Power & Energy
How does a charge pump work?
A charge pump makes a voltage higher than its supply, or below ground, using only capacitors and diodes (or switches), with no coil. Drive a capacitor with a square wave, clamp one side of it with a diode, and its other side rides the wave shifted up; a second diode then catches the top of that into an output capacitor. A ±5 V wave becomes about +8.6 V.
Two halves of each cycle
The square wave drives C1 through a small resistor. D1 runs from C1's far side to ground, and D2 from there to the output capacitor, C2.
- Wave at −5 V: D1 conducts and charges C1 to about 5 V, with its far side held just below ground.
- Wave jumps to +5 V: C1 can't change its voltage instantly, so its far side jumps by 10 V, to nearly +10 V. D2 passes that into C2, and D1 now blocks.
C1 acts like a step stool: it charges up, then the whole capacitor rises with the wave, carrying its charge on top.
Live experiment
Pump the charge up
A ±5 V square wave at 1 kHz drives C1 through 47 Ω; two diodes steer the charge into C2. Watch C1's far side ride the wave, shifted up, and the output climb cycle by cycle to about 8.6 V. Try Schottky diodes, a heavier load, or both diodes turned round.
“Charge pumps” is lesson 6 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 12 before it first.
How much it gives
The first stage, C1 and D1, is a clamp: it shifts the wave up so its bottom sits about a diode drop below 0 V. The second, D2 and C2, is a peak detector that holds the top. So the output is the wave's full swing, peak to peak, less two diode drops:
Schottky diodes, with smaller drops, give about 9.5 V. The output climbs over a few cycles as C2 fills.
Below ground
Turn both diodes round and the pump shifts the wave down instead. D1 now holds C1's far side at or below ground, and D2 pulls C2 to the bottom of the swing: about −8.6 V. That's a handy way to give an op-amp a negative rail from a single supply.
The limits
Each cycle moves only a packet of charge, so a charge pump suits light loads of a few milliamps. Draw more and the output sags: in the experiment, a 1 kΩ load pulls the 8.6 V down to about 6.9 V. Bigger capacitors or a faster drive help; heavy loads need a boost converter with a coil.
The classic MAX232 used charge pumps to make about ±10 V for old serial ports from a single 5 V supply. Flash memory makes its high programming voltages this way, and long ladders of these stages, Cockcroft–Walton multipliers, make thousands of volts.
Free lesson
Now build it yourself
Each lesson comes in three short parts: see the idea, try it on a live board, then build it yourself, with quick checks as you go.
“Charge pumps” is lesson 6 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 12 before it first.
Or start from the very first lesson, no account neededPractice questions
Three quick questions from the lessons. Pick an answer to check it.
Show the answer
Answer: Sags: each cycle C2 loses more charge than C1 can top up. A charge pump only moves a little charge per cycle. Light loads, a few milliamps, suit it; heavy ones need a coil-based switcher.
- Why not “Rises”? Drawing more current never raises the output. Each cycle only refills so much charge.
- Why not “Stays put”? The output only holds while the load takes less than each cycle delivers. Past that it sags.
- Why not “Goes negative”? A doubler's diodes set its sign. A heavier load just pulls it down towards zero.
Show the answer
Answer: It fills at each peak and feeds the load in the dips. The capacitor tops up near each peak, then feeds the load until the next one, so the dips fill in. It can't lift the output above the peak, and with no diodes it just gets pushed back and forth.
- Why not “It turns AC into DC on its own”? On its own a capacitor just follows the AC back and forth. The diodes do the steering.
- Why not “It lifts the voltage above the peak”? A capacitor can only charge to the peak it's fed. It fills dips; it can't lift the output above the peak.
Show the answer
Answer: 10 mC. Q = C × V = 0.001 F × 10 V = 0.01 C = 10 mC, +10 mC on one plate and −10 mC on the other.
- Why not “1 mC”? That's 0.001 F × 1 V. At 10 V it holds ten times more, Q = C × V.
- Why not “100 mC”? That would need 10,000 µF at 10 V. 1000 µF is 0.001 F, and 0.001 × 10 = 0.01 C.
“Charge pumps” is lesson 6 of 6 in Buck Bay, Power & Energy World 2. Free with an account. Lessons unlock in order, so you’ll do the 12 before it first.
Questions people ask
What is a charge pump?
A circuit that moves charge with capacitors and switches, or diodes, to make a voltage higher than its supply, or below ground. It needs no coil, which makes it small and cheap, but it can only supply small currents.
How much voltage does a diode charge pump give?
A clamp-and-peak-detector doubler gives about the drive wave's full swing, peak to peak, less two diode drops. A ±5 V square wave swings 10 V, so the output is about 10 − 1.4 = 8.6 V with silicon diodes, or about 9.5 V with Schottky ones.
How do you make a negative voltage with a charge pump?
Turn both diodes round. The first one then clamps the capacitor's far side at or below ground, and the second pulls the output capacitor down to the bottom of the swing, giving about −8.6 V from a ±5 V wave.
Why does a charge pump's output drop under load?
Each cycle moves only a limited packet of charge, about C × ΔV. Draw more current than that refills and the output sags. Bigger capacitors or a faster drive help; for heavy loads, use a coil-based switcher instead.
Related topics
- How does a boost converter work?A coil fills from the battery, then kicks into the output: how a switch makes more volts than it started with.
- Half-wave vs full-wave bridge rectifiersFour diodes that turn AC into DC, a capacitor that fills the dips, and how to size it for your ripple.
- How does a capacitor work?Two plates and a gap: how a capacitor stores charge and energy, and why it blocks steady DC.
- How does a diode work?A one-way valve for current: anode, cathode, the 0.7 V forward drop and Schottky diodes.