The common approach for generating 5-V and 3.3-V supply rails in portable applications is to convert the main 5-V supply to 3.3 V using a lowdropout (LDO) linear regulator. LDOs offer low cost, small size, low quiescent current, and ease of implementation, but their conversion efficiency is poor—only 67% in this case.
The usual design alternative is the selection of a step-down switching regulator, which can provide up to a 90% typical efficiency. However, it requires an inductor, occupies more space, and is relatively expensive. A second alternative, the charge pump, can bridge this gap between the benefits of the LDO and switching-regulator approaches.
Charge pumps achieve exact multiples of an applied voltage by switching the terminals of external capacitors. They offer low cost, small size, ease of implementation, and conversion efficiencies as high as 95%.
Most charge-pump ICs generate +2 VIN, + 1/2 VIN, or −1 VIN, but the conversion of 5 V to 3.3 V requires a multiple of 2/3 (i.e., 5 V(2/3) = 3.3 V). Obtaining the 2/3 ratio requires two charge pumps, and it’s easily implemented with a combination of +2 VIN and −1 VIN. Those multiples can be obtained from the configuration shown for a dual charge pump (IC1) (Fig. 1). Note that V- connects to ground and GND serves as the output. The entire circuit fits in 0.08 in.2 of board space.
The circuit’s power-conversion efficiency is as high as 95% (Fig. 2). IC1’s output impedance (33 Ω) and typical quiescent current (only 37 µA for this configuration) make it suitable for load currents in the range of 0.1 mA to 5 mA. Because its architecture provides a fixed conversion ratio of (2/3)VIN, VIN must be a regulated 5 V to provide a fixed 3.3-V output. Otherwise, IC1’s very wide input range (2 V to 18 V) allows for a range of dc-dc conversions. In addition, an output of (4/3) VIN is available at V+ (pin 7).
I need some circuits that can really help to improve my knowledge on the charge pump.
Jaisankar -November 25, 2004
In order to understand the working of the charge pump, the component level (capacitor+other discrete component)circuit is required. Please help me.
kiran sv -November 21, 2004
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