Converting an available DC source to a lower or higher voltage is a very common challenge in circuits. One option for the high-to-low conversion is to use a low-dropout regulator (LDO), but how can an LDO easily transform a lower voltage into a higher one?
The answer for AC voltages is well-known: use a transformer, as has been done for over 100 years. Yet, as even every first-year electrical engineering student knows, you can’t use a transformer with DC. The obvious approach is to “chop” the low-voltage DC using an oscillator of some sort, pass the chopped, AC-like waveform through a step-up transformer, and then rectify and filter it at the secondary-side output. This approach can be very successful, and enhanced versions of it are the basis for switching power supplies, used to both increase (boost) and decrease (buck) the voltage between a DC source and a supply rail.
Q: What are the drawbacks of this approach?
A:The key issue is the need for the transformer. This inductive component is relatively large and costly compared to the rest of the power-conversion circuitry it supports. While some power converters prefer or even mandate a transformer due to its inherent galvanic isolation, that benefit is often not needed in low-voltage circuits or localized subcircuits. A transformer-based design’s performance and cost are more suited for dc-dc converters above about 1 to 5 A output, but it is generally not an attractive solution at the low end below a few hundred mA.
Q: What’s the better alternative?
A: Circuit designers have developed a topology called the charge pump, which is difficult to implement with discrete components but very IC-friendly. The charge pump uses capacitors as the energy-storage element.
In the basic execution of this power-conversion technique, current (charge) is alternately switched and directed between two capacitors arranged so the circuit output is twice the input, thus functioning as a voltage-doubling boost converter. For these reasons, the charge-pump converter is also known as a switched-capacitor design.
[Note that a variation of the charge-pump voltage converter is used in phase lock loops (PLL), but that application and its attributes are beyond this power-related discussion.]
Q: How does the charge-pump voltage-doubler work?
A: How is this voltage-doubling boost accomplished? It all starts with a fundamental principle of physics: charge flowing back and forth in a closed circuit is not “lost,” but instead can be transferred via switching between charge-storage elements. In a charge-pump concept, diodes can be used to control the flow of current; in actual practice, the switches are usually switched MOSFETs, and the capacitors are external ceramic or electrolytic devices depending amount of capacitance needed.
Figure 1 shows a two-step charge-discharge cycle in which capacitor C1 charges and then discharges into C2. First, the clock drives the output of inverter 1 low, so D1 is forward-biased, thus charging capacitor C1 to the supply voltage +Vdc; also, D2 is off.
Next, the clock drives the output of inverter 1 high, and the charge on C1 is now in series with +Vdc from inverter 1. As the output of inverter 2 is low, D2 becomes forward-biased and C2 charges to twice Vdc. The voltage thus seen across the load is 2 × Vdc, minus the diode forward-voltage drops and any losses in the inverters.
In practical designs using discrete components, Schottky diodes are usually used instead of conventional diodes because of their lower forward-voltage drop. However, IC-based charge pumps do not use diodes; instead, they use MOSFET switches with low on-resistance RDS(ON). Charge pump efficiency is fairly high, in the 90 to 95% range.
Part 2 examines additional aspects of charge pumps, including their capacitors, non-doubling variations, internal and external clocks, filtering and regulation, and embedded charge pumps.
Chinomso says
I’v been building this 12v to 240v: 2000w inverter circuit. In the circuit, I wanted to embed a chargepump circuit such that the 12v input would be stepped up to drive the power stage of the inverter for maximum output, since 12v from the battery couldn’t do it well. I successfully build the charge pump, however, on powering the circuit, I notice that the chargepump could not stand the strength of the mosfets at the power stage. Please how can this be done?
Borno says
How do you calculate the efficiency of this charge-pump?
Can you explain it by equation?
Thanks
Aimee Kalnoskas says
Thank you for your interest, Borno. Here is the reply from the author, Bill Schweber:
Not surprisingly, the efficiency of charge pump is hard to analyze. It’s a function of diode drop, capacitor ESR, switching losses, duty cycle, internal resistances, and other factors. As a first-cut estimate, 70% efficiency is a good initial number. They are tricky to model and simulate because of their discontinuous waveforms, too.
Possible resources:
–TI app note:
http://www.ti.com/lit/wp/slpy005/slpy005.pdf
–Analog Devices article:
https://www.analog.com/en/technical-articles/clever-circuit-improves-efficiency-of-charge-pump-plus-linear-regulator-solution.html
–Academic article:
https://www.researchgate.net/publication/328218249_Transient_and_Steady-State_Analysis_of_Single_Switched_Capacitor_Converter/fulltext/5bbf494f45851572315f4613/Transient-and-Steady-State-Analysis-of-Single-Switched-Capacitor-Converter.pdf
–Article from Power Electronics Technology aka Power Electronics:
https://www.powerelectronics.com/passive-components/calculating-essential-charge-pump-parameters