Supply uses ac to generate 5V, power-on reset

Edited by Bill Travis

Greg Sutterlin, Maxim Integrated Products, Sunnyvale, CA -- EDN, 5/16/2002

The need often arises for a low-cost logic supply for powering microcontrollers and related circuitry in "white-goods" products, such as industrial controllers and sensors. These applications usually include 24 or 115V ac or higher levels of ac voltage for conversion to 3.3 or 5V dc. The simplest approach to generating low-current logic-supply levels is to apply the rectified and filtered ac input to a high-input-voltage linear regulator. However, power dissipation in the regulator can be considerable, even for modest load currents. A standard shunt regulator also dissipates notable amounts of power in the limiting resistor. A switching regulator minimizes power dissipation, but that type may be impractical for cost-sensitive designs. As an alternative, consider the ac-coupled approach in Figure 1. The circuit suits applications in which 24V ac is available. With proper safety precautions, you can apply the circuit to double-insulated white goods and other products that require a logic supply for control or monitoring functions.

To transfer energy to the regulator with negligible power loss, the circuit uses a coupling capacitor in conjunction with an IC containing a shunt regulator and power-on-reset circuitry. Available with 50-mA maximum output-current capability in 3, 3,3, and 5V shunt-voltage versions, IC₁ also includes a power-on-reset function. Because IC₁ is an active shunt versus a passive zener diode, you must rectify the ac voltage before applying it. Typically, a capacitor follows the rectifier to hold the charge during off cycles. The design in Figure 1 uses a simple half-wave rectifier to save cost. C₁ is the transfer capacitor, and C₂ stores energy. D₁ acts as a half-wave rectifier, and D₂ discharges the transfer capacitor during negative cycles. R₁ limits surge current during the discharge of C₁ and, if applicable, during high-voltage transient testing.

Several simplifications help to approximate the available output current. Assume 0V forward drops in the diodes and 0V shunt voltage in the IC regulator. With, for example, a 60-Hz sinusoidal input of 24V rms amplitude (V_PEAK=33.94V), you calculate as follows:

Peak current in C₁ is I_PEAK(C₁)=C₁(dV_S/dt)=C₁[V_PEAK(dsin(ωt)/dt)]=C₁[ωV_PEAKcosωt))]=C₁ωV_PEAK.

Thus, with C₁=3 µF, ω=377.1 radians/sec, and V_PEAK=33.9V, I_PEAK=38.4 mA. The rms charge current (I_RMS) in C₁ is

Thus, for T=16.7 msec, I_RMS=19.1 mA. By adjusting the value of C₁, you can limit peak-current levels to the value of maximum MAX6330 shunt current, 50 mA, and achieve an output of 20 mA or so. The voltage rating of C₁ must be able to withstand the maximum input voltage. Because peak currents are limited to I_PEAK, practically any small-signal diode can serve as the half-wave rectifier, D₁ . D₂ discharges C₁ during the negative portion of the cycle. The current rating of D₂ depends on the value of V_PEAK and the selected value, 50Ω, of surge-limiting resistor R₁. The maximum reverse voltage on D₁ and D₂ is (V_SHUNT+V_DIODE). C₂ acts as a storage capacitor that maintains load current during the negative portion of the cycle. To calculate its value, use the following approximation based on the allowable level of ripple voltage (V_RIPPLE): C₂=(I_LOAD×T/2)/V_RIPPLE. With V_RIPPLE=150 mV, T/2=8.3 msec, and I_LOAD=10 mA, C₂=550 µF.

Is this the best Design Idea in this issue? Select at www.ednmag.com .