Smoothing, Ripple, and the Reservoir Cap

From throbbing DC to a steady level

The last guide left us with a win and a problem. The win: a half-wave or, better, a bridge rectifier forces the mains AC to flow one way only, so the output never goes negative. The problem: that output still throbs. It is a train of humps that rises to a peak and then falls all the way back to zero, sixty or a hundred-and-twenty times a second. Its long-term average is low, and twice every cycle it briefly sits at nothing at all. As a power rail this is almost useless — a logic chip fed from it would reset on every dip, an audio amp would hum loudly, an LED would visibly flicker.

What we actually want is a flat, steady DC level — the calm voltage a real circuit can run on. The cure is to store energy while each hump is up high and pour it back out into the load while the waveform dips, so the valleys get filled in. We already met the perfect part for the job in the AC rung: a capacitor, the little rechargeable bucket that resists any sudden change in its voltage. Bolt one across the rectifier's output and it does exactly that filling-in, automatically, on every cycle.

The reservoir capacitor: a bucket across the rail

Connect a big capacitor in parallel with the load, right across the rectifier output, and you have a reservoir capacitor (the same part used for filtering is also called a smoothing capacitor). Watch one cycle. As an incoming hump rises, the diodes conduct and charge the bucket up toward the peak. The instant the hump passes its top and starts to fall, the capacitor's voltage is now higher than the falling rectified voltage, so the diodes become reverse-biased and switch off. Cut loose from the rectifier, the capacitor alone takes over feeding the load, draining slowly, until the next hump climbs back above it and tops the bucket up again.

So the capacitor behaves as a peak follower: it grabs the highest point of the waveform and clings to it, sipping out energy in the gaps. The diodes no longer conduct over the whole hump — they wake up only in a brief sliver right around each peak, just long enough to replace the charge the load drew since last time. The throbbing zero-to-peak waveform becomes a nearly flat DC sitting just below the peak. That last fact surprises beginners and matters a lot, so it gets its own note below.

   from rectifier       reservoir    load
   (pulsating DC)          cap
       o-----+-------------+---------+-----o  +Vdc
             |             |         |
            ...           === C     [ R ]
             |             |         |
       o-----+-------------+---------+-----o  GND

   Vout |   ,-.       ,-.       ,-.       recharge to ~Vpk each
   Vpk -|--/   \-----/   \-----/   \--    pulse, then it sags
        | /     \   /     \   /     \     into the load: the
      0 +---------------------------> t   small sag is ripple Vr
           |<- 1/fr ->|   fr = 120 Hz (full-wave on 60 Hz mains)

Ripple: the leftover wiggle

The bucket is not bottomless. Between peaks the load keeps draining it, so the voltage sags a little before the next hump arrives to refill it. That small residual sawtooth — recharge, slow sag, recharge — is the ripple, usually quoted peak-to-peak as Vr. Ripple is simply the part of the AC the reservoir failed to flatten; the whole game of supply design is keeping it small enough that whatever comes next does not care.

We can estimate Vr with a single clean idea. Between top-ups the capacitor discharges into the load for a time t of about one ripple period, while the load draws a roughly constant current I. Charge is Q = C times V, so the voltage it loses is just the charge drained divided by the capacitance: Vr ≈ I times t / C = I / (fr times C), where fr is the ripple frequency. Try real numbers: a 0.5 A load on a full-wave supply (fr = 120 Hz) with C = 4700 uF gives Vr ≈ 0.5 / (120 times 4700 times 10^-6) ≈ 0.89 V peak-to-peak. This is an honest approximation, not a law: it assumes the ripple stays small next to the DC level, so the sag is near-linear and lasts nearly the whole period. When ripple grows large it overestimates a bit, but for design it is exactly the lever you need.

The formula points straight at two ways to shrink ripple: a bigger C, or a higher ripple frequency fr. The second is free, and it is exactly why a full-wave or bridge supply beats a half-wave one for the same capacitor. Half-wave refills the bucket only once per mains cycle (fr = 60 Hz), leaving a long hungry gap; full-wave refills it twice (fr = 120 Hz), so the gap is half as long and the ripple is half as big. Better smoothing for the cost of two extra diodes — usually a bargain.

Sizing the cap — and the price you pay

Design runs the ripple formula backwards. Decide how much ripple you can tolerate, then solve for the capacitor: C ≥ I / (fr times Vr). Say a bridge supply must deliver 0.5 A with no more than 1 V of ripple at fr = 120 Hz. Then C ≥ 0.5 / (120 times 1) ≈ 4170 uF, so you round up to a standard 4700 uF electrolytic. Always round up — a bigger cap only helps ripple — and pick a voltage rating comfortably above the peak rail (a 25 V or 35 V part for our ~16 V example) so it survives a high-mains day.

1. Find the peak rail: take the transformer secondary's RMS, times sqrt(2) for the peak, minus the diode drops (two for a bridge, one for a half-wave).
2. Set the ripple frequency fr: 120 Hz for full-wave/bridge on 60 Hz mains, 60 Hz for half-wave.
3. Pick a ripple budget Vr you can live with — often a volt or two for a rail that a regulator will clean up afterward.
4. Solve C ≥ I / (fr times Vr) at the maximum load current I, then round up to the next standard electrolytic value.
5. Choose the voltage rating well above the peak rail, and confirm the diodes can survive the tall, narrow charging current spikes.

Honest limits: it is still unregulated

Be clear about what a rectifier plus reservoir cap actually is: an unregulated supply. Its output is not a fixed number you can trust. Draw more current and the ripple grows while the average rail droops (look again at Vr ≈ I / (fr times C) — ripple is proportional to load). Let the mains sag or surge by ten percent and the whole rail moves with it. And the ripple never vanishes entirely; it only gets small. That is perfectly fine for a motor, a relay coil, or a heater that does not mind a wobbling rail — but it will not do for a precision reference or a sensitive analog front end.

Cleaning it up is the job of the next pieces. The Zener diode in the very next guide gives a crude fixed reference that clamps the rail to a chosen value; further up the ladder a regulated supply built around a linear regulator actively holds the output rock-steady and scrubs the ripple away, at the honest cost of burning (Vin − Vout) times I as heat. Crucially the reservoir cap does not disappear — it becomes the tank the regulator drinks from. You size it so that even at the bottom of the ripple sag the input still stays a couple of volts above the output the regulator needs, its dropout headroom.

One last practical lever ties back to guide 1. Remember that the diode's drop is a model, not a fixed constant — and a Schottky diode is built to drop far less, around 0.3 V instead of 0.7 V. Swap the bridge for Schottkys and you claw back about 0.8 V across the two conducting diodes, plus they run cooler and switch faster. On a 15 V rail that is noise; on a low-voltage 5 V or 3.3 V supply, where every tenth of a volt and every dropout milliwatt counts, it can be the difference between a design that works and one that does not.