DMT 121ELECTRONIC DEVICES

CHAPTER 2DIODE APPLICATIONS

At the end of this class, students should be able to:-

Understand the concept of load-line analysis and how it is applied to diode networks.

Explain the process of rectification to establish a DC level from a sinusoidal AC input.

Load Line Analysis

The analysis of electronic circuits can follow one of the two paths :1. Actual characteristic or approximate model of the device.2. Approximate model will be always used in the analysis

VD= 0.7 V

Load Line Analysis The load line plots all

possible current (ID) conditions for all voltages applied to the diode (VD) in a given circuit. E / R is the maximum ID and E is the maximum VD.

Where the load line and the characteristic curve intersect is the Q-point, which specifies a particular ID and VD for a given circuit.

Fig. 2.1 Drawing the load line and finding the point of operation

Point of operation of a circuit

Load Line Analysis

The intersection of load line in Fig. 2.2 can be determined by applying Kirchhoff’s voltage in the clockwise direction, which results in:

RIVE

VVE

DD

RD

0

ID and VD are the same for Eq. (2.1) and plotted load line in Fig. 2.2 (previous slide).

Set VD = 0 then we can get ID, where

Set ID = 0 then we get VD, where

0 DVDR

EI

0 DID EV

Fig. 2.2 Series diode configuration

Example For the series diode configuration of Fig. 2.3a,

employing the diode characteristics of Fig. 2.3b, determine VDQ, IDQ and VR

Fig. 2.3 (a) Circuit; (b) characteristics.

Solution

VEV

mAk

V

R

EI

D

D

ID

VD

10

205.0

10

0

0

From the result, plot the straight line across ID and VD.

The resulting load line appears in Fig. 2.4. The Q points occurred at

VDQ 0.78 VIDQ 18.5mA

VR=IRR=IDQR=(18.5 mA)(0.5k)

= 9.25 V

Example

For the series diode configuration of Fig. 2.13, determine VD, VR and ID.

mAk

V

R

VII

VVVVEV

VV

RRD

DR

D

32.32.2

3.7

3.77.08

7.0

Solution:

Example Repeat example 2.4 with the diode reversed

0808

0

0

dIRD

RD

D

VVVVEV

VVE

I

Solution:

Open Circuit

Diode as Rectifier Rectifier: An electronic circuit that converts AC to pulsating DC. Basic function of a DC power supply is to convert an AC voltage

to a smooth DC voltage.

Half-Wave Rectifier

The diode conducts during the positive half cycle.

The diode does not conducts during the negative half cycle.

Sinusoidal Input: Half Wave Rectification

Fig. 2.44 Half-wave rectifier.

Fig. 2.46 Nonconduction

region (T/2 T).

Fig. 2.45 Conduction

region (0 T/2).

Average Value of Half Wave Output Voltage

m

mdcV

VV 318.0

The average value of the half-wave rectified output voltage (also known as DC voltage) is

The process of removing one-half the input signal to establish a dc level is called half-wave rectification

Example

What is the average value of the half-wave rectified voltage?

Solution: Vm/π = 15.9 V

Effect of Barrier Potential(Silicon diode)

Applied signal at least 0.7 for diode to turn on (Vk = 0.7V) Vi ≤ 0.7 V diode in open circuit and Vo = 0V When conducting, Vk=0.7V ,then Vo= Vi – Vk this cause

reduction in Vo, thus reduce the resulting dc voltage level. Now Vdc 0.318 (Vm – Vk)

Example

Draw the output voltages of each rectifier for the indicated input voltages.

Peak Inverse Voltage (PIV)

PIV=peak inverse voltage and is the maximum voltage across the diode when it is not conducting/reverse bias.

Can be found by applying Kirchhoff’s voltage law. The load voltage is 0V so the input voltage is across the diode at tp.

Peak Inverse Voltage (PIV)

Because the diode is only forward biased for one-half of the AC cycle, it is also reverse biased for one-half cycle.

It is important that the reverse breakdown voltage rating of the diode be high enough to withstand the peak, reverse-biasing AC voltage.

PIV=Vm OR accurately PIV (or PRV) Vm• PIV = Peak inverse

voltage• PRV = Peak reverse

voltage• Vm = Peak AC voltage Diode must capable to withstand

certain amount of repetitive reverse voltage

Full-Wave Rectifier A full-wave rectifier allows current to

flow during both the positive and negative half cycles or the full 360°.

Output frequency is twice the input frequency.

VDC or VAVG = 2Vm/π

Full-Wave Rectification

The rectification process can be improved by using more diodes in a full-wave rectifier circuit.

Full-wave rectification produces a greater DC output:Half-wave: Vdc =0.318VVdc =0.318Vmm

=V=Vmm//ππFull-wave: Vdc =0.636VVdc =0.636Vmm

=2V=2Vmm//ππHalf Wave Rectifier

Full Wave Rectifier

Example

Find average value of the full-wave rectified voltage?

Transformer Coupling

Turns ratio, n = Nsec/Npri

V(sec) = nV(pri) (in RMS value)

Vp(sec)=√2 x V(sec)

Full-Wave Rectification

Center-Tapped Transformer Rectifier

Requires

Two diodes

Center-tapped transformer

VDC=0.636(Vm)

Full-Wave Center Tapped Current flow

direction during both alternations. The peak output is about half of the secondary windings total voltage.

Each diode is subjected to a PIV of the full secondary winding output minus one diode voltage dropPIV=2Vm(out)+0.7V

PIV: Full-wave RectifierCenter-Tapped Transformer

PIV can be shown by applying KVL for the reverse-biased diode.

PIV across D2:

27.0

2(sec)(sec) pp V

VV

PIV

VVPIV p 7.0(sec)

VVV

VV

V

outpp

poutp

4.12

7.02

)((sec)

(sec))(

1

2

3

4Substitute 4 to 2:

PIV=2Vp(out) + 0.7 V

Example1. Show the voltage waveforms across each half of the

secondary winding and across RL when a 100V peak sine wave is applied to the primary winding.

2. What minimum PIV rating must the diodes have.

Solution

1.

2. PIV = 49.3 V

Full-Wave Rectification

Bridge Rectifier

Four diodes are required

VDC = 0.636 Vm

Full-Wave Bridge Rectifier The full-wave bridge

rectifier takes advantage of the full output of the secondary winding.

It employs 4 diodes arranged such that current flows in the direction through the load during each half of the cycle.

During positive half-cycle of the input, D1 and D2 are forward-biased and conduct current. D3 and D4 are reverse-biased.During negative half-cycle of the input, D3 and D4 are forward-biased and conduct current. D1 and D2 are reverse-biased.

PIV: Full-wave RectifierBridge Transformer

Vp(out)=Vp(sec) – 1.4 V

PIV=Vp(out) + 0.7 V

ExampleThe transformer is specified to have a 12 Vrms secondary

voltage for the standard 120 V across the primary.

• Determine the peak output voltage for the bridge rectifier.

• Assuming the practical model, what PIV rating is required for the diodes?

Solution

1. Vp(out) = 15.6 V

2. PIV = 16.3 V

Summary of Rectifier CircuitsRectifier Ideal VDC Practical

(approximate) VDC

PIV

Half-Wave Rectifier

VDC = 0.318(Vm) = Vm/π

VDC = 0.318(Vm)-0.7 PIV=Vm

Full-Wave Bridge Rectifier

VDC = 0.636(Vm) =2 Vm/π

VDC = 0.636(Vm)-2(0.7)

PIV=Vm+0.7V

Center-Tapped Transformer Rectifier

VDC = 0.636(Vm) =2 Vm/π

VDC = 0.636(Vm)-(0.7)

PIV=2Vm+0.7V

Vm = peak of the AC voltage = Vp

In the center tapped transformer rectifier circuit, the peak AC voltage is the transformer secondary voltage to the tap.

Power Supply Filters and Regulators In most power supply – 60 Hz ac power line voltage constant dc voltage

Pulsating dc output must be filtered to reduce the large voltage variation

Small amount of fluctuation in the filter o/p voltage - ripple

Power Supply Filters Filtering is the process of smoothing the ripple from the

rectifier.

Power Supply Filters and Regulators – Capacitor-Input Filter

The capacitor input filter is widely used. A half-wave rectifier and the capacitor-input filter are shown.

Power Supply Filters and Regulators Regulation is the last step in eliminating the remaining

ripple and maintaining the output voltage to a specific value. Typically this regulation is performed by an integrated circuit regulator. There are many different types used based on the voltage and current requirements.

A voltage regulator can furnish nearly constant output with excellent ripple rejection. 3-terminal regulators are require only external capacitors to complete the regulation portion of the circuit.

Power Supply Regulators How well the regulation is performed by a regulator is

measured by it’s regulation percentage. There are two types of regulation, line and load.

Line regulation: how much the dc output changes for a given change in regulator’s input voltage.

Load regulation: how much change occurs in the output voltage for a given range of load current values from no load (NL) to full load (FL)

%100

in

out

V

VLine regulation

%100

FL

FLNL

V

VVLoad regulation

Power Supply Filters and Regulators – Capacitor-Input Filter

Surge Current in the Capacitor-Input Filter: Being that the capacitor appears as a short during the initial

charging, the current through the diodes can momentarily be quite high. To reduce risk of damaging the diodes, a surge current limiting resistor is placed in series with the filter and load.

FSM

psurge I

VVR

4.1(sec)

IFSM = forward surge current rating specified on diode data sheet.

The min. surgeResistor values:

Capacitor Input Filter – Ripple Voltage

Ripple Voltage: the variation in the capacitor voltage due to charging and discharging is called ripple voltage

Ripple voltage is undesirable: thus, the smaller the ripple, the better the filtering action

The advantage of a full-wave rectifier over a half-wave is quite clear. The capacitor can more effectively reduce the ripple when the time between peaks is shorter. Figure (a) and (b)

Easier to filter-shorted time between peaks.-smaller ripple.

Capacitor Input Filter – Ripple Voltage

DC

ppr

V

Vr )(

Ripple factor: indication of the effectiveness of the filter

Vr(pp) = peak to peak ripple voltage; VDC = VAVG = average value of filter’s output voltage

•Lower ripple factor better filter [can be lowered by increasing the value of filter capacitor or increasing the load resistance]

[half-wave rectifier]

•For the full-wave rectifier:

)(

)()(

2

11

1

rectpL

AVGDC

rectpL

ppr

VCfR

VV

VCfR

V

Vp(rect) = unfiltered

peak

Example

Determine the ripple factor for the filtered bridge rectifier with a load as indicated in the figure above.

Diode Limiters (Clipper)

Clippers are networks that employ diodes to “clip” away a of an input signal without distorting the remaining part of the applied waveform.

Clippers used to clip-off portions of signal voltages above or below certain levels.

Diode Limiter/Clipper A diode limiter is a circuit that limits (or clips) either the

positive or negative part of the input voltage.

inL

Lout V

RR

RV

1

Example

What would you expect to see displayed on an oscilloscope connected across RL in the limiter shown in above figure.

Solution

VVk

kV

RR

RV in

L

Lout 09.910

1.1

0.1

1

Biased Limiters (Clippers)

The level to which an ac voltage is limited can be adjusted by adding a bias voltage, VBIAS in series with the diode

The voltage at point A must equal VBIAS + 0.7 V before the diode become forward-biased and conduct.

Once the diode begins to conduct, the voltage at point A is limited to VBIAS + 0.7 V, so that all input voltage above this level is clipped off.

A positive limiter

Biased Limiters (Clippers)

In this case, the voltage at point A must go below –VBIAS – 0.7V to forward-bias the diode and initiate limiting action as shown in the above figure.

A negative limiter

Modified Biased Limiters (Clippers)

Example

Figure above shows combining a positive limiter with a negative limiter. Determine the output voltage waveform?

Solution

Summary Limiters (Clippers)In this examples VD = 0

In analysis, VD = 0 or VD = 0.7 V can be used. Both are right assumption.

Summary Limiters (Clippers)

Diode Clampers

A clamper is a network constructed of a diode, a resistor, and a capacitor that shifts a waveform to a different dc level without changing the appearance of the applied signal.

Sometimes known as dc restorers Clamping networks have a capacitor connected

directly from input to output with a resistive element in parallel with the output signal. The diode is also parallel with the output signal but may or may not have a series dc supply as an added elements.

Clamper A clamper (dc restorer) is a circuit that adds a dc level to an ac signal. A capacitor is in series with the load.

Positive clamper – the capacitor is charged to a voltage that is one diode drop less than the peak voltage of the signal.

Vout = Vp(in) – 0.7 V

Negative clamper

Vout = -Vp(in) + 0.7 V

Start with forward-bias!

Diode Clampers

Positive clamper operation. (Diode pointing up – away from ground)

Diode Clampers

Negative clamper operation (Diode pointing down – toward ground)

Diode Clamper

If diode is pointing up (away from ground), the circuit is a positive clamper.

If the diode is pointing down (toward ground), the circuit is a negative clamper

Diode Clamper (Square Wave)

Diode ‘ON’ state Diode ‘OFF’ state

OutputV – Vc = 0 ; Vc = V; Vo = 0.7 V but ideal Vo = 0V

-V - Vc - Vo = 0; Vc = V

Vo = -2 V

Summary of Clamper Circuits

Voltage Multipliers

Voltage multiplier circuits use a combination of diodes and capacitors to step up the output voltage of rectifier circuits.

Voltage Doubler Voltage Tripler Voltage Quadrupler

Voltage Doubler

This half-wave voltage doubler’s output can be calculated by:Vout = VC2 = 2Vm

where Vm = peak secondary voltage of the transformer

Half-Wave Voltage Doubler

Positive Half-Cycle D1 conducts D2 is switched off Capacitor C1 charges to Vp

Negative Half-Cycle D1 is switched off D2 conducts Capacitor C2 charges to Vp

Vout = VC2 = 2Vp

Full-Wave Voltage Doubler

Positif Half-Cycle

• D1 forward-biased → C1 charges to Vp

• D2 reverse-biased

Negative Half-Cycle

• D1 reverse-biased

• D2 forward-biased → C2 charges to Vp

Output voltage=2Vp (across 2 capacitors in series

Voltage Tripler and QuadruplerVoltage Tripler and Quadrupler

Voltage Tripler

Positive half-cycle: C1 charges to Vp through D1

Negative half-cycle: C2 charges to 2Vp through D2

Positive half-cycle: C3 charges to 2Vp through D3

Output: 3Vp across C1 and C3

Voltage Quadrupler

Output: 4Vp across C2 and C4

The Diode Data Sheet

The data sheet for diodes and other devices gives detailed information about specific characteristics such as the various maximum current and voltage ratings, temperature range, and voltage versus current curves (V-I characteristic).

It is sometimes a very valuable piece of information, even for a technician. There are cases when you might have to select a replacement diode when the type of diode needed may no longer be available.

These are the absolute max. values under which the diode can be operated without damage to the device.

The Diode Data Sheet (Maximum Rating)

Rating Symbol 1N4001 1N4002 1N4003

UNIT

Peak repetitive reverse voltageWorking peak reverse voltageDC blocking voltage

VRRM

VRWM

VR

50 100 200 V

Nonrepetitive peak reverse voltage

VRSM 60 120 240 V

rms reverse voltage VR(rms) 35 70 140 V

Average rectified forward current (single-phase, resistive load, 60Hz, TA = 75oC

Io 1

A

Nonrepetitive peak surge current (surge applied at rated load conditions)

IFSM

30 (for 1 cycle)

A

Operating and storage junction temperature range

Tj, Tstg -65 to +175

oC

The Diode Data Sheet (Maximum Rating)

Zener Diodes

The zener diode – silicon pn-junction device-designed for operate in the reverse-biased region

Schematic diagram shown that this particular zener circuit will work to maintain 10 V across the load

Zener diode symbol

Zener Diodes Breakdown voltage – set by controlling the doping level during

manufacture When diode reached reverse breakdown – voltage remains

constant- current change drastically If zener diode is FB – operates the same as a rectifier diode A zener diode is much like a normal diode – but if it is placed in

the circuit in reverse bias and operates in reverse breakdown. Note that it’s forward characteristics are just like a normal

diode.

1.8V – 200V

Zener Diodes The reverse voltage (VR) is increased – the

reverse current (IR) remains extremely small up to the “knee”of the curve

Reverse current – called the zener current, IZ At the bottom of the knee- the zener breakdown

voltage (VZ) remains constant although it increase slightly as the zener current, IZ increase.

IZK – min. current required to maintain voltage regulation

IZM – max. amount of current the diode can handle without being damage/destroyed

IZT – the current level at which the VZ rating of diode is measured (specified on a data sheet)

The zener diode maintains a constant voltage for value of reverse current rating from IZK to IZM

Zener Diodes (Zener Equivalent Circuit)

Since the actual voltage is not ideally vertical, the change in zener current produces a small change in zener voltage

By ohm’s law:

Normaly -Zz is specified at IZT

Z

ZZ I

VZ

Zener impedance

Zener Diodes(Temp Coeff & Zener Power Dissipation and Derating)

As with most devices, zener diodes have given characteristics such as temperature coefficients and power ratings that have to be considered. The data sheet provides this information

Zener Diodes Applications

Zener diode can be used as

1. Voltage regulator for providing stable reference voltages

2. Simple limiters or clippers

Zener Regulation with Varying Input Voltage

As i/p voltage varies (within limits) – zener diode maintains a constant o/p voltage

But as VIN changes, IZ will change, so i/p voltage variations are set by the min. & max. current value (IZK & IZM) which the zener can operate

Resistor, R –current limiting resistor

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 o/p terminal of the zener diode is open (RL=∞)-load current is zero and all of the current is through the zener

When a load resistor (R) is connected, current flow through zener & load RL, IL, IZ

The zener diode continues to regulate the voltage until IZ reaches its min value , IZK

At this point, the load current is max. , the total current through R remains essentially constant.

Zener LimitingZener diode also can be used in ac applications to limit voltage swings to

desired level(a) To limit the +ve peak of a signal voltage to the selected zener

voltage - During –ve alternation, zener arts as FB diode & limits the –ve

voltage to -0.7V(b) Zener diode is turn around -The –ve peak is by zener action & +ve voltage is limited to +0.7V(c) Two back-to-back zeners limit both peaks to the zener voltage ±7V -During the +ve alternation, D2 is functioning as the zener limiter –

D1 is functioning as a FB diode. -During the –ve alternation-the roles are reversed