Figure 2 figure 3 figure 4, Figure 5 figure 6 – Elenco Digital / Analog Trainer in Case User Manual

Page 7

Advertising
background image

-6-

By the addition of a second diode and transformer
winding we can fill in the gap between cycles as
shown in Figure 4. This circuit is called full-wave
rectification. 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 2E. 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
1000

μF capacitors are used, which allows about 5

volts AC ripple when one amp is drawn.

In practice, the current through the diodes is not as
shown in Figure 2C. Because capacitor C1 has a
charge after the first cycle, the diode will not conduct
until the positive AC voltage exceeds the positive
charge 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 diodes
when the anode is more positive than the cathode.
Thus, current will flow in short bursts as shown in
Figure 5.

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 peak
current rating of 10 amps.

REGULATOR CIRCUIT
The regulator circuit in the Model XK-550 power
supply consists of a LM-317 integrated circuit. This
IC is specially designed to perform the regulation
function. Figure 6 shows a simplified circuit of how
the LM-317 IC works.

Transistors Q1 and Q2 form a circuit known as a
differential amplifier. Transistor Q1 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 extremely high.

A) Transformer

Winding

B) Voltage C1

C) Current

through diodes

20V

Peak

20V

Figure 2

Figure 3

Figure 4

Voltage Waveform for Supply

A) Transformer

Winding AB

B) Transformer

Winding BC

C) Output of

diode D1.

D) Output of

diode D2.

E) Total of diodes

D1 & D2.

20V

F) Output of capacitor C1

Ripple depends on load
current (expanded).

Half Wave Rectifier

Full Wave Rectifier

2V

Output

R1

R2

Divider

Q1

Q2

1.5V

Q3

Q4

Q5

Current
Source
Equalized
to 1 Meg.

Figure 5

Figure 6

Advertising