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Design Ideas: June 8, 1995

DC/DC controller generates negative supply

Michael Keagy,
Maxim Integrated Products, Sunnyvale, CA

thumbnail A switch-mode power-supply controller plus a diode-capacitor network generate the modest negative supply current needed for op amps or for LCD bias, without inductors. The circuit in Fig 1 accepts a 2 to 6V input, producing a digitally adjustable output voltage. The diode-capacitor charge pump is driven by the switching action of DHI and DLOW, which normally drive an external MOSFET or pnp transistor in an inductor-based switch-mode power supply.

thumbnail At power-up, the internal 6-bit current-output DAC produces a nominal VOUT of R1x13.33 µA. Then, by holding CTRL high and toggling ADJ, you can adjust VOUT over a 3:1 range in 64 equal steps, according to the value of R1, with VOUT between R1x6.66 µA and R1x20 µA. (If digital adjustment is not required, ground the ADJ pin and connect CTRL to the positive supply.) The maximum current depends on VIN, VOUT, and the number of diode-capacitor stages, each of which consists of two diodes and two capacitors. Maximum output voltage equals VIN-(0.6xnumber of stages), and you can draw more current at a lower output voltage (Fig 2).

The number of stages determines the maximum output current for given input and output voltages. Too few stages will not achieve the desired voltage; having too many stages degrades efficiency because the output current is (approximately) equal to the output current divided by the number of stages. For input voltages less than 5V, the circuit delivers less output current than indicated by Fig 2. For example, a four-stage circuit produces 1 mA at -2V out with 2V in, at -7V with 3V in, at -11V with 4V in, and at -14.5V with 5V in.

The controller changes its behavior when VOUT is much lower than the voltage programmed by R1. Because it is designed for inductor-based circuitry, it compensates for impending dropout by increasing the switching transistor's "on" time at the expense of "off" time. This action normally ramps up the inductor current, but it has an opposite effect for the circuit shown. Because short off times (DHI high) don't allow the capacitors to discharge fully, the available output current actually decreases instead of increases. Thus, you must reduce the load current significantly before regulation can be regained, when VOUT loses regulation due to overload. You should limit the maximum output current drawn at the level where VOUT recovers from dropout, not the higher level at which it enters dropout.

Output ripple is typically less than 1% for the values shown but can be higher if the circuit includes more stages than required by the programmed output voltage. The ripple can be reduced by increasing the output capacitance. Bypass R1 with 220 pF, keep connections between R1 and the IC very short, and place the input-bypass capacitor directly across the IC's input pins. (DI#1710) EDN


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