Elenco Variable Voltage Power Supply Kit - Manual (Page 10)

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The regulator circuit in the Model XP-15K Power
Supply consists of a LM317 integrated circuit. This
IC is specially designed to perform the regulation
function. Figure 6 shows a simplified circuit of how
the LM317 IC works.
Transistors Q1 and Q2 form a circuit known as a
differential amplifier. Transistor Q1’s base is
connected to a stable 1.5V reference voltage. The
base of Q2 is connected to the regulator output
circuit through a voltage divider network. The
collector of transistor Q2 is connected to a current
source. This basically is a PNP transistor biased to
draw about 1mA current. Transistor Q2 sees the
current source as a very high resistor of about 1
meg ohms. Thus, the gain of transistor Q2 is very
high.
Transistor Q5 is called the pass transistor. It controls
the current reaching the output. Transistors Q3 and
Q4 are emitter followers. Their function is to raise
the impedance of the pass transistor. Note that
transistor Q2, Q3, Q4, Q5 and resistor R1 form a
closed loop. Also, note that the feedback to the base
of Q2 is negative, that is, the output at emitter Q5
goes negative. Now, if the 1.25V output voltage
goes down because of current drain at the output,
the base of Q2 will drop, forcing the collector voltage
of Q2 to go higher. This will bring the output voltage
back to 1.25V. This is the basis of all negative
feedback regulators.
Another feature of the LM317 regulator is to protect
the IC against overload and output shorts. If the IC
is overloaded, the junction will overheat. A transistor
will sense this overheating and shut down transistor
Q5.
The LM317 IC is basically a 1.25V regulator. To be
able to vary the output 0 - 15V, we stack the IC on
the negative 1.25VDC voltage as shown in Figure 7.
When R3 equals 0, the output voltage is 0 volts.
REGULATOR CIRCUIT
Each diode conducts when the voltage is positive.
By adding the two outputs, the voltage presented to
capacitor C1 is more complete, thus, easier to filter,
as shown in Figure 2F. When used in 60 cycles AC
input power, the output of a full wave rectifier will be
120 cycles.
Capacitor C1 is used to store the current charges,
thus smoothing the DC voltage. The larger the
capacitor, the more current is stored. In this design,
a 2,200
μF capacitor is used, which allows about 2
volts of AC ripple when one half amp is drawn.
In practice, the current through the diodes is not as
shown in Figure 2E. Because capacitor C1 has a
charge after the first cycle, the diode will not
conduct until the positive AC voltage exceeds the
positive voltage in the capacitor. Figure 5 shows a
better picture of what the current flow looks like,
assuming no loss in the diode.
It takes a few cycles for the voltage to build up on
the capacitor. This depends on the resistance of the
winding and diode. After the initial start-up, there will
be a charge and discharge on the capacitor
depending on the current drawn
by the output load. Remember,
current only flows through the
diode when the anode is more
positive than the cathode. Thus,
current will flow in short bursts
as shown in Figure 5C.
The DC load current may be
one ampere, but the peak diode
current may be three times that.
Therefore, the diode rating must
be sufficient to handle the peak
current. The 1N4001 has a peak
current rating of 10 amps.
Figure 5
A) Transformer
Winding
B) Voltage C1
C) Current
through diodes
20V
Peak
20V Figure 4
Full Wave Rectifier
Figure 3
Half Wave Rectifier
Figure 6
1.25V
Output
R1
R2
Divider
Q1
Q2
1.5V
Q3
Q4
Q5
Current
Source
Equalized
to 1 Meg.
C1
D1
D1
D3
C1
0V - 15V
R2
R3
LM-317
–DC
Figure 7