Welcome back to Electronics. This is Dr.Robinson. In this lesson we're going to look at what are called Half-Wave Rectifiers. In previous lessons, you looked at analyzing circuits containing resistors, diodes and dc sources. And in the process of doing this, you modeled the diodes both as ideal and as non-ideal, considering the forward voltage drop that exist across the diodes. Our objectives for the lesson are to introduce what are call half-wave rectifiers. Well then examine their behavior for sinusoidal inputs and then analyze a particular diode circuit that implements a half-wave rectifier. So here's the definition of the rectifier. A rectifier is a non-linear device that modifies an input voltage such that the output voltage is greater than or less than some threshold value. Now let me illustrate this definition with a picture. So here I've represented the half-way rectifier as a block, and later on in the lesson we'll look at the circuit that is actually inside this block to implement it. For the input to this block, I've applied a sinusoidal varying voltage. This is the zero volt level. So for each half-period, the voltage is positive and for the second half-period, the voltage is negative. Now when we apply this sinusoidal voltage to this half-wave rectifier, from the definition I gave you from the previous slide, there are a number of possible outputs. Let's look at this possible output first. You can see that to obtain the output voltage from the input voltage in this case, when the input voltage is a positive value or greater than the zero-volt threshold, then the output voltages is exactly equal to the input voltage. But, when the input voltage is negative or less than zero volts, the output voltage is set equal to zero volts. Now, this wave form is known as a positive half-wave rectified sine wave. Positive, because it's entirely greater than zero volts, or greater than the zero-volt threshold. And so, it's the half wave rectified sine wave because the output is non-zero only for each half-period of the input waveform. Now, the waveform I've drawn down here is known as the negative half-wave rectified sine wave. Half-wave rectified, again, because the output is non-zero only for every half-period of the input. And negative because it's entirely below the zero-volt threshold. So in this slide I have represented a positive half-wave rectifier in terms of rules that are applied to the input to obtain the output. The input is applied. If the input is greater than the threshold, then the output is equal to the input. But if the input is less than or equal to the threshold, the output is equal to the threshold. Let's look at a circuit that can be used to implement a half-wave rectifier. Now you've seen circuits similar to this one in previous lessons in this course. The circuit consists of an input voltage and series with resister and series with ideal diode. And I have defined this node here as the ground node, or the zero-volt reference node. Now remember, an ideal diode can be thought of as a voltage-controlled switch. When this voltage at the anode is greater than the voltage at the cathode, the diode is forward-biased, or said to be on, and can be thought of as a closed switch or a short circuit. However if the voltage here at the cathode is greater then the voltage at the anode, the diode is set to be reversed biased or off, and can be thought of as an open switch or an open circuit. So in the case where V-in is a positive voltage, that means that V-in is attempting to push current around the loop in this direction from its positive terminal to its negative terminal. However, the direction of the diode does not allow current to flow in that direction. Remember current can only flow from the anode to the cathode of the diode when it's forward biased. So when V-in is positive, current should flow in this direction. However, it's prevented by the diode, the voltage of the cathode is greater than the voltage of the anode, the diode is reversed biased, or off. And we can model this circuit as I've drawn here, with the diode replaced by an open circuit. Now, in this case where V-in is negative, remember if V-in is negative we can think of this side as the positive side, and this side as the negative side. So with V-in negative it's attempting to push current around the loop in this direction. That direction is allowed by the direction of the diode. This voltage is more positive than this voltage. The diode is on and it can be replaced by a short circuit as I've drawn here. So these two circuits in combination model the behavior of your original circuit. This one applies when V-in is negative, this one applies when V-in is positive. On this slide I've redrawn the two circuits. Now, it may be apparent to you the relationship between the output and input voltages for each of these two circuits. But let's go ahead and write some equations to determine those relationships. Because thinking about circuits in this way will help you to analyze more complicated circuits. Now, in this circuit you can see that the output voltage is the voltage between this node and this node. So to get from this node to this node, I go up by a voltage V-in, then I go down by 1IR drop by ohm's law across the resistor. So, we can write that V-out is equal to, we go up from V-in, and then we go down by an IR drop across the resistor to get to V-out. But we know, because of the open circuit, that I is equal to 0, which implies that V-out is exactly equal to V-in. Now, in the second circuit we can see immediately that the out voltage is the voltage measured across a wire, or a resistor of 0 ohm's. Well let's go ahead and write the equation. V-out is equal to I times the resistance of a wire, is equal to V-in divided by R, the current times R wire. But we know that the resistance of a wire, is equal to 0. Which implies that that V-out is equal to 0, for this case. Now remember, these two circuits in combination model the behavior of our original circuit. So I can take these two circuits and form a single piecewise linear equation that represents the behavior of the original circuit. Here I've combined the results of the previous slide into a single piecewise linear equation that describes the behavior of the circuit. The output voltage is equal to the input voltage when V-in greater than 0. The output voltage is equal to 0 when V-in is less than or equal to 0. Now if we apply this sinusoidal input voltage to that circuit, we can determine the output voltage of the circuit by using these rules. For the case where V-in is positive, we can see that the output is exactly equal to the input. So I just copy the input to the output graph when V-in is positive. But when V-in is less than 0, we can see that the output is equal to 0. So from this graph we can see that that circuit that we've analyzed implements a positive half-wave rectifier. Now up to now we assume the diodes were ideal. But what happens if we include the forward voltage drop that exists across a diode when it's turned on? In other words, we model the diode in the original circuit by this circuit here, a battery of Vf, the forward voltage drop in series with an ideal diode. So here I've redrawn the circuit to include the forward voltage drop model. The voltage here is 0 volts, or ground, by definition. To move from this voltage to this voltage, we go down by one voltage drop of Vf volts. Because remember in the battery symbol, the longer side is the more positive side. So 0 minus Vf gets us to this node voltage. So we can label the node voltage here as -Vf volts. Now we can analyze the circuit, just considering the behavior of the ideal diode. Say V-in is a positive voltage such that V-in is greater than minus Vf volts. Then we know the voltage here is greater than the voltage here, which means the diode is off, or reversed-biased, and can be replaced by an open circuit. So, under the condition where V-in is greater than -Vf volts, we redraw the circuit like this. Now for the condition where V-in is less than minus Vf volts, the voltage of the anode would be greater than the voltage of the cathode. The diode would be on and we can represent it as a short circuit. So I've redrawn the circuit here under those conditions, where the output voltage is taken directly across the battery Vf. So lets just analyze the relationship between the output voltage and the input voltage by inspection, rather than writing equations. We can see that in this case the output voltage is exactly equal to the input voltage for the same reasons as it was for the ideal diode case. V-out is the voltage between this node and this node. We go up by V-in and down by one IR drop, but i is equal to 0. Now in this case, we can see that the V-out is the voltage taken directly across a battery of the Vf volts. However the polarities are different. V-out is defined positive on this side but Vf is defined as positive on this side. So we can write that for this case, Vf is equal to -Vf volts. We then combined the two results into a single equation in V-out. In this equation we can see that V-out is equal to V-in when V-in is greater than -Vf volts. And that V-out is equal to minus Vf when V-in is less than or equal to -Vf volts. Now rather than plotting this equation I went to the lab and built on a proto board the circuit we've been analyzing using a one-end 4148 small signal silicon diode. I applied a sine wave to the input and used an oscilloscope to measure the output voltage and input voltage of the circuit. Here's the oscilloscope trace. You can see the input voltage here. This sign is subtly varying voltage. Let me label on this graph -Vf volts, which you may remember for a silicone diode is approximately 0.65 volts. So this is a 2-volt by 200 microseconds squared. So minus 0.65 would be Right there. So we can see that when the input is greater than -0.65 volts, the output is equal to the input and the two curves track each other exactly. But when the input voltage is less than our -0.65 volt level, the output is equal to -0.65 volts, as we would expect from our previous analysis. The circuit we've been analyzing has this particular typology. An input voltage in series with element 1 in series with element 2. Now if we had available to us one resistor and one diode, we can build four different rectifier circuits, assuming that the output voltage is taken across element 2. Now I wanted to leave you with some things to think about at the end of this lesson. For each of these possible circuits, can you perform the same analysis that we did during this lesson? In other words, can you determine the output voltage verses input voltage equation? And additionally, say we took the output across element 1 rather than element 2. How would the output voltage change for each circuit? So in summary, during this lesson we looked at rectification and we looked at a particular type of rectifier known as a half-wave rectifier. In our next lesson we'll look at circuits known as full-wave rectifiers, and determine how they differ from half-wave rectifiers. Thank you and until next time.
