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Topic 3

PN Junction and Diode

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Semiconductor • Semiconductor - materials with characteristics which fall

between insulator and conductors.

– It will not pass current as readily as a conductor nor will it block current as effectively as an insulator.

– E.g. Germanium (Ge), Silicon (Si) and Carbon (C).

• These materials have atomic structures which may be easily altered to obtain specific electrical characteristics.

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• The bandgap energy largely determines the electrical properties of solids:

• EG ≥ 3eV Insulator

• EG ≥ 0.1 - 3eV Semiconductor

• EG = 0 Conductor

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Semiconductor atom

• All atoms have a nucleus where protons and neutrons are located. It carries positive charges, which are surrounded by electrons orbiting in certain shells around the nucleus.

• Electrons occupy energy bands (shells). The 1st shell can only hold 2 electrons, the 2nd 8 electrons, the 3rd 18 electrons, the 4th 32 electrons and so on.

• The outermost shell of an atom is called the valence shell and the electrons orbiting within this shell are called valence electrons.

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• When an atom contains exactly 4 electrons in its outer shell, it does not readily give up or accept electrons.

– Elements which contain atoms of this type do not make good conductors or insulators.

• They are called semiconductor.

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• Each atom shares its 4 valence electrons with 4 neighbouring atoms. – covalent bond.

• Each of the atoms in the structure tries to take an additional electron to fill its valence shell with 8 electrons to achieve stability.

– Therefore each atom tends to be stable and will not easily give up or accept electrons.

• Due to the nature of the covalent bond in semiconductor, semiconductor materials have crystalline structure.

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Conduction in intrinsic (pure) semiconductor

(1) Effect of Temperature• The electrical characteristic of pure semiconductor material is highly

dependent upon temperature.

• At extremely low temperature, the valence electrons are held tightly to their parent atoms. – Therefore, the material cannot support current flow at this temperature. – Pure Ge and Si crystal function like an insulator.

• As the temperature of a Ge and Si crystal is increased, the valence electrons became agitated. – Some with more energy will occasionally break away from the covalent bonds. – They will be free to drift from one atom to the next in a random manner. – These free moving electrons are able to support a small amount of electrical

current if a voltage is applied to the semiconductor. – If they are exposed to extremely high temperature, they conduct as well as an

ordinary conductor.

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(2) Electron-Hole Pair• When an electron receives enough energy, it breaks away from

a covalent bond, a vacancy called hole exist in the bond. – A hole represents the absence of an electron. – Since electron has a negative charge, the hole has the characteristics of

a positively charged particle. – Each corresponding free electron and hole generated is referred to as

an electron-hole pair.

• A hole will exerts an attraction force on an electron and therefore, will be filled by a nearby passing electron.– This process is known as recombination and it causes a continued loss

of holes and free electrons.

• At any given temperature, the rate of recombination of electron-hole pairs is always equal to the generation of new electron-hole pairs so that the total number of free electrons and holes is constant.

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(3) Current flow in pure semiconductor• If a voltage is applied across a pure semiconductor, the free

electrons are attracted to the positive terminal of the voltage source and the holes drift towards the negative terminal. – The movement of particles is shown in the following diagram.

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Extrinsic (impure) semiconductor

• Pure semiconductor materials contain only a small number of electrons and holes at room temperature.

• Impurities (pentavalent or trivalent materials) are added to the semiconductor material through doping to increase the conductivity of semiconductor.

• Doping creates two types of extrinsic semiconductor: n-type and p-type.

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(1) n–type Semiconductor • A pure semiconductor material doped with a pentavalent

materials. – E.g. arsenic (As)– the arsenic atom will replace one of the Ge or Si atoms and share 4 of

its valence electrons with adjacent atoms in a covalent bond. – However, its 5th valence electron is loosely attached to the nucleus of

the arsenic atom and can be easily set free.

Note: no holes are generated when the 5th electron leaves the valence band.

• Since the As atom is called a donor atom.– There are many donor atoms within the semiconductor material, there

are also many additional free electrons within the material.

– Due to the presence of these extra electrons in the doped Ge and Si crystal, the material is called an n-type semiconductor.

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• If a voltage is applied to an n–type semiconductor, the free electrons will flow towards the positive terminal of the battery.

• The free electrons in the n–type come from two sources: – Donor atoms (only electrons are generated, not electron-hole pairs) due

to doping process.– Thermally generated electron-hole pairs by breaking of covalent bond.

• The holes move towards the negative terminal.

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• At normal room temperature the number of free electrons provided by the donor atom will greatly exceed the number of holes and electrons that are produced by the breaking of covalent bonds.

• Therefore the total number of electrons flowing in n–type semiconductor greatly exceeds the number of holes. – Hence, electrons are called the majority carriers while holes are

called minority carriers.

• The material is still electrically neutral.

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(2) p–type Semiconductor • A pure semiconductor material added with a trivalent material.

– E.g. Aluminium (Al). – The Al atom shared its electron with 3 adjacent atoms in the crystal

structure but cannot share with the 4th adjacent atom. – This creates a hole in the covalent bond. – The trivalent atom can take/accept an electron, it is also called acceptor

atom. – A hole created by this doping process is not accompanied by a free

electron.

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• Since many Al atoms are added, a large number of holes will be present in the material. – Holes can be thought of as positive charges because the absence of an

electron leaves a net positive charge on the atom.

– Holes are the majority carriers in p–type material.

– There are a few free electrons that are created when electron-pairs are thermally generated.

– Electrons in p–type material are called minority carriers.

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The pn – junction • A pn–junction is formed at

the boundary between the two regions (n–type and p–type semiconductors) – A diode is created.

– It is the electrical characteristics of the junction rather than the semiconductor sample material that determines the electrical behavior of the entire structure.

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(1) Formation of the Space-Charge Region (Depletion Region)

• At the instant the pn–junction is formed, the electrons near the junction in the n region diffuse (because of high concentration) across the junction into the p region and combine with the holes near the junction.

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• Electrons leaving the n region leaves a layer of donor positive ions.

• When these electrons combine with the holes in the p region, a layer of acceptor negative ions is formed.

• This region is known as space-charge region or depletion region. It is depleted of charge carriers.

• The main purpose of a depletion region is to control current flow.

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(2) Barrier Potential• In the depletion region, the forces between the negative ions

and positive ions form an electric field.

• This electric field prevents the majority carriers from one side to cross over to the other side.

• Therefore, there is no net current flow in pn device. – The potential difference across the electric field is known as barrier

potential.

– It is the amount of voltage required by the majority carriers to get through the electric field. Hence, we need to apply a voltage source across the pn junction that is greater than the barrier potential so that electrons from n region can flow through the device.

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• The barrier potential depends on – the type of semiconductor material, – amount of doping and – temperature.

• The minority carriers are aided by the electric field to cross the junction. – For example, minority carriers in the p region, that is, the electrons

which are near to the depletion region will be swept across by the electric field to the n region and become the majority carrier there.

• By controlling the width of the depletion layer, we are able to control the resistance of the pn junction and thus the amount of current that can pass through the device.– Bias is a potential applied to a pn junction to control the width of the

depletion layer.

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(1) Forward Bias

• A pn–junction is forward biased when the n–type is more negative than the p–type.

• There are two ways in which a pn–junction can be forward biased:

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• In forward bias mode and when VF > barrier potential, the pn–junction begins to conduct.

• The depletion region becomes narrower.

• The majority carriers, for example, electrons, from n side cross the pn–junction and move into the p side with little opposition and conduction occurs.

• There is always a large majority of electrons on the n side and holes on the p side because the battery replenishes them.

IF

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• Once a pn–junction begins to conduct, it provides a slight opposition to current.

• This is the bulk resistance: it is the combined resistance of the n–type and p–type in forward–biased mode.

• Since RB is extremely low (typically, 25 or less), very little voltage drop across this resistance.

• Therefore, IR drop across RB is usually ignored in circuit calculations.

npB RRR

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• When a forward–biased pn–junction begins to conduct, the forward voltage, VF across the junction is slightly greater than the barrier potential.

• When the diode conducts, VF are approximated as VF = 0.7V for SiVF = 0.3V for Ge

• IF is the forward current. • VF is the voltage drop across the diode.

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(2) Reverse Bias • A pn–junction is reverse biased when the n–type is more

positive than the p–type.

• There are two ways in which a pn–junction can be reverse biased:

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• Reverse bias and its effects are illustrated in the following figures.

• Electrons in the n–type will head toward the positive terminal of the source. – This will further deplete the

electrons in the n–type near the junction.

• The same principle applies to the holes in p–type.

• The depletion layer is then widened.

• This increases the resistance of the junction and conduction drops to near zero.

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• The increase in the depletion width is extremely conductive to the minority carriers.

• The electrons in the p–type which are near to the depletion region will be swept across the junction by the electric field into the n region.

• Therefore, there is a current flow - the reverse current, IR.

• It is very small compare to IF.

• IR depends on temperature because the electron-hole pairs (minority carriers) are generated from the breaking of covalent bond which is temperature dependant. – It is not affected by the magnitude of the reverse bias VR.

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PNJUNCTION DIODE • A diode is forward biased even though a resistor, R, is

connected in series with it.– Its p–type is connected to the positive terminal of the source and its n–

type to the negative terminal. – The p–type is guaranteed to be more positive than the n–type.

• But whether the diode conducts or not, will have to depend on the voltage drop across it.

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THE DIODE • There are various types of diode but the ones we will be

studying are the pn–junction diode (normally, we just call it “diode”) and zener diode.

• Diode conducts current in one direction (when it is in forward bias mode) and blocks current in the other direction (when it is in reverse bias mode).

• The symbol is shown here.

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• When reverse biased, the diode conduction drops to nearly zero. What is the voltage drop across the diode?– A diode does not conduct when the symbol points to the more positive

of the diode potentials.

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DIODE MODELS

• A model is a representation of a component.

• It is used to show the characteristics of the component when the component is used in a particular application.

• The diode has 3 models. Each has its particular use in a particular application.– The ideal diode model. – The practical diode model.– The complete diode model.

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(1) The Ideal Diode Model

• This model represents the diode as a switch: – close (forward bias) or open (reverse bias).

• When the diode is forward bias (closed switch), we can say that:– The diode has no resistance: it acts as a short circuit.– There is no voltage drop across the diode.

• When the diode is reverse bias (opened switch), we can say that:– The diode has infinite resistance; therefore, it acts as an open

circuit.– The applied voltage drops across the diode terminals.

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• Normally, the ideal model of the diode is used in the initial stages of circuit troubleshooting.

– We are only interested in whether the diode is acting as a one-way conductor.

– If it is not, then the diode is faulty and needs to be replaced.

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Example

SolutionSince D1 is reverse biased, it acts as an open circuit. Therefore,

IT = 0A

VD1 = VS

VR1 = 0A

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Example

SolutionSince D1 is forward biased, it acts as a short circuit. Therefore,

VD1 = 0V

VR1 = 5V

IT = VR1 / R1 = 5 / 1000 = 5mA

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(2) The Practical Diode Model

• It is a bit more complex than the ideal diode model.

• The model includes characteristics – forward voltage VF,

– peak reverse voltage VRRM,

– average forward current IO,

– forward power dissipation PD(max))

which are useful for mathematical analysis to help determine which diode will be used for a given circuit.

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• VF is the voltage drop across the diode. • The following figures show the diode characteristics curve.

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• IF suddenly increases when the voltage across the diode VF VK (knee voltage). – Knee voltage is a general term to describe the situation where current

suddenly increase/decrease.VK 0.7 V (Si)

– In an actual circuit, VK = 0.7V to 1.1V, depending on the current through the device.

– As long as the diode is conducting, even if the applied voltage VS is increased further, the voltage drop across the diode will remain VF = VK.

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Example

SolutionD1 is forward biased. It acts as a short circuit.VD1 = 0.7V

VR1 = 5 – 0.7 = 4.3V

IT = VR1 / R1 = 4.3 / 1000 = 4.3mA

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Peak Reverse Voltage VRRM • When the diode is reverse-biased, the

maximum reverse voltage that will not force the diode to conduct is called VRRM.

• When VRRM is exceeded, the depletion layer breaks down and the diode conducts in the reverse direction. – Any insulator will conduct if an applied

voltage is high enough to cause the insulator to break down.

• The diode conducts in the reverse direction when the VR > VRRM. Normally, when a diode is forced to conduct in the reverse direction, the device is destroyed.

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• The reverse current in this case is known as the avalanche current.

• Avalanche breakdown is the result of collisions of high – energy minority carriers with atoms to break the covalent bond.– When the reverse bias is large enough, some of the minority carriers

will gain sufficient energy and when these carriers collide with the atoms, covalent bonds are broken.

– So, more electron – hole pairs are generated and will gain high – energy (due to the large reverse bias) to cause further broken covalent bonds.

– As the process is repeated, the saturation current quickly increases and breakdown takes place.

• VRRM is a very important limiting parameter.

• In practice, the VRRM should at least be 20% greater than the maximum voltage that the diode is expected to block. – The reason for 20% is because in practical applications, there will be

minor variations in the source amplitude.

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Example During the negative alternation of the input, the peak value Vpk (50 V) is dropped across the reverse-biased diode.

SolutionVRRM = 1.2 VR(pk) = (1.2) (50) = 60V (minimum)

The VRRM rating for D1 must be 60 V or greater to ensure that the component isn’t damaged by the voltage source. As long as VRRM rating is 20% greater than the maximum reverse voltage in a circuit, we are not concerned about the diode’s exact value.

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Average Forward Current IO

• This rating indicates the maximum allowable value of dc forward current.

• Again, IO must be taken as 20% greater than the forward current IF.

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Example Determine the minimum average forward current rating that would be required for the diode shown.

SolutionIF = (VS – 0.7) / RL = (50 – 0.7) / 200 = 246.5mA

Therefore, D1 must have an IO > 246.5mA.

IO = 1.2 IF = (1.2) (246.5) = 295.8mA (minimum)

Note: when dealing with ac circuits, you will need to determine the average (dc equivalent) current for the circuit.

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Forward Power Dissipation PD(max)• This rating indicates the maximum possible power dissipation

of the device when it is forward biased.

ExampleCalculate the minimum forward power dissipation rating for any diode that would be used in the circuit shown.

SolutionIF = (VS – 0.7) / RL = (10 – 0.7) / 100 = 93mAPF = IFVF = (93mA) (0.7) = 65.1mWPD(max) = 1.2 PF = (1.2) (65.1) = 78.12 mW (minimum)

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(3) The Complete Diode Model

• The complete diode model most accurately represents the true operating characteristics of the diode.

• It includes characteristics e.g. – bulk resistance RB, – reverse current IR

that are meant for circuit development, high-frequency analysis, etc.

• The model is used to explain many of the differences between predicted and measured circuit values.

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Bulk Resistance RB• The effect due to RB is that VF become non-constant.

– VF is constant in the practical diode model.

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Diode Behaviour

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• The diode equivalent circuit looks as follows.

Note: VB and RB only exist when the diode is in forward bias, not reverse bias.

• As IF increases, IFRB increases too. – The total voltage across a diode varies with IF. – The measured values of VF generally vary between 0.7 V and 1.1 V. – In a low-current circuit, the VF will tend to be closer to 0.7 V because

lesser voltage drop across RB. – In a high-current circuit, the VF will tend to be closer to 1.1 V.

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Reverse Current IR

• In reverse bias, the diode is supposed to stop current flow.

• However, there is a small current IR that flows because of the minority carriers.

• IR is made of two components

IR = IS + ISL

where IS = the reverse saturation current (It varies with temperature)

ISL = the surface-leakage current (It varies with the reverse voltage).

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• Since ISL << IS, it is safe to assume that IR IS. – This makes IR dependent on temperature as well.

– The voltage developed across any series resistance by IR is usually insignificant. However, knowing it helps us to explain:

• Any voltage across the series resistance may increase dramatically with increases in operating temperature.

• We often do not measure the full applied voltage across a reverse-biased diode.

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Diode Capacitance • When a diode is reversebiased,

it forms a depletion layer (insulator) between two semiconductor materials.– It has some measurable amount of

junction capacitance.

• Junction capacitance is very important in high-frequency operation of a diode. – There is a type of diode designed

to make use of its junction capacitance: it is called a varactor.

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Temperature Effects on Diode Operation

• An increased in temperature means increased thermal activity and decreased diode resistance. – This is true for both forward bias

and reverse bias.

• As temperature increases, at a fixed value of VF, IF increases.

• As temperature increases, at a fixed value of IF, VF decreases.

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• As temperature increases, IR increases for all values of VR (< VRRM).

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ZENER DIODE

• A pn–junction diode operated in this region is usually destroyed by the excessive reverse current and the heat it produces. – But the zener diode is designed to

work in the reverse breakdown region.

• The reverse voltage across the diode VR remains near-constant voltage (called zener voltage, VZ) despite a drastic change in current flow through the diode.

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• Zener diode is useful as a voltage regulator. – A voltage regulator is a circuit designed to maintain a relatively

constant voltage despite anticipated variations in load current or input voltage.

• Zener diode is useful in the reverse bias mode, not in the forward bias mode.

• There are two types of reverse breakdown: avalanche breakdown and zener breakdown.

• The materials of the zener diode are designed in such a way that the zener breakdown occurs at relatively low VR.

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• Zener breakdown is a type of reverse breakdown that occurs at relatively low reverse voltages.

• The n–type and p–type materials of zener diode are heavily doped, resulting in a narrow depletion layer.

• Therefore, the depletion layer can break down at a lower VR than the depletion layer in a pn–junction diode.

• Zener diodes with low VZ ratings experience zener breakdown, while those with high VZ ratings will experience avalanche breakdown.

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Zener Operating Characteristics • A Zener diode maintains a

near-constant VR for a range of IR values.

• Zener knee current IZK is the minimum current required to maintain voltage regulation (constant voltage).

• To use the zener as a voltage regulator, the current through the diode must never be lesser than IZK.

VZT

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• The maximum Zener knee current IZM is the maximum amount of current the diode can tolerate without being damaged.

• Zener test current IZT is the current level at which the VZ rating of the diode is measured. – It is an important parameter in circuit design. – If you need to have a zener voltage that is as close to the rated

value of VZ as possible, you need to design the circuit to have a current value equal to IZT.

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• Zener impedance ZZ is the zener diode’s opposition to any change in current.

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• There are two equivalent circuits for the zener diode. • The ideal model considers the zener to be a voltage source equal

to VZ. – When placed in a circuit, this voltage source opposes the applied circuit

voltage.

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• The practical model includes a series resistor, ZZ. • This model is mainly used for predicting the response of the

diode to a change in circuit current.

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Example A 1N4736 zener diode has a ZZ of 3.5. The data sheet gives VZT = 6.8V at IZT = 37mA and IZK = 1mA. What is the voltage across the zener terminals when the current is 50mA? When the current is 25mA?

SolutionFor IZ = 50mA,ΔIZ = IZ − IZT = 13mAΔVZ = ΔIZZZ = (13mA) (3.5) = 45.5mV

The change in voltage due to the increase in current above the IZT value causes the zener terminal voltage to increase. The zener voltage for IZ = 50mA isVZ = 6.8 + ΔVZ = 6.8 + 0.0455 = 6.85V

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Example A 1N4736 zener diode has a ZZ of 3.5. The data sheet gives VZT = 6.8V at IZT = 37mA and IZK = 1mA. What is the voltage across the zener terminals when the current is 50mA? When the current is 25mA?

SolutionFor IZ = 25mA,ΔIZ = IZ − IZT = −12mAΔVZ = ΔIZZZ = (−12mA) (3.5) = −42mV

The change in voltage due to the decrease in current below IZT value causes the zener terminal voltage to decrease. The zener voltage for IZ = 25mA is

VZ = 6.8 − ΔVZ = 6.8 − 0.042 = 6.76V

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Zener Regulation with a Variable Load • The zener diode maintains a nearly constant voltage across RL as long as the zener

current is greater than IZK and less than IZM.

– When the output terminals of the zener regulator are open (RL = ), the load current IL is zero and all current is through the zener.

– When a load is connected, part of the total current is through the zener and part through RL.

– When RL decreases, IL increases and IZ decreases.

• The zener diode continues to regulate the voltage until IZ reaches its minimum value IZK. At this point IL is at maximum and the load is called a full load.

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Negative Series Clipper

• It eliminates the negative alternation of the ac input. – When D1 is forward bias, VL = Vin – 0.7.

– When D1 is reverse bias, VD1 = Vin, VL = 0V.

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COMMON DIODE APPLICATIONCLIPPERS (LIMITERS)

• A clipper (or limiter) is a diode circuit that is used to eliminate some portion of a waveform. – For example, the half-wave rectifier is a clipper that

eliminates one of the alternations of an ac signal.

• There are two basic clippers:– Series clipper: series clippers contain a diode in series with

the load.– Shunt clipper: shunt clippers contain a diode in parallel with

the load.

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(1) Series Clippers • The half-wave rectifier is actually a series clipper. • The series clippers provide an output when the diode is

forward biased and no output when the diode is reverse biased. – Two types: the negative and the positive series clippers.

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Positive Series Clipper

• Works the same way as negative series clippers with only the following differences:– The output voltage polarities are reversed.– The current directions through the circuit are reversed.

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(2) Shunt Clippers

• The shunt clipper provides an output when the diode is reverse biased and shorts the output signal to ground when the diode is forward biased.

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VR(S) = Vin + 0.7 V

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Example

Determine the value for VL for each for the input alternations.

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Solution

• When the input is positive, the diode is forward biased. Thus, VL = 0.7 V.VR(S) = Vin – 0.7 = 10V – 0.7 = 9.3 Vpk

• When the input is negative, the diode is reverse biased. Thus,VL = (RL / RL + RS) Vin

= (1.2k / 1.4k)(–10V) = − 8.45Vpk

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• What is the use of RS?

• It is used as a current-limiting resistor.

• During forward biased, the diode shorts the signal source to the ground during positive alternation of the input signal. – The diode and the components in the signal source may be

damaged by the excessive forward current.

• Note: Make the RS much lower than the RL.

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(3) Biased Clippers

• The level of ac voltage in a series or a shunt clipper is limited to approximately zero.

• In a biased clipper, the level to which an ac voltage is limited can be adjusted by adding a bias voltage VB in series with the diode.

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• In practice, a potentiometer is used to provide an adjustable value of VB.

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(4) Clipper Applications

• They are generally used to perform one of the two functions:– Alter the shape of a waveform– Provide circuit transient protection

• E.g. the half-wave rectifier is an example where the circuit alters the shape of an ac signal, changing a sine-wave to a pulsating dc.

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Transient Protection• A transient is an abrupt current or voltage spike that has an

extremely short duration. – A current surge is an example of a transient. – Transients can damage circuits whose input must stay within certain

voltage or current limits.

• For example, digital circuits have inputs that can handle only voltages that fall within a specified range. – A voltage transient that goes outside the specified voltage range cannot

be allowed to reach the input. – A clipper can be used in this case to prevent such a transient from

reaching a digital circuit.

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Digital circuits

The transient occurs on the input line

It will forward bias D2, causing the diode to conduct

With D2 conducting, the transient will be shorted to ground, protecting the input to circuit A.

Input square wave goes above 5V, D1 will be FB.

It shorts any input greater than 5V back to the +5V supply.

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• The figure shows a clipper designed to protect the output of circuit B from transients produced when a square wave is used to drive a speaker.

• A speaker is essentially a big coil (inductor). – When a square wave is used to drive the coil in a speaker, that coil produces a counter emf. – This counter emf is produced each time that the output voltage makes the transition from +5V to 0V.

Its magnitude is greater than the original voltage and opposite in polarity.

– Each time circuit B tries to drive the output line to 0V, the speaker tries to force the line to continue to some negative value.

– This counter emf could destroy the driving circuit B. – The clipping diode shorts the counter emf produced by the speaker coil to ground before it can harm

the driving circuit.

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AM Detector • Another common application for the diode clipper is in a

typical AM receiver. • Purpose: produce dc output voltage that varies with the peak

of the input signal.

• The input signal has an average value of 0V. – This is because each positive peak has an equal and opposite negative

peak. • The clipping diode is used to eliminate the negative portion of

the input waveform. • With the negative portion eliminated, the capacitor charges

and discharges at the rate the peak input varies.

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CLAMPERS (DC RESTORER)

• A clamper is a diode circuit that is used to shift a waveform either above or below a given reference voltage without distorting the waveform.

• There are two types: – positive clamper and – negative clamper.

• A positive clamper shifts its input waveform so that the negative peak of the waveform is approximately equal to the clamper dc reference voltage.

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• In the figure above, the clamper dc reference voltage is at 0V. • The peak-to-peak value of the output waveform remains at

20Vpp but the 10Vpk becomes 20Vpk and the −10Vpk becomes the clamper dc reference, 0V.

Note: The dc value for the output waveform is no longer the same as the input waveform: Vin(dc) = 0 V and Vout(dc) = 10 V.

• In summary, the Vdc normally changes but the Vpp does not when the waveform goes through the clamper circuit.

• For sine-wave, the dc (average) value falls halfway between the peak to peak.

• A negative clamper does the opposite to that of the positive clamper.

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(1) Clamper Operation • The clamper works on the basis of switching time constants. When the

diode is forward biased, it provides a charging path for the capacitor as shown.

• C1 is completely charged when TC = 5RD1C1

where RD1 is the bulk resistance of the diode.

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• When the diode is reverse biased, C1 discharge through RL. C1 is completely discharged when TD = 5RLC1.

• For the clamper circuit to work properly, TD >> TC.

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Consider a clamper has values RD1 = 10 , C1 = 1 F and RL = 10 k. Therefore, TC = 5RD1C1 = 5(10)(1 X 10-6) = 50 µs

TD = 5RLC1 = 5(10000)(1 X 10-6) = 50 ms

A square-wave is applied to the input of the clamper circuit.

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• When the input is at 5V, D1 is forward biased.

• C1 charges quickly to 5 V. The voltage drop across RL, VL = 0V.

• When the input is at −5V, D1 is reversed biased.

• C1 will discharge through RL.

• The voltage drop across RL is the sum of Vin and VC1 which is −10 V.

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Note: choose C1 so that it will discharge very slowly, to maintain a constant −10 V.

• A clamper provides an output square wave that varies between 0 V and −10 V.

• When designed properly, clampers work for a variety of input waveforms, not just square-wave.

Input

Output

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(2) Biased Clampers

• Biased clampers shift a waveform so that it falls above or below a dc reference other than 0 V.

• Figure shows a negative-biased clamper. – If VB is −20 V and R1 is set so

that the potential applied to D1 is −10 V.

– The input waveform would be shifted so that its positive peak voltage is approximately −10 V.

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• The positive-biased clamper shifts its input waveform so that the negative peak of the output waveform is approximately equal to the potential applied to D1.

• The Zener diodes are used in Zener clampers to set the dc reference voltages of the circuits.– If let’s say a 10 V Zener is

used, the dc reference voltage is approximately 10V.

92

• Let’s say a 10 V Zener is used in a positive zener clamper, the dc reference voltage is approximately −10 V.

• The operation of any clamper depends on the one–way conduction of the pn–junction diode, not the Zener diode. – Zener diode conducts in the forward

and reverse bias mode. Therefore, it cannot be the diode that performs the one-way conduction.

– During the positive alternation of the input, the diode branch conducts, clamping the positive peak to (VZ + 0.7 V).

– During the negative alternation of the input, D1 is reverse biased, blocking conduction through the diode branch.

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• Zener clampers come only in two varieties:

– Negative clampers with positive dc reference voltages– Positive clampers with negative dc reference voltages

• We cannot have a positive zener clamper with a positive dc reference voltage.

• If D1 is reversed, there is nothing to prevent the diode branch from conducting during the negative alternation of the input signal.

• The clamping action is therefore eliminated. This is not desirable.

94

VOLTAGE MULTIPLIERS • A voltage multiplier produces a dc output voltage that is a

multiple of an ac peak input voltage.– For example, a voltage doubler provides a dc output voltage that is

twice its peak input voltage.

• They are not power generators. – This is because the input peak current is decreased.

• The device has a low current capability.– The output current capability is reduced every time the voltage is

increased.

• Therefore, it is usually used in high-voltage, low-current applications. – For example, the cathode-ray tube in a television.

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(1) Half-Wave Voltage Doublers • The half-wave voltage doubler consists of two diodes and two

capacitors.

• Electrolytic capacitors are normally used because of their high capacity ratings.

• In practice, multiplier circuits are used in power supply applications.

96

• During the negative alternation of the input cycle, D1 is forward biased, D2 is reverse biased.

• C1 charges to the peak value of input signal.

• C2 discharges through RL.

97

• During the positive alternation of the input cycle, D1 is reverse biased, D2 is forward biased.

• VS and C1 are effectively connected in series with C2.

• C2 charges to approximately 2VS(pk).

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• The time constant of the above circuit is such that C2 discharges very little across RL during the negative alternation of the input.

• During the positive alternation of the input, C2 charges to approximately 2VS(pk).

• In this way, the output waveform closely resembles that of a filtered half-wave rectifier

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• The circuit has a dc output voltage and ripple voltage.

• The dc output voltage is approximated as Vdc 2VS(pk).

• The output ripple from the multiplier is calculated in the same manner as for a filtered half-wave rectifier.

• The amount of ripple depends on the values of the capacitors and the current demand of the load. – High capacitor values and low current demand reduce the amount of

ripple.

• To get a negative voltage doubler, reverse the direction of the diodes.

• The circuit output is a negative voltage, -2VS(pk).

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(2) Full-Wave Voltage Doublers • During the positive half-cycle

of the input, D1 is forward biased, D2 is reverse biased. – C1 charges to VS(pk).

• When the input polarity is reversed, D1 is reverse biased, D2 is forward biased.– C2 charges to VS(pk).

• Since C1 and C2 are in series, the total voltage across the two components is Vdc 2VS(pk).

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BASIC POWER SUPPLY CIRCUITS

• The power supply is a group of circuits that convert the ac energy provided by the wall outlet to dc energy for our daily use.

102

• C3 is a filter capacitor. It is used to reduce the ripple at the output. – A full-wave voltage doublers is used as a dual-polarity power supply.

• A basic linear power supply can be broken into three circuit groups:– Rectification– Filtering– Voltage regulation

103

TRANSFORMER • Transformer is not a solid-state device, but they are closely

related to the operation of power supply.

The above figure shows the schematic for a transformer. • Transformers are made up of inductors that are in close

proximity to each other, yet are not electrically connected.• An alternating voltage applied to the primary induces an

alternating voltage in the secondary. • Thus, a transformer provides ac coupling from primary to

secondary while providing physical isolation between the two circuits.

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• Three types of transformer:– Step-up.– Step-down.– Isolation: the output voltage is equal to the input voltage.

• The turns ratio is the ratio of the number of turns in the primary to the number of turns in the secondary. – The turns ratio is equal to the voltage ratio of the component.

where NS = the number of turns in the secondary,NP = the number of turns in the primary,VS = the secondary voltage, andVP = the primary voltage

p

S

p

S

VV

NN

105

• Ideally, transformers are 100% efficient. This means 100% of the input power is transferred to the output.

PS = PP

• Therefore, VSIS = VPIP

• From the formula,– For a step-down transformer, IP < IS

– For a step-up transformer, IP > IS

• Current varies in the opposite way that voltage varies.

P

S

S

P

VV

II

S

P

p

S

II

NN

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RECTIFIER

• A rectifier is a diode circuit that converts the ac to pulsating dc.

• There are 3 basic types of rectifier circuits:– Half-wave rectifier– Full-wave center-tapped rectifier– Full-wave bridge rectifier (the most commonly used)

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(1) Half-Wave Rectifier

• Half-wave rectifier is a simple diode.

• It is connected in series between a transformer and its load.

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• The diode is used to eliminate either the negative or positive alternation of the input. – A sine-wave at the input, we should get a half-wave rectified output

waveform.

• The output will drop across RL.– Therefore we can observe (using a scope) or measure (using a

multimeter) the output across RL, not the diode. The diode direction determines which half-cycle is eliminated.

110

Basic Circuit Operation

• During positive cycle of the input, the diode is forward biased and provides a path for the current.

• Therefore, VD1 0V. VS will drop across RL (VL = VS).

111

• During negative cycle of the input, the diode is reverse biased and acts as an open.

• Therefore, VD1 = VS.

• No current flow in the circuit, so, no voltage developed across RL, VL = 0V.

112

Load Voltage and Load Current • What we have discussed so far is based on an ideal diode. Take

VF into account, the peak load voltage, VL(pk) is given as

VL(pk) = VS(pk) – VF

• We can get VS(pk) from the transformer turns ratio formula.

NS / NP = VS(pk) / VP(pk)

• Usually, source voltages are given as rms values. Convert to the peak value using

Vpk = Vrms / 0.707

• The peak load current is found asIL(pk) = VL(pk) / RL

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Example

What is the peak load current for the circuit shown in the figure with a 120 Vac input?

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Solution

• VP(pk) = VP(rms) / 0.707 =

• VS(pk) = (NS/NP) VP(pk) =

• VL(pk) = VS(pk) – VF =

• IL(pk) = VL(pk)/RL = 5.59mA

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Average Voltage and Average Current

• The average voltage Vave is the dc equivalent value of the ac waveform. – In most cases, Vave and Vdc are used to describe the same value.

• For half-wave rectifier:Vave = Vpk /

Iave = Ipk /

116

ExampleDetermine the dc load voltage and current for the rectifier shown. What role does the value of Iave play in component substitution?

117

SolutionVS(pk) = 24 / 0.707 =

VL(pk) = VS(pk) – VF =

Vave = VL(pk) / =

IL(pk) = VL(pk) / RL =

Iave = IL(pk) / = 528.39 μA

The maximum dc forward current that can flow through a diode is equal to the average forward current IO rating of the device. A faulty diode from a rectifier may need to be substituted. Therefore, Iave can be calculated as estimation or measured, so as to make sure that Iave < IO.

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Peak Inverse Voltage, PIV

• It is the maximum amount of reverse bias that will be applied to a diode in a rectifier circuit.

• For a half-wave rectifier, the PIV is given as PIV = VS(pk)

– When the diode is reverse biased, no voltage drop across RL. – All of VS is dropped across the diode. – Thus, PIV determines the VRRM of the diode.

119

(2) Full-Wave Center-Tapped Rectifier• This type of rectifier circuit requires the transformer to be

a center-tapped transformer. • The transformer has a lead connected to the center of the

secondary winding. – For a 24V center-tapped transformer, voltage from the center tap

to each of the outer winding terminals is 12V.

Center-tapped

120

121

Basic Circuit Operation • The load voltage is approximately equal to half the secondary

voltage because the transformer is center tapped. VL(pk) VS(pk)/2

122

Load Voltage and Load Current • The peak load voltage is,

VL(pk) = [VS(pk)/2] – 0.7

• The full-wave rectifier produces twice as many output pulses (per input cycle) as the half-wave rectifier.

• The average load voltage for a full-wave rectifier,Vave = 2(Vpk/)

123

Example

Determine the dc load voltage, the peak current and also the dc load current for the circuit shown.

124

Solution

• VS(pk) = 30 / 0.707 =

• VL(pk) = ½VS(pk) – VF =

• Vave = 2(VL(pk)/) =

• IL(pk) = VL(pk)/RL =

• Iave = Vave/RL = 2.56mA

125

Peak Inverse Voltage PIV • When one of the diodes is reverse biased, the voltage across that

diode is approximately equal to VS = 34 Vpk.

• The secondary voltages are +17 Vpk and 17 Vpk.

• Assuming D1 is ideal, then VD1 = 0V and the cathode of D1 is at +17 Vpk.

• Since the cathode of D1 is connected directly to the cathode of D2, the cathode of D2 is also at +17 Vpk.

• Since D2 is off, it will cause an open in the circuit. Therefore VD2 = +34 Vpk.

• Therefore, the reverse voltage across either diode is PIV = 2 VL(pk) = VS(pk)

126

Full-Wave versus Half-Wave • For a 24Vac (rated) transformer, the dc output voltages for the

two circuits are calculated asVave = 10.58 Vdc (half-wave rectifier)Vave = 10.36 Vdc (full-wave rectifier)

• The two circuits produce nearly identical dc output voltages the same values of transformer secondary voltage. – Question: why bother with the full-wave rectifier when it has the same

dc output values as the half-wave rectifier?

• The reasons are– If VL(pk) for the two circuits are equal, the full-wave rectifier will have

twice the dc load voltage and power efficiency of the half-wave rectifier. We need to use a step-up transformer with turns ratio 1:2.

– The full-wave rectifier has twice the output frequency of the half-wave rectifier, which has an impact on the filtering of the rectifier output.

127

(3) Full-Wave Bridge Rectifier

• It has the following advantage:– It does not require a center-tapped transformer and therefore, can be

directly coupled to the ac power line.– It produces nearly double VL(pk) of the full-wave center-tapped rectifier.

So, we get a higher dc output voltage from the supply.

128

• The current direction through the load does not change, nor has the resulting polarity of the load voltage.

129

Load Voltage and Load Current • The center-tapped transformer is essential for the center-

tapped rectifier to work.

• But the output voltage in a center-tapped rectifier is reduced to half the secondary voltage by the center tap on the transformer secondary.

• The bridge rectifier does not require a center tap transformer to work. It can be coupled directly to the ac line input, just like the half-wave rectifier.

• The output voltage, across the load, isVL(pk) VS(pk) (ideal diode)

• For practical diode model,VL(pk) = VS(pk) – 2(0.7)

130

Example

Determine the dc load voltage and current values for the circuit shown.

?

?

??

?

131

Solution• The peak secondary voltage is VS(pk) = 12 / 0.707 =

• The peak load voltage is VL(pk) = VS(pk) – 2VF =

• The dc load voltage is Vave = 2(VL(pk) / ) =

• The dc load current is Iave = Vave/RL = 825.8µA

132

Example• A center-tapped rectifier has a 12Vac transformer and a 12k

load. Determine the dc load voltage and current values for the circuit.

• Use the values for comparison with the bridge rectifier.

• The table below summarizes the results of the two examples above:

Value Bridge Rectifier Center-tapped Rectifier

VL(pk) 15.57V 7.79V

Vave 9.91V 4.96V

Iave 825.8µA 413.3μA

133

Peak Inverse Voltage PIV• Ideal diode :

PIV = VS(pk)

• Practical diode :PIV = VS(pk) – 0.7

Why not VS(pk) – 2(0.7)?

134

FILTER• Aim of power supply –

produce a constant dc output voltage.

• Filter: reduce variations in the rectifier output signal.– A little bit of voltage variation

still exist after filtering ripple voltage, Vr.

• It is not desirable• In audio amplifier, ripple can

produce an annoying “hum” at 60 Hz or 120 Hz.

• Ripple voltage depends on 1. Type of rectifier2. Filter component values3. Load resistance

Vr

Half-wave rectifier

Filter

135

Basic Capacitive Filter• Most basic type and most commonly used.• It is a capacitor connected in parallel with the

load resistance.

Diode is forward-biasedCapacitor charged to VS(pk)

Voltage decrease to zero (diode is FB) & increase in negative direction (Diode is RB).Capacitor discharges through RL.

136

• The capacitor charges through the diode.– Let’s say the bulk resistance of diode is 5 , it takes 5RDC = 2.5 ms

to fully charge the capacitor.– The discharge path for the capacitor is through the load resistor. – For the capacitor to fully discharge, it takes 5RLC = 500 ms.

• In other words, the capacitor charges instantaneously and discharges very slowly.– For a 60 Hz signal, the capacitor discharges for 16.7 ms before it is

provided with another charging voltage.

137

• Vr can be minimized by increasing the discharging time.

– Use a high capacitance filter together with a high resistance load.

• However, the load of a power supply may consist of any kind of circuit, each with its own input resistance characteristics.

– It is impractical to vary RL in to control Vr.

• Choosing the suitable capacitance depends on1. The amount of surge current Isurge that the rectifier diodes

can stand.2. The cost of using a larger-than-required capacitance.

138

Surge Current• When the power supply is first turn on, the filter

capacitor acts as a short circuit.

• The current through the diode is only limited by the resistance of the transformer secondary winding, RW, and the bulk resistance of the diode, RW, which are very low.

• The surge current,Isurge = VS(pk) / [RW + RB]

139

Example• If the input of the circuit shown in the the previous slide is

170 Vpk, a turns ratio of 2:1, RW = 0.8 , RB = 5 , what it the initial value of surge current for the circuit?

SolutionVS(pk) = (NS / Np) VP(pk) =

Isurge = VS(pk) / [RB + RW] = = 14.66 A

This value may damage the diode. If the surge current is too large for the diode, a resistor can be added to limit the current as shown in the next slide.

140

• For the circuit above, its surge current is given as Isurge = VS(pk) / [RW + 2RB + Rsurge ]

• Adding the surge resistor limits the current to a safe level for the diode to operate. However, this reduces the voltage drop across the output.– The value of the surge resistor, Rsurge, must be small

compare to the load resistor, RL.• Another option is to use a lower capacitance

filter. It charges faster thus reducing the duration of the surge.– The trade off is greater ripple voltage.

141

Filtered Output Voltages• Output from a filter has peak, average (dc) and

ripple voltages.

142

• The dc output voltage can be calculated asVdc = Vpk (Vr / 2)

• The ripple voltage is Vr = (ILt) / C

where IL = Vdc / RL

IL = the dc load currentt = the time between charging peaks.

• Note the equations are in a loop. To solve this problem, assume Vdc Vpk.

• The ripple factor, r, is an indication of the effectiveness of the filter.

r = Vr / Vdc

143

144

Example• Determine the value of Vdc for the circuit shown.

The input voltage has a frequency of 60 Hz.

Vdc = 16.16 Vr = 0.015

145

Filter Effects on Diode PIV• The filter does not have significant effect on the

PIV across each diode of the full-wave rectifier.

• However, the PIV across the diode for a half-wave rectifier is twice the secondary voltage.

PIV = 2VS(pk)

146

ExampleThe zener diode shown has values of IZK = 3 mA and IZM = 100 mA. What is the minimum allowable value of RL?

RL(min) = 241

147

Voltage Regulation• Voltage regulator is used to maintain a constant power supply

output voltage.– There are many types, we will look at Zener diode.

• The zener current must be kept within the rangeIZK < IZ < IZM

148

• The current through RS IT = (Vin VZ) / RS

• The current through RL IL = VZ / RL

• The current through D1 IZ = IT IL

• To maintain zener regulation, IZ > IZK IL(max) = IT IZK

• The minimum load resistance RL(min) RL(min) = VZ / IL(max)

• Likewise, the maximum load resistance RL(min) RL(min) = VZ / (IT IZM)

149

• Zener diode reduces the amount of ripple voltage present at the filtered output.

• To the ripple waveform, there is a voltage divider present in the regulator. The voltage divider is made up of RS series with ZZ||RL.

• The ripple output of the regulator is

rSLz

Lzoutr V

RRZRZV

||

||)(

150

ExampleThe filtered output from a full-wave rectifier has a peak-to-peak ripple voltage of 1.5 V. What will the ripple at the load equal?

0.129 Vpp

151

• Zener regulator is rarely used. It wastes a large amount of power.

• Ideally, a voltage regulator is able to maintain a constant dc output voltage even though there are changes in the input voltage of the load current demand.– In practice, a change in the input voltage does cause a change in the

regulator’s output voltage.

• Line regulation indicates how much change occurs in the output voltage for a given change in the input voltage.

Line regulation = Vout / Vin

– A 2 V/V means that the output voltage will change by 2 V for each 1 V change in the regulator’s input voltage.

• The smaller the line regulation value, the closer the regulator is to the ideal case (0V / Vin).

152

• A practical voltage regulator experiences slight change in output voltage where there is a change in load current demand.

• Load regulation specifies the change that occurs in the output voltage over a certain range of load current values, from minimum current (no load) to maximum current (full load).

Load regulation = (VNL VFL) / IL= Vout / IL

– A 10 V/mA means that the output voltage will change by 10 V for each 1 mA change in load current.