Switched-capacitor voltage multiplier achieves 95% efficiency
You can use low-on-resistance analog switches to improve efficiency.
Marián Štofka, Slovak University of Technology, Bratislava, Slovakia; Edited by Martin Rowe and Fran Granville -- EDN, June 10, 2010
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A capacitor that you charge through a resistor operates at 50% efficiency; hence, many engineers avoid using switched-capacitor dc/dc converters. That efficiency figure holds true only for capacitors with no initial voltage, however. If you decide to switch a precharged capacitor, you can transfer energy to an output with a power-efficiency approaching 100%.
CIN and COUT are filtering capacitors at the input and output, respectively. RP is a protective resistor, which limits the inrush current to the capacitors at power-on. As the output voltage rises, it closes IC5 and shorts out this resistor. Schottky diodes D1 and D2 allow you to power the ICs using the input voltage until the output rises, at which time the higher output voltage powers the ICs. CDC is a storage and decoupling capacitor for this power bus. The higher power-supply voltage is necessary for proper operation, and it lowers the on-resistance of the analog switches. The 0.4Ω on-resistance of IC1 results in low circuit losses and high-efficiency operation. IC1, IC2, and IC3 exhibit a break-before-make operation, which is essential in this case.
For a 50%-duty-cycle clock, you can calculate the theoretical power-efficiency of the converter, according to the following equation:
If the value of COUT is equal to the value of C, the power loss due to charging of the three capacitors is about two-thirds of the power loss during the discharging phase. The power consumption of the control circuit reduces the efficiency of this calculated value. For CMOS circuits, the power consumption rises linearly with the operating frequency. By choosing the operating frequency, you can optimize the efficiency of the circuit. The optimum frequency is inversely proportional to the load resistance, RL. Fortunately, the efficiency maximum is flat, so you can achieve efficiencies higher than 90% over a wide range of values for RL. You can attain 94% efficiency driving a 120Ω load over clock frequencies of 100 to 400 kHz. If you set a 229-kHz operating frequency, an input of 2.2V yields a 2.87V output at an efficiency of 95.9%. The optimum clock frequency shifts to lower values at lighter loads. You can drive a 269Ω load at 100 kHz and achieve an output of 2.88V.
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You are welcome Tony
Marian
Marian Stofka - 2010-17-6 00:55:16 PDT -
You are quite right Marian. So long as the output ripple is small, so that the capacitor voltages do not change much (relative to input voltage) the possible efficency can be quite high. I'd missed the fact that the circuit relies on SMALL capacitor voltage changes.
Thank you for pointing this out.
t
Tony Hooley - 2010-16-6 04:01:53 PDT -
Continuation of response to Tony:
As dV is much smaller than V; and this is the case of the DI, you can see that the capacitor's energy increment dW is much higher than the loss on resistance of the switch.
Marian Stofka - 2010-14-6 06:44:47 PDT -
Dear Tony,
Your words of "Unfortunately,...quoted." are just a verbal expression, which is not backed by any mathematical theory. If you made a rigorous analysis, you probaly got the same formula for the efficiency, as is given in the DI. What is even more important is, that the numerical data at the end of DI, including the over-95% efficiency, come from an experiment.
Perhaps the following will help you to absorb the idea:
Let assume a capacitor C, which has been precharged to voltage V. If you charge it further to a voltage V+dV;
the capacitor's energy increment will be: dW=C.V.(dV); "d" denotes her Greek "delta". Simultanously, the loss on resistor will be (1/2).C.(dV).(dV). If you now evaluate a "loss/(energy increment)" ratio, you get: (1/2).(dV)/V. As is dV
Marian Stofka - 2010-14-6 06:19:23 PDT -
Unfortunately this circuit cannot operate as quoted. Charging the three series-connected capacitors from the power line is exactly equivalent to charging one single capacitor of 1/3rd the capacitance of each from the power line, through the power line's internal resistance, which is never zero. This will gurantee that 50% of the energy consumed during the charging cycle is dissipated in the PSU's internal resistance, and 50% stored in the capacitors. You cannot charge a capacitor losslessly (even ideally) without using an inductor.
Tony Hooley - 2010-14-6 01:29:10 PDT
