Power Supply Unit 5
Power Supply Unit 5
Power Supply Unit 5
POWER SUPPLIES
5.1 RECTIFIERS
Diodes are referred to as non-linear circuit elements because of the above characteristic curve.
For most applications the non-linear region can be avoided and the device can be modeled by piece-wise
linear circuit elements. Qualitatively we can just think of an ideal diode has having two regions: a
conduction region of zero resistance and an infinite resistance non-conduction region. For many circuit
applications, this ideal diode model is an adequate representation of an actual diode and simply requires
that the circuit analysis be separated into two parts: forward current and reverse current. Figure 5.1 shows
a schematic symbol for a diode and the current-voltage curve for an ideal diode.
Figure 5.1 a)Schematic symbol for a diode and b) current versus voltage for an ideal diode
Figure 5.2
Equivalent circuit model of a junction diode
A diode can more accurately be described using the equivalent circuit model shown in figure 5.2. If a
diode is forward biased with a high voltage it acts like a resistor (Rf) in series with a voltage source (VPN).
For reverse biasing it acts simply as a resistor (Rr). These approximations are referred to as the linear
element model of a diode.
Figure 5.3 shows a half-wave rectifier circuit. The voltage source VS is an AC source
……..(5.1)
of frequency radians per second, and Vm is the maximum or peak voltage. Note that
where f is the frequency in Hz. Generally, the source vS is the secondary winding of a transformer.
The diode is the component which does the rectification, since it permits current flow in one
direction only. The resistor RL represents the resistance of the load drawing the power.
Let's analyse this circuit assuming the diode is ideal. When vS > 0, the diode is forward biased,
and so switched on; therefore vout = vS. But when vS < 0, the diode is reverse biased, i.e. switched off, and
hence vout = 0 V. This is illustrated in Figure 5.4.
.
Figure 5.4 Half-wave rectifier waveforms
The load voltage waveform vout is always positive, and so has a non-zero DC component, the average
value VAVG which we calculate as follows:
……(5.2)
If we use the practical diode model to take into account the diode voltage drop, then we need to reduce Vm
by 0.7 V when forward biased:
…..(5.4)
This means that the average voltage is reduced by the forward bias voltage drop across the diode.
The peak inverse voltage (PIV) is defined as the peak voltage across the diode when reverse biased: (note
that with zero current the voltage drop across RL is zero)
PIV = Vm……….(5.5)
The diode must be capable of withstanding this voltage.
In the half-wave rectifier the voltage is zero for half of the cycle. Full-wave rectifiers are designed using
two or more diodes so that voltage is produced over the whole cycle.
Figure 5.5 shows a full-wave rectifier designed using two diodes and a center-tapped AC supply (i.e.
center-tapped transformer).
The waveforms are shown in Figure 5.6. The center tapping implies that the two source voltages v1 and v2
are a half cycle out of phase. We see that diode D1 conducts when source v1 is positive, and D2 conducts
when v2 is positive, giving the waveform vout.
.
Figure 5.6 Full-wave rectifier waveforms
……..(5.6)
since the waveform is non-zero twice as much as in the half-wave case. Here, Vm(out) equals Vm if we
regard the diodes as ideal, and equal to Vm -0.7 V if we use the practical model.
Figure shows a bridge rectifier built from four diodes and a single AC source. The waveform
of vout is the same as for the center-tapped full-wave rectifier.
Figure 5.7 Bridge full-wave rectifier
The average voltage for the bridge rectifier is the same as in HWR but the peak inverse voltage is
…….(5.8)
It can be seen from Figures that the waveform vout is not very smooth. For many applications it is desired
to have a much smoother DC waveform, and so a filtering circuit is used. We will consider the filtered
half-wave rectifier of Figure, and leave the filtered full-wave rectifiers up to you to work out (not hard-see
lab).
Figure 5.7 Filtered Half-wave rectifier.
The waveform produced by this filtered half-wave rectifier is shown in figure 5.8 illustrating the ripple.
Here, ripple is defined as the difference between the maximum and minimum voltages on the waveform,
Figure (i.e. peak-to-peak).
Figure 5.8 Filtered Half-wave rectified waveform showing Vrpp
and VDC.
…..(5.9)
where Vrpp is the peak-to-peak ripple voltage and VDC is the DC component of the ripple waveform. T is
the period of the AC source voltage: T=1/f, . For f=50 Hz (the frequency of the AC supply in
Australia), T= 20 ms.
We now explain how to calculate (approximately) Vrpp and VDC. Think of the ripple waveform as being
approximated by a triangular waveform so that
…..(5.10)
Using symmetry. Suppose that at the beginning of a cycle the capacitor is fully charged to Vm(out), and that
the capacitor is large enough so that the time constant RL C is much larger than T. The rate of change of
vout at the beginning of the cycle t=0 is
…….(5.11)
…….(5.12)
approximately (straight line approximation). This allows one to design C for a given load and desired
ripple.
…….(5.13)
β = R2 / (R1+R2) …….(5.14)
This reduces the output voltage Vo of the difference amplifier due to the 180‟ phase
difference provided by the amplifier.
Vo is applied to the base Q1 which is used as an emitter follower.
Vo follows Vo‟ i.e. Vo also reduces.
Hence the increase in Vo is nullified.
Similarly reduction in output voltage also gets regulated.
Advantages:
1. Low cost
2. High reliability
3. Reduction in size
4. Excellent performance
Characteristics:
The regulated output voltage is fixed at a value specified by the manufacturer. For e.g.
78XX series has a output voltage at 5, 6, 8 etc.,
The unregulated input voltage must be at least 2V more than the regulated output voltage
i.e. | Vin | >= Vo +2 volts. For example, if 7805 regulator has Vo = 5V then Vin = 7V
The load current Iomax may vary from 0 to rated maximum output current.
Thermal shunt down : the IC has a temperature sensor (built-in) which turns off IC when
it becomes too hot. The output current will drop and remain there until the IC has cooled
significantly.
The three terminal regulators discussed earlier have the following limitations
1.No short circuit operation
2. Output voltage is fixed.
These limitations have been overcome in the 723 general purpose regulator, which can be
adjusted over a wide range of both positive or negative regulated voltage. This IC is
inherently low current device, but can be boosted to provide 5 amps or more current by
connecting external components.
Limitation:
It also has no short circuit current limits.
It has no in-built thermal protection.
Figure 5.11 723 general purpose regulator
Ii ( as IQ is small )
I B Ii I R1
VEB(on)
IO
R1
The maximum current Io(max) for a 7805 regulator is 1A from the data, assuming
Veb(on)= 1 V& = 15 we get from equation (8),
LIMITATION:
It also has no short circuit current limits.
It has no in-built thermal protection.
Figure 5.13 a. current fold back characteristic curve b. current fold back
The voltage at terminal CL is divided by R3-R4 network.
The current limit transistor Q2 conducts only when the drop across the resistance Rsc is
large enough to produce base emitter voltage of Q2 to be atleast 0.5 V.
As Q2 starts conduction, transistor Q1 begins to turn off and the current IL decreases .
This reduces the voltage v1 at the emitted of Q1 and also the output voltage vo .
Voltage at the base of Q1= v1R4/(R3+R4)
Thus the voltage at the CL terminal drops by a smaller amount compared to the drop in
voltage at Cs terminal.
This increases Vbe of Q2 thereby increasing the conduction of Q2 which in turn reduces
the conduction of Q1 i.e., the current IL further reduces.
This process continue till Vo=0 V & V1 is just large enough to keep 0.5 v between CL &
Cs terminal.
This point is Isc & has been reduced by lowering both IL & Vo.