Lab 3: Semiconductor Diodes

Christopher AgostinoLab Partner: MacCallum Robertson

April 6, 2015

Introduction

In this lab we will gain a working knowledge of the non-linear circuit element known as a diode andthe interesting tasks we can achieve using them in our circuits. In particular, we can use them to filterout the negative parts of AC signals and thus, with capacitors and resistors, convert the signal fromAC to DC. In addition to this, we will explore LEDs as well as Zener diodes, which allow us to createmuch more effective voltage dividers. Perhaps most importantly, however, is the knowledge we will gainregarding graphical analysis, particularly load-line analysis as well as graphical perturbation analysis.We will also explore the causes of the AC component to a DC signal and what factors may affect that.

Lab 3: Semiconductor Diodes

3.1

We obtained a 1N4448 diode and connected the DMM to the diode. We measured the resistance to be623 Ω in the forward direction and it reads exactly 0 Ω in the reverse direction indicating that the DMMis unable to measure any current through the DMM.

3.2

We obtained a plastic-stick-mounted 1N4448 diode and measured the resistance in the forward directionto be 622 Ω. It is not exactly the same as the diode we used in 3.1 but is quite similar. This resistancevalue indicates that it has a forward voltage drop of 622Ω · 1mA = .622V. We then placed our fingers onthe diode and noted that the resistance began to drop. It eventually settled to 596 Ω, corresponding toa voltage drop of .596 V. Next, we placed said diode on a stick in a cup of liquid nitrogen. We measuredthe resistance to be 1.047 kΩ which corresponds to a voltage drop of 1.047 V.

3.3

We played around with the offset adder and noted that it has a range from -8.64 V to 8.71 V. We thenmeasured different currents at relatively low resistances as shown in Figure 1.

1

Figure 1: Output impedance of the breadboard’s offset adder

I determined a line of best fit using the polyfit function in the numpy python package. The lineof best fit’s slope indicates the output impedance of the offset adder which as we can see in Figure 1is 24.016 Ω. This is relatively low compared to the output impedance of the signal generator which isusually set to be 50Ω so the offset adder is indeed a relatively stiff output impedance.

3.4

We created the circuit as shown in Figure 2.

Figure 2: Offset Adder with Ammeter and Diode

We then varied the applied voltage using the offset adder and measured the voltage and current acrossthe diode. We plotted them on a linear graph in Figure 3 and on a log-linear graph in Figure 4.

2

Figure 3: Measured Characteristic Curve for our Diode

Figure 4: Measured Characteristic Curve on log-linear scale

3

3.5

We used the curve tracer to find the characteristic curve for our diode as well as the isat value and thevoltage coefficient for the plastic stick which our diode was on and I plotted it in Figure 5.The isat value we found was 1.13∗10−9A and the voltage coefficient was 22.76. We can use the saturationcurrent value to calculate the diode constant, n.

I(V ) = Isat ∗ (eV ∗enkT )

I’ve left out the factor of 1 as it is negligible for voltages above .1 V.

Figure 5: Characteristic Curves measured by Curve Tracer and with measured values

Looking at the data provided by the curve tracer, we can pick any voltage-current pair to solve for n.I’ll use the pair (1.002 V, 1.385 ∗ 10−1 A). We then plug in Isat and note that kT at room temperatureis approximately 1/40 eV.

I(1.002V ) = 1.385 ∗ 10−1 = 1.13 ∗ 10−9 ∗ e40∗1.002/n

ln(1.385 ∗ 10−1

1.13 ∗ 10−9) = 40 ∗ 1.002/n→ n =

40 ∗ 1.002

ln( 1.385∗10−1

1.13∗10−9 )= 2.15204299168

The lab manual suggests that this value is usually 2 for discrete diodes. Our answer is within 10% of thesuggested value which, for tabletop electronics, seems to be pretty good. We then plotted the data givenby the curve tracer on the same plot as the data we took in 3.4 There is an incredible mismatch betweenour measurements and the measurements given by the curve tracer. This could occur for a variety ofreasons.We then placed the diode on a stick into the liquid nitrogen and performed the same analysis. In theliquid nitrogen, the saturation current was 4.47 ∗ 10−7 A and the voltage coefficient was 2.12. Thecharacteristic curve of the diode in the liquid nitrogen is shown in Figure 6.

4

Figure 6: Characteristic Curve for diode on stick immersed in liquid nitrogen

3.6

We set up the circuit as shown in Figure 7 with the -12 V power supply from the breadboard. The actualvoltage drop across the whole circuit was -12.28 V. We measured the voltage drop across the resistor tobe -11.73 We can use Kirchoff’s loop law to find the current through the diode, assuming that the diodeon a stick has the same resistance as it did in previous parts of the lab, 623 Ω.

−12.28V = −11.73V + I ∗R→ I ∗R = −.55V → I = −.55V/623Ω = .883mA = Idiode

.

Figure 7: -12 V supply with 1 MΩ resistor in series with a reverse biased diode.

3.7

We built the have-wave rectifier as shown in Figure 8.

5

Figure 8: Half-wave rectifier

The scope trace produced by the signal can be found in Figure 9.

Figure 9: Scope trace for Half-Wave Rectifier

The alternating signal shown in the scope trace is the input signal whereas the signal which has onlypositive components is the output signal. The output signal contains only positive components becausethe diode only passes positive signals. The diode will not pass currents from negative applied voltagesunless it exceeds the breakdown voltage at which point it will source as much negative current as possible.

3.8

We added a 1 µF capacitor in parallel with the 47kΩ resistor in the circuit from Figure 8. We noticedthat the signal in Figure 10 was even more rectified than in Figure 9.

6

Figure 10: Scope Trace for Half-Wave Rectifier with Capacitor in parallel

This is because the capacitor combined with the resistor works to smooth the waveform by filteringout parts of the AC signal.

(a) We used the signal generator to double the input frequency and found that the output signal,shown in Figure 11, was even more rectified than that in Figure 9 and 10.

Figure 11: Scope Trace of Output signal with doubled input frequency

This is likely because higher frequency signals have less time to dissipate and therefore smooth thewaveform even more.

(b) We doubled the capacitance of the half-wave rectifier. The output signal is shown in Figure 12 andagain we saw a similar affect as we did in part a.

7

Figure 12: Scope Trace of Output signal when capacitance is doubled

This is because the time constant of the circuit has increased and therefore the discharging of hecapacitor takes a longer time so there is a more constant voltage output in the circuit.

(c) We doubled the load resistor for the half-wave rectifier and the output signal can be found in Figure13.

Figure 13: Scope Trace of Output signal when resistance is doubled

Coupled with the doubled frequency and capacitance from parts a and b, we saw an fairly straight,nearly DC, output signal. This is because the increased resistance decreases the output current ofthe AC signals and they are thus attenuated.

From this, we can clearly see a pattern between the amplitude of the AC component of the signal andthe circuit components

3.9

Using the diode on a stick, we built the circuit as shown in Figure 14 below with the signal generatordisconnected.

8

Figure 14: Load-Line Circuit. Credit: Lab Manual

We measured the output voltage of the circuit across the diode at different voltages as shown in Tables1 and 2.. I then plotted the data given from the curve tracer for the characteristic curve in Figures 15and 16 and used .25 V,, .5 V, .7 V, 1.0 V, 2.0 V as the source voltages, Vo, to plot the load lines for the10kΩ and 1kΩ resistors. Using the formula given in the lab manual,

I(V ) =Vo − VR

Vin(V ) Vout (V) with 10k Resistor Predicted Value from Load-line Analysis (V) Percent Error.250 .225 .249 9.6%.5 .393 .416 5.5%.7 .44 .471 6.59 %1 .473 .503 5.96%2 .525 .55 V 4.5%

Table 1: Voltage across diode with various input voltages

Vin(V ) Vout (V) with 1k Resistor Predicted Value from Load-line Analysis (V) Percent Error.250 .272 .241 12.9%.5 .477 .472 1.1%.7 .541 .55 1.6 %1 .587 .598 1.8 %2 .633 .653 V 3.06 %

Table 2: Voltage across diode with various input voltages

9

Figure 15: Characteristic curves with load-lines for .25 V,.5 V,.7 V, 1 V, and 2 V with 10k resistor

Figure 16: Characteristic curves with load-lines for .25 V,.5 V,.7 V, 1 V, and 2 V with 1 k resistor

3.10

We reconnected the signal generator to the Offset Adder and then connected the Offset Adder directlyto the oscilloscope, removing the diode and the resistors from before. We messed with the offset adder,varying the voltages. From this, it is clear that the offset adder does exactly what its name suggests inthat it simply adds a certain voltage to the input signal and outputs that new combined signal.

3.11

We recreated the circuit in Figure 14 as we did in 3.9 but this time we connected the signal generatorto the input of the of the Offset Adder. We adjusted the signal generator to output a .1 V peak to peak

10

sine wave and adjusted the offset adder to add an extra +1 V DC to the signal. It simply looked like asine wave shifted up by 1 V, as shown in Figure 17.

Figure 17: Graph of .1 V peak-to-peak sine wave offset by 1 V with offset

It is not a rectified signal as we see in 3.7 where the negative half of the signal is cut off. This isbecause the offset adder increases the voltage by +1 V so the whole signal increases by 1 V. The reasonthe signal was rectified in 3.7 was because the diode was not allowing negative parts of the signal to passthrough but in this case, there are no negative parts to the signal as the sine wave oscillates from .9 Vto 1.1 V so the whole signal is able to pass through the diode.

3.12

We then changed the scope settings to measure the AC component of the output signal for the circuitin 3.11 as well as the AC component of the input signal.

Figure 18: AC Component for the DC Offset for the circuit in 3.11 with 10k resistor.

11

Figure 19: AC component for the DC Offset for the circuit in 3.11 with 1k resistor.

As the graphs in Figure 18 and Figure 19 show us, as we increase the Voffset, the peak of the ACcomponent of the signal decreases, in what seems to be, an exponential fashion. It decays more quicklyfor the 10kΩ resistor than for the 1kΩ resistor. In both cases, the Vpeak of the AC component of theinput signal seems be constant, regardless of the voltage offset, which we should expect seeing as howthere are no other circuit components which could really affect the signal and because the larger offset.

3.13

We grabbed an LED and placed it in series with a resistor and a 5 V power supply. First we placed itin reverse and measured the voltage across it with the DMM and found it was 0 V. We then measuredthe voltage drops across the various resistors (100Ω, 300Ω, 3300Ω, 33000Ω, 330000Ω) and across the LEDwith the different resistors. I’ve plotted the the voltage drops for each in Figure 20 against current. Isolved for the current by dividing voltage drop across the resistors by their respective resistances and byKirchoff’s current law, the same current will be present throughout the entire series circuit, so the LEDvoltage drop associated with that resistance will have the same current. I present the results in Table 3.

Figure 20: LED and resistor Voltage drops in series

12

3.14

We used the Curve Tracer to find the characteristic curve for the LED which is plotted in Figure 21.I performed a load-line analysis on this curve,which can be found in Figure 21, to find the operatingpoints for each of the resistances we tested in order to compare them to the LED voltage drops we foundin 3.13. I’ve compared them in Table 3. All of our measurements were within 5 percent of the predictedvalues from the load-line analysis.

Figure 21: LED Characteristic Curve with load lines for different resistors

Resistance (Ω) Vdrop across LED (V) Predicted Voltage (V) Percent Error100 3.674 3.70 0.7%300 3.391 3.39 0.029 %3.3k 3.05 3.06 0.030 %33k 2.695 2.74 1.6 %330k 2.45 2.35 V 4.26 %

Table 3: Voltage across diode with various input voltages

3.15

An approximate expression for the peak to peak voltage of the ripple in the rectifier built in 3.8 can beshown, as was done in lecture, to be

Vpp ≈Vin

Rload · f · Cload

where Vpp is the peak to peak voltage of the AC signal, Vin is the input voltage, Rload is the resistanceof the load, f is the frequency of the input signal, and Cload is the capacitance of the load. This equationmakes sense because as you increase the frequency, you decrease the amount of time that the signal hasto dissipate, making the output signal flatter. Similarly, increasing the resistance and the capacitance ofthe load decreases the peak to peak voltage because it increases the time constant of the circuit whichmeans that it takes longer for the capacitor to charge and discharge, thereby decreasing the amount offluctuations possible. Finally, it makes sense that it is proportional to the input voltage because with alarger input voltage, there is a larger possibility for variations so it makes sense that it would contributeto the peak to peak amplitude of the AC component of the signal.

3.16

I performed a graphical perturbation analysis for .7 V, 1 , and 2 V from 3.12 to find what the peak ofthe AC amplitude should be in 3.12. These can be found in Figures, 22, 23, 24

13

Figure 22: Graphical Perturbation Analysis for load-line with source voltage .7 V

Figure 23: Graphical Perturbation Analysis for load-line with source voltage 1 V.

14

Figure 24: Graphical Perturbation Analysis for load-line with source voltage 2 V.

For .7 V, the perturbation analysis tells us that the amplitude of the A.C. component of the signalshould be 7.5 mV, whereas we measured 20 mV,which is 166% off from the predicted value. For the 1V input, it was predicted to be 4.1 mV, where we measured 14 mV, 241% off from the predicted value.Finally, the perturbation analysis suggests that for the 2 V input, the amplitude of the AC signal shouldbe 1.6 mV, but we measured 8 mV, 400% off from the expected value.It is clear from these errors that the predictions do not take into account the full story, suggesting thatthere is likely another source of electrical noise which we are not considering, likely noise that has leakedin from the offset adder or from the signal generator.

3.17

In this problem we will try to calculate output AC signal amplitudes using small signal impedances ofthe diode. We recall that the impedance of any element is given by the equation

Z =∂V

∂I

We can look at some small interval on the characteristic curve for our diode and approximate it as aline. Thus our equation for the impedance can be approximated as a linear one where

ZDiode =∆V

∆I

We then consider differences between the operating points at positive and negative perturbations aboutsome point along the characteristic curve.

ZDiode =V+ − V−I+ − V−

For any equilibrium point, we can thus analyze our circuit linearly, as a voltage divider where

Vout =ZDiode

Zoffsetadder +R+ ZDiode· Vin

15

However, we found earlier that the impedance of the offset adder is small relative to other circuitcomponents, so we can rewrite the equation as being

Vout ≈Vin · ZDiode

R+ ZDiode

and the values for V+out and V−out perturbed positively and negatively from some DC and AC inputvoltages are given by the equations

V+out =VDC + VAC

R+ ZDiode· ZDiode

V−out =VDC − VAC

R+ ZDiode

The peak to peak output AC voltage is then defined as the difference between V+out and V−out whichcan be shown to be

Vout peak−peak AC =Zdiode

R+ Zdiode· Vin peak−peak AC

where in 3.16 we used .05 V as the peak AC Signal, so it makes sense that the peak-to-peak input ACcomponent is 0.1 V for this analysis. I performed this analysis for load-lines with .5 V,.7 V, 1.0 V and± .05 V at the intersections. Doing so finds the same intersections as in part 9 but looking at the points±.05 V away yields an impedance of 806 Ω for the 1 V load-line, 1668 Ω for the .7 V one, and 4887 Ω forthe .5 V. Using .1 V as our Vin peak−peak AC we can solve for the Vout peak−peak AC , giving us a bettervalue for the AC component’s amplitude.For the different load-lines, we use our equation

Vout peak−peak AC =Zdiode

R+ Zdiode· Vin peak−peak AC

and input our values of 10kΩ for R and 806Ω for the Zdiode. This data is thus displayed in Table 4 aswell as plotted as the thicker line in Figure 18.

Vin(V ) Zdiode(Ω) AC Amplitude (mV) Measured Amplitude (mV).5 V 4887 7.4 33.7 V 1668 14.3 201.0 V 806 32.8 14

Table 4: Estimating AC component using small signal impedance

From this analysis it seems clear that the larger the impedance the larger the output voltage, causingthese estimations to act sort of as a voltage divider which overestimates the voltage output. Thus itoverestimates the AC component of the voltage for higher resistances.

3.18

We used the curve tracer to get the characteristic curve for the Zener diode and I have plotted this curvein Figure 25.

16

Figure 25: Characteristic Curve for Zener Diode

We then built a circuit that would reduce the 12 V input to a 6.2 V output, using a resistor in serieswith a Zener diode which the circuit in Figure 26 shows.

Figure 26: Zener Diode used to create voltage divider

We checked various resistances until we found one that created the proper voltage drop. We used theequation I = (Vo − V )/Rs and used 15 mA for the current. The power supply actually puts out 12 V sothe voltage drop across the resistor should be 12 V-6.2 V so that the output of the circuit will be 6.2 V.R = 5.8V/15mA = 386.67Ω. Since this isn’t a normally available resistance, we grabbed a 390 Ω resistorand used that in our circuit. We measured the output voltage of the circuit to be 6.102 V, which is 1.5

17

% off from the intended value, which is pretty good for our purposes.

3.19

We opened up MultiSim and opened the BiasDiod schematic. We set VAC to .00001 and ran an ACAnalysis on the circuit. The graph of the input (larger) and output(smaller) signals is shown in Figure27.We then changed the VAC to 1 VAC and reran the simulation. When looking at the Fourier Analysis,we saw frequency doubling, in that we saw a spike in the second harmonic of the waveform. The signalsare plotted in Figure 28 and the gross distortion of the output signal is very obvious. As we increasedthe VAC, the amplitudes of the higher harmonics increased and at 2 V, the amplitude of the secondharmonic was about 1/2 of that of the first one. We decided that significant frequency doubling occurswhen the second harmonic has an amplitude of at least 1/5 that of the 1st one, which we saw occur inthe Fourier Analysis when the voltage was approximately .5 VAC.The harmonics are never stronger than the fundamental. I think that frequency doubling occurs becausethe diode is a non-linear circuit component and reacts in a way that alters the input frequencies andadds in different frequencies and we are able to see the ones which correspond to harmonics becausethey are the harmonics of the original waveform. It thus makes sense that higher alternating voltagescan generate higher amplitude second harmonics. Current responds exponentially to a diode’s Voltage.If we look at the Taylor Series expansion for the exponential function

ex = 1 + x+ x2/2 + x3/6 + ...+ xn/n!

we can see that, for small x, or Voltages, only the first order terms will play a role. However, as voltagesbecome larger, the square of the voltage which, in this case, is the input sine wave, starts to play a largerrole. One could decompose the input wave into a series of sines and cosines using a Fourier Series, whichlikely is how the signal is generated in the first place and note then that when squaring the signal, youget higher order terms and cross terms which generate the different harmonics that one sees when usingthe Fourier Analyzer.

Figure 27: .00001 VAC input for BiasDiod circuit AC Analysis

18

Figure 28: 1 VAC input for BiasDiod circuit AC Analysis

Conclusion

In this lab we have learned a great deal about diodes. We have learned how to use a curve tracer to findthe characteristic curve of a diode and how to use load-lines to compare our own measurements to thevalues predicted by the characteristic curve. In addition to this, we have learned that, contrary to myown intuition, diode resistance actually increases when temperature decreases. We learned briefly aboutLEDs and about Zener diodes which allow us to make very effective voltage dividers where the outputvoltage is essentially independent of the load resistance. We also learned briefly about how to changea signal from AC to DC and how to smooth the waveform by tuning the resistance, capacitance, andinput frequency. We also learned about offset adders and what exactly they do. We also learned how toapply graphical perturbation analysis to the characteristic curves. It was an interesting introduction tonon-linear circuit elements and the sorts of techniques used to analyze them.

19