introduction and Basic example of Power Supplies

A Transformer Designed for AC

introduction and Basic example of Power Supplies…………………..

Anything electronic essentials a source of power. This may be as easy as a battery, but often will involve reducing high-voltage AC to the normal 1–12V DC that most electronics work. Sometimes you require to generate higher voltages from a low-voltage battery. This may be stepping up the supply from a single 1.5V AA battery to a 6 or 9V, or it may be to generate much higher voltages for applications such as the 400V to 1.5kV supply needed by Geiger-Müller tubes.

≡ Convert AC to AC Power Supplies

A Transformer Designed for AC
A Transformer Designed for AC

This has two identical primary coils and two identical secondary coils, all wound on the same former. This provides some flexibility in how the transformer used. For example, US electricity 110V whereas much of the world
works 220V AC. If you are designing a product and require the same low-voltage AC out‐put whether the input 100V or 220V, then two primary coils allow you to do this if you power the primaries in parallel for 110V and series for 220V. The dual secondaries provide similar flexibility to wire in series for double the output voltage.

Transformers are simply rated to handle a certain amount of power. The resistance of the windings causes them to heat as current flows through them, and if things get too hot the insulation on the wire will break down and the transformer will break. A transformer will usually rated as so many VA. For most loads connected to a transformer, 1VA the same as 1Watt, but if large inductive loads like motors are being driven, then the current and voltage become out of phase and the apparent power will be lower than the VA rating.

≡ Convert AC to DC Power Supplies

Utilize an AC step-down transformer and then rectify and smooth the output. shows one schematic for the simplest method of

AC-to-DC Power Supply
AC-to-DC Power Supply

achieving this. The output voltage is shown as 9V. This will be 1.42 (square root of 2) of the stated output voltage of the transformer. The diode D rectifies the low-voltage AC from the transformer. The capacitor C then smooths this out to a constant DC that theoretically will be the peak voltage from the cathode of the diode. The diode will prevent the capacitor from discharging back through the transformer and so once it is at the peak voltage it will stay there.

A Basic Unregulated AC-to-DC Power Supply
Figure A Basic Unregulated AC-to-DC Power Supply

The schematic of Figure A Basic Unregulated AC-to-DC Power Supply missing any kind of load. The capacitor doing charged to the peak voltage, but nothing using that charge. If you add a load to the output as shown in Figure A Basic Unregulated AC-to-DC Power Supply the capacitor will still receive its kicks of voltage from the diode and transformer, but now it will be discharged through the load at the same time.

When working power supplies you will hear the application of the word “load,” which is the thing that is being powered by the power supply. If thinking about how the power supply will operate, you can only think of this load as resistance. In A Basic Unregulated AC-to-DC Power Supply, the load is represented by R. The characteristic voltage drop during each cycle due to the load discharging C is called the ripple voltage.

The size of this ripple voltage can calculated from the load current and also from the size of the capacitor according to this formula.

Vripple = I/ 2 f C

wherever I is the current in amps, f is the AC frequency (60Hz), and C is the capacitance in Farads. As an example, if the load current is 100mA, a 1000µF capacitor would reduce the ripple voltage to:

Vripple = I/ 2 f C =  . 1A /( 2x60Hz ×  . 001F )=  . 833V

See that the ripple voltage is proportional to the load current, so in the preceding example, a current of 1A would result in a ripple voltage of 8.3V! This situation of affairs can improve by utilizing a full-wave rectifier, which effectively doubles the frequency f in the preceding formula, halving the ripple voltage.

≡ Convert AC to DC with Less Ripple Power Supplies

The half-wave rectification is not so efficient. There is two option to this:

  1. If you have a center-tapped or dual secondary transformer, you just need two diodes for full-wave rectification.
  2. If you only have a single secondary winding, you can use four diodes as a bridge rectifier
Full-wave Rectification using Dual Secondary Windings
Figure Full-wave Rectification using Dual Secondary Windings

Full-wave rectification with dual windings.

shows how you can make use of both the positive and negative halves of
the AC cycle. The secondary windings are in serial, with the central connection between the windings becoming the ground of the DC output. At the end of the winding connected to D1 is at its maximum, the end of the winding connected to D2 will be at its minimum. D1 and D2 will take it, in turn, to provide a positive voltage to the capacitor and the whole AC cycle can contribute to charging the capacitor.

Using a Bridge Rectifier

Using a bridge rectifier
Figure. Using a bridge rectifier

If you only have a single secondary winding, when you can still achieve full-wave rectification by using an arrangement of four diodes called a bridge rectifier. Figure Using a bridge rectifier shows an unregulated DC power supply using a bridge rectifier. To understand how this works, imagine that point A is positive and B is negative. This will enable current to flow from A through D2 to charge the capacitor. If there is a load on the power supply, then the current from the negative side of the DC output will be able to flow back through D4 to B. If the AC cycle flips polarity and it is B that is positive and A negative, then the positive DC output is supplied through D3 and returns to A through D1.

≡ Convert AC to Regulated DC Power Supplies

Use a linear voltage regulator IC after your unregulated DC power supply.

A Regulated DC Power Supply
Figure A Regulated DC Power Supply

Figure A Regulated DC Power Supply shows such a circuit. As you can see, the first stage of the project is just an unregulated power supply. linear voltage regulator IC such as the very common 7805 regulators has three pins:

  1.  GND
  2. Vin—unregulated DC input
  3. Vout—regulated DC output

In the case of the 7805, the output voltage fixed at 5V, so as long as the input voltage is above about 7V the output will not have any significant ripple voltage because the regulator chip needs about 2V more than its output voltage of 5V to successfully regulate the voltage. At first, sight, since C1 and C2 are in parallel, it is not immediately obvious why you would require both.

But, the reason is that C1 will be a large electrolytic capacitor (470 F or more) and C2 will be a smaller MLC (typically 330nF) with a low ESR that should be positioned as close as possible to the voltage regulator. C2 along with C3, which is typically 100nF, needed by the voltage regulator IC to ensure that the regulation remains stable.

In choosing values for C2 and C3, it is best to consult the datasheet for the voltage regulator IC that you are applying. This will usually specify values for the capacitors as well as tell you other important properties of the voltage regulator such as:

  1. The maximum current that can be drawn from the output (1A for a 7805)
  2. Maximum input voltage (35V for a 7805)
  3.  Dropout voltage—the number of volts by which the input should exceed the output for normal operation (2V for a 7805)

Some linear voltage regulators are defined as low-dropout (LDO). These have a much smaller dropout voltage than the 2V of the 7805. Some only require as little as 150mV more at the input than the output. If the input voltage of the regulator does fall below the dropout voltage, the output voltage will slowly decrease as the input voltage drops.

These regulators such as the LM2937, which have a dropout of 0.5V, are great if your input voltage is unavoidably close to the desired output. Maybe you have a 6V battery and need a 5V output. Following the input voltage close to the output voltage also helps to reduce the heat made by the regulator. Voltage regulators are prepared in various packages from tiny surface-mount devices to larger packages such as the TO-220 package of the 7805, which are created bolted to a heatsink for the higher current applications.

Most voltage regulators including the 7805 have protection circuitry that monitors the device’s temperature and if it gets too hot, it drops the output voltage and therefore limits the output current so that the device is not destroyed by the heat. Voltage regulators also have a mounting hole designed to be connected to a heatsink. This is not needed for low currents, but as the current gets higher a heatsink will become necessary.

The 05 part of the 7805 refers to the output voltage and, as you might expect, you can also find 7806, 7809, 7812, and other voltages up to 24V. There is also a parallel 79XX range of negative voltage regulators should you need to provide regulated positive and negative power supplies as you do for some analog applications. The 78LXX range of regulators are a lower-power (and smaller package) version of the 78XX regulators.

≡ Converting AC to Variable DC Power Supplies

Using an LM317 Voltage Regulator
Figure Using an LM317 Voltage Regulator

Figure Using an LM317 Voltage Regulator shows a typical schematic for a variable output voltage regulator using the LM317. The output voltage of the LM317 varies according to the formula

Vout = 1 . 25( 1 + R 2/ R 1)

This only holds true as long as R2 is relatively low resistance (less than 10kΩ). So, if R1 is 270Ω and R2 is a 1kΩ pot, then at one end of the pot’s travel, the output voltage will be:

Vout = 1 . 25 (1 + R 2/ R 1) = 1 . 25 (1 + / 270) = 1 . 25V

When the pot’s knob is rotated fully the other way, the output voltage will be:

Vout = 1 . 25 (1 + R 2 /R 1 )= 1 . 25 (1 + 2000 /270 )= 10 . 5V

As R2 increases, the output voltage also grows, but as with a fixed-output regulator, the regulator needs some spare voltage to do the regulating with. In the case of the LM317, this is about 1.5V less than the input voltage. Adjustable voltage regulators like the LM317 are available in a number of different sized packages and ready to handle different amounts of current before their
temperature-protection circuitry kicks in to prevent damage to the IC.

≡ Regulate DC Voltage Efficiently

Use a switching voltage-regulator IC like the one in the schematic The LM2596 can provide regulated power supply at 3A

A Switching Voltage Regulator Using the LM2596
Figure. A Switching Voltage Regulator Using the LM2596

externally needing a heatsink. The FB (feedback) pin allows the regulator to monitor the output voltage and adjust the width of the pulses to keep the voltage constant. The EN (enable) pin can be used to turn the IC on and off. Linear regulators as explained in suffer from the disadvantage that they simply burn off the excess voltage as heat, which means they get hot and also waste energy.

Switching regulator designs such as the buck regulator operate at 85% efficiency around independently of the input voltage. By comparison, a linear regulator with a high input voltage and a low output voltage may only have an efficiency of possibly 20 to 60%.

In a related manner to PWM a bull converter uses a transistor to switch power to an inductor that collects the energy of each pulse at a high frequency (150kHz for the LM2596). The longer the pulse the higher the output voltage. A feedback mechanism changes the pulse length in reply to the output voltage, ensuring a constant output. It offers little sense to design a switching power supply from discrete components.

if there are ICs like the LM2596 that will do the whole thing so well. There are many switching voltage-regulator ICs ready and generally, they’re data‐ sheet will include a well-designed application circuit and sometimes even a circuit board layout. If you are making a one-off project, then it is worth considering buying a ready-made switching-regulator module, either from eBay or suppliers like Adadruit and Sparkfun that specialize in modules.

≡ Convert a Lower DC Voltage to a Higher DC Voltage Power Supplies

Use a boost-converter IC as the foundation for a design like the one shown in Using a TPS61070 Boost Converter

 Using a TPS61070 Boost Converter
Figure Using a TPS61070 Boost Converter

This design will produce an output voltage of 5V from an input voltage of within 0.9V and 5V at 90% efficiency. The SW (switch) output of the IC used to send pulses of current through the inductor L1 generating high-voltage spikes that charge C1.

The output VOUT monitored by the FB (feedback) pin via the voltage divider formed by R1 and R2 that sets the output voltage.

≡ Convert DC to AC Power Supplies

Use an inverter. These devices use an oscillator to drive a transformer and generate high-voltage AC from low-voltage DC.

An Inverter Schematic
Figure An Inverter Schematic

This design utilizes the CD4047 timer IC as an oscillator. This IC provides a similar function to the much more famous 555 timers but has the advantage that it has both natural and inverted outputs.

These two outputs each drive an NPN Darlington power transistor to energize one half of the primary coil in turn.

The transformer used here is the kind used in recipes to step down an AC voltage, but in this case, it used to step the voltage up. The windings that would normally apply as the secondary side of the transformer driven by the transistors and the output from the transformer will be taken from the coil that would usually connect to AC.

To drive the circuit from 12V DC at 60Hz to produce a 110V AC output you want a transformer with a 12-0-12V secondary and a 110V primary. The power to the CD4047 is supplied through a 100Ω resistor (R5) and a 100µF decoupling capacitor across the IC’s power supply.

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