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Unregulated Power Supply

Unregulated Power Supply

An Unregulated Power Supply is the simplest of power supplies to construct.

Nearly all electronic devices and circuits require some form of a DC power source for their operation supplied from either from a battery, solar cell or mains connected unregulated power supply.

While batteries have the advantage of being small, portable and ripple free, they need replacing (or recharging) frequently and are also expensive as compared to a conventional DC power supply.

Since in our homes, schools and workplaces we have a convenient, reliable and economical source of electrical power, it makes sense to use the domestic mains AC supply to power our circuits. However, the mains AC supply is a lot higher (usually 220-250 V rms) than the much smaller DC voltage provided by a battery. The process of converting this higher AC voltage into a much lower DC voltage is called Rectification.

Rectification is the process of converting AC power into DC power. In the Diodes tutorials we saw that a diode conducts current in just one direction (from Anode to Cathode) and not in the reverse direction. A diodes ability to switch current in one direction only makes it ideal for converting a two directional alternating current into a constant direct current or DC supply as shown.

Diode Rectifier

diode rectifier

 

We can see that the AC input to the diode is a sinusoid which alternates between the positive and negative half cycles, while the output from the diode is rectified DC having a waveform which goes only positive to zero volts, the negative half cycles being blocked. This type of output waveform is called “half wave pulsating DC”.

Half Wave Unregulated Power Supply

The purpose of a power supply is to provide the required amount of electrical power at a specified voltage and current level, for example +9 volts at 500mA. The electrical characteristics of any power supply will depend on the circuit or circuits being powered, but generally all unregulated power supplies consist of a transformer to step down the AC mains voltage to the required level as well as providing electrical isolation and a diode rectifier to provide an unstabilised output voltage.

Consider the half wave unregulated power supply circuit below.

Half Wave Unregulated Power Supply

half wave unregulated power supply

 

The mains input is applied to the primary winding of the mains transformer, T1 with the transformers secondary winding supplying low voltage AC to the rectifier diode D1. The resulting output waveform contains a DC voltage level which is approximately equal to 1/π or 0.318 of the peak voltage.

So for example, if the sinusoidal peak voltage is 10 volts, the equivalent DC output would therefore be: 0.318 x 10 = 3.18 volts. Then it is important that you choose the right voltage transformer for your unregulated power supply.

As we have seen above, the output waveform from the diode is pulsating DC. Obviously this pulsating DC voltage is not suitable to power most electronic circuits as not only does the supply voltage vary considerably and rapidly compared to an ideal DC battery supply, there is no supply voltage at all for 50% of the time during the negative half cycle.

Very often when rectifying an alternating voltage we want to produce a steady state direct voltage such as we would obtain from a battery supply and free from the waveform variations mentioned above. One way to overcome this problem is to add a smoothing capacitor across the output terminals effectively connecting it in parallel with the load.

We know that a capacitor has the ability to store an electrical charge on its plates and we can use this ability to help smooth out some of the pulsating waveform. The capacitor, C1, usually called a smoothing capacitor or reservoir capacitor, becomes charged up by the current flowing through the forward biased diode during the positive half cycle. The amount of charge on the capacitors plates depends upon the peak positive output voltage from the transformer, T1 and the value of the capacitor as charge, Q equals V x C (volts x capacitance).

As the output voltage from T1 starts to reduce to zero, the charged up capacitor now takes over supplying current to the load. At some point the output voltage from T1 crosses over zero and supplies the negative half cycle which reverse biases the diode into cut-off. During this half cycle, capacitor C1 is supplying all the current to the load and discharges itself at a rate determined by the loads time constant.

On the next positive half cycle, transformer T1 takes over control again supplying power to the load and continues to do so until the output voltage from T1 returns to its positive peak value once again. During this period C1 recharges again and provides the output current to the load when the the voltage from T1 drops away again until the next positive peak voltage from T1 as shown.

Half Wave Rectifier Waveforms

half wave rectifier waveforms

 

As the capacitor, C1 cannot have an infinite value, it cannot provide a perfectly smooth DC output supply which in some cases can take on the form of a sawtoothed waveform. The variations in the output waveform due to the capacitors inability to maintain a steady output is called “Ripple” and ripple is produced for every complete cycle of the AC input. In other words for a half wave rectifying circuit the amount of pulsating DC ripple frequency will equal the input AC frequency.

The amount of ripple present on the output waveform depends upon the characteristics of the load but for a given capacitor value, a greater load current (smaller load resistance) discharges the capacitor more and so increases the ripple content of the output waveform.

You may think, why not use a bigger value capacitor to reduce the ripple content, but there are limits to using large smoothing capacitors (usually electrolytic’s) with regards to cost, size and increasing their value beyond a point will not reduce ripple significantly. Also using large value smoothing capacitors can require very large charging currents to be supplied through the diode bridge. However, it is possible to improve the ripple content present in the output voltage supplied by an unregulated power supply by adding more capacitors of different values in parallel across the output terminals.

Full Wave Unregulated Power Supply

We have seen that the output voltage from a half wave unregulated power supply can be difficult to filter to a smooth DC level because the output voltage and current are applied to the load for only half of each input cycle. Also another disadvantage of a half wave unregulated power supply is the relatively long periods between the capacitors charging pulses supplied by the transformer necessitating the use of a comparatively large electrolytic type smoothing capacitor.

However, if we added a second rectifying diode to the circuit so that every half cycle of the input, instead of every other half cycle, contributes to the rectified output waveform, the amount of ripple content will be greatly reduced and this can be achieved by using a Full Wave Unregulated Power Supply

A full wave unregulated power supply varies from its half wave cousin by using a mains transformer with a center tapped secondary winding and two rectifying diodes as shown.

Full Wave Unregulated Power Supply

full wave unregulated power supply

 

We can see that the two halves of the secondary winding are effectively feeding to separate half wave rectifying circuits of the type described above, with the two outputs being combined together and smoothed by the common smoothing capacitor, C1.

The two diodes, D1 and D2 operate in a push-pull type arrangement because the transformer secondary is grounded (0V) to create a 180o phase difference between the upper and lower half secondary windings. Then the upper half provides a positive going voltage and the lower half a negative going voltage.

When the AC input waveform is positive, a positive voltage is developed across the upper half of T1 secondary forward biasing diode, D1 turning it “ON”, while the corresponding negative voltage developed across the lower winding of T1 secondary reverse biases diode, D2 turning it “OFF”. Then current is supplied to the load through diode D1 only.

When the AC input waveform swings negative, a negative voltage is developed across the upper half of T1 secondary turning diode D1 “OFF”, while a positive voltage is developed on the lower half of T1 secondary forward biasing and turning “ON” diode, D2. Then current is supplied to the load through diode D2 only.

Then the two diodes and the center-tapped transformer switch the two directional AC current developed across the secondary winding alternately through the load. The resulting output waveform contains a DC voltage level which is approximately equal to 2/π or 0.636 of the peak voltage.

So for example, if the sinusoidal peak voltage is 10 volts, the equivalent DC output would therefore be: 0.636 x 10 = 6.36 volts, twice that for the half wave rectifier as shown.

Full Wave Rectifier Waveforms

full wave rectifier waveforms

 

The advantage of this full wave unregulated power supply circuit is that it requires a smoothing capacitor of about half the value of the one required for the half wave circuit because it is charged up twice as frequently in a full wave circuit as it is in a half wave circuit and the amount of discharge for a given load current is therefore less.

Also, because two half cycles appear across the smoothing capacitor for every one cycle at the input, the ripple content will be lower and the ripple frequency will be twice that of the input frequency. For example, if the sinusoidal input frequency is 50Hz, then the ripple frequency will be 100Hz. As a result this higher ripple frequency is easier to smooth out any fluctuations.

Unregulated Power Supply Summary

One of the main disadvantages of an unregulated power supply is that its output voltage is affected significantly by changes in the mains voltage and also by changes in the load current. As the load draws more current, the DC terminal voltage decreases.

Also the output waveform produced by a half wave unregulated power supply has a DC level of approximately 0.318 x Vpeak together with a large AC variation resembling a sawtoothed waveform. This output waveform is generally known as a pulsating DC voltage.

In order to remove some of the AC content a smoothing capacitor is used allowing the DC content to pass and reducing the AC to a small ripple. A half wave rectifier produces a ripple frequency the same as the input frequency.

One way to increase the DC output voltage, reduce the waveforms ripple content and improve efficiency is to use a full wave rectifier which consists of two diodes and a center-tapped transformer to generate two equal and opposite waveforms across each half of the secondary winding. The main disadvantage of a full wave unregulated power supply is that it requires a bigger transformer for a given output power.

Half wave unregulated power supplies are cheap and simple to construct that convert AC power to pulsating DC power. We have seen that smoothing capacitors can be used to change this pulsating DC from the rectifier either half wave of full wave to a reasonably smooth and ripple free DC supply to power electronic circuits or to charge batteries.

33 Comments

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  • ABENAUIA

    I have enjoyed it and learnt alot from it

  • vg-trade

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  • Joe W.

    So is the advantage of using a full-wave bridge rectifier circuit with 2 more rectifier diodes that you don’t need a center-tapped transformer?

    • Wayne Storr

      Not necessarily. You could still use a center-tapped transformer with a bridge rectifier circuit, just do not connect the center-tap. A center-tapped transformer full-wave rectifier is in effect two half-wave circuits back-to-back as the two halves of the secondary feed two rectifier circuits which operate in a push-pull arrangement. The advantage of a full-wave bridge rectifier is that it is more efficient as the polarity of the single secondary winding is reversed every half cycle so it provides two positive pulses per cycle. Also the untapped transformer could be smaller, and as two extra diodes are cheaper and less bulky than transformers, a full-wave bridge rectifier is generally smaller and cheaper than a standard full-wave rectifier.

  • Cornellie K Ronoh

    This is awesome

  • SUKANT NATH

    Sir can we attain a particular type of current and voltage through this such as 12 v 2 amp DC.As we are using this for survelliance system input .

    • Wayne Storr

      You can design a dedicated power supply to produce any combination of voltages or currents, within reason

  • Ansar shaheen

    Implementation of non regulated DC power supply

  • Muhammad Anas

    Write

  • Kebede Tasew

    that is great

  • Delower Hossen

    Good

  • TSCA Electrical Group of Company Incorporated

    Thanks for sharing your story, Very nice!

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  • Isaac Sunday

    Diagram of unregulated power supply

  • Md Fahad uddin

    Thanks to give this study materials

  • Bongani M.

    Well explained…tutorials….
    Very clear…well understood.

  • JOSEPH

    This page is very useful

  • JITENDRA SURANA

    i have a quetion that what happens when we use non filtered dc supply in a coil load, as example tesla coil…

  • 0782376970

    i real , enjoyed fully. thank you alot

  • Foudama

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  • sanjaya

    great work. your articles are explanatory. keep it up.

  • Ramakrishnan.V

    I am fascinated by the tutorials and has kept all these tutorials in my PC so that I can read them and understand during my free time. I wish if you can post some tutorials on MCUs such as 8051, Arduino, ATMEGA and PICs so that I can learn them which is my dream. Will you please post theses tutorials especially C coding dedails of these MCUs.
    Thanks,
    Regards,
    V.Ramakrishnan.

  • Rock

    As the author points out ” You may think, why not use a bigger value capacitor to reduce the ripple content…” Yes, I agree. In fact my understanding is that depending on the design load, the input capacitor value is critical as increasing the value beyond a point will not reduce ripple significantly and only decreases the capacitor charging time every cycle thereby increasing the instantaneous charging current, potentially stressing the rectifier and transformer unnecessarily. I am actually here looking for the math or formula to determine the optimum input cap value.
    As the author also points out that adding more caps in parallel will reduce ripple, I believe they also need to have series resistors or chokes in series between them.