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Part 4: AC-DC Rectification By Andrew Levido Power Electronics Part 4: AC-DC Conversion with Rectifiers Power electronics as we know it today started with the invention of the mercuryarc rectifier in 1902. Rectifier-type AC-DC converters therefore predate the DC-DC converters we have been studying so far, by many decades. This month, we will look into these deceptively simple circuits. B efore the invention of the mercury-arc rectifier, the conversion of alternating current to direct current at scale required the use of rotating machines (eg, an AC motor driving a DC generator). I think their early emergence is the reason that most power electronics textbooks and courses start with rectifiers. I have taken a different approach because I believe that rectifier circuits are more challenging to analyse than DC-DC converters. In rectifier circuits, many of the quantities are sinusoidal or partly-­ sinusoidal instead of square or triangular, as they are in DC-DC converters, so calculating averages and RMS values is more difficult. Regardless of this extra difficulty, we will take a very similar approach to analysing rectifier circuits as we did with DC-DC converters. That means we will start with the simplest possible configuration; in this case, a single-­phase half-wave rectifier feeding a resistive load, as shown in Fig.1(a). Similar to the DC-DC converter analysis, we have a source voltage and current on the left, and a load voltage and current on the right. I’ll use the same conventions for AC, DC and average quantities as I have previously. We will also use the same average value analysis technique we learned in the first article in this series. As a reminder, average value analysis is based on the fact that under periodic steady-state conditions (PSS), the average voltage across any inductor is zero and the average current through any capacitor is zero. A single-phase half-wave rectifier We will initially assume the voltage source and diode are ideal components. The source voltage is an AC Practical Electronics | July | 2026 quantity given by the expression vs = Vs(pk)sin(ωt). This Vs(pk) term describes the amplitude of the sinusoidal voltage, and the sin(ωt) part just describes a unit sinewave (a sinewave with an amplitude of one). The resulting voltage is shown in red on the top graph. It is also common to describe AC quantities in terms of their ‘root mean square’ or RMS value. This is calculated by squaring the signal, taking the average over one cycle, and working out the square root of the result. For sinusoidal signals like our source voltage, the RMS value is just the peak value divided by √2. The horizontal axis is also worthy of a few comments. Instead of using time as we have done so far, it is conventional to use the closely related units of phase angle (ωt, pronounced omega-t – yes, ω is the lower-case version of ω) since trigonometric functions like sine operate on angles. If ωt is an angle and t is time, then ω must be an angular velocity, expressed in radians per second. Angular velocity in this context is just another way of describing frequency. You can see from the graph of vs that the length of one full cycle is 2π radians, so it should be apparent that one cycle per second (1Hz) is equal to 2π rad/s. Don’t get hung up on this if it does not seem intuitive to you – it is not critical to understanding what follows. Given that the diode can only conduct when the voltage across it is positive, the voltage seen at the load is just the positive half-cycles of the input. Since the load is resistive, the load current will have the same shape, and the source current will be the same as the load current. The average load voltage ‹v l› is harder to calculate than for the DC-DC converter where all the waveforms were square, since we have to calculate the area under the half-sine curve and divide it by the length of one cycle (2π). This requires a fairly straightforward integration, but I will spare you the details and simply give you the result, which turns out to be ‹vl› = Vs(pk) ÷ π. Adding a filter The output voltage is not exactly smooth DC, so to improve things, we could add an LC filter just as we did for the DC-DC converter. Before we do that, I want to add just the inductor, as in Fig.1(b), to demonstrate a point Fig.1(a): half-wave rectifiers are simple but have several disadvantages. Their input current has a DC component, they have a poor power factor, and they require more filtering than their full-wave counterparts. 69 Power Electronics series relating to switches like diodes that can’t be switched off externally. With the inductor in circuit, when the source positive half-cycle reaches zero at phase angle π, the inductor current is still flowing, so the diode must remain in conduction. If a current is flowing in the diode, the voltage across it must be zero. The voltage vx must therefore go negative, following the source voltage, until the diode current falls to zero. The load current (and source current since they are the same) is therefore ‘smeared’ past the end of the half-cycle. The average voltage ‹vx› will be lower than for the unfiltered rectifier because vx is negative for a period after the zero crossing. The average load voltage ‹vl› will be equal to ‹vx› since the average inductor voltage must be zero due to our steady-state analysis rules. If we were to try to increase the inductance to the point where the load (and therefore source) current is continuous, the diode would have to conduct continuously, forcing the voltage at vx to follow the input voltage for the whole cycle. Under these circumstances, the average voltage at vx would be zero, so the current would also be forced to zero. Fig.1(b): a series inductor provides smoothing, but introduces new problems. This circuit clearly will not work to create a smooth DC current in the load. We have already seen the solution in the DC-DC examples; introduce a freewheeling diode, as shown in Fig.1(c). We can now make the inductor as large as we like, since the inductor current now has a path to flow when the source voltage reverses. In a practical circuit, we would probably add a capacitor to the output, making an LC filter, but it can be ignored for the purpose of our analysis. After all, if the inductor is large enough to eliminate current ripple, no current will flow into or out of it. So now the voltage at vx will be a half-sinusoid, which we know from above has an average value of Vs(pk) ÷ π. Due to the periodic steady-state rules, the average load voltage will be the same. Ohm’s law dictates that the average load current must be Vs(pk) ÷ πR, and since there is no current ripple, this is also equal to the DC current, Il. The input current will therefore have a rectangular shape, with an amplitude of Il and a duty cycle of 50%, as shown in blue in the upper graph. The current though D2 (lower graph) will be similar, but out of phase by half a cycle. Commutation Fig.1(c): a second 'freewheeling' diode solves most of those problems. Fig.2: the non-zero source inductance L2 causes commutation, where both diodes conduct simultaneously for a short period, impacting power factor and regulation. 70 We have assumed a perfect voltage source up to this point, but we know that won’t be the case in real life. If our rectifier is powered from the mains, it will see a source voltage with an inductive component. If it comes from the mains via a transformer, there will be even more inductance, due to the transformer’s leakage inductance. In fact, it is hard to think of a situation where there won’t be any source inductance. Adding source inductance to the circuit of Fig.1(b) makes no difference to its operation, because it is effectively in series with L1. However, things get more complicated if we add it to the single-phase rectifier with a freewheeling diode, as shown in Fig.2. When D1 is conducting and D2 is off, the current through L2 is the DC load current, Il, so the voltage across it is zero and vx tracks the source. However, a problem arises when the source voltage goes negative. D1 must remain conducting while current is flowing in L2, and the voltage vx cannot go negative due to D2, so Practical Electronics | July | 2026 Part 4: AC-DC Rectification the voltage across L2 begins to rise until the energy stored in L2 is all transferred to the load via D1. Therefore, the current through D1 does not drop instantly at the end of the half cycle; instead, it tapers off, as shown in the figure. This means D2 does not take over the current instantly either, and it ramps up in a complementary manner, because the total current flowing through L1 to the load remains fixed. D1 and D2 are therefore both in conduction for a (hopefully) short period. The same thing happens in reverse when the positive half-cycle starts. The input current is now zero, so it takes a finite time for it to ramp up through L2 and D1 to the level of the load current. During this period, D2’s current ramps down to maintain the constant load current. While both diodes are conducting, the voltage at vs is held at zero, truncating the beginning of the voltage half-cycle and slightly reducing the average output voltage. This process, where current is transferred between the switches, is called commutation. The effect of commutation is to reduce the load regulation and change the shape of the source current. The effect on regulation is equivalent to adding a resistor of value XL2 ÷ 2π in series with the output. The effect on the input current is to make it more trapezoidal, meaning that the RMS source current for a given level of load current is higher with commutation than without. Power factor This segues nicely into the topic of power factor, which has become increasingly important in power electronics. All the rectifiers we have looked at so far have a source current waveform that is non-sinusoidal. This means that the apparent power entering the rectifier, calculated as the source voltage times the source current, is higher than the real power delivered to the load, calculated as the average over a cycle of the instantaneous voltage times the instantaneous current. How can this be? Consider the simple example in Fig.3. At the top, we have a sinusoidal source voltage (red trace) feeding some converter that produces a distorted (non-sinusoidal) current waveform (blue trace). In this case, the source current iS the Practical Electronics | July | 2026 sum of two sinusoidal currents: iS1, at the same frequency as the source voltage, and iS2, which is of lower magnitude but twice the frequency. The lowest chart shows the instantaneous product of the source voltage vs with the fundamental component of the current iS1 (in dark green) and the product of the source voltage vs with the second harmonic component of the current iS2 (in light green). The average power available from the fundamental component of the current is positive, but the average power available from the harmonic component is zero. This turns out to be true for all harmonics. From Fourier theory, we know that any periodic current waveform can be decomposed into a series of sinusoids, including a fundamental component and its harmonics. However, if the source voltage is sinusoidal (like the mains), only the fundamental component of the current contributes any useful power to the load. This is the ‘real power’, which we designate ‹p›, in units of watts. While the harmonic components of the current do not contribute to real power, they do contribute to the RMS value of the current, and therefore to the ‘apparent’ power – the product of RMS source voltage and RMS source current. We use S to describe apparent power, which has units of VA (volt-amps), and which will always be greater than or equal to the real power. The ratio of real power to apparent power, ‹p› ÷ S, is the definition of power factor. It is a unitless quantity that varies between zero and one. A power factor of one means that the real and apparent power are equal, as would be the case for a resistive load, for example, and implies that the current is purely sinusoidal and in phase with the voltage. A power factor less than one means that the real power doing useful work is lower than the apparent power being consumed. Power factor is important because the electrical supply system has to be dimensioned for apparent power. For example, a UK domestic power outlet is rated to deliver 230V at 13A for a nominal apparent power of 2990VA. If you applied a unity power factor load (like a resistive heater), you can expect to get 2990W of usable power from such an outlet. Fig.3: with a sinusoidal voltage source, only the fundamental component of the current waveform contributes to real power. The average power of any current harmonics is zero. Fig.4: if the current is sinusoidal but out of phase with the voltage, the average power available will be reduced. However, if the load produces a nonsinusoidal current, the usable power will be lower. If the power factor were 0.75, for example, you would only be able to get 2243W of useful power from the circuit, even though the RMS current sourced from the outlet would be 13A. Clearly, we can get the most out of the power distribution system by keeping the power factor high. Mains-connected power electronics has become a major contributor to poor power factor in electricity distribution systems around the world. A variety of techniques are available to improve or ‘correct’ power factor. 71 Power Electronics series We will cover some of these circuits in the next instalment of this series. For completeness, I should point out that just eliminating current harmonics won’t get you to unity power factor if the current is out of phase with the voltage. In fact, when I studied electrical engineering (many decades ago), the only discussion of power factor related to phase. The distortion component was skipped altogether because switch-mode supplies were far less common then. Fig.4 shows why phase matters. A sinusoidal source voltage feeds a device that draws a current that is also purely sinusoidal, but slightly out of phase with the voltage. The green trace shows the instantaneous power obtained by multiplying the two. If you compare this to the green power trace in Fig.3, you will see that instead of riding on the zero line, this trace is shifted down, reducing the average power compared to the inphase case. The vertical dashed lines show that as the phase shift increases, the zerocrossings of the voltage and current sinusoids diverge, so the power curve must drop to keep its zero crossings aligned. When either the voltage or cur- Fig.5: the full-wave bridge rectifier with inductor overcomes the disadvantages of half-wave rectifiers. The input current has no DC component, energy is delivered to the load on every half-cycle, and the power factor is much better. Fig.6: we typically use a capacitor filter instead of an inductor. Circuit (A) is simple and cost-effective, but has a fairly poor power factor. 72 rent is zero, the instantaneous power must also be zero. If the phase shift reaches ±π/2 radians (±90°), the average power, and therefore the power factor, drops to zero. If voltage and current are pure sinusoids, cos(ø), where ø is the phase shift, is a shorthand way to calculate power factor. The power ratio equation is the one to use in power electronics as it works for both phase- and distortion-­ related power factor or any combination thereof. Full-wave rectifiers The half-wave rectifiers that we described above are rarely used in practice since they suffer from three major drawbacks. First, the average input current is non-zero. This means there is a DC component to this current, which will not play nicely with transformers in the source network (and can accelerate conductor corrosion in some cases). Secondly, they require large filtering components to achieve low voltage ripple because energy is supplied only during every other half-cycle. Lastly, they have a poor power factor. We can confirm this pretty easily. Consider the half-wave rectifier with freewheeling diode in Fig.1(c). We saw that the average load voltage was ‹vl› = Vs(pk) ÷ π. We know the load current Il is DC, so we can calculate the average power dissipated in the load (and therefore supplied by the source) to be ‹p› = Vs(pk)Il ÷ π. The RMS input voltage is Vs(pk) ÷ √2 and the RMS input current is √Il² ÷ 2 = Il ÷ √2, so the apparent power must be S = Vs(pk)Il ÷ 2. Dividing ‹p› by S cancels the voltage and current terms, leaving a power factor for this topology of 2 ÷ π, which is about 0.64. Not great. Fig.5 shows a full-wave bridge rectifier and its associated waveforms. You can think of this circuit as two single-phase rectifiers with freewheeling diodes – in positive half-cycles, D1 conducts and D3 is the freewheeling diode, while in negative half-cycles, D2 conducts and D4 is the freewheeling diode. The full-wave rectifier addresses all the problems of the half-wave rectifier. The input current swings positive and negative alternately, so has an average value of zero and therefore no DC component. Power is supplied Practical Electronics | July | 2026 Part 4: AC-DC Rectification every half-cycle, doubling the output frequency and thus requiring less filtering to achieve a given level of voltage ripple. The power factor is also much better. The average load voltage is twice that of the half-wave rectifier, so the average power is ‹p› = 2Vs(pk)Il ÷ π. The RMS input voltage is the same, but the RMS current is now just I l, giving an apparent power of S = Vs(pk) Il ÷ √2. Dividing ‹p› by S cancels the voltage and current terms, as before, but leaves us with a power factor of 2√2 ÷ π, which is about 0.90. This is much better. Capacitive filters Of course, all of this assumes the presence of a large inductor to smooth the current, but this is not how we usually build rectifier-filter circuits. For the most part, we simply add a capacitor directly after the half- or full-bridge, as shown in Fig.6. The circuit acts like a peak detector, with the capacitor charging to Vs(pk) via the diodes at the crest of the half- or full-wave rectified waveform (shown dotted). The capacitor discharges via the load until the next peak. The voltage ripple can be (roughly) approximated by assuming that the voltage takes on a sawtooth profile (ie, it charges instantly at the crest and that the discharge is linear). For a half-wave rectifier, this gives Vl(pk-pk) = Il ÷ f C, where f is the source frequency and C is the capacitance. For a full-wave rectifier, the ripple is half of this, ie, Vl(pk-pk) = Il ÷ 2f C. This approximation is quite pessimistic for small supplies where the source impedance is relatively high, as we shall see. This topology means that current Fig.7: I built and simulated this simple transformer/rectifier circuit. It showed we could get a maximum power of about 14W from this 20VA transformer due to the limited power factor. only flows into the capacitor for a short period, resulting in a current waveform that is made up of narrow spikes of current aligned with the crests of the input voltage. The width of the current spikes is related to the ripple (the lower the ripple, the narrower the spikes), the source impedance and the capacitor’s ESR. All of this is difficult to calculate, but is a great candidate for simulation and experimentation. A practical example I had a 20VA, 240V to 12+12V toroidal transformer in my junk box, so I decided to build the simple full-bridge AC-to-DC converter shown on the righthand side of Fig.7 to see how it performed. Notice that the transformer is specified for apparent power. The transformer windings are connected in series to get a nominal 24V RMS, which is rectified by four chunky 6A10 (6A, 1000V) diodes I had lying around, and filtered by a 1000µF 63V electrolytic capacitor. First, I measured the open-circuit voltage of the transformer (28.6V RMS), the DC resistance of the secondary windings (1.3W each), the transformer leakage inductance (250µH) and the filter capacitor’s ESR (0.06W). This allowed me to build the simulation model shown in Fig.7. The simulation results are shown in Fig.8. The average output voltage is 31.6V, with a peak-to-peak ripple of 3.4V. The average output power is therefore 16.0W. The input current is shaped as we would expect, but the input voltage shows a flattened top. This is due to the voltage drop across the source impedance when the current pulses occur, and is typical for supplies of this size. The simulator calculated the RMS input voltage and current to be 26.2V and 0.95A, respectively, for an apparent power of 24.9VA (a little higher than our transformer’s rating). The power factor is therefore 0.64. The simulation compares well with the measured results below. The average load voltage is 30.6V and there is 4.8V peak-to-peak ripple. The output power is therefore 15.3W. These measurements were taken Fig.8: the experimental results agree fairly well with those obtained by simulation. Practical Electronics | July | 2026 73 Power Electronics series using a Current Probe and two Differential Probes, described in the November 2025 and January 2026 issues of this magazine, respectively, so the appropriate scaling factors need to be applied. The RMS input voltage and current are 26.0V and 0.824A for an apparent power of 21.9VA (still a touch too high for the transformer in the long term). The power factor is therefore 0.69, slightly better than the simulation, but nothing to get excited about. The important thing to note here is that the relatively low power factor puts an upper limit on the real power you can get from a given transformer. For this 20VA toroid, it is about 14W. I should also point out that this type of rectifier/filter results in fairly high 100Hz current ripple in the capacitor, which raises its internal temperature and potentially shortens its life. In this circuit, the capacitor ripple current is 0.8A RMS. Electrolytic caps usually come with a maximum 100Hz ripple current specification, so it is worth checking you are not exceeding this limit. The ripple current rating is one of the reasons you almost always see large filters made up of multiple parallel capacitors. If the capacitors are identical the ripple current rating of 5-year collections 2019-2023 £49.95 2018-2022 £49.95 2017-2021 £49.95 2016-2020 £44.95 Purchase and download at: www.electronpublishing.com 74 the bank is the sum of the ripple current rating of each capacitor. The ‘peaky’ current waveform has an impact on the output voltage you will achieve with this circuit. There are two diode drops between the peak value of the transformer secondary voltage and the voltage across the filter capacitor. These will likely be higher than the nominal 0.6-0.7V you might expect because the capacitor only charges when the current is at its peak. The diode data sheet should provide a curve called “instantaneous forward characteristic” or similar, which relates forward voltage drop to peak forward current. In the example of the 6A10 diodes I used, this curve shows a forward voltage of 0.8V at the peak current we see in the simulation, for a total drop of 1.6V. It is not at all unusual for the voltage drop to approach 2V if larger currents are involved. This drop can eat up a significant portion of the available voltage in low-voltage applications. As an aside, this is why active rectifiers are becoming more popular (see our November 2024 & September 2025 designs); they involve very little voltage loss and so improve efficiency. Inrush current This topology also comes with potential inrush current concerns. When power is first applied, and the capacitors are fully discharged, the inrush current is limited only by the supply impedance and the capacitor ESR. This is rarely a problem with lowvoltage supplies fed by relatively small transformers like this one, but can be a big problem for off-line rectifier/filters and very large capacitor banks, as you might find in a highpower audio amplifier, for example. The Variable Speed Drive for Induction Motors (December 2025 & January 2026) used a bank of five 330μF 400V capacitors to filter the rectified mains. A simulation made at the time showed that without inrush limiting circuitry, the peak inrush current would be around 200A, almost certainly tripping the supply circuit breaker and maybe damaging the rectifier diodes. In that case, we used a special inrush-­l imiting thermistor with a cold resistance of 10W to limit these peaks to less than 35A. The thermistor’s resistance drops as current passes through it, and the model we used is rated to conduct 15A continuously. These low-cost inrush limiters are available in various sizes and packages and are very common in off-line converters of all sizes. There was even one in the flyback converter we looked at last month. You can use a resistor to limit inrush if you short it out with a relay or similar after it has done its job, but you need to be sure that the resistor can withstand the short pulse of power that occurs during inrush. In the case of the Variable Speed Drive, this peak power is well over 100W for a few milliseconds. Some power resistors are specified for pulse power, but many are not, so be careful. Other topologies There are many variants of the full-wave rectifier, and a couple of the more common ones are shown in Fig.9. The load voltages shown assume the diode voltage drops are negligible. At the top is a centre-tapped variant that is a little more efficient for lowvoltage supplies than the full bridge, since the current passes through only one diode instead of two. This comes at the expense of a more complex transformer, but in reality, dual secondary windings are common in small off-the-shelf transformers. The middle rectifier uses a dualwinding transformer to produce a symmetric split (±) power supply – very useful for audio amplifiers or op amp circuits. The final circuit is a full-wave voltage doubler that is effectively two halfwave rectifiers in series. During positive half-cycles, the upper capacitor is charged almost to the peak of the secondary voltage, and during negative half-cycles, the lower capacitor is charged to a similar voltage. The result is an output voltage twice what could be expected from the same transformer with a full bridge rectifier. Phase control No discussion of rectifiers would be complete without introducing phase-controlled rectifiers. This is a technique that is used less these days than it used to be, but is still relevant in industrial applications where very high currents must be controlled. Practical Electronics | July | 2026 Part 4: AC-DC Rectification Fig.10: using a thyristor in place of the diode allows the output voltage of this half-wave rectifier to be controlled. Fig.9: here are some variations on the full-wave rectifier theme that might come in handy. The classical phase-control switch is the thyristor (sometimes called the silicon controlled rectifier, or SCR). You can think of a thyristor as a diode that will not conduct in the forward direction until the appropriate gate signal is applied. When the gate is positively biased with respect to the cathode while the anode-cathode voltage is positive, the thyristor switches on and remains on, even if the gate bias is removed, until the current drops to zero. In this sense, it is a latching device. In fact, the anode current must drop to zero for a short time (tq) for the thyristor to recover its forward blocking capability. This time is in the order of tens to hundreds of microseconds, limiting the application of thyristors to fairly low-frequency applications. Thyristors are very robust devices and are available in voltage ratings up to 6kV and current ratings up to 4kA. They have very good overload performance and, unlike most semiPractical Electronics | July | 2026 Fig.11: the full-wave phase-controlled rectifier is capable of inversion, where energy is transferred from the load to the source. conductor switches, can be protected by fast fuses. A more modest example, the 800V TN1605H-8G, is rated for a current of 16A RMS (8A average) and can withstand a non-repetitive half-cycle (10ms) surge of 160A. Just as a matter of interest, this thyristor requires a gate current of 1.5mA to switch on (you would usually drive it at around 5mA to be sure), and has a tq of 25µs at 25°C, rising to 85µs at 150°C. Fig.10 shows a half-wave phase-­ controlled thyristor rectifier. In this case, the thyristor is switched on at a phase angle (sometimes called firing angle or delay angle) of θ. You can see that if θ is zero, the output will be identical to the half-wave rectifier (ie, ‹vl› = Vs(pk) ÷ π), but as the phase angle increases, the output voltage drops until it is zero when θ = π. The relationship between phase angle and average voltage is not linear, due to the truncated sinusoidal shape of the voltage waveform. It can be shown that the output voltage is ‹vl› = (Vs(pk) ÷ 2π) × (1 + cos[θ]). Full-wave phase-controlled rectifier The full-wave phase-controlled rectifier (Fig.11) has some very interesting properties, as we shall see. The thyristors are gated on at phase angle θ as before, with SCR1 and SCR4 on in the positive half cycle and SCR2 and SCR3 on in the negative one. Let’s have a close look at what happens over a couple of cycles. As we come to the end of a positive half-cycle (at phase angle π, let’s say), SCR1 and SCR4 are conducting, but SCR2 and SCR3 have not yet been gated on to take over the constant current through the inductor. This means the current continues to flow through SCR1 and SCR4, so they must remain conducting past phase angle π, and the voltage at vx follows the input voltage negative. When SCR2 and SCR3 switch on at phase angle π + θ, the current 75 Power Electronics series Fig.12: an AC-DC converter can, in theory, operate in any one of these quadrants. Inversion occurs in quadrants II and IV. Fig.14: a three-phase rectifier has excellent low output ripple and a very good power factor, thanks to energy being delivered to the load six times per cycle. the uncontrolled bridge rectifier in Fig.5, but now its phase can shift from being in phase with the input voltage when θ = 0 to 180° out of phase when θ = π. Let’s pause and take in what this means. On the load side, a positive DC current with a negative average voltage means negative power is ‘dissipated’ in the load! The same is true on the source side; at phase angles greater than π/2, the average input power is also negative, so power is delivered to the source. This means that for phase angles greater than π/2, this circuit will transfer energy from the load to the source. This is known as inversion, and it can be very useful in practice. All four quadrants Fig.13: a three-phase source or load can be configured in star (Y) or delta (∆), since no current flows in the Neutral wire if the phases are balanced. commutates from SCR1 and SCR4 to SCR2 and SCR3, so the voltage flips positive abruptly. The same thing happens as the negative half-cycle comes to an end at phase angle 2π, although this time, it is SCR2 and SCR3 that remain on. The upshot of this is that the average voltage, ‹vx›, and hence the average load voltage, can go negative. The input current is a square wave with a 50% duty cycle, as it was for 76 I will give a couple of examples where this can be useful, but first it is worth looking at the general case illustrated in Fig.12. Here, I have shown a generic ‘four-­quadrant’ AC-DC converter powering a DC motor. The output voltage and current of this converter can each be positive or negative. This gives four possibilities, illustrated by the four quadrants in the V/I chart. The quadrants are denoted by Roman numerals counter-clockwise from the top right. In quadrants I and III (shaded green), both the voltage and the current have the same sign, so the power is positive. In quadrants II and IV, the voltage and the current are of opposite polarities, so the power is negative. I have positioned the current arrows on the motors so that they are always on the most positive terminal, but they are consistent with the upper diagram in that positive current flows down and negative current flows up. You can see that in quadrants I and III, the current flows into the most positive terminal, so the motor will be consuming power and driving the load (possibly in opposite directions, depending on the motor type). In quadrants II and IV, the current emerges from the most positive terminal of the motor, so the motor must be behaving as a generator and exporting power. Getting back to the full-wave phasecontrolled rectifier in Fig.11, we can see that this operates in two quadrants (I and IV, because the current is always positive). If the load on this converter was a DC motor with a highinertia mechanical load, quadrant I could be used to drive the load, and quadrant IV could provide regenerative braking. The reversed voltage applied to the motor creates a retarding torque that brings the motor and load to a stop quickly, since the energy stored in the rotating mass is transferred to the source. The motor will come to a stop much faster than it would if left to coast, losing its stored energy only to friction and windage. I have also seen this circuit used to quickly switch off (and therefore de-magnetise) a large electromagnet, moving the energy stored in the magnet’s inductance to the supply much faster than it would if we relied on the freewheeling effect of a full bridge rectifier. This allowed the electromagnet, which was picking up and Practical Electronics | July | 2026 Part 4: AC-DC Rectification moving steel components, to release them promptly on command. We should note here that the inversion described here is not sufficient to create a DC-AC converter. The AC source must be present for this circuit to work, and power can only flow back to the source while it is present. I will cover DC-AC inverters in a future article. Three-phase systems The electromagnet example I mentioned above was actually fed from a three-phase supply. I won’t go into multi-phase rectifiers in any great detail, since they are really only used in industrial applications, but I will touch on them for completeness. A three-phase voltage source is just three sinusoidal voltages with equal amplitude and frequency, but shifted in phase by one third of a cycle. I have drawn these three sources and loads (at the top of Fig.13) in a slightly unconventional manner, but you can probably see why this arrangement is called a star or Y configuration. The centre point of the star is the Neutral connection. The phase-­toNeutral voltages are shifted from each other by 2π/3 radians (or 120°), and this brings some very useful benefits. Firstly, if the load in each phase is balanced, the sum of the three line currents is zero, and no current flows in the Neutral wire, which is why I have shown it dotted. In fact, many three-phase loads such as motors don’t even have a Neutral terminal. On top of this, for balanced loads, any current harmonics that are multiples of three (the 3rd, 6th, 9th etc) also cancel to zero so, they don’t contribute to apparent power, meaning you get a power factor advantage for free. Given that the Neutral is unnecessary if the load is balanced, we could think about line-to-line voltages rather than line-to-Neutral voltages, and redraw the circuit as shown at the bottom of Fig.13. Again, I have drawn it unconventionally, but this configuration is called a delta (∆) arrangement because the elements are arranged in a triangle. You can have a star-connected load with a delta-connected source and vice versa. The line-to-line (or phase-to-phase) voltages are the sum of the two relevant line-to-Neutral voltages. I will leave the maths out, but it is easy enough to Practical Electronics | July | 2026 show that the line-to-line voltages are larger than the line-to-Neutral voltages by a factor of √3, and displaced from them in phase by π/6 radians (30°). The nominal line-line voltage for domestic three-phase supplies in the UK is therefore 400V, corresponding to √3 times the 230V nominal phaseto-Neutral voltage. Before you rush to correct me, I am well aware the typical line-to-line voltage is closer to 415V in many areas of the country, as we continue to transition from the old 240V/415V standard to the newer 230V/400V standard. In any case, 240V/415V is within the allowable range of the 230V/400V standard and vice versa. www.poscope.com/epe A three-phase full-wave rectifier Fig.14 shows a three-phase fullwave rectifier with an inductor. The upper graph shows the U-phase lineNeutral input voltage and the U-phase line current, with the V-and W-phase line-Neutral voltages shown dotted. The lower graph shows the voltage at vx, with the six half-cycles that contribute to it shown dotted. The average voltage at vx and hence the average load voltage ‹vl› = 3vll(pk) ÷ π or approximately 0.96 times the peak line-line voltage. The load voltage ripple is obviously much lower than for the single-phase rectifier, since there are now six ‘pulses’ of voltage each cycle instead of two. The power factor for the three-phase rectifier is also better than the single-­ phase case, as you might expect by observing that the line current waveform looks more sinusoidal with its ‘stepped’ shape. The power factor also turns out to be equal to 3/π (0.96) for this topology, so quite close to unity. You can also use thyristors in place of the diodes to create a phase-­ controlled version of this rectifier. It behaves in much the same way as its single-phase counterparts with phase angles above π/2, producing negative output voltages. Like the singlephase case, it can operate in quadrants I and IV. That’s all for this month. Next month, we will continue to look at AC-DC converters, with a focus on power factor correction. While we are on the topic of being responsible with regard to the power grid, we will also touch on the basics of electromagnetic PE interference (EMI) filtering. - USB - Ethernet - Web server - Modbus - CNC (Mach3/4) - IO - PWM - Encoders - LCD - Analog inputs - Compact PLC - up to 256 - up to 32 microsteps microsteps - 50 V / 6 A - 30 V / 2.5 A - USB configuration - Isolated PoScope Mega1+ PoScope Mega50 - up to 50MS/s - resolution up to 12bit - Lowest power consumption - Smallest and lightest - 7 in 1: Oscilloscope, FFT, X/Y, Recorder, Logic Analyzer, Protocol decoder, Signal generator 77