ECE_208_Lab3

ECE_208_Lab3 - 3 A SMALL SIGNAL ATTENUATOR INSTRUCTIONAL...

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Unformatted text preview: 3 A SMALL SIGNAL ATTENUATOR INSTRUCTIONAL OBJECTIVES 1. Given a diode, you should be able to make DC measurements of the diode current versus voltage over a specified current range. 2. From the collected ID—VD data, you should be able to use graphical and analytical methods to determine the reverse saturation current (lo) and the ideality factor (n). 3. Using the ideal diode equation, lo, and n, you should be able to compute the dynamic resistance (rd) of the given diode for specificvalues of diode current. 4. You should be able to select the operating points (VD, ID) of a diode in a variable attenuator circuit to obtain various amounts of small signal voltage attenuation and compare theoretical and actual results. 5. You should be able to specify waveform amplitude (small signal) limitations such that the output signal of the attenuator circuit is not observably distorted. 6. You should document your experimental investigation, design and analysis of the results in a clear and concise manner. PRELAB l. The following data has been obtained by measuring diode current vs. voltage using the method given in this experiment. Table 1. Sample ID VS. VD Data l VD (volts) 1 ID (amps) rd—theoretical 0.312 1.30x 10“6 1D 0.375 7.22 x 10—6 %.:th 0.463 4.71 x 10—5 + VD _ 0.515 1.89 x10”: 0.587 1.00 x 10— (a) Plot the data points (VD, ID) on the included piece of semilog paper. Draw the “best” straight line through the points. (b) Calculate nVT and 10 using the technique discussed in Section 2.0 of this experiment. (c) Calculate the values of dynamic resistances, rd, corresponding to the operating points given in Tablel. Enter these values in the table. 2. Using the small signal equivalents of the attenuator circuit as shown in Figure 3(b), derive an expression for rd in terms of R1, R2, and the attenuation ratio, Vow/Vin. 3. Using the circuit of Figure 3(a) and the diode of Table 1, what value of VCC is required if R2 = 3.3K and if VD = 0.587 volts? EE208, Electronic Devices and Design Laboratory 3—1 1.0 INTRODUCTION AND DEFINITIONS In this experiment you will explore the concepts of small signal analysis through the design of a voltage attenuator circuit. Sections 1.1 through 1.3 present the voltage attenuation circuit and describe how a diode may be used to provide variable attenuation. Section 2 provides the theoretical background and equations useful in designing the circuit. Section 3 leads you through the design and testing of your actual attenuator circuit. WHAT IS SMALL SIGNAL ANALYSIS? Small signal analysis is a technique used to analyze non—linear circuits which have reasonably linear operating regions. Small signal analysis of a device assumes: — the device is caused to operate at an operating point in a reasonably linear operating region; — a signal input is applied to the device; — a signal output results; — the output signal is linearly related to the input signal. HOW LARGE IS A SMALL SIGNAL? Small signals may be any size —— so long as the output signal is linearly related to the input signal. HOW CAN YOU TELL IF THE INPUT AND OUTPUT SIGNALS ARE LINEARLY RELATED? A good test for linearity is human observation: - If the input signal closely resembles the output signal, then the signals are proportional and we say that the network is linear for signals of that magnitude. In this case, small signal analysis may be applied effectively. — If the magnitude of the input signal is increased until the output signal no longer resembles the input signal, we say that the output signal is distorted. In this case, small signal analysis is not effective because a sinewave input of that magnitude does not produce a sinewave output. E13208, Electronic Devices and Design Laboratory 3—2 1.1 THE VOLTAGE DIVIDER AS AN ATTENUAT OR Voltage attenuation is a process whereby the magnitude or amplitude of an electrical signal is decreased. An example of a voltage attenuator is shown in Figure l where R1 is fixed and R2 is a variable resistor. Vin Fig. 1. Simple Voltage Attenuator. Analyzing the simple voltage divider gives v = R2 Out R1+R2 Vin (1 ) Notice that Vout is attenuated by different amounts as R2 is varied and that Vout can never exceed Vin. EE208, Electronic Devices and Design Laboratory 3—3 1.2 THE SEMICONDUCTOR DIODE AS A VARIABLE RESISTOR In many attenuator applications it is desirable to change the amount of attenuation electrically rather than mechanically (i.e. by turning the potentiometer knob). The necessary component to implement this concept is a “resistor” whose value can be changed electrically in a known manner. One device which can be used as a variable “resistor” is the diode. To see this, consider the diode current vs. voltage (I—V) plots of Figure 2. Fig. 2 Diode l—V Characteristics (a) Linearized Model (resistor and battery) (b) Diode Equation, iD : Io[exp(vD/nVT) — 1] Figure 2(a) is a sketch of the linear diode model used in a previous experiment and Figure 2(b) contains a plot of the diode equation, ip = Io[exp(vD/nVT) — 1]. Note that VD and ID are DC values whereas VD and iD represent the total (DC + AC) diode voltage and current. Two important concepts of this model are: (1) The choice of a diode model depends upon the current range. That is, the values of R and E depend on the value of current (iD) flowing through the diode. (2) The resistor value, R, of the diode model was found by dividing the change in voltage across the diode (AVD) by the corresponding change in current through the diode (AiD). Thus AVD R: ‘ AiD (2) EE208, Electronic Devices and Design Laboratory 3—4 Fairly large current swings (eg. 0 to 8 mA) were used when displaying the diode I—V characteristic. A line was then fit to the curve and the resistor value was calculated as the inverse slope of the line. Now consider the situation depicted in Figure 2(b). A DC voltage, VD, is applied to forward bias the diode. This voltage results in a DC current flow, ID. Then, a small AC voltage signal, vD(t), is added to this bias. The total voltage, VD, across the diode is VD = VD + VD“) (3) DC AC This small voltage perturbation causes the diode voltage to vary about VD. This in turn causes the current to vary around ID. Refer to AVD and AiD of Figure 2(b). Notice that, in a small region about the Operating point (VD, ID), the curve approximates a straight line. As a result, a SMALL CHANGE in voltage, causes a PROPORTIONAL SMALL CHANGE in current. This example defines the small signal concept. The small change in stimulus, and the proportional small change in response, are referred to as small signals. The small signal concept is used to analyze non—linear devices which have reasonably linear operating regions. Let’s allow the amplitude of the AC voltage to get smaller and smaller. AVD and AiD will both decrease. In the limit, the ratio of the AVD to AiD becomes a derivative . Av dv _ , lim _D = ID VD,ID 2 rd 2 dynamic resrstance (4) AvD—>0 AID dlD The dynamic resistance, rd, is defined as the inverse slope of the ip vs. VD characteristic at a bias point (VD, ID). Notice what happens as Point Q, the DC operating point, (VD, ID) is varied. See Figure 2(b). If VD is decreased, the slope of the curve decreases. Conversely, the dynamic resistance will decrease if VD increases since the slope will increase. The diode is the variable “resistor” that we were looking for! EE208, Electronic Devices and Design Laboratory 3—5 1.3 A VARIABLE VOLTAGE ATTENUATOR The variable attenuator circuit that makes use of the dynamic resistance of the diode is shown in Figure 3(a). Fig. 3. (a) Diode Attenuator Circuit (b) Small Signal Equivalent of (a) In Figure 3(b), it has been assumed that vS is a small signal. The amount of voltage attenuation (i.e. Vent/Vin) can be controlled by selecting the proper operating point (VD, ID) for the diode. 2.0 BACKGROUND: DYNAMIC RESISTANCE AND DIODE PARAMETERS 2.1 DERIVAT ION OF rd FROM THE IDEAL DIODE EQUATION. The ideal diode equation relates the diode current, ID, to the voltage across the diode, VD. ID = Io[exp(qVD/nkT) — l] (5) where IO 2 reverse saturation current q 2 electronic charge : 1.6 x 10'19c n = dimensionless ideality factor k = Boltzmann’s constant = 1.38 X 10'23 J/K T = temperature in degrees Kelvin (K) The expression is often written as ID = IO[CXP(VD/T]VT) — 1] (6) where VT 2 thermal voltage 2 kT/q (7) E13208, Electronic Devices and Design Laboratory 3—6 The values VD and ID above are to be interpreted as DC quantities. They specify the operating point of the diode, (VD, ID). If a small signal voltage, vd, is added to VD, the total voltage applied to the diode is VD = VD + Vd (8) The varying voltage results in a varying diode current, id, and the total instantaneous current is 1D = ID + id (9) Including these small signal variations into the ideal diode equation gives in = Io[CXp(VD/HVT) — 1] (10) The dynamic conductance, gd is a small signal parameter and is defined as the ratio of the differential current to the differential voltage at a specific operating point (VD, ID). Thus, di 1 gd 2 d D ZVIO €Xp(VD/nVT) VD (VDJD) “n T (VDJD) 1 :~~——10exp(VD MVT) (11) WT rearranging Equation 10 we find that To exp(vD/1]VT) : iD + Io (12) Then gd (13) (NW) The dynamic resistance, rd, is defined as 1/ gd or rd 2 dYD (14) (1113 (VDJD) V rd = n T (15) ID + IO By definition, rd is the inverse slope of the ID — VD diode characteristic at the operating point (VD, ID). It is the small signal resistance of the diode at a given DC bias. EE208, Electronic Devices and Design Laboratory 3—7 2.2 CALCULATION OF nVT AND 10 FROM ID VS. VD DATA The diode parameters nVT and 10 can be determined by using static DC measurements of diode current versus diode voltage. Note that n and the calculation of Io are dependent on the current range used. Recall the ideal diode equation in terms of DC values ID : Io[exp(VD/nVT)—1] (16) For VD >> nVT, exp(VD/nVT) >> 1 and the equation can be reduced to ID = 10 CXP(VD/nVT) (17) Taking the natural log of both sides 1 v +1 1 18 M D 11(0) ( ) ln(ID) = This equation can be thought of as an equation of a line y : ax + b, where y 2 map), a : 1 WT pOifltS (V131, 1131) and (Vm, 1oz) , x : VD, and b : ln(Io). Since any line can be defined by two points, consider the DC 1 l ln(IDl) = “TVDI + ln(IO) mam) : WVDZ + ln(IO) (19) T F subtracting and rearranging gives V — V an : D1 D2 lnIDl — lnTDZ Thus, the nVT product can be obtained using two (VD, ID) data points. Once nVT has been determined, the ideal diode equation and a (VD, ID) data point can be used to solve for 10: T I0 = ——*D (21) CXPWD NWT) “1 EE208, Electronic Devices and Design Laboratory 3—8 2.3 SEMI-LOG GRAPHICAL CONSIDERATIONS In the previous section, nVT was determined using just two (VD, ID) data points. The problem with “this technique is that some error is present in any measurement and if one or both of the (VD, ID) points is incorrect,” the nVT calculation could be way off. A better method would be to obtain a number of data points and use some sort of average to minimize any possible error. This “averaging” can be accomplished by plotting the points on semi—log paper and getting two (VD, ID) points graphically. Semi—log graph paper has one axis which is scaled linearly and one which has a logarithmic (base 10) scale. Since the magnitude of ID varies much more than VD, it is appropriate to plot ID on the logarithmic axis. Recalling Equation (18) ln(ID) = -1-——VD +1n(10) (22) WT it can be seen that (VD, ID) data points should plot as a straight line, the slope of which is l/ nVT. Due to measurement error the points won’t be exactly linear and a best “fit” line will have to be drawn. After the (VD, ID) points are plotted and a best “fit” line is obtained, any two points on the line can be used find nVT. Equation 22 can not be used in its present form since it contains natural logarithm terms (base C) while the log axis of the semi—log paper is base 10. The following conversion is required. log10 (x) (23) 1n(X) : 10g10(e) Then Equation 20 becomes VD _VD nVT = log10(e)[————‘——————Z—--~] (24) 10g10 (IDI ) —— 10g10 (1132) Note that this equation is simplified if the (VD, ID) points are chosen such that the ID values are a decade (ie. a factor of 10) apart. Io can be determined using nVT calculated above and any (VD, ID) point on the line. See Equation 21. EE208, Electronic Devices and Design Laboratory 3—9 3.0 EXERCISES 3.1 Determine Diode Parameters A. Acquire Diode Current vs. Diode Voltage Data. Using the circuit shown in Figure 4, adjust VCC to obtain diode currents of from 10—6 to 10“3 amperes. Choose 10 points over the range with one point near each decade (for example: lxlO‘S). Record VD for each ID. ID should be determined by measuring the voltage across R with a digital multimeter. Don’t forget to measure the exact value of R with the digital multimeter. +VCC IDl R210k§2 Fig. 4. Circuit used to measure ID— VD data. B. Plot data and calculate parameters. Plot the ID — VD data collected in the previous section on the sheet of semi—log paper included at the end of this experiment. Draw the “best” straight line through the data points that you can. Then, calculate values of nVT and lo using the method presented in the “Diode Parameters” sections (2.1 & 2.2) of this experiment. C. Theoretical rd values. Using the values of nVT and 10 determined above along with the theoretical dynamic resistance expression (Equation 15), calculate rd values corresponding to each of the measured ID — VD data points. What are the maximum and minimum values of rd for 1D in the range of 1 MA to 1 mA? 3.2 Design the Attenuator The schematic of the variable attenuator is given in Figure 3(a). Let the frequency of the source be 5 kHz and R1 : 10 K9. Choose a value for C1 such that its impedance is negligible at this frequency. Using the maximum magnitude of rd obtained in 3.1.C, determine R; such that the attenuation ratio (Vent/Vin) is 1/2. Then, calculate values of rd to obtain attenuation ratios of 1/5 and 1/10 keeping R2 fixed. Compute the operating points and VCC required to obtain these three values of attenuation. EE208, Electronic Devices and Design Laboratory 3'10 3.3 Determine a Maximum “Small Signal” Input, vim. Use a triangle wave to establish the maximum input signal, Vin, which can be considered a “small signal” over the entire attenuation range (1/2 to 1/10). See section 1.0 for definition of “small signal”. Without changing the amplitude, switch the input waveform to a sinewave. Does the output signal appear to be undistorted by the attenuator over the attenuation range: 1/2 to 1/ 10? Measure the rms value of this input sinewave using the DMM. This will be the rms value of the small signal used to test the attenuator in Section 3.4. 3.4 Test the Attenuator for Small Signals Using the small signal sinusoid determined in 3.3, set the diode current to the currents designed in 3.2 and compare the actual attenuations to the predicted attenuations (1/2, 1/5, and 1/10). 3.5 Investigate distortion of large signals Adjust the source voltage, vs, so that vin is a 10 volt peak—to—peak triangle waveform (5 KHZ). Make a sketch of vout vs. time for. an attenuation ratio of 1/10. Discuss the distortion and its cause. EE208, Electronic Devices and Design Laboratory 3—11 ...
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ECE_208_Lab3 - 3 A SMALL SIGNAL ATTENUATOR INSTRUCTIONAL...

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