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The and its Usage

Consider a two-stage amplifi er system, as shown in Fig. 1 . Each amplifi er provides an increase of the . This effect is

referred to as the power , A p , of the amplifi er. This means that the signal output power from an amplifi er is greater than its signal input power. This may be expressed as:

P1 P2 P3 AP1 AP2 i/p o/p power power

F i g . 1

P P A out or o p P P in i

For the system shown, if the input power P 1 is 1 mW, and the gains of the two amplifi er stages are 10 times and 100 times respectively, then

the fi nal output power, P 3 , may be determined as follows.

P2 App1 ;soPA2 11 P P1 3 therefore, P2 1010 10 mW P3 Ap2 ; so PA322p P watt P2 therefore, P 100 10 103 1 W 3

From these results it may be seen that the output power of the system is 1000 times that of the input. 1

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In other words, overall signal power gain,

P 1 A 3 1000 p P 103 1

or, overall signal power gain,

AAA ppp12 In general, when amplifi ers (or other devices) are cascaded in this way, the overall gain (or loss) is given by the product of the individual stage gains (or losses). Note: The effi ciency of any machine or device is defi ned as the ratio of its output power to its input power. However, this does NOT mean that an amplifi er is more than 100% effi cient! The reason is that only the signal input and output powers are considered when quoting the power gain. No account is taken of the comparatively large amount of power injected from the d.c. , without which the amplifi er cannot function. In practice, small signal amplifi ers will have an effi ciency fi gure of less than 25%. Power amplifi ers may have an effi ciency in the order of 70%. It is often more convenient to express power gain ratios in a logarithmic form, known as the Bel (named after ). Thus a power gain expressed in this way is:

P A logo Bel p P i

where P i input power; Po output power; and to the base 10 are used. The Bel is an inconveniently large unit for practical purposes, so the decibel (one tenth of a Bel) is used.

P Hence, A 10 log o decibel (1) p P i

The unit symbol for the decibel is dB. For the two-stage amplifi er system considered, the power gains would be expressed as follows:

Ap1 10log 10 10 dB Ap2 10log 100 20 dB and A 10log 1000 30 dB p

Note that the overall system gain, A p , when expressed in dB is simply the sum of the individual stage gains, also expressed in dB.

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Worked Example 1 Q A communications system, involving transmission lines and amplifi ers, is illustrated in Fig. 2 . Each section of attenuates (reduces) the signal power by a factor of 35.5%, and each amplifi er has a gain ratio of 5 times. Calculate the overall power gain of the system as (a) a power ratio, and (b) in .

i/pA AA o/p

F i g . 2

A

P (a) For each line, o 0. 355 Pi so, total loss 0..... 355 0 355 0 355 0 355 00. 1 59

P For each , o 5 Pi so, total gain 5551 25 therefore,, overall gain 1125 0. 0 59 2 times Ans

(b) For each line, is

10log 0 . 355 4 . 5 dB* so total attenuation45. 4 18 dB For eeach amplifier, gain105 log 7 dB so total gain7321 dB

Hence, overall gain of the system 21 1 8 3 dB Ans

* Note that a loss or attenuation expressed in dB has a negative value; whereas a gain has a positive value. A further point to note is that if the gains and of the system had originally been expressed in dB, then the calculation would simply have been as follows:

Overall system gain (3 7) (4 4.5) 3 dB Ans

The above example illustrates the convenience of using decibel notation, since it involves only simple addition and subtraction to determine the overall gain or attenuation of a system. It has also been shown that a gain of 2 times is equivalent to a power gain of 3 dB (more precisely, 3.01 dB). It is left to the reader to confi rm, by using

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a calculator, that an attenuation of 2 times (i.e. a gain of 0.5) is equal to 3 dB. This fi gure of 3 dB will frequently be met when dealing with the curves for amplifi ers and other frequency- dependent circuits, such as series and parallel tuned circuits. In order to gain a ‘ feel ’ for power gains and losses expressed in dB, the following list shows the corresponding power gain ratios.

⎪⎧10 000 times 40 dB ⎪ ⎪ 130 000 times dB ⎪ ⎪ 100 times 20 dB gains⎨ ⎪ 10 ttimes 10 dB ⎪ ⎪ 23 times dB ⎪ ⎩⎪ 10 times dB

⎪⎧ 05. times 3 dB ⎪ ⎪ 01. times 10 dB ⎪ ⎨ losses⎪ 001. times 20 dB ⎪ ⎪ 000. 1130 times dB ⎪0. 0001 times 40 dB ⎩

Worked Example 2 Q (a) Convert the following gain ratios into dB. (i ) 250, ( ii ) 50, ( iii ) 0.4 (b) Convert the following gains and losses into ratios. i ) 25 dB, (ii ) 8 dB, (iii ) ؊ 15 d B)

A

(a) (i) Ap 10 log 250 24 dB Ans (ii) Ap 110507 log dB Ans (iii) A 10044 log . dB Ans p

(b) (i) 2510 log (ratio) dB so 2.5 log (ratio)

therefore, (ratioo) antilog 2.5361 times Ans (ii) 810 log (ratio) dB 08.o log (ratio) therefore, (ratio) antilog 0.863. times Ans (iiii)1150 log (ratio) dB 11.5 og (ratio) therefore, (ratio)antilog (1..5 ) 0 032 Ans

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Although the decibel is defi ned in terms of a power ratio, it may also be used to express both voltage and current ratios, provided that certain conditions are met. These conditions are that the resistance of the load is the same as that of the source, i.e. the conditions for maximum power transfer. Consider such a system, whereby the two resistance

values are R , the input voltage is V 1 , and the output voltage is V 2 volt. Let the corresponding currents be I 1 and I2 .

V 2 V 2 P 1 watt and P 2 watt 1 R 2 R ⎛ ⎞ ⎜V 2 R ⎟ gain10 log⎜ 2 ⎟ ⎜ 2 ⎟ ⎝ R V1 ⎠ ⎛ ⎞2 ⎜V2 ⎟ 10 log ⎜ ⎟ ⎝⎜ V ⎠⎟ 1

V2 hence, voltage gain, A v 20 log dB V1 2 2 Also, using the fact that PIR11 watt and PIR2 2 watt it is left to the reader to verify that:

I current gain, A 20 log 2 dB (2) i I 1

As shown in Further Electrical and Electronic Principles , Chapter 2, a tuned circuit used as a pass-band or stop-band fi lter has a bandwidth. The same concept also applies to a.c. amplifi ers. In the latter case, the frequency response curve for a voltage amplifi er would be similar to that shown in Fig. 3 . The bandwidth is defi ned as that range of frequencies over which the voltage gain is greater than, or equal to,

Avm /2 , where A vm is the mid-frequency gain. A similar response curve for the amplifi er current gain could also be plotted.

Now, power gain voltage gain current gain

or, A p A v Ai

gain

Avm

A vm− √2

Bandwidth

f1 f2 f(Hz)

F i g . 3

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The cut-off frequencies, f 1 and f 2 , defi ne the bandwidth, and at these frequencies, the current and voltage gains will be:

Avm Aim Av and Ai respectively 22 A AAvm im pm thus Ap 222 These points on the response curve are therefore referred to as either the cut-off points, the half-power points, or the 3 dB points.

Worked Example 3 Q An amplifi er is fed with a 50 mV, 200 µ A signal. The amplifi er has a voltage gain of 75 times and a current gain of 150 times. Determine (a) the voltage, current and power gains, expressed in dB, and (b) the output voltage, current and power. You may assume that input and output resistances are the same value.

A

3 6 V1 50 10 V; I1 200 10 A; A v 75; Ai 150 (a) A v 20 log 75 dB 37. 5 dB Ans

Ai 20 log 1 50 dB 43 . 5 dB Ans Ap AAAvi75111 50 250 times so, A 1110405 log 250 dB. dB AAns p (b) VVA volt 50 1 03 75 3. 75 V Ans 2 1 v 6 II2 1 Ai amp 20011 0 55030 mA Ans PVI watt375.. 00311 25 . mW Ans 222

Worked Example 4 Q An amplifi er has a voltage gain of 10 times at a frequency of 150 Hz; 60 times between 2 kHz and 12 kHz; and 15 times at 35 kHz; Determine (a) the voltage gain, expressed in decibel, for each case, and (b) the voltage gain at the limits of its bandwidth.

A (a) at 11 50 Hz: A v 20 log 0 20 dB Ans at mid-frequencies:A v 20 ll o g 6 035. 6 d B Ans at 35 kHz:A 20 log 1 5 23. 5 dB Ans v

(b) The limits of the bandwidth occur when A v is 3 dB down on the mid- frequency value, or AAvvm/2 . Therefore, AAvvm33563326.. dB Ans

A vm 60 or, A v 42. 4226 times 2 2 hence, A 20 log 42.426 32. 6 dB Ans v

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The decibel is defi ned in terms of a power ratio. Thus, the power output of a device cannot be quoted directly in dB. For example, to say that an amplifi er has an output power of 20 dB is completely meaningless. In this case it only makes sense to say that the amplifi er has a power gain of 20 dB. However, actual values of output power may be expressed in decibel form provided a known reference power input is used. In electronic work, this reference power is one milliwatt (1 mW), and powers expressed in this way are designated as dBm. Considering such an amplifi er, it could be said to have a power output of 20 dBm. This would mean that its actual output power was 0.1 W. This fi gure is obtained thus:

P 10 log 2 20 dBm P1 P log 2 2 P1 P 2 antilog 2 100 P 1

3 and since P 1 1 mW, then P 2 100 10 0.1 W or 100 mW.

Supplementary Worked Example 1 Q An circuit has input and output of 8.5 V and 2.4 V respectively. Determine the attenuation, expressed in decibels.

A VVio8.5V; 2.4 V V attenuation 20 logO decibel Vi 24. 20 log 20 log 0.2824 85. so, attenuation11 dB Ans

Supplementary Worked Example 2 Q An amplifi er has a power gain of 40 dB. Express this gain as a power ratio.

A

401 0 log (ratio) decibel 4 log (ratio) antilog 4 ratio 4 so, raatio10 Ans

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Supplementary Worked Example 3

Q For the attenuator pad shown between terminals AB and CD in Fig. 4, calculate (a) the output (V O) and input (V i ) voltages, (b) the voltage attenuation in decibels, and (c) the output power.

A R1 C

270 Ω

r 40 Ω

R V R2 V L i R O Ω 330 Ω 3 100 Ω E1 25 V 330

B D F i g . 4

A

(a) Using Thévenin ’ s theorem and looking in at terminals CD the circuit will appear as shown in Fig. 5 .

E C

R1 270 Ω

R3 Ro 330 Ω r R2 40 Ω 330 Ω

D F

F i g . 5

rR RR 2 ohm EF 1 rR2 40 330 270 40 330  REF 305. 68 RR 305. 68 330 R EF 3 ohm O REF R3 305. 68 330  RO 158. 7

To obtain the Thévenin generator emf EO , consider Fig. 6

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A I R1 C

270 Ω

r 40 Ω R V 2 R3 AB Eo 330 Ω 330 Ω

E1 25 V

B D

F i g . 6

RR() R Rr 231 ohm RRR23()1 Total resistance, 330 600 40 330 600 R 252. 9  E 25 I 1 amp0. 0989 A R 252. 9 VErI volt 25 ( 0. 0989 40 ) AB 1 V 21.005 V AB

Resistors R 1 and R3 act as a potential divider with V AB applied across them, so

R 330 E 3 V volt2051 . OABRR 600 3 1

EO 11.575 V, and the Thévenin equivalent circuit is as Fig. 7

C

Ro 158.7 Ω Ω Vo RL 100

Eo 11.575 V

D

F i g . 7

R 100 V L E volt 11.575 O O RRLO 258. 7 VO 4. 474 V Ans

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Looking in at terminals AB of the original circuit, the resistance ‘ seen ’ R i is illustrated in Fig. 8 , and may be obtained as follows:

RR RR 3 L ohm GH 1 RR3 L 3301 00 270 3301 00  RGH 346. 744 R R 346. 7 330 R GH 2 ohm i RRGH 2 346. 7 330  Ri 169. 08

A G

Ri 270 Ω

Ri R2 330 Ω

R3 RL 330 Ω 100 Ω

B H F i g . 8 So the circuit is equivalent to that shown in Fig. 9 , from which:

A

r 40 Ω Ri Ω Vi 169.08

E1 25 V

B

F i g . 9

R 169. 08 V i E volt 25 i 1 Rri 209. 08 so,. V 20 21 7 V Ans i

V (b) attenuation 20 log O decibel Vi 4. 474 20 log 20. 21 7 attenuatioon13 dB Ans

22 VO 4. 474 (c) PO watt RL 100 P 02. W Ans O

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