Skip to main content

Basic Regulated Power Supply

regulated power suppply


The ac from the transformer secondary is rectified by a bridge rectifier D1 to D4 which may also be a block rectifier such as WO4 or even four individual diodes such as 1N4004 types. (see later re rectifier ratings).

The principal advantage of a bridge rectifier is you do not need a centre tap on the secondary of the transformer. A further but significant advantage is that the ripple frequency at the output is twice the line frequency (i.e. 50 Hz or 60 Hz) and makes filtering somewhat easier.

As a design example consider we wanted a small unregulated bench supply for our projects. Here we will go for a voltage of about 12 - 13V at a maximum output current (IL) of 500ma (0.5A). Maximum ripple will be 2.5% and load regulation is 5%.

Now the rms secondary voltage (primary is whatever is consistent with your area) for our power transformer T1 must be our desired output Vo PLUS the voltage drops across D2 and D4 ( 2 * 0.7V) divided by 1.414.

This means that Vsec = [13V + 1.4V] / 1.414 which equals about 10.2V. Depending on the VA rating of your transformer, the secondary voltage will vary considerably in accordance with the applied load. The secondary voltage on a transformer advertised as say 20VA will be much greater if the secondary is only lightly loaded.

If we accept the 2.5% ripple as adequate for our purposes then at 13V this becomes 13 * 0.025 = 0.325 Vrms. The peak to peak value is 2.828 times this value. Vrip = 0.325V X 2.828 = 0.92 V and this value is required to calculate the value of C1. Also required for this calculation is the time interval for charging pulses. If you are on a 60Hz system it it 1/ (2 * 60 ) = 0.008333 which is 8.33 milliseconds. For a 50Hz system it is 0.01 sec or 10 milliseconds.

The formula for C1 is:

C1 (uF) = [ ( IL * t ) / Vrip ] X 106
C1 = [ ( 0.5A X 0.00833 ) / 0.92V ] X 106
C1 = 0.00453 X 106 = 4529 or 4700 uF

Remember the tolerance of the type of capacitor used here is very loose. The important thing to be aware of is the voltage rating should be at least 13V X 1.414 or 18.33. Here you would use at least the standard 25V or higher (absolutely not 16V).

With our rectifier diodes or bridge they should have a PIV rating of 2.828 times the Vsec or at least 29V. Don't search for this rating because it doesn't exist. Use the next highest standard or even higher. The current rating should be at least twice the load current maximum i.e. 2 X 0.5A or 1A. A good type to use would be 1N4004, 1N4006 or 1N4008 types. These are rated 1 Amp at 400PIV, 600PIV and 1000PIV respectively. Always be on the lookout for the higher voltage ones when they are on special.

TRANSFORMER RATING - In our example above we were taking 0.5A out of the Vsec of 10V. The VA required is 10 X 0.5A = 5VA. This is a small PCB mount transformer available in Australia and probably elsewhere. This would be an absolute minimum and if you anticipated drawing the maximum current all the time then go to a higher VA rating.

source: http://my.integritynet.com.au/purdic/power1.html

Comments

Popular posts from this blog

LM317T Voltage Regulator Circuit with Pass Transistor

This is the schematic diagram of voltage regulator circuit with pass transostor. The regulator is based regulator IC of LM317T. The LM317T output current can be raised by utilizing an additional power transistor (on circuit, it is 2N2955) to share a portion of the total current. The amount of current sharing is established with a resistor placed in series with the LM317 input and a resistor placed in series with the emitter of the pass transistor. In the above scheme design , the pass transistor will start conducting when the LM317 current reaches about 1 ampere, due to the voltage drop across the 0.7 ohm resistor. Current limiting happens at about 2 amperes for the LM317 which will drop about 1.4 volts across the 0.7 ohm resistor and make a 700 millivolt drop across the 0.3 ohm emitter resistor. Thus the total current is limited to about 2+ (.7/.3) = 4.3 amperes.

6V to 12V DC Voltage Doubler

Here is the 6V to 12V DC voltage doubler circuit design, it also called DC voltage miltiplier or DC to DC converter. This dc voltage doubler circuit will need about 2A from the 6V input supply to produce the full 800mA at 12V for the power output. This circuit is very useful to generate higher voltage from a low power source, but this circuit will deliver low output current. So it should only be used for low current driven applications. Also, the output voltage may be unstable, so a voltage regulator (IC78XX) of proper rating can be used regulation and smooth output. But voltage regulator IC itself consume some current, and reduce the deliverable current.

Sealed Lead Acid (SLA) 12V Battery Charger with Current Limiting

This is the circuit design of Smart Sealed Lead Acid (SLA) 12V Battery Charger featuretwith Current Limiting. The charger uses a two step process for charging SLA batteries – a current limited ‘fast’ mode followed by a constant voltage ‘float’ mode. Maximum charging current is 1A. An onboard LED indicates when the charger is in ‘fast’ mode. When the LED goes out the battery is charged and the charger has switched to ‘float’ mode.