A switch-mode power supply (SMPS) offers a convenient and efficient way to reduce a large input voltage (the ac mains voltage) to power lower-voltage loads. The heart of the switching power supply is a pulse width modulation (PWM) controller. This integrated circuit draws the energy it needs to operate from an auxiliary winding once steady-state operation is reached.
However, before this happens, there is no energy source in the circuit to crank the power supply, even if the ac mains is suddenly applied. To energize the PWM chip, a resistor is wired from the bulk capacitor (the high-voltage rectified rail) to charge up a VCC capacitor. During this startup period, the controller draws as little current as it can. Some controllers draw 500 µA, while others consume less than 50 µA, depending on whether the IC was fabricated in bipolar technology or the lower power CMOS technology.
When the voltage across the VCC capacitor reaches a given threshold, the controller starts to pulse, and its consumption dramatically increases. At that point, the startup current delivered by the resistor becomes negligible and the VCC capacitor solely delivers energy to the controller. If everything is well designed, after a few milliseconds the auxiliary winding connected to VCC via a diode takes over and self-supplies the controller.
The VCC capacitor selection is made on the hold-up time needed from this capacitor to supply the controller until the auxiliary takes over. If the VCC voltage drops below a level called undervoltage lock-out (UVLO) — for example, in the presence of an overload — the controller stops its pulses.
Usually, the UVLO level is selected to ensure a proper bias of the internal logic circuit and to enhance the MOSFET with a sufficient Vgs voltage. Keep in mind that the final value of Vgs,in current-mode SMPS is the driving voltage minus the sense resistor drop, usually 1 V. Fig. 1 depicts a typical controller configuration with a UC3843.
There are various ways to calculate the startup resistor value. One uses the startup time, which is the time allowed for the power supply to deliver its nominal voltage value. This time can be split into two intervals: startup1 is the time sequence described previously (controller crank-up), and startup2 is the time needed by the power supply to ramp up VOUT to its final value.
Typically, startup2 is much shorter than startup1, so we'll focus on this latter parameter. Suppose startup2 equals 10 ms and we need a startup1 time of 250 ms from an 85-Vac input. Our controller startup current is 50 µA and its VCC turn-on voltage is 15 V. When the controller starts to pulse at VCC = 15 V (and consumes current because it drives a capacitive load, the MOSFET), the VCC capacitor must hold enough voltage at least until the auxiliary voltage takes over and before VCC reaches the UVLO level (startup2 parameter). The value of capacitance required for the VCC capacitor can be calculated from:
CVcc > (ICC1 × startup2) / (VCC(turn-on) - VUVLO) (Eq.1)
where ICC1 is the total controller consumption with the selected switching frequency and MOSFET. Given a startup1 of 250 ms, a startup2 of 10 ms, a VCC(turn-on) of 15 V, a VUVLO of 8 V and an ICC1 of 2.5 mA, we can calculate the VCC capacitor required to hold 7 V during 10 ms when delivering 2.5 mA of total current:
CVcc>(2.5 mA × 10 ms)/(15 V - 8 V) > 3.6 µF or a 10-µF normalized capacitor.
Now that we have selected a 10-µF capacitor, we can determine the needed startup current (IST) from the startup1 design constraint:
IST > (VCC(turn-on) × CVcc)/ startup1 = (15 V × 10 µF) / 250 ms IST > 600µA
To this result, we should add IC, the current consumed by the controller below VCC turn-on voltage. If an IC of 50 µA is specified, then the startup resistor must provide 650 µA from the high-voltage rectified rail Vbulk, which has a minimum value VIN(min) of 85 × 1.414 = 120 Vdc. The value of the startup resistor (Rstartup) can then be calculated as:
Rstartup = (VIN(min) - VCC(turn-on))/ (IST + IC) = (120 V - 15 V) / 650 µA = 161-kV resistor (Eq. 2)
Despite the fact that this resistor must be a costly high-voltage type, the problem comes from its permanent connection to the high-voltage rail. When this rail will start to ramp up to 374 Vdc (265 Vac), the consumption will then be (374 V)2 /161 kΩ or approximately 800 mW, if we neglect the VCC level. With 800 mW of permanently wasted consumption, forget about low standby power.
An interesting alternative is the half-wave connection, whereby the startup resistor is wired to the mains and takes power from the half-rectified input voltage rather than Vbulk. However, this change merely reduces power consumption by a factor of π / 4, which in this case reduces startup resistor losses to 630 mW.
Bipolar Startup Source
Rather than connecting “power” resistors to the bulk, we can create a small power supply that will charge the VCC capacitor and then turn off when the auxiliary winding takes over. Fig. 2 offers a solution built around ON Semiconductor's MPSA44 and a Zener diode.
When Vbulk is applied, the resistor R biases the Zener diode and its reference voltage, diminished by two forward drops — Vbe and D — charges the VCC capacitor. When the controller has started to pulse, the auxiliary voltage rises. If a proper design has been made (for example, a final voltage above the Zener level minus the two drops), then D blocks and the MPSA44 turns-off, leaving only the low-value biasing current through R. Now, you can try to boast low standby power. D prevents a lethal E-B avalanche.
Direct Connection to High-Voltage Rail
Recent progress in high-voltage technologies has enabled almost lossless startup sources. The principle is similar to that of Fig. 2, except that the source duty is left to a lateral MOSFET. Fig. 3 depicts the internal circuitry of the NCP1203 controller, directly supplied from the high-voltage rail.
At power-on, the device injects current (3 mA, typical) into the VCC pin and charges up the VCC capacitor. When the VCC voltage reaches 12 V, the current source opens and the controller starts to pulse. As usual, the capacitor supplies the controller until the auxiliary winding takes over. Once the startup sequence is done, the current source being off, it is fully transparent and no longer consumes leakage, except a 35-µA current. At 330 Vdc, power consumption becomes 11.5 mW. Low no-load standby power becomes possible.
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