Demo 3.2: Turning AC into DC
Mains electricity alternates. Almost every circuit you will ever build wants a steady voltage that does not. Bridging that gap is the single most common thing diodes are used for, and the whole of it rests on the one-way behaviour you spent the previous demonstration sweeping a slider across.
A rectifier is a circuit that turns alternating current into direct current. The word direct is doing less work than you might hope: it means only that the current never reverses, not that it holds still. Getting from a sine wave to something a circuit can actually run on takes three steps, and each one exists because the step before it was not good enough.
The three steps
Section titled “The three steps”One diode passes the positive half of the cycle and blocks the negative one. The output never goes negative, so it is direct current by definition, and it is at zero volts for more than half of every cycle.
Four diodes in a bridge recover the wasted half. Nothing about a negative half cycle is inherently useless: the current is simply flowing the wrong way, and a bridge gives it two different routes so that it arrives at the load from the same side both times.
A capacitor across the load fills in the gaps. It charges up on each peak and supplies the load on its own in between, so the output stops falling all the way to zero. What is left over is called ripple, and how much of it there is turns out to be what a power supply is really judged on.
How to use it
Section titled “How to use it”Half-wave has a cursor you drag through two full cycles. The circuit beside it shows the diode conducting or blocking at that instant, and the readouts show what the load is getting.
The bridge is the same idea with four diodes. Drag the cursor across a zero crossing and watch the conducting pair swap over, with the diodes numbered as they are in the lecture: D1 and D2 on one half cycle, D3 and D4 on the other. The thing to watch is the arrow through the load, which does not move.
Smoothing adds the reservoir capacitor. You can change the capacitance over four decades, change the load, and switch between one diode and the bridge with everything else held still, which is where the real argument for full-wave rectification shows up.
Ways to get it wrong covers five faults, four of which leave the supply apparently working.
Rectifiers, Half-Wave and Full-Wave
One diode, then four, then a capacitor, and the ways each of them goes wrong.
Walkthrough
Section titled “Walkthrough”Step 1: Watch one diode do the obvious thing
Section titled “Step 1: Watch one diode do the obvious thing”Open Half-wave and drag the cursor through a full cycle.
On the positive half the diode conducts and the load gets the input less one forward drop. On the negative half it blocks and the load gets nothing. The output is the shape everyone expects.
Now look at the numbers rather than the shape. With 17 V peak in, which is what a transformer labelled 12 V rms actually delivers, the peak output is 16.22 V and the average is about 5.04 V. That average is the useful DC output and it is less than a third of the peak, because the circuit is at zero for more than half of every cycle.
Read the caution underneath. The diode does not start conducting at the zero crossing, it starts once the input has climbed past its forward drop, so the conduction figure is 48.8 per cent rather than a clean 50. That part is a curiosity. The part that matters is the voltage: every point of the output sits a forward drop below the input, which costs about 5 per cent of the peak at 17 V and nearly a quarter of it at 3 V. Drop the peak input slider to 3 V and watch the whole trace shrink towards the axis.
Step 2: Recover the wasted half
Section titled “Step 2: Recover the wasted half”Open The bridge and drag the cursor slowly across a zero crossing.
Above zero, the left input node is positive, D1 and D2 are forward biased and carry the current, and D3 and D4 block. Below zero the situation reverses exactly: D3 and D4 carry it and D1 and D2 block. That is the pairing from the lecture slides, and the demonstration is wired to match.
Now watch the arrow through the load while you drag. It does not move. That is the entire trick and it is worth being precise about what is happening: the bridge is not turning a negative half cycle into a positive one, which would be impossible. It is redirecting it, so that current which was about to return through one path is guided through the load in the direction it went the first time.
Right at the zero crossing all four diodes are off, because the input has to climb past two forward drops before anything conducts.
Step 3: Count the cost and the benefit
Section titled “Step 3: Count the cost and the benefit”Look at the four readouts.
The peak output is 15.45 V rather than the 16.22 V the single diode managed, because current now passes through two diodes in series on its way round. The rule of thumb is that a bridge costs 1.4 V, and at a real load current it is a little more: here it is 1.55 V.
The average is 9.35 V against 5.04 V for half-wave. The demonstration puts that at 1.86 times, not two, and the missing fraction is precisely the second diode drop. It is worth noticing, because the textbook comparison of 2Vp/π against Vp/π says exactly two and assumes perfect diodes.
Then count the humps on the output trace: four across two cycles, where the half-wave circuit gave two. From 50 Hz mains, a bridge ripples at 100 Hz. That sounds like a detail and it is the main reason bridges are used, for a reason that only becomes visible on the next tab.
Step 4: Fill in the gaps
Section titled “Step 4: Fill in the gaps”Open Smoothing. The capacitor charges up as each hump rises and supplies the load on its own as it falls away, so instead of dropping to zero the output sags a little and is topped up again.
Set the capacitor to 100 µF and then to 1000 µF and watch the ripple shrink from about 1.3 V to about 0.15 V. Then read the estimate on the page:
V_ripple ≈ I / (f_ripple × C)
That comes straight from the definition of capacitance. A capacitor losing a steady current I for a time t falls by I·t/C, and the time here is one ripple period. The demonstration measures slightly less than the estimate predicts, because the capacitor is actually being topped up for part of each cycle rather than discharging for all of it, and the two converge as the ripple gets smaller.
Read the formula for what it tells you to do. Ripple falls if you increase the capacitance, draw less current, or rectify at a higher frequency. The first is why power supplies are full of large electrolytics. The third is why a switch-mode supply running at 100 kHz needs a capacitor a thousand times smaller than a mains-frequency one, and therefore why modern power supplies are so much smaller than the ones they replaced.
Step 5: Find the real argument for a bridge
Section titled “Step 5: Find the real argument for a bridge”With the capacitor at 470 µF, switch between One diode and Four in a bridge with nothing else touched.
Ripple goes from about 0.65 V down to about 0.30 V, a factor of a little over two. The reason is entirely the ripple frequency: the bridge’s gaps are half as long, so the capacitor has 10 ms to discharge instead of 20 ms.
That is the argument worth carrying away. The higher average output is nice, but a bigger transformer could also buy you that. What three extra diodes buy you and nothing else will is that the same smoothing capacitor does roughly twice the job.
Step 6: Learn the faults that do not stop the supply
Section titled “Step 6: Learn the faults that do not stop the supply”Open Ways to get it wrong.
Diode in backwards still rectifies, and the output is negative. Nothing is damaged and nothing announces the mistake, but any polarised electrolytic downstream is now reverse biased, and those fail loudly.
One bridge diode open is the classic field failure and the one worth being able to spot. The pair that diode belonged to can no longer conduct, so the bridge quietly becomes a half-wave rectifier: the average falls from 9.35 V to 5.04 V and the ripple frequency drops from 100 Hz back to 50 Hz. The equipment does not stop. It hums, runs warm and misbehaves under load.
The diagnosis is on the oscilloscope and it is about frequency rather than voltage. A full-wave supply from 50 Hz mains must ripple at 100 Hz. If it is rippling at 50 Hz, one half of the cycle is missing, whatever the multimeter reads, and a multimeter alone will never tell you because it reports the average and the average merely looks low.
One bridge diode shorted is the dangerous one. It puts the transformer secondary across a near short circuit for half of every cycle, and a mains transformer will deliver a great deal of current into that before anything gives way. If a supply blows its fuse the instant it is switched on, this is the first thing to test for.
Capacitor, but no load is the one that catches designers rather than repairers. With nothing to discharge it, the capacitor sits at the peak, and because almost no current is flowing the diode drops shrink too. A supply built around a 12 V transformer reads about 15.8 V unloaded. A capacitor rated at 16 V has almost no margin left, and it will be sitting at that voltage every time the equipment is switched on before the load draws anything.
Check your understanding
Section titled “Check your understanding”A bridge rectifier is fed from a transformer whose peak output is 17 V. What is the peak voltage across the load?
A full-wave supply running from 50 Hz mains is measured on an oscilloscope and its ripple is at 50 Hz. What has most likely happened?
Match each part to what it contributes
You want to halve the ripple in a mains-frequency supply. Which of these would do it? Select all that apply.
A supply built around a transformer labelled 12 V rms is measured with no load connected. What would you expect to read?
Wrap-up
Section titled “Wrap-up”Five things to take away.
- A rectifier makes current one-directional, not steady. Half-wave gives an average of about Vp/π and is at zero more than half the time.
- A bridge sends both half cycles through the load the same way. The output very nearly doubles, and the shortfall from exactly double is the second diode drop.
- Two diodes are always in the path of a bridge, so it costs about 1.5 V rather than 0.7 V. On a 5 V supply that is a third of everything you have.
- The capacitor makes the supply usable, and ripple is roughly I/(f·C). More capacitance, less current or a higher frequency all reduce it.
- Full-wave ripples at twice the supply frequency, so the same capacitor does about twice the job. That, rather than the higher average, is what the extra diodes are really buying.
The output is now usable and it is still not fixed. It ripples, it sags as the load increases, and it rises and falls with whatever the mains is doing. Anything that needs a genuinely constant voltage follows this with a regulator, and the simplest regulator is a resistor and a diode operated in the one region the previous demonstration deliberately left alone: reverse breakdown. That is a Zener diode, and it comes next, along with the light emitting diode, which is the same device again with a wider band gap.
© 2026 Derek Molloy, Dublin City University. All rights reserved.