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*Introduction:*

Several semiconductor devices or systems (such as car stereo systems) need a negative dc source or both negative and a positive dc source. For the sake of simplicity, we will examine only positive dc power supplies. Though, a positive dc supply can be converted in a negative one by simply reversing two leads in same way as we reverse polarity of the dry cell.

Consider the following circuits:

1. Half-wave rectifier

2. Full-wave rectifier

3. Full-wave bridge circuit.

*Half-Wave Rectifier:*

The alternating secondary voltage is applied to the diode connected in series with load resistor. Let equation of alternating secondary voltage be:

V_{S} = V_{sm}ωt

Working:

During positive half-cycle of input ac voltage, diode D is forward-biased (ON) and conducts. While conducting, diode serves as a short-circuit so that circuit current flows and therefore, positive half-cycle of input ac voltage is dropped across. It comprises output voltage VL. Waveform of load voltage comprises of half-wave rectified sinusoids of peak value V_{LM}.

During the negative input half-cycle, the diode is reverse-biased (OFF) and so, does not conduct i.e. there is no current flow. Hence, there is no voltage drop across RL. The negative input half-cycle is suppressed i.e. it is not utilized for delivering power to the load. As seen, the output is not a steady dc but only a pulsating dc wave having a ripple frequency equal to that of the input voltage frequency. This wave can be observed by an oscilloscope connected across . When measured by a dc meter, it will show some average positive value both for voltage and current. Since only one half- cycle of the input wave is used, it is called a half-wave rectifier. It should be noted that forward voltage drop across the diode has been neglected in the above discussion. We have, in fact, assumed an ideal diode (having zero forward resistance and infinite reverse resistance).

Average and RMS Values:

Let

V_{m} = maximum value of transformer secondary voltage, V_{s} =rms value of secondary voltage,

V_{LM} =maximum value of load voltage, V_{sm} = diode drop - secondary resistance drop

V_{L} = rms value of load voltage, I_{L} = rms value of load current, V_{L}(dc) =average value of load voltage

I_{L}(dc) = average value of load current, I_{LM} = maximum value of load current, R_{L} = load resistance

R_{S }= transformer secondary resistance, r_{d} = diode forward resistance

Now,

R_{0} = R_{S} + r_{d}

I_{LM} = V_{sm} - V_{B} /(R_{S} + r_{d}) + R_{L}

= V_{sm} -V_{B}

V_{LM} = I_{LM} . R_{L}

V_{L}(dc) = V_{LM}/π = 0.318V_{LM}

I_{L}(dc) = I_{LM}/π = 0.318I_{LM}

I_{L} = I_{L}M_{n}/2 = 0.5ILM = 0.5V_{LM}/R_{L}

Efficiency:

Efficiency of rectification is given by ratio of output dc power to total amount of input power supplied to circuit. It is also known as conversion efficiency.

η = P_{dc}/P_{in} = power in load/input power

Now,

P_{dc} = IL_{(dc)}^{2 }(R_{L}) = (I_{LM}/ π)^{2}.R_{L} = I_{LM}^{2}.R_{L}/ π^{2}

P_{in} = I_{L}^{2} (R_{L} + R_{0}) = (I_{LM}/2)^{2} (R_{L} + R_{0}) = I_{LM}^{2} (R_{L} + R_{0})/4

Thus:

η = P_{dc}/P_{in} = (4/π2)R_{L}/(R_{L} + R_{0}) = 0.406/(1+ R_{0}/R_{L}) = 40.6%/(1 + R_{0} /R_{L})

If R_{0} is neglected η = 40.6%. Obviously, it is maximum possible efficiency of half-wave rectifier.

Though rectified voltage is not filtered voltage, it however has a dc component and ripple component. For the half-wave rectified signal, output dc voltage is:

V_{dc} = 0.38Vm

Rms value of ac component of output signal can be computed to be

V_{r}(rms) = 0.385Vm

Percent ripple of half-wave rectified signal can then be computed as:

γ = V_{r}(rms)/V_{dc} x 100% = 0.385Vm/0.318Vm x 100% = 121%

Peak Inverse Voltage (PIV):

It is the maximum voltage which takes place across rectifying diode in reverse direction. The diode is reverse-biased during negative half-cycle and maximum voltage applied across it equals maximum secondary voltage i.e. V_{sm}

Transformer Utilization Factor (TUF):

While designing any power supply, it is essential to find out rating of transformer. It can be achieved provided TUF is known. Value of TUF depends on amount of power to be delivered to load and kind of rectifier circuit to be utilized.

TUF = dc power delivered in load/ ac rating transformer secondary

P_{dc}/P_{ac} rated = P_{dc}/P_{in} rated

At first sight it might seem as if above ratio is same as conversion efficiency. Really, it is not so because rating of transformer secondary is different from actual power delivered by secondary.

The rated voltage of transformer secondary is V_{sm}/√2 but actual current flowing through secondary is I_{L }= I_{LM}/2 (and not I_{LM}/√2) as it is half-wave rectified current.

TUF = (V^{2}_{sm}/πR_{L})/(V^{2}_{sm}/2√2R_{L}) = 2√2/π = 0.287

Though, because of saturation effects produced by flow of direct current through transformer secondary, value of are further reduced to 0.2.

**Equivalent Circuit of a HW Rectifier:**

Diode has been replaced by equivalent circuit. Transformer secondary has been replaced by ac sinusoidal generator having peak value of Vsm. Resistance Rs represents transformer secondary resistance.

Obviously,

I_{LM} = (V_{sm} - V_{B})/(R_{s} + r_{d}) + R_{L} = (V_{sm} - V_{B})/(R_{0} + R_{L})

η = (4/π^{2})(R_{L}/(R_{s} + r_{d}) + R_{L})

Voltage regulation is provided by

V_{R} = (V_{NL} - V_{FL})/V_{FL} x 100%

Under no-load condition i.e. when no output current flows, voltage has maximum value.

When rectifier is completely loaded i.e. when output current flows, there is drop over. Therefore, output voltage is decreased by this much amount.

V_{FL} = V_{NL} (R_{L}/(R_{0 }+ R_{L}))

Substituting value in equation, we get:

V_{R }= R_{0}/R_{L}

*Full-Wave Rectifier:*

In this case, both half-cycles of input are used with help of two diodes working alternately. For full-wave rectification, use of a transformer is necessary (although it is optional for half-wave rectification).

Full-wave rectifier circuit utilizing two diodes and center-tapped transformer is shown below. Center-tap is usually taken as ground or zero voltage reference point.

Working:

When input ac supply is switched on, ends M and N of transformer secondary become +ve and -ve and alternately. During positive half-cycle of ac input, terminal M is +ve, is at zero potential and N is at -ve potential. Therefore, being forward-biased, diode D_{1} conducts (but not D_{2} that is reverse-biased) and current flows. Consequently, positive half-cycle of voltage appears across RL.

During negative half-cycle, when terminal N becomes +ve, then D_{2} conducts (but not D_{1}) and current flows. Current keeps on flowing through R_{L} in same direction (i.e. from A to B) in both half-cycles of ac input. It signifies that both half- cycles of input ac supply are used. Also, frequency of rectified output voltage is twice the supply frequency. Of course, this rectified output comprises of dc component and several ac components of diminishing amplitudes.

Average and RMS Values:

V_{L}(ac) = √VL_{2} - VL_{2}(dc)

Likewise,

I_{L} = I_{LM }= 0.7.7I_{LM}

I_{L}(dc) = 2I_{LM}/π = 0636I_{LM}

I_{L}(ac) = I^{2}L - I2L(dc)

Incidentally, I_{L(ac)} is same thing as I_{r(rms)}

Efficiency:

P_{in} = I^{2}_{L}(R_{0} + R_{L}) = (I_{LM}/√2)(R_{0} + R_{L}) = 1/2 I^{2}_{LM}(R_{0} + R_{L})

η = P_{dc}/P_{in} = (8/π^{2})(R_{L}/R_{0} + R_{L}) = 0.812/(1 + R_{0}/R_{L}) = 81.2%/(1 + R_{0}/R_{L})

It is twice the value for half-wave rectifier for simple reason that full-wave rectifier uses both half-cycles of input ac supply. Ripple Factor

For a full-wave rectified voltage, dc value is

V_{dc} = 0.636Vm

Rms value of ac component of output signal can be computed to be:

V_{r}(rms) = 0.308Vm

Percent ripple of full-wave rectified signal can be computed as:

γ = V_{r}(rms)/V_{dc} x 100% = 0.308V_{m}/0.636V_{m} X 100% = 48%

*Full-Wave Bridge Rectifier:*

It is the most often-utilized circuit for electronic dc power supplies. It needs four diodes but transformer utilized is not center-tapped and has maximum voltage of V_{sm}. Full-wave bridge rectifier is available in three distinct physics forms.

1. Four discrete diodes

2. One device inside the four-terminal case

3. As part of array of diodes in an IC.

Working:

During positive input half-cycle, terminal M of secondary is positive and N is negative as shown. Diodes D_{1} and D_{3} become forward-biased (ON) whereas D_{2} and D_{4} are reverse- biased (OFF). Therfore, current flows along producing the drop across R_{L}.

During negative input half-cycle, secondary terminal becomes positive and M negative.

Therefore, current keeps flowing through load resistance R_{L} in same direction AB during both half-cycles of ac input supply. As a result, point A of bridge rectifier always serves as anode and point C as cathode. Output voltage across is RL

Its frequency is twice that of the supply frequency.

Average and RMS Values:

These are same as for center-tapped full-wave rectifier.

Efficiency %η = 81.2/(1 + 2r_{d}/RL)

Ripple Factor:

It is same as for full-wave rectifier i.e. γ = 48%

PIV:

PIV rating of each of four diodes is equal to -entire voltage across secondary. When Secondary and Diode Resistances are considered

I_{LM} = (V_{sm} - 2V_{B})/(R_{s} + 2r_{d}) + R_{L} = (V_{cm} - 2V_{B})/(R_{o} + R_{L})

η = (8/π^{2})(R_{L}/(R_{0} + 2r_{d} + R_{L})) = (8/π^{2})(R_{L}/(R_{0} + R_{L}))

V_{R} = (R_{s} + 2r_{d})/R_{L }= R_{0}/R_{L}

Advantages:

After the start of low-cost, highly-reliable and small-sized silicon diodes, bridge circuit has become much more popular than center- tapped transformer full-wave rectifier. Main reason for this is that for bridge rectifier, a much smaller transformer is needed for same output as it uses transformer secondary continuously unlike 2-diode full-wave rectifier that uses two halves of secondary alternately.

So, benefits of bridge rectifier are:

1. No center-tap is needed on transformer

2. Much smaller transformers are needed

3. It is appropriate for high-voltage applications

4. It has less PIV rating per diode.

Disadvantage is need for twice as many diodes as for center-tapped transformer version. But ready availability of low-cost silicon diodes has made it more economical despite its need of four diodes.

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