OverviewBackground1. Diode Reverse Current2. Zener Diode3. Forward Diode Current4. Diode Rectifier5. Diode ClampPhysics 375, Laboratory 5DiodesOverviewThe purpose of these experiments is to measure the current-voltage curve of a signal diode andzener diode, and to observe the behavior of diodes in circuits.BackgroundA diode is made from the junction of a p-type and n-type semiconductor. The internal potentialsof the semiconductors make the diode have a current to voltage relationship of:For negative voltages the current approaches a small current −I0. For positive voltages the currentgrows exponentially with the voltage. This has the effect of looking like the diode permits currentto flow without resistance for positive voltage (forward direction) but blocks current for negativevoltage (reverse direction). The diode becomes very useful in letting current flow in only oneway. A zener diode is designed to run in reverse direction but the diode’s barrier breaks down at awell defined voltage, so the zener diode can be operated to produce a fixed voltage across thezener diode.1. Diode Reverse CurrentConnect the variable power supply, a silicon signal diode (1N914), resistor, and DMM:Figure 1: Diode Reverse Current MeasurementMake sure the negative terminal of the power supply is connected to ground. Measure the voltagedrop across the resistor for supply voltages of 0.1, 0.2, 0.5, 1.0, 2.0 and 5.0 V. Calculate the cur-rent through the resistor which equals the reverse current through the diode, and the voltageacross the diode which is the difference between the supply voltage and the voltage across theresistor. Plot the current vs. voltage for the diode.II0eeV kT⁄1–()=R =1MΩDMMV+−1N914PHYS375Lab5,p.22. Zener DiodeBuild the circuit in figure 1 with a zener diode (1N750) instead of the signal diode and a 1kΩresistor instead of the 1 MΩ resistor. Set the power supply voltage so that the current is 1.0 mAthrough the resistor (I = V/R), and calculate the voltage drop across the zener diode. Repeat forcurrents of 2 mA, 5 mA, 10 mA and 15 mA. What is the percentage change in VZfrom IZ=5mAto IZ=15mA?3. Forward Diode CurrentDisconnect the ground terminal of the power supply from the negative terminal. Reverse thedirection of the diode, change the resistor and reconnect the ground to the point between the diodeand resistor as shown in figure 2.Figure 2: Diode Forward Current MeasurementConnect the DMM common terminal to ground and the measure the voltage across the resistor .Adjust the power supply so that VR= 10 mV. From Ohm’s law, I = −VR/R. measure the voltageacross the diode VD. Repeat for VR= 50 mV, 100 mV, 200 mV, 500 mV and 1 V. Plot the cur-rent vs. voltage for the diode.R = 100 ΩDMMV+−1N914VDCOMVRPHYS375Lab5,p.34. Diode RectifierThe following circuit is a extension of a differentiating circuit.Figure 3: Diode RectifierSet the function generator for square waves of 1 kHz and 1 V amplitude. Use both channels of theoscilloscope to compare the input signal with the output signal. How does the output comparewith the pure differentiating signal? Remove the 220 kΩ resistor and observe the output again.Try to explain what the 220 kΩ resistor does, using a couple of other values to see a pattern.5. Diode ClampConstruct the following diode clamp circuit.Figure 4: Diode ClampSet the function generator to produce sine waves at 1 kHz with a 10 V amplitude. Observe theshape and peak voltage of the output. Now add a 680 Ω resistor between the diode and the powersupply and observe the output. This is equivalent to a 15 V supply going through a voltagedivider of 2 kΩ and 1 kΩ.33 kΩscopevin0.001 µF220 kΩ1N914scope1kΩ+5