Capacitive DC-DC converters: how a charge pump works.

One supply, but you need a different voltage.

Your circuit runs from 5 V, but one part of it needs 10 V. Another wants −5 V. Classic examples are the ±10 V an RS-232 line driver needs from a 5 V logic supply, the bias for an LCD, or the higher voltage a flash memory uses while it writes. Converting one DC voltage into another is called DC-DC conversion.

Most people meet DC-DC conversion through inductor-based buck and boost converters. There is a simpler family that uses no inductor at all: capacitive converters, also called charge pumps or switched-capacitor converters. A few switches and capacitors, a clock, and the right sequence produce a doubled or inverted voltage.

This guide builds both a voltage doubler and a voltage inverter in Simvolta. Every result below is either calculated, and labelled as a prediction, or measured in the desktop solver, and labelled as a result.

A capacitor is a bucket of charge you can carry.

A capacitor stores charge Q in proportion to the voltage across it: Q = C · V. Charge it to 5 V, disconnect it, and it keeps those 5 V between its plates, wherever you move it. That is the whole trick. A charge pump fills a capacitor from the supply in one phase, then reconnects it somewhere else in the next phase.

The capacitor that moves is the flying capacitor, C1. If its negative plate is moved from ground to the 5 V supply, its positive plate rises to 5 V + 5 V = 10 V. Nothing in C1 changed; its reference point did. A second capacitor, the output or reservoir capacitor C2, collects these transfers and keeps the output steady between them.

Use the buttons to switch phases. The highlighted path is where conventional current flows in that phase; the voltages are the ideal, unloaded values.

Q = C · VE = ½ · C · V²

The voltage doubled; the energy did not. Each coulomb that reaches the output passes through the 5 V supply twice: once while it charges C1, once while C1 is stacked on top of the supply. The supply therefore delivers twice the output current, and input power is at least the output power. A charge pump changes voltage and current in opposite directions, just like a transformer or a boost converter.

Four switches, two capacitors and one clock.

The doubler in the bundled example uses Simvolta's voltage-controlled switches as ideal electronic switches, the job MOSFETs do inside a charge-pump chip. A PWM generator is the clock: 0 V or 5 V, 10 kHz, 50 % duty.

S1 and S2 close while the clock is high: that is phase 1, which charges C1 across the supply. S3 and S4 close while the clock is low: phase 2, which stacks C1 on top of the supply and connects it to the output. To make S3 and S4 work in the opposite phase from the same clock, their control inputs are wired the other way round, clock to Ctrl− and ground to Ctrl+, and their threshold is set to −2.5 V. When the clock is low, their control voltage is 0 V, which is above −2.5 V, so they close.

Step by step: charge, then stack.

Phase 1, clock high. S1 connects C1's top plate to 5 V and S2 connects its bottom plate to ground. C1 charges to 5 V. Meanwhile C2 is disconnected from C1 and alone supplies the load, so the output droops slightly.

Phase 2, clock low. S3 lifts C1's bottom plate to 5 V. C1 still holds about 5 V, so its top plate sits near 10 V, and S4 connects it to the output. C1 tops up C2 and supplies the load. The clock then goes high again and the cycle repeats 10,000 times a second.

The plot below is real solver output from the example, two clock periods after it has settled. Watch C1's plates move together: the bottom plate jumps between 0 V and 5 V, and the top plate follows it from 5 V to about 9.9 V, so the voltage across C1 hardly changes.

Predict the output before you run it.

With no load, C1 and C2 eventually charge to exactly 10 V and the converter stops moving charge. With a load, every coulomb the load takes must first be carried by C1, and that costs voltage. In the slow-switching limit, where each phase is long enough for the switch currents to finish, the charge pump behaves like an ideal 2 × source in series with a resistance of 1/(f · C1). Here that is 1 / (10 kHz × 10 µF) = 10 Ω.

So the predicted output is 10 V × 1 kΩ / (1 kΩ + 10 Ω) ≈ 9.90 V, the load current about 9.9 mA and the input current twice that: about 19.8 mA. The efficiency of this ideal-switch circuit is then VOUT / (2 × VIN) ≈ 99 %. The ripple estimate comes from the droop of C2 alone in phase 1, plus C2 and C1 together in phase 2.

R_out ≈ 1 / (f · C1) = 1 / (10 kHz × 10 µF) = 10 ΩVOUT ≈ 2 · VIN · RL / (RL + R_out) = 10 V × 1000 / 1010 ≈ 9.90 VI_in = 2 · I_out η = VOUT / (2 · VIN) ≈ 0.99ΔV ≈ I_out · (T / 2) · [1 / C2 + 1 / (C1 + C2)] ≈ 9.45 mV

• The switches are ideal apart from 10 mΩ on and 1 MΩ off: no gate charge, dead time or switching loss.
• Each phase lasts 50 µs, more than 250 times the time constant of the switches and capacitors (about 0.2 µs), so every transfer finishes: the slow-switching limit applies.
• The output is close to constant (C2 ≫ C1), and the result is read after start-up has finished.

Build it in Simvolta.

The video shows the whole build in the real desktop app, from an empty canvas to the result. Or open the finished circuit: choose Open Example… and pick Charge Pump Voltage Doubler.

Watch · 1:32: video, Build a charge-pump voltage doubler

1. 01 · 0:03Start with a 5 V supply
2. 02 · 0:07Add four voltage-controlled switches
3. 03 · 0:16C1 moves charge, C2 holds it, R1 is the load
4. 04 · 0:25Add a 10 kHz PWM clock and grounds
5. 05 · 0:33Press W, then wire the power path
6. 06 · 0:50Wire the clock to every switch
7. 07 · 1:04S3, S4: threshold −2.5 V, the other phase
8. 08 · 1:08Type C1 = 10u and C2 = 100u
9. 09 · 1:13Run a 10 ms transient
10. 10 · 1:15Probe VOUT: 5 V in, 9.9 V out
11. 11 · 1:20Change one thing: C1 = 1u
12. 12 · 1:22Run 100 ms: slower, and lower: ≈ 9.1 V

Read the transcript

1. Start with a 5 V supply.
2. Add four voltage-controlled switches.
3. C1 moves charge, C2 holds it, R1 is the load.
4. Add a 10 kHz PWM clock and grounds.
5. Press W, then wire the power path.
6. Wire the clock to every switch.
7. S3, S4: threshold −2.5 V, the other phase.
8. Type C1 = 10u and C2 = 100u.
9. Run a 10 ms transient.
10. Probe VOUT: 5 V in, 9.9 V out.
11. Change one thing: C1 = 1u.
12. Run 100 ms: slower, and lower: ≈ 9.1 V.

1. Place a DC voltage source (5 V is the default) and a ground. Add four voltage-controlled switches, two capacitors, a resistor and a PWM generator (its defaults are 0–5 V, 10 kHz, 50 %).
2. Wire the power path: the source to S1 and S3; S1, C1's top plate and S4 together; S3, C1's bottom plate and S2 together; S4 to C2 and the load; S2, C2 and the load to ground.
3. Wire the clock to Ctrl+ of S1 and S2 and to Ctrl− of S3 and S4. Connect each switch's other control pin to ground. Press W for the Wire tool; clicking empty canvas adds a bend, and ending on a wire makes a junction.
4. Select S3 and S4 and set Threshold to -2.5. Set C1 to 10u and C2 to 100u. Keep the load at 1k.
5. Choose Transient, 10 ms with a 1 µs output step, and press Run. Press P and click the load's wire to plot VOUT.

Run it: what the simulation shows.

The output climbs in small stairs. Each step is one phase-2 transfer: the first one shares C1's charge with an empty C2, and every later one closes about a tenth of the remaining gap, because C1 is a tenth of C2. That gives a start-up time constant of roughly eleven clock periods, about 1.1 ms. After 1 ms the output is at 6.12 V, after 5 ms at 9.82 V, and it then settles at 9.90 V.

The predicted and simulated outputs differ by about 1 mV. The slow-switching formula ignores the 20 mΩ of the two closed switches and the small ripple, which accounts for the difference. The input current shows the energy bookkeeping: about 20 mA from 5 V in, 9.9 mA at 9.9 V out.

Change one thing: a smaller flying capacitor.

Make one prediction before you rerun the circuit. What happens to the output if C1 shrinks from 10 µF to 1 µF?

The same 100 Ω appears if you keep C1 at 10 µF and lower the clock to 1 kHz: the simulation settles at 9.089 V. The output resistance depends on the product f · C1, the amount of charge C1 can carry per second for each volt it droops. The ripple tells the two experiments apart. With the slower clock, C2 has to hold the output for ten times longer, so the ripple grows from about 9 mV to 86 mV. A smaller C1 at 10 kHz keeps it near 9 mV.

This is how a charge pump is designed in practice: choose f · C1 for the output resistance you can accept, and f · C2 for the ripple you can accept.

Experiment B: make a negative voltage, in slow motion.

An inverter uses the same four switches, rewired. In phase 1, C1 charges to +5 V exactly as before. In phase 2, S3 connects C1's positive plate to ground, so its negative plate now sits about 5 V below ground, and S4 connects that plate to the output. The output settles near −5 V. This is the topology of the classic ICL7660 voltage converter.

The bundled Live: Charge Pump Inverter (slow motion) example runs at 1 Hz so you can watch each transfer. A 1 kΩ resistor in series with C1 stretches the current pulses to a time constant of 0.1 s, the role a surge-limiting resistor plays in a relay-built charge pump. Go Live and turn on the current-flow overlay. In phase 2 the current flows up from ground, through the load and C2, into the output node: that is what a negative supply looks like.

Watch · 0:45: video, Watch a charge-pump inverter in slow motion

1. 01 · 0:03Open Example… → Charge Pump Inverter
2. 02 · 0:07Go Live at 1×: the clock ticks once a second
3. 03 · 0:09Press P: probe VOUT and C1's lower plate
4. 04 · 0:15Dots are current: charge C1, then transfer
5. 05 · 0:20VOUT steps down toward −4.5 V
6. 06 · 0:24Close S5: an extra 10 kΩ load. VOUT sags
7. 07 · 0:33Open S5: the pump catches up

Read the transcript

1. Open Example… → Charge Pump Inverter.
2. Go Live at 1×: the clock ticks once a second.
3. Press P: probe VOUT and C1's lower plate.
4. Dots are current: charge C1, then transfer.
5. VOUT steps down toward −4.5 V.
6. Close S5: an extra 10 kΩ load. VOUT sags.
7. Open S5: the pump catches up.

Slowing the clock makes the converter weak. At 1 Hz with 100 µF, 1/(f · C1) is 10 kΩ; with the 1 kΩ resistor included, the two-phase formula gives about 10.1 kΩ. With RL = 100 kΩ the prediction is −4.54 V. Close S5 to add a 10 kΩ load in parallel, and the prediction falls to −2.36 V.

What the slow inverter measures.

The plot shows both loads from start-up, simulated as 20-second transients of the same example: S5 open, then S5 closed from the start.

The two simulators agree to the millivolt, but both outputs are smaller in magnitude than the formula predicts: by 0.6 % at the light load and 3 % at the heavy one. That is a model assumption failing, not a numerical error: the formula assumes the output voltage stays constant during a cycle. At 1 Hz with a heavy load, C2 droops by 0.73 V each cycle, so the assumption no longer holds.

The inverter also passes each coulomb through the supply only once, so its input current equals its output current, where the doubler needed twice the output current. And efficiency halves as the output sags: 46 % at −2.29 V. Regulating a charge pump by letting it sag throws away the difference, like a linear regulator.

Where the losses come from.

Charging a capacitor from a voltage source always loses energy in whatever resistance is in the path. When charge flows from a source at V1 to a capacitor at V2, the loss is ½ · C · (V1² − V2²) per transfer, no matter how small the resistance is: less resistance only makes the current spike shorter and taller. Charge pumps are efficient when C1 changes its voltage very little per cycle, which is why the doubler above reaches 99 % with a large C1 and loses efficiency as soon as the output is pulled away from 2 × VIN.

The idealized switches hide several real effects, and the simulator shows only what its models contain:

• Switch resistance. At high frequency the capacitors stop mattering and R_out levels off at about 4 × the resistance in each phase's current loop (8 × Ron for this four-switch doubler). Raising the clock beyond that point only adds switching loss. Real charge-pump ICs such as the ICL7660 have about 55 Ω of output resistance, mostly from their switches.
• Capacitor ESR. It adds to R_out too: the ICL7660 datasheet multiplies C1's ESR by four.
• Diodes instead of switches. Many discrete charge pumps use diodes. Each conducting diode subtracts its forward drop, so a diode doubler from 5 V gives noticeably less than 10 V even with no load.
• Switching and gate-drive loss, dead time and parasitic plate capacitance. None of these are in the voltage-controlled switch model; it is a logic-level switch, not a MOSFET.
• Start-up surge. At the first transfer, C1 stacked on 5 V meets an empty C2 through 20 mΩ: the simulated input current spikes to 500 A for a fraction of a microsecond. Real chips limit it with switch resistance and soft-start; relay-built charge pumps use a series resistor.

Keep model limits separate from numerical accuracy. The solver reproduces the ideal circuit to within a millivolt of an independent simulator; that does not make the ideal switch a real one. In the current build, very fine output steps (below about 1 µs) on this doubler with the default 100 MΩ switch off resistance can stop at the first step with an “inconsistent system” error. That is a solver robustness limit, not physics: use a 1 µs or longer output step, or a 1 MΩ off resistance as in the example.

Charge pump or inductor converter?

Both families switch at high frequency and store energy temporarily. They differ in what stores it and in how they handle a change of ratio.

Test your understanding.

Answer each one before you open it.

Continue learning.

Charge pumps tie together several ideas from the other guides: RC charging and time constants, the difference between a model and a real component, and choosing between a transient run and Live simulation. The doubler example page repeats the build video with the circuit file name and the key result. A good next experiment: give each switch 1 Ω of on resistance and watch R_out move from 1/(f · C1) toward the fast-switching limit.