Assume again that you have a supply voltage of 12 Volts, but you need 3.3 Volts. Your load consumes 1 Ampere.
A linear regulator acts as a resistor which drops the unneeded 8.7 Volts. In the process it converts 8.7 Watts into heat. 3.3 Watts are used by the load; a rather dismal efficiency.
Enter the switching regulator: instead of creating a resistance between input and output, it connects an inductor between the two for short periods of time.
The switch, S1, is driven by a pulse generator (PWM, or
pulse-width modulator). The
pulses are rapid, so that the inductor value can be small.
The inductor, together with C1, smoothes out the switching pulses.
When the switch is closed, the left node of the inductor is at Vcc (assuming the switch has no resistance), but when the switch opens, this voltage jumps abruptly to a large negative value, created by the energy stored in the inductor. It is the purpose of D1 to catch this negative spike so it does no harm to the switch and provides a path for the current during the off period.
Figure 14-20 shows the resulting waveform at the output, for duty cycles of 10%, 20% and 40%. The average output voltage is simply proportional to the duty-cycle, but there is a noticeable ripple, the remains of the switching frequency (100kHz).
There is also an overshoot, which becomes more pronounced as the duty-cycle increases. This
undesirable behavior is due to the LC filter (L1, C1).
With ideal components the voltage conversion is 100% efficient.
Out D1 PWM S1 L1 27u 4.7u C1 RLoad 10 Vcc
Fig. 14-19: Reducing a supply voltage with a series switch and inductor.
Fig. 14-20: The resulting waveform at the output. Time/uSecs 100uSecs/div 0 100 200 300 400 500 Output / V 0 1 2 3 4 5 6 7 8 40 20 10
But when you add some resistance to the switch and inductor and a forward voltage drop for the diode, the efficiency drops. For example, with a total
resistance of just 50mΩ
and a diode drop of 0.3Volts (a Schottky diode) the efficiency is 94%.
The circuit of figure 14-19 is not a regulator; we have to add feedback to make the output voltage immune to supply fluctuations. This is accomplished by amplifying the difference between a fraction of the output voltage (R1, R2) and a reference voltage in an error amplifier. S1, an abstract simulation symbol, is now used as both a switch and a comparator (with the on/off thresholds set just a few millivolts apart). The output of the low-pass filter (R3, C2) following the error amplifier is thus compared with a triangle wave (100kHz, 2Vpp). In this way the regulator finds the duty cycle which gives the desired output voltage. Such a circuit is generally called a Buck
Regulator.
There are a few items to consider, which are peculiar to a switching regulator:
First, an actual switch is not a perfect device; you will have to make a painful compromise, weighing voltage drop and speed: the lower the voltage drop the more current it takes to drive the device. For example, a discrete MOS transistor with an "on"
resistance of 100mΩ at 1 Ampere
has a total input capacitance of about 1nF. At a switching frequency of
100kHz you will need to turn the device on and off in less than 50nsec, otherwise the dissipation during switching becomes significant. This means the output of the comparator (the driver stage) has to provide 100mA to
Vcc Vreg 4 7 n C2 Error Amp. R2 2 7 k RLoad 10 4.7u C1 L 1 27u S1 Triangle 1 0 0 kHz D1 R1 3 2 k 1.2 Vref R3 100k
Fig. 14-21: "Buck" regulator.
Time/mSecs 200µSecs/div 0 0.2 0.4 0.6 0.8 1 Probe1-NODE / V 0 0.5 1 1.5 2 2.5 3
charge and discharge 1nF. If you push the switching frequency to 500kHz, this current increases to 0.5 Amperes.
Second, the current level that the switching transistor needs to handle is always larger than the average output current. If you use a small inductor, the peak current can exceed the average by a factor of three or more; with a large inductance this factor is between 1.1 and 1.4.
Third, the voltage drop (and switching speed) of the diode is just as important as that of the switching transistor, their peak currents are roughly equal.
Fourth, the output LC filter (L1, C1) form a pole, which makes frequency compensation (R3, C2) more challenging.
We can step up the voltage by using the induced voltage in an inductor. Switch S1 connects the inductor L1 across the power supply (here assumed to be 1 Volt). The current flowing through the inductor is given by:
I V t
L = ∗
As soon as the switch is turned off, a positive voltage appears at the anode of the diode, created by
the stored current. This voltage is averaged by C1.
Output RLoad 25 D1 L1 10u PWM 47u C1 S1 Supply
Fig. 14-23: By using inductive charge the output voltage can be made higher than
the supply voltage.
Time/mSecs 1mSecs/div 0 1 2 3 4 5 D1-cathode / V 1 1.5 2 2.5 3 3.5 4 70 Percent 50 Percent 30 Percent
Fig. 14-24: Output voltage for three different duty cycles.
Time/mSecs 5µSecs/div 1.435 1.44 1.445 1.45 1.455 1.46 mA 0 50 100 150 200 250 300 Switch Current Output Current
Fig. 14-25: Currents through switch and load.
The magnitude of the output voltage depends on how long the inductor is charged (i.e. what peak current is reached). Thus, by changing the duty cycle, the output voltage is altered. Note that in this configuration, too, the current the switching device must handle is considerably larger than the output current.
Add feedback and we have a Boost Regulator. As before, the switch symbol represents both the switch and a comparator (i.e. the switch turns on and off within a few millivolts of the differential input signal). But be aware that, in this configuration, the feedback circuitry must have some specific characteristics: the output of the error amplifier must be constrained so that it stays within the amplitude of the triangle wave-form, otherwise the regulator can hang up at either zero or full output.
The frequency compensation network (R3, C2) also provides a "soft start", i.e. the output voltage builds up gradually, without much of an overshoot.
The same principle of using the inductive "kickback" voltage is also used to regulate larger voltages (such as a 110V or 220V line input). The inductor becomes a transformer, with a secondary winding delivering a lower voltage (isolated from the line).
Feedback to the switching device is effected through an optical link (an LED and a phototransistor) to also provide isolation.
Some of the devices in such a line regulator, including the switching transistor, need to operate at high voltage; you need to be aware that this increases device size considerably (see panel).
Fig. 14-26: Boost switching regulator.
R3 50k Vcc (1 Volt) D1 L1 10u S1 R1 32k 47u C1 RLoad 25 R2 27k 1.2 Vref Triangle Error Amplifier 100n C2 Output (2.5V) Time/mSecs 2mSecs/div 0 2 4 6 8 10 12 14 Output Voltage / V 0 0.5 1 1.5 2 2.5
Fig. 14-27: Soft-start of the boost regulator.