In the previous tutorials we saw that a semiconductor signal diode will only conduct current in one direction from its anode to its cathode (forward direction), but not in the reverse direction acting a bit like an electrical one way valve.

A widely used application of this feature and diodes in general is in the conversion of an alternating voltage ( AC ) into a continuous voltage ( DC ). In other words, Rectification.

But small signal diodes can also be used as rectifiers in low-power, low current (less than 1-amp) rectifiers or applications, but were larger forward bias currents or higher reverse bias blocking voltages are involved the PN junction of a small signal diode would eventually overheat and melt so larger more robust **Power Diodes** are used instead.

The power semiconductor diode, known simply as the **Power Diode**, has a much larger PN junction area compared to its smaller signal diode cousin, resulting in a high forward current capability of up to several hundred amps (KA) and a reverse blocking voltage of up to several thousand volts (KV).

Since the power diode has a large PN junction, it is not suitable for high frequency applications above 1MHz, but special and expensive high frequency, high current diodes are available. For high frequency rectifier applications **Schottky Diodes** are generally used because of their short reverse recovery time and low voltage drop in their forward bias condition.

Related Products: Diodes, Transistors and Thyristors | Discrete Misc

Power diodes provide uncontrolled rectification of power and are used in applications such as battery charging and DC power supplies as well as AC rectifiers and inverters. Due to their high current and voltage characteristics they can also be used as free-wheeling diodes and snubber networks.

Power diodes are designed to have a forward “ON” resistance of fractions of an Ohm while their reverse blocking resistance is in the mega-Ohms range. Some of the larger value power diodes are designed to be “stud mounted” onto heatsinks reducing their thermal resistance to between 0.1 to 1^{o}C/Watt.

If an alternating voltage is applied across a power diode, during the positive half cycle the diode will conduct passing current and during the negative half cycle the diode will not conduct blocking the flow of current. Then conduction through the power diode only occurs during the positive half cycle and is therefore unidirectional i.e. DC as shown.

Power diodes can be used individually as above or connected together to produce a variety of rectifier circuits such as “Half-Wave”, “Full-Wave” or as “Bridge Rectifiers“. Each type of rectifier circuit can be classed as either uncontrolled, half-controlled or fully controlled were an uncontrolled rectifier uses only power diodes, a fully controlled rectifier uses thyristors (SCRs) and a half controlled rectifier is a mixture of both diodes and thyristors.

The most commonly used individual power diode for basic electronics applications is the general purpose 1N400x Series Glass Passivated type rectifying diode with standard ratings of continuous forward rectified current of 1.0 amp and reverse blocking voltage ratings from 50v for the 1N4001 up to 1000v for the 1N4007, with the small 1N4007GP being the most popular for general purpose mains voltage rectification.

A rectifier is a circuit which converts the *Alternating Current* (AC) input power into a *Direct Current* (DC) output power. The input power supply may be either a single-phase or a multi-phase supply with the simplest of all the rectifier circuits being that of the **Half Wave Rectifier**.

The power diode in a half wave rectifier circuit passes just one half of each complete sine wave of the AC supply in order to convert it into a DC supply. Then this type of circuit is called a “half-wave” rectifier because it passes only half of the incoming AC power supply as shown below.

During each “positive” half cycle of the AC sine wave, the diode is *forward biased* as the anode is positive with respect to the cathode resulting in current flowing through the diode.

Since the DC load is resistive (resistor, R), the current flowing in the load resistor is therefore proportional to the voltage (Ohm´s Law), and the voltage across the load resistor will therefore be the same as the supply voltage, Vs (minus Vf), that is the “DC” voltage across the load is sinusoidal for the first half cycle only so Vout = Vs.

During each “negative” half cycle of the AC sinusoidal input waveform, the diode is *reverse biased* as the anode is negative with respect to the cathode. Therefore, NO current flows through the diode or circuit. Then in the negative half cycle of the supply, no current flows in the load resistor as no voltage appears across it so therefore, Vout = 0.

The current on the DC side of the circuit flows in one direction only making the circuit **Unidirectional**. As the load resistor receives from the diode a positive half of the waveform, zero volts, a positive half of the waveform, zero volts, etc, the value of this irregular voltage would be equal in value to an equivalent DC voltage of 0.318 x Vmax of the input sinusoidal waveform or 0.45 x Vrms of the input sinusoidal waveform.

Then the equivalent DC voltage, V_{DC} across the load resistor is calculated as follows.

Where V_{max} is the maximum or peak voltage value of the AC sinusoidal supply, and V_{S} is the RMS (Root Mean Squared) value of the supply.

Calculate the voltage across V_{DC} and the current I_{DC}, flowing through a 100Ω resistor connected to a 240 Vrms single phase half-wave rectifier as shown above. Also calculate the DC power consumed by the load.

During the rectification process the resultant output DC voltage and current are therefore both “ON” and “OFF” during every cycle. As the voltage across the load resistor is only present during the positive half of the cycle (50% of the input waveform), this results in a low average DC value being supplied to the load.

The variation of the rectified output waveform between this “ON” and “OFF” condition produces a waveform which has large amounts of “ripple” which is an undesirable feature. The resultant DC ripple has a frequency that is equal to that of the AC supply frequency.

Very often when rectifying an alternating voltage we wish to produce a “steady” and continuous DC voltage free from any voltage variations or ripple. One way of doing this is to connect a large value Capacitor across the output voltage terminals in parallel with the load resistor as shown below. This type of capacitor is known commonly as a “Reservoir” or *Smoothing Capacitor*.

When rectification is used to provide a direct voltage ( DC ) power supply from an alternating ( AC ) source, the amount of ripple voltage can be further reduced by using larger value capacitors but there are limits both on cost and size to the types of smoothing capacitors used.

For a given capacitor value, a greater load current (smaller load resistance) will discharge the capacitor more quickly ( RC Time Constant ) and so increases the ripple obtained. Then for single phase, half-wave rectifier circuit using a power diode it is not very practical to try and reduce the ripple voltage by capacitor smoothing alone. In this instance it would be more practical to use “Full-wave Rectification” instead.

In practice, the half-wave rectifier is used most often in low-power applications because of their major disadvantages being. The output amplitude is less than the input amplitude, there is no output during the negative half cycle so half the power is wasted and the output is pulsed DC resulting in excessive ripple. To overcome these disadvantages a number of **Power Diode** are connected together to produce a Full Wave Rectifier as discussed in the next tutorial.

Error! Please fill all fields.

Fantastic discussion , I loved the insight , Does someone know if my assistant could get a fillable AO 442 version to edit ?

What are common cathode diodes? And how to use these diodes in power switching circuits?

Common cathodes means that all cathodes of the diodes are connected together and at the same potential.

what is a power diode ? I still don’t get it

hello everyone,

I have made full wave bridge rectifier circuit using IN4007 diodes. As per the theory we all know if my input voltage is below the threshold of the diode it will not conduct but in my case I’m using signal from function generator if I give 4V rectifier is working very well but it is also conducting when supply is 1V only. I don’t the reason pls help me out from this problem.

can you please help me.

One disadvantage of half-wave rectification is that one [ ] of the source waveform is not used, resulting [ ] only 50% of the available input power being [ ] available at the output!

i cant get how to calculate the power consumption of rectifier

Anne, diodes have a non-linear voltage/current relationship so when reverse biased they don’t dissipate any power and when in conduction have a characteristic forward voltage drop which doesn’t vary much with changes in current.

The voltage drop varies according to diode type and might be as low as 0.15V for Schottky diodes found in low voltage/high current switch mode power supplies or as high as 15V for the rectifier you might find in a microwave oven.

In the example above, with a resistive load taking 1.08A, I’d expect the diode to be dissipating about 0.65V x 1.08A = 0.7 Watts; the 0.65V being the forward voltage drop across a typical silicon rectifier diode.

Where the rectifier is loaded by a capacitor the arithmetic looks horrific at first sight because the current only flows in a brief gush as the AC voltage rises above the stored voltage remaining on the capacitor and ceases as soon as the capacitor’s stored voltage is above that of the AC waveform. In fact, that doesn’t matter, it’s the average current that determines the power dissipation, so again it’ll be 0.65V (or whatever) times the current taken by the load.

Dear Webmaster,

I am an EE student as of now. I would like to know if the smoothing capacitor is the only filter that would work on a half-wave rectifier. Would an LC filter work as well? If so, how would its diagram and waveform differ from the C filter.

Thanks and regards,

Prince

Rectifier smoothing circuits can use a parallel capacitor filter, series inductor filter, LC filters or pi filters, etc. The ideal and simplest arrangement for filtering is to connect a capacitor in parallel with the load so that it presents a high impedance for the DC and a low impedance at the frequency being used. A series inductor (usually called a choke in this type of application) can be used which will present a low impedance to the DC current and is more effective for heavy current than for light current loads.

A further improvement can be achieved by using a LC filtering nework. At the ripple frequency, the inductor exhibits a high value of inductive reactance, XL while the capacitor exhibits a low value of capacitive reactance, Xc. The effect of this LC filtering is to minimizing the ripple across the load while having little effect on the DC voltage.

Really nice !!!!

How can I download these content?

Dear Webmaster,

With reference to Example No.1, power dissipated by the resistive load of 100Ω is 116 Watts, many textbooks and websites use a similar approach when talking about both full-wave as well as half-wave rectified DC.

When the load is a non-linear component such as a diodes or a LED that has its own characteristic voltage drop, using average voltage of the rectified sinusoidal wave-form for the resultant output DC voltage is appropriate since average current will be used for the calculation of power dissipation of non-linear components as in P = V * I.

However, when the load is resistive or a linear component like a resistor in the example, Vrms of the respective wave-form of either full-wave or half-wave rectified should be used for calculating the power dissipation of the resistor. The reason is that instead of using the equation P = V * I in the case of non-linear component, P = (I^2)(R) or P = (V^2) / R will be used for resistive load.

Therefore, in Example No.1, the resultant voltage should be, V = Vpeak / 2

= SQRT(2) * 240 Vrms /2 = 169.71V,

and power dissipated by the resistor, P = V^2 / R

= (169.71)(169.71) / 100 = 288W.

If the power supply is 240Vrms sinusoidal wave, power dissipated by the same resistor will be, P = (240)(240) / 100 = 576W, which doubles the value of the rectified half-wave DC output that has exactly half the wave-form of the AC mains.

Regards

John Lam

Hello John, Thank you for your detailed explanation. My example of 108V, 1.08A and 116W was calculated using the equivalent DC values of the rectified waveform for a half wave rectifier. In this example, your values of 169V 1.69A and 288W are based on the instantaneous RMS values for half of the output waveform only.

While I agree that your calculations are correct, they are only valid for one half of the waveform as the other half is blocked due to the diode resulting in zero voltage, current and power output half of the time. However your calculations are perfectly valid for the DC equivalent values of a full wave rectified output as both halves of the input waveform are presented across the 100 Ohm load be it in rectified form.

Thanks for your helpfuj articles.