In telecom applications, the most common power-supply input voltage is −48 V. For years, telephone company central offices have supplied −48 V to operate standard phones, and they continue to do so. So for standardization purposes, the dc-dc converters supplying logic (+5 V) and analog (+12 V) voltages often use −48 V as the input supply.
The higher input voltage—higher than a +5-V supply, for example—has a substantially lower input current for a given output power. This significantly reduces wire losses and costs. Since phone lines span long distances, the power lost in the form of wiring losses can be significant.
To convert the higher and inverted input voltage of −48 V to lower, multiple positive voltages, flyback converters are often the best approach at low to medium powers. In a flyback converter, the output voltage can be configured to have any polarity and amplitude with respect to the input voltage. Also, multiple outputs are readily available just by adding secondary windings to the transformer.
The circuit shown in Figure 1 yields a low-cost solution for converting a −48-V supply to the dc voltages needed for a TV set-top box. The magnetic element is not a transformer but a coupled inductor, although additional windings can be added to it just the same. A transformer by definition does not store energy. Rather, it transforms one voltage or current level to another. For simplicity, the term "transformer" will be used to describe the multiwinding structure.
The coupled inductor is set up so the −48 V is impressed across the entire primary winding during the on-time of the primary switch. The secondary output voltages are tapped off the primary winding, and current is supplied to the output during the off-time.
As a result, the primary and secondary share the same winding by virtue of configuring the ground of the secondary as the positive terminal of the primary. This has the advantage of fewer windings over an all-isolated transformer structure. The transformer, then, is cheaper. Also, and perhaps more importantly, the leakage inductance between the input and output are drastically reduced. This reduction yields smaller voltage spikes on the primary switch at turn-off, providing less stringent snubbing requirements. Additionally, a lower-voltage MOSFET can be used, improving efficiency while reducing cost.
An isolated winding supplies the bias current for the SC1101 pulse-width-modulation (PWM) controller IC. This low-current winding needs to be isolated since its return is −48 V, not ground. In other words, the −48-V rail serves as the chip's ground.
When the main switch (Q2) is on, current flows in the primary winding for a time determined by the duty cycle and switching frequency. Current flows from ground, or the return of the secondary, to the −48-V rail through MOSFET Q2, storing energy in the transformer. Once the current ceases, the voltage at Q2's drain increases to meet the voltage required by the coupled inductor's outputs.
The voltage at pin 1 of the transformer is clamped to 10 V (plus a diode drop) during Q2's off-time (Fig. 1, again). The number of turns in the transformer is:
Pins 1-8 = 9 (10-V winding to Q2 drain
Pins 1-2 = 6 (6.5-V winding to 10-V winding)
Pins 2-6 = 9 (6.5-V winding to ground)
Pins 6-3 = 24 (−17-V winding to ground)
Pins 4-5 = 9 (bias winding to −48 V, isolated)
From this turns information, a primary versus secondary turns ratio (TR) can be derived. The derived TR is equivalent to having a standard transformer with separate primary and secondary windings. Remember that in this structure, the primary and secondary windings are the same. This reduces leakage, number of turns, and cost:
TR (10-V winding) = 9/(9 + 6 + 9) = 0.375
TR (6.5-V winding) = 6/(9 + 6 + 9) = 0.25
TR (−17-V winding) = 24/(9 + 6 + 9) = 1
TR (bias winding) = 9/(9 + 6 + 9) = 0.375
The duty cycle (D) of the flyback converter now can be calculated. Since the loop is closed around the highest power winding for best coupling, the 10-V, 1.5-A winding with a TR of 0.375 determines the duty cycle:
D = VOUT/(VOUT + TR × VIN) = 10.6 V/(10.6 V + 0.375 × 48) = 0.37
For the main output of 10 V, 0.6 V is added for the forward drop of the rectifier diode.
This is a suitable duty cycle since at high input voltages, input currents are lower, and narrow duty cycles do not result in excessive peak switch currents. Also, since the output ac ripple current is calculated using the off-time (1 − D), low values of D result in longer off-times and lower peak output currents, which reduces output-capacitor size and cost:
IOUT (avg) = IOUT (dc) = IOUT (pk) × (1 − D)
or:
IOUT (pk) = IOUT (avg)/(1 − D)
The nine additional turns between the drain of Q2 and the 10-V winding are used to optimize the duty cycle. If those additional turns were reduced to zero, D would be reduced to 0.22 at VIN = 48 V. This would yield too low a duty cycle at maximum input voltages of 60 V or higher, increasing the peak input currents.
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