (f) = Rg = RL

Consider the real part :

Rg = RL C = 1 RL – Rg ½ 1 + 2RL

2C2 2f0RL Rg

Then consider the imaginary part :

L - R L2 C = 0

1 + 2RL2C2

Replacing the value of C in the equation above :

L = 1 Rg( RL – Rg ) 2f0

b) Calculations :

If operating frequency fo is 30 kHz,

L = R g ( RL – Rg )2f0

= 600 ( 1800 – 600 ) 2 (3.142) 30x103

= 4.49 mH

C = 1 RL – Rg ½ 2foRL Rg

= 1 1800 – 600 ½2( 30x103 ) 600

= 5.894 x 10-9 nF

d) Measurements and Calculation of the Power Transfer Ratio

Given Vo = 6 volts.

Frequency, f(kHz)

V2(f)(volts)

PL(f)(mW)

t(f)

5 4.52 11.4 0.7610 4.65 12.0 0.8015 4.77 12.6 0.8420 4.92 13.5 0.8925 5.08 14.3 0.9630 5.13 14.6 0.9835 5.0 13.9 0.9240 4.69 12.2 0.8245 4.18 9.7 0.6550 3.63 7.3 0.49

e) Graph t(f) against f is shown below :

DISCUSSION

(i) The Attenuator circuit consists of Resistors only because it absorbs power has no effect on frequency. So therefore, it will not affect the input signal.

(ii) Design of a variable T-network Attenuator having 0 to 50 dB and Ro = 600 .

(iii) The Impedance Transformer consists of L and C only because it does not absorb power, therefore the circuit will have impedance matching, to get the maximum power transfer.

CONCLUSION

In this experiment, we have studied about T-network attenuator, variable T-network attenuator and variable -network attenuator. In both networks, the resistors are arranged symmetrically so that the resistance looking in the circuit equals the resistance loading circuit. We also learned about variable network of T and - network by combining more of these network together.

For impedance transformer, we learned that the purpose of impedance transformer is to have impedance matching so that maximum power transfer can take place.