NOISE INTERNAL NOISE. Thermal Noise

Transcription

1 NOISE INTERNAL NOISE Thermal Noise Shot Noise Frequency dependent noise THERMAL NOISE Resistors in series Resistors in parallel Power Spectral Density EQUIVALENT NOISE RESISTANCE SIGNAL-TO-NOISE RATIO (S/N) TANDEM CONNECTIONS NOISE FACTOR AND NOISE FIGURE MEASUREMENT OF NOISE FIGURE FRIIS FORMULA Noise in a Multi-stage (Amplifier) system Amplifiers in cascade NOISE TEMPERATURE Relationship between noise figure and noise temperature....7 SINAD RATIO QUESTIONS & PROBLEMS ON NOISE & S/N Noise is any undesirable signal energy that interferes with a desired signal. Sources include: Environmental noise: due to cosmic and solar radiation, lightening discharges etc. Man made noise: from electric arc discharges from spark plugs, drills, motors, switching lights on and off, mains hum, etc. Internal noise: This type of noise is inherent in all electrical and electronic circuits, and will be considered in some detail because ultimately it places a limitation on the performance of electronic circuits from the point of view of limiting the smallest desired signal that can be distinguished from noise. A random variable has a finite rms. value but its average (or dc) value is zero. All formulas referring to random noise apply only to the rms. value of such noise, not to its instantaneous value which is unpredictable, although unlikely to be more than 10 times the rms. value. INTERNAL NOISE There are three major types of internal noise: Thermal noise, Shot noise, Frequency Dependent noise. Thermal Noise Due to the random thermal interaction between the free electrons and vibrating ions in a conductor. It is dependent on temperature and called thermal noise (or Johnson noise after the first thorough investigator in 1928). Resistors are the major contributors to thermal noise, but it also exists within all other electronic devices. The maximum available power of this generated noise is given by P noise = ktb watts, where k = Boltzmann s constant = 1.38 * j/k T = The resistor temperature in degree Kelvin (K) = 0 C Eqn1 /tmp/jfile.doc 1 11/04/2008

2 B = The bandwidth (in Hz) of frequencies involved This simple law is based on observation. It means that a conductor can be considered as a generator of electrical energy by virtue of its being at a finite temperature. The power spectrum density, which is the average noise power per Hertz of bandwidth is constant and equal to kt watts/hz. (This is the noise power per unit bandwidth available at the terminal of a resistance R and temperature T). The same noise power exists in a given bandwidth regardless of the centre frequency i.e. gives the same noise per unit bandwidth anywhere in the spectrum and is called white noise (in optics, white light contains all colour frequencies). The noise power is directly proportional to the bandwidth, so it is advisable to limit the bandwidth of a receiver to the smallest usable value. In communications the impact of noise is of special importance at receivers where generally the strength signal is very small and comparable to the noise. (e.g. noise power of 4 * W and the received signal power at the same point of 5 * W). The noise power can be represented by a noise voltage source having an rms. voltage of v n. Using the equivalent circuit of a resistor as a noise generator this resistor noise voltage may be calculated. R Assume the load R L is noiseless and is receiving the maximum noise power generated by R. For maximum power transfer, R L must be equal to R and, P noise = v n2 /4R Eqn2 Equate Eqn1 and Eqn2 to find the rms. noise voltage v n generated by a resistor R Noise v n = 4kTBR Eqn3 Source The thermal noise associated with any device is a direct result of this inherent resistance and to their composition. Eqn3 applies to copper wound resistors, with all other types exhibiting somewhat greater noise voltages. Different types of resistors of equal resistance value produce different noise levels, which give rise to the distinction low noise resistor. Standard carbon resistor are the least expensive but the most noisy. Metal film resistors offer a good compromise in the cost/performance comparison and can be used in all but the most demanding low-noise designs. The ultimate noise performance (lowest noise generated, given by Eqn3) is obtained with the most expensive and bulkiest resistors, the wire wound resistors. But these might not be suitable for RF applications. The thermal noise contribution of a transistor is found by considering the input resistance of the transistor (h ie in the case of a common emitter configuration) in parallel with all other resistors at the input of the stage which the source sees, i.e. R source //R bias //h ie equivalent noise resistor. As a general rule equation Eqn3 is used as a guide line without taking into account the type of resistor, unless stringent design requires otherwise. Example An Amplifier operating over the frequency range from 18 to 20 MHz has a 10 kohm input resistance. What is the rms. noise voltage at the input to this amplifier if the ambient temperature is 27 degrees C? Answer v n = 4kTBR = 18.2 microvolts. It would be futile to expect this amplifier to handle signals unless they were considerably larger than 18.2 microvolts. Shot Noise R L The major contributor of transistor noise is called shot noise, it exists in all amplifying devices and virtually all active devices. When driving a speaker excessive shot noise sounds like a shower of lead shot falling on a metallic surface. /tmp/jfile.doc 2 11/04/2008

3 Shot noise is due to the particle nature of the current carriers in all semiconductors. The current carriers, even with DC where the current is assumed to be constant, are not moving in exactly steady continuous flow since the distance they travel is somewhat different for each carrier due to their random motion. Although the average dc current is constant minute variations occur which appear as a randomly varying noise current superimposed on the DC. Except for diodes, there is no formula to calculate the exact shot noise current for a transistor (or other device) - the shot noise is caused by the currents in the e-b and c-b diodes. The user must refer to the device data sheet for the shot noise characteristics. These often specify an equivalent noise resistor, i.e. a resistor value for a fictitious resistor that produces the same amount of thermal noise as the device shot noise when applied to v n = (4kTBR) This equivalent input-noise resistor (assumed noiseless) together with its noise source are then included as noise sources at the input of the device. For a diode the shot noise current is calculated exactly by i n = (2eI dc B) where, i n = rms. shot noise current e = charge of an electron = 1.6 * C I dc = dc diode current B = Bandwidth of system Shot noise increases with dc bias current except in MOSFETs where it is independent of DC bias. Frequency dependent noise There are two components: At frequencies below 1kHz, excess noise or flicker noise or 1/f noise occurs. It is inversely proportional to frequency and directly proportional to temperature and dc current levels. It is not very well understood and occurs in bipolar junction transistors and FETs. At high frequencies device noise increases rapidly near the device high frequency cut-off. These High and Low frequency effects are fairly unimportant in receivers, as the critical stage (front end) is usually working well above 1 khz and below the device high frequency cut-off. The overall noise versus frequency curve for semiconductor devices have a bath-tub shape Semiconductors possess inherent resistances, they generate thermal noise in addition to shot noise. The noise characteristics provided in manufacturers data sheet usually takes into account the effect of both these types of noise. THERMAL NOISE Equations 2 & 3 show that the thermal noise properties of a resistor (or conductor) may be represented by a voltage source equivalent circuit or a current-generator equivalent circuit for thermal noise. Resistors in series Consider two sources of thermal noise in series, for example two resistors. Since two resistors in series can be replaced by a single resistor R = R1 + R2 it follows that v n 2 = 4kTBR = 4(R1 + R2) ktb = v n1 2 + v n2 2 This shows that the resultant noise voltage squared is equal to the sum of the squares of the individual noise voltages i.e. v n = (v n12 + v n22 ). Note that just summing the noise voltages v n1 + v n2 would have given the wrong result; for this reason it is usually best to work in terms of noise voltage squared - or noise power. The argument can be extended to any number of resistors in series, giving the resultant noise voltage squared by /tmp/jfile.doc 3 11/04/2008

4 v n2 = 4(R 1 + R 2 + R 3 + ) ktb Resistors in parallel Here it is easier to work in terms of the conductance G and the current-generator equivalent circuit. Since two conductances in parallel can be replaced by single conductance G = G1 + G2, then I n 2 = 4G p ktb = 4(G 1 + G 2 )ktb = I n12 + I n2 2 The argument can also be extended to any number of resistors in parallel. If it is desired to obtain the voltage-generator equivalent circuit for resistors in parallel, the equivalent parallel resistor R p is used where 1/R p = 1/R 1 + 1/R 2 and V n2 = 4R p ktb Example Two resistors, 20 kohm and 50 kohm, are at room temperature (290 K). Calculate, for a bandwidth of 100 khz, the thermal noise voltage (a) for each resistor (5.66µv & 8.95µv) (b) for the two in series (10.59 µv) (c) for the two in parallel (4.78µv) Power Spectral Density Power spectral density (PSD) = kt W/Hz Example: At room temp (17 degrees C), T = = 290 K so that we have PSD = 1.38 * * 290 = 4 * W/Hz For a band width of 1 MHz the available noise power is PSD * BW = ktb = 4 * * 10 6 = 4 * Watts Consider a received signal of 1µ Volt in a 50 Ohm impedance. Available power = V 2 /4R = (10-6 ) 2 /4 * 50 = 5 * Watts. This is the same order as noise power S/N = 1.25 Any amplification after this will not help, since it will amplify both noise and signal. EQUIVALENT NOISE RESISTANCE R g Generator Outout Impedance R i Equivalent Noise Source V n R n To Amplifier Amplifier Input Impedance the output impedance of the signal source or generator. Noise originating in a device is represented at the input of the device by means of a virtual resistance R n which would generate the same noise at room temperature. The actual device is then considered noiseless. The noise is treated as if it is fed into the input of the device after the noise which has been introduced by the input impedance of the device and For example using: R n = 400 Ω, R g = 50Ω, R i = 600 Ω, B = 10 khz, T = 290 K, Vg = 1 * 10-6 V gives v n = 4kTB(R th + R n ) = 0.267µV (this thus contains noise from Rg//Ri (= R th ) with Rn in series) The amplifier noise resistance, Rn, should be part of the manufacturer s data for the device. SIGNAL-TO-NOISE RATIO (S/N) S/N = 10 log 10 (P signal /P noise ) also S/N = 20 log 10 (V signal /V noise ) (because we are looking at the same impedance point) /tmp/jfile.doc 4 11/04/2008

5 TANDEM CONNECTIONS If M is the number of communications links in tandem through which a signal of power Ps is being transmitted, the noise powers P n1, P n2, P n3,,p m, will add. Overall S/N = 10log 10 If all P n s are identical then S/N = 10log 10 = 10 log 10-10log 10 (M) If M = 3 links, and each link has S/N = 60 db, overall S/N = 60 -log 10 (3) = db If the S/N ratio of any one link is much worse than any of the others, that link will determine the overall S/N. Example M = 3 links, and two of them have S/N = 60 db, and the third 40 db. Calculate the overall S/N. Which Link must you improve to improve the overall S/N? Answer P n1 = P n2 = P s * 10-6 and P n3 = P s * 10-4 Total noise = P n1 + P n2 + P n3 = P s (2 * ) P s * 10-4 overall S/N = 10log 10 [P s /(P s * 10-4 )] = 40 db NOISE FACTOR AND NOISE FIGURE Noise factor or noise ratio (F) for any network is defined as F = Since any network will add noise, then (S/N) output < (S/N) input, i.e. worse at the output than the input i.e. F > 1. F is the factor by which the amplifier degrades the available S/N at the input. Noise Figure (NF) for any network is defined as NF = 10log 10 F in db Example An amplifier of power gain 20 db has an input consisting of 100 µw signal power and 1 µw noise power. If the amplifier contributes an additional 100µW of noise, determine:(a) the output signal-to-noise ratio (b) the noise factor (c) the noise figure Solution G = 10log 10 (g) = 20 db, i.e. g = log -1 (2) = 100 S out = S in * g = 100 * 10-6 * 100 = 0.01 W N out = N in * g + noise generated by amplifier = 1 * 10-6 * * 10-6 = 200 * 10-6 W F = (S in /N in ) / (S out /N out ) = (100 * 10-6 /1 * 10-6 ) / (0.01/200 * 10-6 ) = 100/50 = 2 NF = 10log 10 (F) = 10log 10 (2) = 3 db In general NF = 10 log 10 = 10 log 10 (S/N) input - 10 log 10 (S/N) output If voltages are used NF = 20 log 10 = 20 log 10 (S v /N v ) input - 20 log 10 (S v /N v ) output MEASUREMENT OF NOISE FIGURE = input (S/N) in db - output (S/N) in db The procedure for determining the noise figure for an amplifier (or other device) is: 1. Measure the source resistance R s and temperature T. Determine the bandwidth BW and calculate N vi = 4kTR s B 2. Adjust the input signal voltage so that it is ten times the noise voltage, i.e. S vi = 10N vi, and measure the output voltage (with meter or oscilloscope). The output reading will be S vo + N vo S vo (because the signal is much bigger than the noise) 3. Set S vi = 0 and measure N vo. 4. NF = 20 log 10 (S v /N v ) input - 20 log 10 (S v /N v ) output = log 10 (S v /N v ) output If the amplifier used has a very narrow BW (a few Hertz) then a spot noise figure is obtained, whereas if the filter BW is large then a broadband or integrated noise figure is obtained. /tmp/jfile.doc 5 11/04/2008

6 FRIIS FORMULA Consider an amplifier with a power gain g. g = S o /S i where, S O and S i are the signal power at the output and input of the amplifier, respectively. The amplifier Noise Factor, F, is given by F = (S i /N i )/ (S o /N o ) = (S i * N o ) / (S o * N i ) = N o /gn i Thus N o = FgN i = gn i + (F - 1) gn i = noise from input + noise introduced by amplifier Here we see that the output noise is the amplified input noise increased by the factor F over what it would be if the amplifier was noiseless. So the fraction of the total available noise at the input, contributed by the amplifier is equal to (F - 1) gn i / FgN i = (F - 1)/F Example The noise figure of an amplifier is 11 db. What fraction of the noise power at the output is contributed by the amplifier? Solution NF = 10log 10 (F) db 11 = 10log 10 (F) i.e. F = log -1 (1.1) = Amplifier contribution of noise power = (F-1)/F = ( )/12.59 = 0.92 i.e. 92% of the power in the output is produced by the amplifier. Example The noise factor of a radio receiver is 15. What is its noise figure? If the input S/N ratio is 35 db, what will be the output S/N ratio? Solution Noise Figure = 10log 10 (15) = db Si/Ni = 35 db, i.e. Si/Ni = 3162 F= (Si/Ni) / (So/No) = 15, thus So/No = 210 or 23.2 db Noise in a Multi-stage (Amplifier) system into each amp. NF 1 NF 2 n 1 n 2 g 1 g 2 Total noise output noise of amplifier 1 = F 1 g 1 n 1 Total noise input to amplifier 2 Consider the diagram which consists of a cascade of two amplifiers with respective available power gains and noise figures of g 1 (= G 1 db), NF 1 db and g 2 (= G 2 db) and NF 2 db. n 1 and n 2 are the equivalent noise powers input = Actual noise from previous stage + equivalent noise input to this stage Noise output from amplifier 2 = Amplified input noise + noise contributed by the amplifier = F 1 g 1 n 1 + ( n 2 ) The overall gain = g 1 g 2 so that the overall noise factor, F t is = F 1 g 1 g 2 n 1 + (F 2-1) g 1 n 2 F t = N o /g 1 g 2 n i = (F 1 g 1 g 2 n 1 + (F 2-1) g 1 n 2 ) / g 1 g 2 n i thus (using the fact that n 1 and n 2 are equal, if the two amplifiers are similar) F t = F 1 + (F 2-1)/g 1 From this it is clear that almost all of the system noise in the output is due to the first stage. In a two stage amplifier it is most important that the input stage introduce as little noise as possible. In many systems a great deal of effort and expense is put into the first stage (using better components, best circuit layout, maximum isolation from sources of interference) to reduce the noise figure. It is not necessary that the later stages be of such a high quality. /tmp/jfile.doc 6 11/04/2008

7 Example A mixer stage has a noise figure of 20 db and is preceded by an amplifier with a noise figure of 9 db and power gain of 15 db. Calculate the overall noise figure. Answer NF 1 = 9 db i.e. F 1 = 7.9 and NF 2 = 20 db i.e. F 2 = 100 G 1 = 15 db i.e. g 1 = 31.6 F t = F 1 + (F 2-1)/g 1 F t = (100-1)/31.6 = overall noise figure, NF t = 10log 10 (F t ) = db Example: Interchange the noise figures of the two stages in the previous example so that the high noise stage is first. What is the new overall noise figure? Answer: 20 db Amplifiers in cascade F t = F 1 + (F 2-1)/g 1 + (F 3-1)/g 1 g 2 + (F 4-1)/g 11 g 21 g 3 + This shows that the first stage is still the most important in determining the overall noise figure of a system. Most effort must go into getting 1 st stage noise as small as possible. NOISE TEMPERATURE Radar receivers in the 1000 MHz range had noise figures in the 1950's ranging from 10 to 15 db, i.e. F = 10 to 30. Most of the noise was thus developed in the system = (F-1)/F. Modern amplifiers have F down to 2 to 3. These devices require low temperatures and cooling systems for their operation. In these cases it is not internal noise that is significant (hence noise figure is not used in these cases) but the external noise introduced at the antenna that provides the major problem. In these cases it is the concept of noise temperature that is relevant. The noise temperature of any given system is defined as the effective temperature of a thermal noise source which if placed at the system input, would produce the same noise power at the system output, the system being considered noiseless. If this effective temperature T e is much less than the equivalent temperature T s of actual noise sources at the system input, then the system introduces effectively no noise of its own. The noise power generated by a source at temperature T s is N s, where N s is given by: N s = kt e B Watts The total noise output, N o, output from an amplifier of gain G is the sum of the amplified noise at the input, GkT s B, plus the internal noise in the amplifier, N n : N o = GkT s B + N n Suppose that N n is represented by a thermal noise source of temperature T e at the system input, then: N n = GkT e B and N o = GkT s B + GkT e B = Gk(T s + T e )B The above equation defines T e, the effective temperature of the system. Relationship between noise figure and noise temperature If the noiseless model is used then the following equation is obtained for the noise figure: F = ( total o/p noise power)/(o/p noise power due only to i/p noise power) F = GkB(T T ) s + e Te = 1 + GkBT T s s /tmp/jfile.doc 7 11/04/2008

8 SINAD RATIO This is used to compare the quality of radio receivers. SINAD stands for Signal plus Noise and Distortion. The distortion is introduced by non linearities in the receiver. All circuits introduce distortion, and so it can be treated along with the noise. SINAD = To obtain the SINAD figure an RF signal modulated by a 400 Hz signal is fed into the input of the receiver. The output power (signal + noise + distortion) is measured. Then a very selective band stop filter is used to filter the 400 Hz signal from the output, and the output power is measured (noise + distortion). The SINAD figure is calculated. QUESTIONS & PROBLEMS ON NOISE & S/N 1. Explain how thermal noise power varies with (a) temperature (b) frequency 2. Thermal noise from a resistor is measured as 4 * W for a given bandwidth and at temperature of 20 o C. What will the noise power be when the temperature is changed to (a) 50 o C? (B) 77 o K? 3. Calculate the rms. noise voltage appearing across a 10-kΩ resistor at room temperature and for an effective noise bandwidth of 10 khz. 4. Three 5-kΩ resistors are connected in series. For room temperature (kt = 4 * j) and an effective noise bandwidth of 1 MHz, determine (a) the noise voltage appearing across each resistor, and (b) the noise voltage appearing across the series combination (c) What is the rms. noise voltage which appears across the same three resistors connected in parallel under the same conditions? 5. The noise generated by a 1000-Ω resistor can be represented by a rms. current source of 4.0 na in parallel with the 1000-Ohm Resistor. Show that this is equivalent to having the noise represented by a root-mean-square noise voltage source of 4 µv in series with the resistor. 6. Explain why inductance and capacitance do not generate noise. Explain also how they control the noise in a circuit. A parallel tuned circuit is resonant at a frequency of 1 MHz and has a Q factor of 50. The tuning capacitor is 200 pf. Calculate the effective noise bandwidth of the circuit. Also calculate, at room temperature, the rms. noise voltage appearing across the circuit. 7. Write brief notes on the sources of noise, other than thermal, which arise in electronic equipment. In particular, describe how the power spectrum density varies with frequency for the various sources. 8. An amplifier has an actual input resistance of 5000Ω and an equivalent noise resistance of 1000Ω referred to the input. What is the total effective input resistance of the amplifier? What is the rms. noise voltage, at room temperature, at the input of the amplifier for a noise bandwidth of 1 khz? 9. A signal generator of internal resistance 50Ω and internal EMF 10 µv is connected to the input of an amplifier that has an effective noise resistance of 1200 Ω and an input resistance of 75Ω. Calculate the S/N ratio at the input for a noise bandwidth of 1 khz and at room temperature. 10.In previous problem, what will be the S/N ratio at the input when the signal source has an internal EMF of 1 µv? 11.Three telephone circuits, each having an S/N ratio of 44 db, are connected in tandem. What is the overall S/N ratio? A fourth circuit is now added, what has an S/N ratio of 34 db. What is the overall S/N in this case? 12.The noise factor of a radio receiver is 15 : 1. What is its noise figure? If the input S/N ratio is 35 db, what will be the output S/N ratio? /tmp/jfile.doc 8 11/04/2008

9 13.Given that the noise figure of an amplifier is 11 db, what is the fraction of the total available noise, at the input, contributed by the amplifier? 14.A mixer circuit having a noise figure of 16 db is preceded by an amplifier having a noise figure of 9 db and an available power gain of 25 db. What is the overall noise figure of the combination? 15.Explain what is meant by (a) the equivalent noise temperature of an amplifier input, and (b) the equivalent noise temperature of a lossy passive network. Given that the equivalent noise temperature of an amplifier is 25 o K, what is its noise factor? 16.The mixer stage of a microwave receiver has a noise figure of 11 db and is preceded by a low-noise amplifier having an available power gain of 20 db and an equivalent noise temperature of 33 o K. Calculate the effective noise temperature of the combination referred to the input of the amplifier. Acknowledgements: The major part of this document is based on the work of Adel Farrag, Donal O Toole and Brian Keogh of the Institute of Technology, Tallaght. /tmp/jfile.doc 9 11/04/2008