A Very Functional Transistor Circuit to Demonstrate Biasing, Voltage and Current Gains, and Frequency Response

Transcription

1 A Very Functional Transistor ircuit to Demonstrate iasing, Voltage and urrent Gains, and Frequency Response Robert J Scoff, P 1 Abstract - Over the last four years The ngineering Technology Department at The University of Memphis developed a very functional and easy to use transistor circuit to demonstrate almost every aspect of transistor circuit operation It is a common emitter amplifier with a standard four resistor bias network with easy to calculate base, emitter, and collector voltages All the resistor values are common, off the shelf values Almost any NPN small signal transistor will work A 2N3904 was used, but the transistor selection is not critical When the circuit is built, the measured bias voltages always compare favorably to the calculated voltages The calculated and measured voltage and current gains at mid band frequency also always compare favorably The gains are high enough to be meaningful, but low enough to not cause oscillations The ode plot of the frequency response is also easy to do and verify There are two low frequency cutoffs which are almost equal and one high frequency cutoff When this circuit is used, the students get to analyze and test it very thoroughly From biasing, to mid band frequency gains, to frequency responses, students get to work with a circuit that really shows all these things y learning how to analyze and build one circuit very well the students gain a firm foundation on which to build their understanding of solid state circuitry Keywords: Transistor, ircuit, Frequency, Gain, iasing INTRODUTION LT S LOOK AT TH IRUIT 75KΩ 300 Ω +24 VD 75Ω 1K Ω 2N3904 β = Ω 1000 uf 01uf V out Notice that this circuit is a standard base biased circuit with all of the resistors and the capacitors being standard values The coupling capacitors can be 2 uf, which causes a slight change in the low frequency cutoff points The 01 uf capacitor is added to give a definite high frequency cutoff The 1000 uf emitter bypass capacitor can be any value large enough to not allow the 75 ohm resistor to affect the cutoff frequency due to the capacitors There are at least two ways to determine the D bias points One way will be shown The circuit is first stripped of all components that do not affect the D voltage levels This is shown in figure # 2 Figure # 1 asy to Use Transistor ircuit 1 Robert J Scoff, P is an assistant professor in The ngineering Technology Department at The University of Memphis, Memphis, Tennessee, <rscoff@memphisedu>

2 FINDING TH D PARAMTRS At this stage the students are required to write the series of equations shown in Figure # 2, and solve for the emitter, base, and collector voltages 75K Ω I VD I 1 + I b I c V I b I e 1K Ω V 2N3904 β =150 V 24 Ω 75 Ω If V = 07 volts, V = V + 07 volts Also, I c = β*i b and I e = I b + β* I b = ( β+ 1) * I b Then an equation can be written for each resistor as follows: ( β+ 1) * I b = V / (24 +75) or 1 I b = V / (99 * ( β+ 1)) 2 I 1 = (V + 07) / (10 * 10 3 ) 3 I 1 + I b = (24 - (V + 7)) / (75 *10 3 ) And for the collector circuit: 4 β*i b = (24 - V ) / 10 3 Figure # 2 D iasing onditions and quations Instead of memorizing the equations for the unknown transistor circuit voltages and currents, the students are expected to write an equation for each resistor in the circuit by applying Ohm s Law Then, any mathematically correct method can be used to solve for the unknown quantities rrors can be easily introduced if an effort is made to determine the currents I 1 and I b If equations 1, 2, and 3 are looked at, it can be seen that equation 1 is an expression for I b and equation 2 is an expression for I 1 If the expressions for I 1 and I b from the first two equations are substituted for I 1 and I b in equation # 3, an expression containing only V as an unknown is derived as shown elow V / (99 * ( β + 1)) + (V + 7) / (10 * 10 3 ) = (24 - (V + 7)) / (75 * 10 3 ) Notice th at there is one equation and one unknown Notice, also, that it was easy to derive This makes life a little easier, as the unknown currents are not even found The value for V comes out to be 133 volts for this circuit This makes V equal to 203 volts (V + 7) Now the only thing left to determine is V Something to observe here is that Ohm s Law, Kirchoff s Voltage Law, and urrent Law were the only concepts that were used efore finding V, it is helpful to determine the D emitter current And it is simply V divided by the total emitter resistance It comes out to be 1343 * 10-3 amps This helps in two ways First, it is necessary to know an expression for I b (which is I e divided by (β+1) ) to find collector current, and I e will be used later in the A analysis Now look at equation number 4 in figure # 4 It can be seen that I b = I e / (β + 1); then: β * I / ( β + 1) = (24 - V ) / 10 3 e And if I e = 133 / 99 and β = 150, then the above equation simplifies to: (150 *133) / (151 * 99) = (24 - V ) / 103 Although the data sheets show that the value of β can be from 30 to 300, a value of 150 was chosen because experience has shown that most of the commercially purchased transistors have a β of about 150 Also, for the purposes of this paper, the D and A values of β are made equal This also eliminates confusion for the student because the student doesn t have to be concerned with another variable, β, or a different value of β for the D and the A models As always, the intention is to make things understandable Solving for V, V = 1065 volts Once the three transistor voltages are known, the circuit can be checked to see if it is in the linear operating range In this case, V is 24 volts, V is 1065 volts, and V is 133 volts This makes V equal to 932 volts Therefore, the circuit is an amplifier with a satisfactory dynamic range When this circuit is built in the lab, the D operating points come close to the above calculated values At this point, it is interesting to note that the base current is never

3 found This is important because the value I b is so much smaller than the rest of the unknowns, that errors can be introduced if extreme care is not taken with the actual calculations A ONDITIONS MID-AND VOLTAG AND URRNT GAINS To determine mid-band voltage and current gains, the T parameter equivalent circuit is used This is shown in figure # 3 β Figure # 3 onverting to T-Parameters i b r e' It is necessary to find the A parameter, r e The formula is derived elsewhere as: r = 25 mv / I (ma) 1 e This is approximately the slope of the forward biased diode curve The base emitter junction is a forward biased diode, so this expression for r e does give a good approximation of the A resistance of the transistor emitter in the T-parameter model Using V = 133 volts and R = 99 ohms, r e is found to be: r e = 25mv / I where I = 133volts / 99 Ω, which gives r e to be 186 Ω Now, to use this T-Parameter transistor equivalent circuit, the circuit is redrawn with the transistor replaced with its T-parameter equivalent The resistors are left in the circuit Since this is an A equivalent circuit that is being determined, something has to be done with the D power supply This is no problem if it is remembered that all D sources look like short circuits to A Then, since it is mid-band voltage and current gains that are being determined, all the capacitors can be replaced by their impedance at the mid-band frequency At the mid-band frequency, the coupling and low frequency bypass capacitors, by definition, have approximately zero impedance There is also a high frequency by pass capacitor in the circuit It is chosen to have a very high impedance at midband ecause of this, it can be removed from the circuit The new V out equivalent circuit, with the transistor replaced by its model, the coupling i in 1KΩ and low frequency by pass capacitors β replaced by shorts, and the high 300Ω *i b frequency by pass capacitor replaced β*i by an open is shown in figure # 4 i b i out efore proceeding, there is an V b 75KΩ 1 r important concept that needs to be e' = 186Ω introduced What does the transistor base circuit look like to the rest of the circuit It can be seen that there is an 24Ω input current, i b, flowing into the transistor There is also a voltage, i 1 ( β+1)*i b V 1, across the base circuit And by Ohm s Law, V 1 equals (β + 1) times i b times (24 + r e ) Now, the Figure # 4 A quivalent ircuit for Mid-band Frequency impedance, Z, is equal to V/I Therefore:

4 Z =V / I = ( β + 1) * (24 + r e' ) * i b i b If the value of β = 150 is used, Z = 3905 Ω This value of Z, which is called the input impedance of the base, will be useful later in this paper It is also written as Z in VOLTAG AND URRNT GAIN ALULATIONS The next step is to determine mid-band voltage and current gains This circuit is simple enough so that finding these quantities is fairly easy, but also shows all the concepts that are necessary to understand how a transistor amplifier works To proceed, the output circuit in figure # 4 is looked at The current β * ib flows through the parallel combination of 1KΩ in parallel with Therefore V out = -β * 667 * i b The negative sign is generated because of the assumed direction of V out Then the input voltage to the transistor, V 1 is looked at Since V = I * R, V 1 = ( β + 1) * ( ) i b Therefore the voltage gain from V 1 to V out is: V V = out - β * i b * 667 = gain V 1 ( β + 1) * (24 + r e' ) * i b This is equal to It can be seen that this is not the complete gain of the circuit, only the gain from the base to the collector To get the gain from the source, Vin, to the base of the transistor, it is necessary to look at the input circuit and replace the transistor with its equivalent impedance This is shown in figure # Ω i in i b V 1 75KΩ 3905 Ω Now it is a simple matter to replace the three parallel resistors with their equivalent resistance of 2707 Ω The circuit now looks like figure # 6 and has a gain of 09 The total gain is then equal to the product of the two individual gains Therefore: V gain (total) = 09*(-2562) = Figure # 5 quivalent Input ircuit of Transistor Amplifier 300Ω V out The next step is to determine the current gain Looking at figure # 4, it can be seen that there three components of current gain The first, and most obvious one is β (in this example, 150), because the collector current is β times i b, and the base current is i b Then, looking at the collector circuit in figure # 4, which is two resistors in parallel, the current gain is found by using Kirchoff s current Law to be: 2707 Ω I gain (load) = 1KΩ / (1KΩ + ) = 1/3 Now, to find the current gain of the input circuit, look at figure # 5 The only current that is amplified is i b y using Kirchoff s urrent Law again, The current gain is: Figure # 6 Simplified Input ircuit Of Transistor Amplifier I gain (input circuit) = 8825KΩ / (8825KΩ KΩ) = 0693

5 The 8825KΩ is the parallel combination of and 75KΩ The total current gain is then the product of the three individual current gains or: Igain (total) = 0693 * 150 * 0333 = 346 FRQUNY RSPONS Up to now, there was only fleeting mention of frequency That was the mid-band frequency Now it is time to look at the frequency response of this circuit Part of the real beauty of this circuit is the way that the 3 cutoff frequencies fall into an easily calculated and measured range Note that the emitter bypass capacitor is so large that it has no measurable effect on the low frequency cutoff This is done on purpose to make the circuit easier to understand Note, also, that if the emitter bypass capacitor were to be taken into account, the voltage gain would change between two limits as the effective emitter resistance changed from 24 Ω to 99 Ω V gain = -667 / 2586 = and V gain = -667 / = -661 This unnecessarily complicates the analysis of the voltage and current gains of the circuit Remember that the original intent of this circuit was to simplify the analysis of a transistor circuit Another intent was to have a circuit that could be easily tested and built As will be seen, the low and high frequency cutoffs are frequencies that are easy to implement with almost any general purpose audio frequency function generator and oscilloscope This continues with the idea of making all the calculations easy and the circuit practical and easy to build and test Knowing that the emitter bypass capacitor is large enough that it will have no effect on frequency analysis, The effect of the two capacitors will be shown DTRMINING TH LOW FRQUNY UTOFF First, it is necessary to determine the cutoff frequency of the input and output coupling networks The first thing to do is to add the coupling capacitors to the network of figure # 4, as shown below in figure # 7: 300 Ω 75KΩ Figure # 7 A quivalent ircuit of Transistor ircuit Showing oupling apacitors i V b 1 r e' β*i b = 186Ω 24Ω ( β+1)*i b 1KΩ β*i b V out The next thing to do is to only look at the frequency effects caused by the input coupling capacitor This circuit and its simplified version are shown in figure# 8 Note that the transistor has been replaced by its equivalent impedance of 3905 Ω Now one of the things that make this such an easy circuit to analyze will be shown When the three parallel resistors in the original model are replaced with its equivalent resistance of 2707 Ω, the new equivalent circuit shows 3007 Ω in series with the capacitor If the capacitive reactance is now set equal to 3007 Ω, the first low frequency cutoff is found to be 24 cycles per second as shown in the following equation: Setting X = -j3007 results in j3007 = -j / (2 * π *f * ) Solving for f (the only unknown) results in f = 24 cycles per second

6 300 Ω K 75K Ω Ω Ω V out 300 Ω Vin 2707Ω V out Figure # 8 quivalent ircuit of the Input of the Transistor Amplifier of Figure # 1 An important thing to do now is to look at the effect of the other capacitor The equivalent circuit of the output circuit is shown in figure # 9 1K Ω β * i b r e' 186Ω 1KΩ V th 24 Ω Figure # 9 quivalent ircuit of the Output of the Transistor Amplifier of Figure # 1 To convert the original model to the simplified model only requires that the Thevinin equivalent circuit be found for everything in the original circuit to the left of the capacitor Since current sources are opened when finding the Thevinin equivalent resistance, the Thevinin resistance is the 1 K Ω collector resistor A quick inspection reveals that the capacitor is in series with 3000 Ω Therefore the cutoff frequency due to the second capacitor is almost exactly the same as for the first capacitor So both cutoff frequencies are approximately 24 cycles per second When both low frequency cutoffs are the same frequency, the actual cutoff frequency shifts upward to 37 cycles per second The relationship that shows this is: (2) f cl f' cl = 2 1/n - 1 Another way of looking at this is to think that the two frequency responses multiply to cause the output at 24 cycles per second to be or ½ This is true because each of the half power points is where the output voltage is down to 707 of the mid band voltage Therefore 24 cycles per second is no longer the half power point, but the quarter power point The half power point is shifted to 37 cycles per second

7 DTRMINING TH HIGH FRQUNY UTOFF The only other capacitor that needs to be taken into account is the01 uf connected across the 2 K Ω load resistor It can just as easily be connected from the collector of the transistor to ground because the impedance of the coupling capacitor is almost zero at the high frequency cutoff point This is shown on the circuit model in figure # 10 β * i b r e' 186Ω 24 Ω 1KΩ 01 uf R th = 667 Ω Figure # 10 High Frequency quivalent ircuit of the Output ircuit and Its Thevinin equivalent V th 01 uf Note that the coupling capacitor has been replaced with a short, and the two resistors are now in parallel Then the Thevinin equivalent circuit was determined, with the 01 uf capacitor being considered the load For finding the cutoff frequency, the value of V th is irrelevant Opening the current source leaves the Thevinin resistance to be the two resistors in parallel When the capacitive reactance of the 01 uf capacitor is set equal to 667 Ω s, The cutoff frequency becomes cycles per second as shown below: Xc = -j667 = -j / (2* π *f*) which gives f = cycles / second The thing to notice about this circuit is that all the analytical results are easy to work with and reproduce The circuit also gives accurate, reliable results when tested The gains are big enough to see easily, and yet small enough that oscillation is not a problem FRQUNY RSPONS The last thing to do with this circuit is to run a frequency response The frequency range of 10 to 100,000 cycles per second will be used because it covers the frequency range where all the calculated cutoff frequencies occur This frequency range works well in practice, although some difficulty in measuring voltages on the oscilloscope occur at frequencies less that 40 cycles per second To do a frequency response, the frequencies that are used are 10, 20, 40, 80, 100, 200, 400, 800, 1000, 2000, 4000, 8000, 10000, 20000, 40000, 80000, and cycles per second Only 17 measurements are required When the calculations are done with a calculator and pencil, many of the numbers are the same and differ by a factor of 10 This helps the students to think about what is actually going on For this paper, the calculations and graphing are done by use of an xcel spreadsheet For the students, it is arranged that the calculations have to be done by hand the first time that a frequency response is generated If xcel is used, the following chart can be generated This is just an example ecause of the versatility of xcel and other graphing programs that are available, there are many ways to do this same thing Freq Xc Xc Ang A1 Ang Ang Hi Hp Lo Tot Phase Total D Abs 22uf 01uf 1 deg 2 2 Pass *Hp pass Out shift gain gain Total Rad rad deg out out Gain

8 hart # 1 xcel Generated hart t o Gra ph Frequency and Phase Shi ft Response s Now that the chart for doing the frequency response is generated, graphs of gain, D gain, and phase shift will be made The graphing function of xcel was used to generate the following three graphs D Voltage Gain Versus Frequency for Transistor Amplifier Frequency Graph # 1 Voltage Gain Versus Frequency for the Transistor Amplifier of Figure # 1 D D Gain Versus Frequency for Transistor Amplifier Frequency Graph # 2 D Gain Versus Frequency for the Transistor Amplifier of Figure # 1

9 Phase Shift in Degrees Phase Shift Versus Frequency for Transistor Amplifier Frequency Graph # 3 Phase Shift Versus Frequency for the Transistor Amplifier of Figure # 1 As shown above, the gain and phase shift versus frequency curves are easy to generate from the circuit parameters What can not be so easily shown is the experimental results that the students obtain When the amplifier is tested and the experimentally obtained frequency response is compared to the analytical results there is usually a close correlation The goal of having the students run an experiment and get results that compare favorably to the calculated results is not only a morale booster, but it usually helps their understanding In this case their understanding of R networks and transistor amplifiers is reinforced Robert J, Scoff, P The University of Memphis 203 Technology uild ing Memphis, TN mail: rscoff@memphisedu Telephone ell Phone: References: (1) Floyd, lectronic Devices, onventional urrent Version Seventh dition, 2005 Pearson, Prentice Hall, Upper Saddle River, New Jersey, Page 936 (2) Floyd, lectronic Devices, onventional urrent Version Seventh dition, 2005 Pearson, Prentice Hall, Upper Saddle River, New Jersey, Page 940