Designing a Capacitive Charge Pump Power Supply Jim Hagerman, IBiS Networks Circuit

Transcription

1 esigng a apacitive harge Pump Power Supply Jim Hagerman, BiS Networks ircuit The basic circuit is shown below and is the classic non-isolated A to capacitive converter. t works just fe and provides a simple and low-cost solution for many products. There s just one thg every application note ve read explas it wrong! ectification The circuit employs half-we rectification. Other application notes show how the put A voltage sewe A gets rectified to somethg like weform B. Well, that s just pla correct. This is a current mode converter, not voltage! The portion of put sewe actually beg rectified is shown weform. The risg voltage jects a current to the put via capacitor. Not only that, but current reaches a maximum as the put voltage crosses through zero, not at the voltage peaks. erivation For the purposes of understandg the method of operation, let s ignore and everythg to the right of it. We can also ignore the fuse F and resistor, as they serve no purpose under normal conditions.

2 As the put voltage rises, a current is jected to defed as The impedance is defed by 1 π f As we are lookg to determe the erage put current ( will store charge and smooth the put voltage), we start by obtag the erage put voltage. For half we rectification, this is given by p π π This equation remas true even though we are usg weform. One thg we need to note is that the negative and positive peak voltages across are not the same. urg the off half of the cycle, current is pulled of. Therefore, the negative peak voltage is reduced by the amount of, with forward biased. For the on half of the cycle, current pushes to resultg a positive peak reduced by. ombg the above equations, we get π + 1 π f π f ( ) This is the full cycle erage put current. We can take advantage that the peak A put voltage is many times greater than any practical zener voltage (typically we want somethg like a 5 put), which simplifies the formula to f e-arrangg this we solve for f

3 For an application runng at 10 and Hz this simplifies to (with given µf) 50 should be a good quality film capacitor that can handle the bipolar ripple currents and voltages. Polyester is a good choice; polypropylene is even better. oltage The put current is turned to an put voltage via the zener. The circuit will fall of regulation if the current is too high, and must be less than the erage put current. Under these conditions, maximum put voltage is is used to filter the discontuous put current to a smooth put voltage, where the ripple is dependent on the current. The value of this filterg capacitor is determed by i t f For a Hz system with a voltage ripple of 1, this simplifies to egulation Most applications will require a regulated put voltage. An LO type is perfect for this design, as we don t want to burn any extra unnecessary power across the regulator. For a 5 put with 1 ripple on, then should be at least 6. Takg to account the forward voltage drop across, the mimum zener voltage should be 6.8. Protection Sce this supply is powered directly from A mas, the capacitor should he an appropriate safety ratg. n other words, it needs to fail as an open circuit, not a short. When such a capacitor cannot be found the required value, addg a fuse will sidestep the issue. That is why F is shown the circuit. Similarly, is only there for surge protection. Under normal conditions it is not needed, but when there is a voltage surge (ES, lightng, etc.), the peak current through and may exceed their ratgs. To solve this the resistor is added. But what

4 value should it be? m proposg a value that is 5% of the impedance of. This will provide some level of protection with significantly alterg circuit performance π f For a Hz system this simplifies to (with given µf) 130 The power dissipation this resistor is determed by the put current. However, is given for a full cycle erage. Sce the entire charge is pumped durg half a cycle (half-we rectification), that value is actually double. And it is the same, but the opposite direction for the other half of the cycle (current that isn t beg used except to charge ). Therefore, the power dissipation the resistor is P ( ) n a practical application, the resistor should he a ratg of at least 3x this to keep temperatures reasonable. Example Let s say we he a need to power a 5 circuit at up to 75mA. Step one is to set slightly higher, to prevent drops. Let s chose 80mA. We then get 50 50(0.08) 4 uf Let s select a value of 4.7µF to buy us some headroom agast low put voltage and component tolerance. Aga our choice for zener voltage is 6.8. The mimum filter capacitor needs to be (0.075) 1,500µ The series resistor calculates to F Ω 4.7 Power dissipation the resistor is

5 P ( ) ( 0.08) 7 0. W 7 Similarly the power dissipation the zener diode can be as high as 0.54 watts, if the current drops to zero. n summary, for a practical design we get omponent alue atg 4.7µF 50A 7Ω W 15,000µF W ual Supply As we noted earlier, the off portion of the cycle burns just as much put current as the on portion. We can use this! By addg just a few more components we can add a negative supply. This is often useful, especially if one desires to power analog circuitry. The key to dual supplies is to stall back-to-back zener diodes. The put voltage is now equal to the zener voltage. Everythg else calculates the same. Usg dual puts doubles the overall conversion efficiency of the supply.