A DC to DC voltage doubler is a switched-capacitor circuit that uses oscillating switches and capacitors to step up a DC input voltage to exactly twice its amplitude without using an inductor or transformer. In a real circuit, this topology changes the design paradigm by allowing engineers to power high-voltage components—like RS-232 transceivers, LCD bias rails, or high-side MOSFET gates—directly from single-cell lithium or 5V USB rails without adding bulky, expensive magnetic components. Beginners frequently confuse switched-capacitor charge pumps with inductive boost converters; while both step up voltage, boost converters use inductors to store energy in magnetic fields and can achieve arbitrary step-up ratios, whereas doublers rely purely on electrostatic charge transfer and are strictly limited to integer multiples of the input.

The Mechanics of Switched-Capacitor Charge Pumps

At the bench, a voltage doubler operates by rapidly reconfiguring a "flying" capacitor between parallel and series connections relative to the input supply and a "reservoir" capacitor. This is typically driven by an internal oscillator and a set of MOSFET switches inside an integrated circuit like the classic ICL7660 or MAX1044.

Phase 1 (Charge): Switches connect the flying capacitor ($C_{fly}$) directly across the input voltage ($V_{in}$). It charges to $V_{in}$.
Phase 2 (Transfer): Switches flip $C_{fly}$ so its positive terminal connects to $V_{in}$ and its negative terminal connects to the reservoir capacitor ($C_{res}$). The reservoir now sees $V_{in} + V_{Cfly}$, effectively charging to $2 \times V_{in}$.

Because energy is transferred in discrete packets rather than continuously, the output is not a perfectly flat DC line. It contains a sawtooth ripple dictated by the switching frequency, the capacitance values, and the load current. Furthermore, the physical switches have on-resistance ($R_{ds(on)}$), and the capacitors have equivalent series resistance (ESR), which combine to give the doubler a non-zero output impedance.

Worked Numeric Example: 5V to 10V Gate Driver

Let's look at a common bench scenario: you need to drive a high-side N-channel MOSFET (like an IRFZ44N) from a 5V Arduino. To get a low $R_{ds(on)}$ and prevent the MOSFET from overheating, the gate-to-source voltage ($V_{gs}$) needs to be at least 10V. We will use an ICL7660A charge pump IC to double the 5V rail.

Circuit Parameters:

  • Input Voltage ($V_{in}$): 5.0V
  • Switching Frequency ($f$): 10kHz
  • Flying Capacitor ($C_{fly}$): 10µF (Low-ESR ceramic)
  • Reservoir Capacitor ($C_{res}$): 10µF (Low-ESR ceramic)
  • Average Load Current ($I_{load}$): 20mA (gate drive charging current)

Calculating Output Voltage Drop:
The theoretical output impedance ($R_{out}$) of a basic charge pump doubler is approximated by the formula:

$$R_{out} \approx \frac{1}{f \times C_{fly}} + 2R_{sw} + ESR$$

Plugging in our numbers for the capacitive reactance term: $1 / (10,000 \times 0.00001) = 10\Omega$. Adding roughly $5\Omega$ for the internal MOSFET switch resistance and $1\Omega$ for capacitor ESR, our total effective $R_{out}$ is about $16\Omega$.

Under a 20mA load, the voltage drop is:

$$V_{drop} = I_{load} \times R_{out} = 0.02A \times 16\Omega = 0.32V$$

Final Output Voltage: $(2 \times 5.0V) - 0.32V = \mathbf{9.68V}$. This is more than enough to fully enhance the IRFZ44N gate.

Calculating Output Ripple:
The peak-to-peak ripple voltage ($\Delta V$) on the reservoir capacitor is:

$$\Delta V = \frac{I_{load}}{f \times C_{res}} = \frac{0.02}{10,000 \times 0.00001} = \mathbf{0.2V \text{ (peak-to-peak)}}$$

A 200mV ripple on a 9.68V DC gate drive is entirely negligible for power switching applications.

Where You Meet This In Practice

While discrete doublers are great for learning, you will mostly encounter this topology integrated into silicon for specific subsystem tasks:

  • RS-232 Transceivers: The legendary MAX232 IC uses an internal charge pump to generate the $\pm$10V rails required by the RS-232 standard from a single 5V supply.
  • High-Side Gate Drivers: Motor driver ICs (like the TI DRV8701) use a bootstrap charge pump to generate a voltage higher than the motor supply rail, allowing them to turn on high-side N-channel MOSFETs efficiently.
  • LCD Bias Rails: Generating the negative contrast voltage (e.g., -5V) or positive bias rails for character and graphic LCDs from a 3.3V microcontroller rail.
  • EEPROM Programming: Generating the 12V to 15V "Vpp" programming pulse required to write to older parallel EEPROMs from a standard 5V logic system.
When to Abandon the Doubler: If your load requires more than 50mA to 100mA of continuous current, or if you need a step-up ratio other than exactly 2x (or 3x with a tripler stage), abandon the switched-capacitor topology. The output impedance scales poorly at high currents, and the physical size of the required low-ESR capacitors will exceed the size of a small inductor. Switch to an inductive boost converter (like the MT3608 or TPS61230) for high-current, high-ratio step-up needs.

DC to DC Voltage Doubler FAQ

How much current can a DC to DC voltage doubler supply?

Practical integrated charge pump doublers like the ICL7660A or MAX1044 are limited to roughly 20mA to 40mA of continuous output current. Higher-capacity variants (like the TPS60310) can push 100mA to 150mA, but they require physically larger ceramic capacitors to keep the output impedance low. Beyond 150mA, the capacitive energy transfer becomes highly inefficient compared to magnetic (inductive) transfer.

Why is my DC to DC voltage doubler output dropping under load?

Voltage sag under load is the hallmark of high output impedance in a charge pump. This is almost always caused by using standard aluminum electrolytic capacitors instead of low-ESR ceramic or tantalum capacitors for the flying and reservoir caps. The ESR of a cheap electrolytic cap can easily add 2 to 5 ohms of resistance per cycle, multiplying the voltage drop. Swap to X7R ceramic capacitors and ensure your PCB traces between the IC and the caps are short and wide.

Can I use a DC to DC voltage doubler instead of an inductive boost converter?

You can, but only for low-current, noise-tolerant applications. A doubler wins when you need a cheap, low-component-count solution (just the IC and two caps) and want to avoid the electromagnetic interference (EMI) and layout complexities of an inductor. However, an inductive boost converter wins when you need high efficiency, high output current, or a specific output voltage that isn't an exact integer multiple of your input.

Does a charge pump voltage doubler generate EMI noise?

Yes, but differently than an inductive switcher. Because there is no inductor to generate radiated magnetic fields, the EMI is primarily conducted noise injected back into the input supply rail via the high di/dt switching currents charging the flying capacitor. This can corrupt sensitive analog-to-digital converter (ADC) readings on the same microcontroller. To mitigate this, always place a dedicated 1µF to 10µF ceramic bypass capacitor directly across the $V_{in}$ and GND pins of the doubler IC, and consider adding a small ferrite bead on the input line if powering sensitive analog circuitry.