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Basic Power Supplies
The Power Supply converts the 120 Volts AC into various DC
voltages that the game actually runs on. A transformer steps the
voltage down to a voltage close to what is desired. We still have
AC coming out of the transformer. To convert that to DC we use a
Rectifier. Various schemes provide different means of
rectification, each with benefits and draw backs. The output of
the rectifier is Pulsating DC. These need to be filtered out to a
smoother DC using a large filter capacitor. The output at this
point is smoother DC, but unregulated. To regulate the voltage to
a specific voltage we use a Voltage Regulator.
Now that we know where we are heading, lets build one, shall
we?
First we select a
Transformer. The output
voltage for our simple
application must be higher
than the voltage we want
to end up with after we
regulate it. Since most of
our circuits work on +5
Volts, we need a voltage
higher than 5V coming out
of the transformer. 6.3 V
is popular and available
even at Radio Shack, so
let’s select that. The
current rating of the
secondary must be able to
supply our desired
current. Most of our exercises can be done with ¼ Amp, or less, so
let’s select one with at least that much current capability.
Next we need to Rectify this
6.3 V AC coming out of the
transformer to a DC level. We have
at least three basic designs to
choose from. The Half Wave
rectifier lets us use only one
side of the AC. Since we need
positive voltage we can use this
circuit.
An alternative would be a full wave rectifier. This allows us
to use both sides of the AC signal. The advantage is that we can
use a smaller filter
capacitor. With a half wave
rectifier we have a 8 ms gap
between positive pulses. Our
filter capacitor must hold a
charge during this time,
meaning we need a larger
filter capacitor. With a
full wave rectifier, we have
a sequence of 8 ms pulses
(our rectified AC signal)
with no gaps. This means our
filter cap can be half the
size as with half wave
rectification.
The disadvantage is that our transformer must be twice as
large. Since we are using half of our secondary on each half
cycle, we need twice the voltage out, or 12.6 V, with a center
tap.
A third alternative is to use a circuit called a bridge
rectifier.
This allows us to use both sides of our AC signal using only
a 6.3 volt transformer. The disadvantage is that we loose twice as
much voltage dropped across the rectifier diodes. Two of the four
are conducting at any given time.
Full wave requires less filtering, so that’s a good idea for
us in our application. Cost of a transformer is less for a 6.3
volt transformer, compared to a 12.6 volt transformer, so it looks
like a Bridge rectifier design is best for this application. The
question to us is can we afford the loss of voltage across the two
diodes at one time.
Lets take a close look at the design we are talking about.
The voltage at the rectified output will be up to the peak value
of our rectified AC line. 6.3 (AC RMS) x 1.414 = almost 9 Volts.
So we will have a maximum of 9 Volts once we rectify the AC. Less
our loss across the diodes (0.6 V each x 2) equals 1.2 Volts lost
in rectifying. 9 – 1.2 = 7.8 volts. We have a 2.8 Volt difference
between the 5V we want at the end, and the 7.8 Volts coming out of
the rectifier and filter capacitor.
This gives us a DC voltage, but it is unregulated. As our
line voltage varies, the output of our power supply will too. As
current in the load varies, so will our voltage. We need a
regulator.
A Zener
diode provides a
steady voltage
across it, but
will only provide
10’s of milliamps
of current. A
transistor
amplifies
current, so let’s
put the two
together.
If we want 5.0 volts out we need to choose a Zener diode that
is 0.6 Volts above that. Note the direction our current is flowing
through the regulating transistor, from emitter to collector. The
base voltage will be 0.6 Volts more positive than the emitter. We
have +5 Volts at the emitter, so we want to set the base to 0.6
volts above the emitter, or 5.6 V. It just so happens that this is
a standard value for a Zener diode, so we are in luck.
What resistance do we need for R1? We will have a maximum of
9V at the collector. Actually only 7.2 Volts we calculated
including loss across the diodes, but we have to make our
calculations for “worst case” conditions. When the transistor is
off we draw no current, the maximum we will see at the collector
is 6.3 x 1.414, or 8.8 volts. We will calculate for 9 volts since
our line voltage may vary.
9 V minus the 5.6 volts across the Zener makes 3.4 Volts,
maximum, dropped across the resistor.
How much current will be passing through R1? If we have our
expected 250 mA we want to supply, our emitter current will be 250
mA. If the gain of the transistor is 100, we will have
1/100<sup>th</sup> of that 250 mA, or 2.5 mA of base current. To
this 2.5 mA we will have the 20 mA through the Zener, according to
the design specs of the Zener. So we have 2.5 mA, plus 20 mA, or
22.5 mA through R1.
With 3.4 volts across it, and 22.5 mA through it, we need a
150 ohm resistor. This calculates out to less than 100 mW, so a ¼
W resistor is sufficient.
Selection of filter capacitors is loose. With a bridge
rectifier we need about 1,000 uF per Amp of current. With ¼ Amp of
current, we need about a 250 uF (or higher) capacitor, rated at 10
Volts or higher. Looking at our shelves, we find a drawer full of
1,000 uF, 25 V, capacitors. That should work just fine. No more
elaborate calculations are needed. C1 is now chosen to be a 1,000
uF 25 V capacitor.
This gives us a working design, but has the disadvantage of
not limiting our current. If we accidently short the output to
ground the transistor will conduct itself into destruction trying
to keep the output at 5 Volts. We need to limit the output current
to 250 mA. Beyond that we want the power supply to just shut down.
Adding current limiting to our power supply is a simple task.
We only need
to add R2 and Q2 to
our design. When
the current through
R2 gets to the 250
mA level we develop
0.6 Volts across
it, turning Q2 on,
shorting out the
emitter – base
circuit of Q1,
turning Q1 off.
What resistance do we want for R2? 0.6 volts across it. 250 mA
through it. That comes out to be 2.4 ohms. That’s a standard
value. 0.6 V x 250 mA comes out to 150 mW, so it also can be a ¼
Watt resistor.
This gives us the safety of current regulation with only a
small sacrifice to voltage regulation. Our output voltage will be
a little less since we have the loss across R2 to consider. At
lower currents we may only drop tenths of a volt across R2. As we
approach 250 mA we will get a maximum of 0.6 volts across it.
Beyond that our current limiter should activate and shut down the
power supply.
When the power supply turns off the voltage at the collector
will raise to the full 9V level. That’s why we used 9 V when
making calculations for C1 and R1.
The whole of our circuit may be shown in discrete components,
as above, or as a block diagram, and just call it a Voltage
Regulator.
This is such a standard design, why don’t we add a circuit
for thermal safety, and a few more bells and whistles and stuff it
all inside a transistor case? Great idea. That bring us to our
next lesson.
Regulated Power Supplies
The next step in developing our power supply is to turn the
pulsating DC, unregulated, in to a specific, well regulated,
current limited voltage. Most logic circuits run on +5 Volts. This
must be within 5% of 5 Volts under varying load conditions.
In order to get a stable, predictable, output voltage under
any predictable
current level we
need to add a
regulator. In
today’s circuits,
this is usually a
simple threeleaded component.
The voltage
regulator puts out
a specific voltage
over a certain
range of operating
current.
The LM7805T is a very popular device in the gaming industry.
It provides a regulated +5 Volts at any current level from about
0.050 Amps to over 1 Amp. The part number can be broken down to
make sense of the device. “LM” is the manufacturer’s prefix for
analog devices, and is an industry standard. Other manufacturers
may use another prefix. The 78xx family of devices are all
positive voltage regulators. 7805 is a 5 Volt device. 7812 is a 12
volt device. 7815 is a 15 volt device. 7824 is a 24 volt device.
The T suffix indicated the case the device is built in, and
therefore the current that we may draw from the
regulator. T indicates a TO-220 device, rated at 1 Amp. “K”
indicates a TO-3,
rated at 1.5 Amp. “L” indicates a TO-92 device, rated at 100 mA,
(0.100 Amp). We can read the part number and tell what the device
does, and its ratings.
The 79xx family is a similar device made for negative
voltages.
These devices provide a regulated output voltage, as well as
current limiting. If we try to draw more than the rated current
value the device shuts itself off, preventing circuit damage and
blown fuses. An excessive current drain just turns the device off.
The internal operations of these devices will be demonstrated as
we get into electronic components.