Elenco Digital / Analog Trainer - Manual (Page 8)

AC TO DC CONVERTER
The AC to DC converter consists of diodes D1, D2 and capacitor C1. Transformer T1 has two secondary
windings which are 180 degree out of phase. The AC output at each winding is shown in Figure 2A and 2B.
Diodes are semiconductor devices that allow current to flow in one direction. The arrow in Figure 3 points to the
direction current will flow. Only when the transformer voltage is positive will current flow through the diodes. Figure
3 shows the simplest possible rectifier circuit. This circuit is known as a half-wave rectifier. Here the diode conducts
only half of the time when the AC wave is positive as shown in 2C. Use of this circuit is simple but inefficient. The
big gap between cycles require much more filtering to obtain a smooth DC voltage.
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
µ
F capacitors are used, which allows about 1
volt AC ripple when .25A 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 .25A but the peak diode current may be three times
that. Therefore, the diode rating must be sufficient to handle the peak current.
The IN4001 has peak current rating of 10 amps.
REGULATOR CIRCUIT
The regulator circuit in the Model XK-150 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.
Figure 2
Figure 3
Figure 4
Figure 5
Figure 6
-7-
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
A) Transformer
Winding
B) Voltage C1
C) Current
through diodes
20V
Peak
20V
2V
Output
R1
R2
Divider
Q1
Q2
1.5V
Q3
Q4
Q5
Current
Source
Equalized
to 1 Meg.