Semiconductor theory predicts that the current through a diode is given by

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1 3 DIODES 3 Diodes A diode is perhaps the simplest non-linear circuit element. To first order, it acts as a one-way valve. It is important, however, for a wide variety of applications, and will also form the starting point for understanding transistors. This lab will require one day. Reading: HH Sections (pgs ) 3.1 Current vs. Voltage Semiconductor theory predicts that the current through a diode is given by I = I s [exp(v/v 0 ) 1], where V is the voltage across the diode, I s is a current scale, and V 0 is a voltage scale on the order of kt/e for absolute temperature T, Boltzmann constant k and electron charge e. We can verify this prediction using the ELVIS 2-wire analyzer. Obtain a 1N914 diode from the supply cabinet. Note that one end of the diode is marked with a bar; this end is the cathode (see Fig. 1). Place the diode into the DUT contacts, with the cathode end in the negative terminal. Run the analyzer from -10 V to 10 V with a 0.5 V step and low gain. Describe the response. Getting a more detailed plot takes some extra work, because of the large range of currents involved. In particular, the low gain setting on the analyzer doesn t give an accurate value for small currents, but the high gain setting doesn t let the current grow large enough to see everything we want. So to get a clear picture, break the plot into three sections: First, run the analyzer from -10 V to 0 V with 1 V steps and high gain. Export this data and paste into your report. Second, run the analyzer from 0 V to 0.5 V with a 0.1 V increment and high gain. Paste this data into your report as well. Finally, run the analyzer from 0.5 V to 1.0 V with a 0.05 V increment and low gain. Paste this data along with the others in your report, and plot all three together on one graph. To compare to the theory prediction, we need to know I s and V 0. To get these, make another plot with just the third section of data, with V > 0.5 V. If you plot this on a semilog scale, it should appear as a straight line. Excel can fit this for you: with this chart selected, go to the Chart menu and select Add Trendline. Chose an exponential function, and then Figure 1: The diode circuit element. 3-1

2 3.2 Diode Drop 3 DIODES go to the Options tab and select Display equation on chart. Click OK, and Excel should produce a line through the data. From the parameters in the equation, you can read out I s and V 0. Record these in your report. Is V 0 on the order of kt/e as expected? Now that you have I s and V 0, construct a theoretical prediction for I(V ) and plot it along with the full range of your data. How well does the prediction agree? 3.2 Diode Drop Most often, it is not necessary to use the full theoretical model for the diode. Instead, we can approximate the IV relation as { 0 if V < V D I = if V > V D. In practice this means that sufficent current will flow to reduce the voltage across the diode to V D, which is termed the diode drop. Clearly this is an imprecise model, since the IV curve you measured above is not an ideal step function. However, it is quite steep, and the diode drop model is usually adequate. Estimate the diode drop here by determining the voltage drop at which I = 1 ma. (You can do this by reading it off your plot, rather than by measuring it directly.) How much would this estimate change if a current of 2 ma or 5 ma were used instead? The diode drop can also be measured using the diode setting on your DMM. What value does it give for the 1N914 diode, and what current does this correspond to? In general, the diode-drop model is accurate to a few tenths of a volt; if an application requires more precision than that, the better model should be used. 3.3 Power Diodes According to its datasheet, the 1N914 diode can conduct a forward current of about 75 ma before it overheats and breaks. Larger diodes can carry much larger currents. For instance, the 1N4001 diode is rated for 1 A. There is a tradeoff to using higher power diodes, however, because they generally have a slower response. The limit to a diode s response speed is primarily due to the effective capacitance of the diode junction, and higher power diodes have a larger capacitance. This causes high-frequency signal components to leak through the junction as if it weren t there. This capacitance can t be measured directly because of the forward conduction. The circuit of Fig. 2 solves this problem by measuring a pair of diodes in series. Wire up this DUT+ DUT- Figure 2: Circuit for measurement of diode capacitance. 3-2

3 3.4 Rectifiers 3 DIODES circuit using 1N914 diodes. What capacitance is obtained? Compare to the minimum resolvable capacitance, which can be observed by running the meter with nothing attached. Now set up the circuit using 1N4001 diodes instead. What do you obtain for the capacitance of a single 1N4001 diode? 3.4 Rectifiers A common use for diodes is rectification, in which an ac signal is convered to dc. For instance, rectification is needed to convert an ac power line to a dc power supply. Fig. 3 shows the simplest type of rectifier, often referred to as a half-wave bridge. Assemble this circuit using R load = 2.2 kω. Observe the output. Are the amplitude and polarity of the peaks what you expect? Can you see the effect of the diode drop? The full-wave bridge of Fig. 4 is more efficient (and thus more common) than the half-wave bridge. Before building anything, think through the circuit and describe in your report what you expect the output signal to look like. Then wire up the circuit (again with R load = 2.2 kω) and see if you got it right. 3.5 Filtering and Ripple We can reduce the remaining ac component in the output of a rectifier using a capacitor as a filter. Place a 15 µf capacitor in parallel with the load resistor in Fig. 4. Notice that this type of capacitor is polar, meaning that the terminal marked with a negative sign should be held at a more negative voltage than the positive terminal. Be sure to orient the capacitor in your circuit correctly. What is the average output level? The output should also have a small ac ripple; what is the peak-to-peak amplitude of this ripple? What changes if you replace the load with a 1k resistor? The ripple in the output should have a characteristic sawtooth shape. We can understand where it comes from with a simple model: During each ac cycle, the capacitor is alternately charged by the supply and then discharged through the load. In the limit of small ripple amplitude, the voltage is nearly constant during the discharge, so the discharge current is nearly constant as well and is simply given by I discharge = V/R load for dc output voltage V. The ripple amplitude can then 1N4001's 120 Vac 6.3 Vac out load Figure 3: Half-wave bridge. Figure 4: Full-wave bridge. 3-3

4 3.6 Zener Diodes 3 DIODES be estimated using I discharge = C dv dt C V t where V is the variation in V (the ripple amplitude) and t is the discharge time. The charging time of the capacitor is typically short, since the diodes and transformer have a low output impedance. The discharge time is therefore approximately equal to the signal period. Using your circuit s values for R, C, V and t, what ripple amplitude V is predicted, and how does it compare to your observations? Replace the 15 µf capacitor with a 470 µf capacitor, and again compare the ripple to the prediction. This circuit should now be a respectable dc power supply. However, a real power supply would include a voltage regulator to improve its stability in response to changing loads. Diodes can be used for this purpose as well, as explained below. 3.6 Zener Diodes Obtain a 1N746A (an equivalent 3.3 V) zener diode from the cabinet, and install it in the IV analyzer with the cathode on the DUT- terminal. If you try to run the analyzer from -10 V to 10 V, you will immediately get an error. Run it instead from 0 V to 1 V: are there any obvious differences from the 1N914 diode? Now reverse the diode so that the cathode is in the DUT+ terminal, and run the analyzer from 0 V to 10 V on the low gain setting. How is the result different from what you would expect with the 1N914 diode? The conduction observed at around 3 V is termed reverse breakdown. All diodes will exhibit reverse breakdown if a large enough reverse voltage is applied. For regular diodes, this voltage is normally 50 V or more, but in a zener diode, it is designed to be a specified low value. An important use for a Zener diode is as a voltage reference. For example, wire up the circuit of Fig. 5 and measure the output voltage with a VPS setting of 10 V. Due to the nonlinearity of the diode, the output is relatively insensitive to the input voltage and output currents. To demonstrate this, change the input supply voltage to 8 V and note the output change. Also, attach a 1 kω resistor from the output to ground as a load. Based on the observed voltage drop, what is the output impedance of the regulator circuit? This type of circuit is useful any time you need a fixed reference voltage. A fancier version with a temperature stabilized zener diode is available as the LM399 integrated circuit. Figure 5: Voltage reference using zener diode. 3-4

5 3.7 Diode Clamps 3 DIODES Figure 6: 5 V clamp. Figure 7: Zener clamp. Figure 8: Limiter. 3.7 Diode Clamps Another use of diodes is for circuit protection. Various configurations of diodes can limit a signal to a specified range, and thus prevent accidental damage to more sensitive and expensive components. For instance, the the circuit of Fig. 6 prevents the output signal from exceeding 5 V. Construct the circuit and drive it with a 10 V pp sine wave from the function generator. Vary the DC offset from -5 to +5 V and verify that it performs as claimed. Two other types of voltage clamp are shown in Figs. 7, and 8. You don t need to build these circuits, but think about them and explain what they do in your report. 3-5