LABORATORY 8 DIODE CIRCUITS

Transcription

1 LABORATORY 8 DIODE CIRCUITS A solid state diode consists of a junction of either dissimilar semiconductors (pn junction diode) or a metal and a semiconductor (Schottky barrier diode). Regardless of the type, the circuit symbol for a diode is as shown in Fig. 8.1a, and the corresponding device in Fig 8.1b Fig. 8.1a Fig 8.1b If V is positive, the diode is forward-biased. Then, the diode can conduct a significant positive current I even though V is a small voltage of typically 0.7 V for the most common diode (silicon diode). If V is negative, the diode is reverse-biased. This negative current is so small that it is often considered to be zero. Thus, the usual function of a diode is to allow current to flow in the direction of the arrow (the forward direction) for positive V s, but not allow any current to flow in the reverse direction for negative V s. Only a small forward bias (positive V) is required to cause a diode to conduct a significant current I, and the less this voltage, the better. Ideally, this voltage would be zero volts. Also, ideally, a diode can conduct any value of current I in the forward direction, with this value being determined not by the diode, but by other components in the circuit in which the diode is connected. Also, ideally, a diode conducts zero amperes for a negative V, regardless of the voltage magnitude. Put another way, an ideal diode is a short circuit for a voltage V that tends to be positive (but it cannot be more than 0 V). Also, an ideal diode is an open circuit for a negative V. Thus, an ideal diode acts like a switch that is closed for current flow in the direction of the arrow in the diode circuit symbol, and open otherwise. Essentially, it is an electronically operated switch. This ideal approximation is satisfactory for analyzing many circuits that contain diodes, provided that the voltage levels are much greater than 0.7 V. Fig. 8.2 shows the I-V characteristic for an ideal diode. Fig. 8.2

2 Fig. 8.3 shows the I-V characteristic of an actual, physical diode. The part of the curve in the first quadrant is the forward characteristic, and the part in the third quadrant is the reverse characteristic. The current I s is called the reverse saturation current. For a reverse voltage V B, the diode breaks down and draws a large reverse current. Fig. 8.3 The diode forward characteristic is shown on an expanded scale in Fig Observe the turn-on voltage V T. For forward voltages less than V T, a diode conducts very little current. Also, in the normal forward operating range, the diode voltage is approximately V T, almost irrespective of the current value. For the common silicon diode, V T is approximately 0.7 V. Fig. 8.4 Except for the reverse-breakdown region, the I-V characteristic of a diode may be expressed analytically as I = I s (e 40V - 1) at room temperature (20 C). For a silicon diode, the saturation current I s is of the order of 1 na. Although the forward characteristic is exponential, because of the large factor 40 in the exponential exponent, the characteristic appears to be almost vertical for a forward voltage slightly greater than V T, as can be seen in Fig Usually, a stripe on the diode casing designates the cathode (-) end. If there is any doubt, a DMM ohmmeter can be used to measure the diode resistance in both directions. A diode has a small resistance in the forward direction, which is for current flow from anode to cathode. It has almost an infinite resistance in the reverse direction, for current flow from cathode to anode.

3 Measurement of Diode I-V Characteristics An oscilloscope display of the diode I-V characteristic can be obtained using the circuit of Fig. 8.5 where the device is a diode. However, this circuit configuration can be used to displace the I-V characteristic of any two-terminal device. It is important to note this method but in this lab, we will use one of the built-in panels available in VI ELVIS. Fig. 8.5 Procedure: The Two-Wire Current-Voltage Analyzer panel is a stand-alone instrument that is a basic two-wire I-V curve tracer. It is capable of measuring four quadrant IV signals within ±10V and ±10mA. The procedure for obtaining an I-V graph is as follows: 1. Select a diode and place it on the DMM readout of the ELVIS protoboard. Connect one end of the device to CURRENT HI and the other end to CURRENT LO as shown in Fig Set the current limits to ±10 ma and the voltage sweep range from 0V (start) -1.2V (stop) with increments of 0.10V. 3. Run the tool. 4 Comment on how the characteristic obtained in Step 1 compares to that of an ideal diode. Fig. 8.6

4 Diode Circuits Diodes are used in many types of circuits. For example, they are used in rectifier circuits to convert AC to DC. Also, they are used in clipper circuits to select for transmission that part of a waveform that is either greater or less than some reference value. The Half-Wave Rectifier Consider the circuit of Figure 8.7. Assume that v i represents a DC voltage source V i in series with the diode and the resistor. We monitor the voltage V 0 across the resistor. If we consider the diode to be ideal (V T = 0) then current will flow in the diode only if V s is positive, the diode will act as a short circuit and V 0 = V i. For V i < 0, the diode will be reverse biased, allowing no current flow and V 0 will be zero. If the diode is not ideal, then for the current to flow, V i must be equal to or greater than V T. When the current flows in the diode, there is a voltage drop V T across it and the voltage across the resistor is given by V 0 = V i V T. Again, for V i < 0, V 0 = 0. Now reconsider the circuit of Figure 8.7, where we have an AC voltage source v i (t) of angular frequency ω (Period T = 2π/ω) and amplitude V p. v i (t) = V p sin(ωt) During the time 0 t T/2, v i (t) is positive and the diode, if ideal, will act as a short circuit, resulting in v 0 (t) = v i (t). However, in the second half of the period, v i (t) is negative and the diode will remain open during this time, resulting in v 0 (t) = 0. If the diode is not ideal, then the diode current will remain zero during the time v i remains less than V T. However, when v i > V T, v 0 (t) = v i (t) V T. Prelab: 1. Consider the circuit of Figure 8.7. Let v i (t) = 5.0sin(377t) V and V T = 0.7 V. Sketch the waveform v 0 (t) and show the times when v 0 is zero. Does your answer depend on the value of the resistor? 2. Consider the circuit shown in Figure 8.8. Let v i (t) = 5.0sin(377t), V T = 0, V r =2.0V. Sketch the output waveform v 0 (t) and show the maximum and minimum values of the output voltage. 3. Consider the circuit shown in Figure 8.9. Let v i (t) = 5.0sin(377t), V T = 0 and V r1 = 3.0 V and V r2 = 2.0 V. Sketch the output waveform v 0 (t) and show the maximum and minimum values of the output voltage.

5 Procedure: 1. Construct the half-wave rectifier circuit shown in Fig Use a 5-V peak, 1- khz sinusoidal voltage for V i. Again, select the Large Amplitude Waveforms tab of the function generator otherwise the generator is limited to 2.5V. The output of the function generator will be from DAC0. 2. Observe the output voltage V o on an oscilloscope and sketch it. Be sure that the AC-GND-DC switch (input coupling) for CH-A and CH-B are both set to DC to avoid shifting of the waveforms in a vertical direction. Can the diode turn-on voltage be determined by looking at the output voltage waveform? If so, what is it? Fig. 8.7 The Diode Clipper Circuits 1. A clipping circuit is shown in Fig Let V R = 2 V and the input voltage V i be a 5-peak, 1 khz sinusoidal voltage. Observe and sketch the output voltage V o. This circuit transmits that part of the v i waveform that is more negative than V R + V T. Fig. 8.8

6 2. Reverse the direction of the diode in the circuit of Fig Then, observe and sketch the output voltage v o. How does this output voltage compare to that in step 2? 3. Diode clippers may be used in pairs to perform double-ended limiting at two independent levels. Fig. 8.9 shows a double-diode clipper that limits at two independent levels. Set V R1 = 3 V and V R2 = 2 V. Test this clipping circuit with a sinusoidal input that has a peak amplitude of 5 V and a frequency of 1 khz. Observe the output voltage V o and sketch it. Please note that this will require the use of both the positive and negative variable DC supplies. These outputs are labeled SUPPLY+ and SUPPLYon the protoboard. CHECK YOUR CONNECTIONS BEFORE POWERING THE CIRCUIT. 4. Comment on the diode circuits studied in this experiment. Where might they be used in? Fig. 8.9