C hapter 2 diode applications

DIODE

APPLICATIONS

CHAPTER 2

MDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN

TOPIC OUTLINES LOAD LINE ANALYSIS

SERIES DIODE CONFIGURATIONPARALLEL AND SERIES-PARALLEL CONFIGURATION

AND/OR GATESHALF WAVE RECTIFICATIONFULL WAVE RECTIFICATION

CLIPPERSCLAMPERS

ZENER DIODESVOLTAGE MULTIPLIER CIRCUITS


Load Line Analysis

• The load line plots all possible current (ID) conditions for allvoltages applied to the diode (VD) in a given circuit. E/R is themaximum ID and E is the maximum VD.

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


How to determine the Q point of asystem?

• Identify diode model

• Using Kirchoff’s Law :– Set VD = 0V (horizontal line)– Set ID = 0A (vertical line)– Obtain VDQ, IDQ from the graph intersection ( Q-point)


Diode Approximation

Approximate model notationIn Forward Bias:Silicon Diode: VD = 0.7VGermanium Diode: VD = 0.3V

In Reverse Bias:Both diodes act like opens VD = source voltage and ID =0A

Ideal model notationVD = 0V and ID = 0A

Si

Ge


Diode in DC Series Circuit:Forward Bias

The diode is forward bias.• VD = 0.7V (or VD = E if E <0.7V)• VR = E – VD

• ID = IR = VR /R


The diode is reverse biased.• VD = E• VR = 0V• ID = IR = IT = 0A

Diode in DC Series Circuit:Reverse Bias


• An open circuit can have any voltage across itsterminals, but the current is always 0A.

• A short circuit has a 0V drop across its terminals,but the current is limited only by the surroundingnetwork.

• Source notation :


For the series diode configuration below, employing the diodecharacteristics of figure below, determine VDQ, IDQ and VR.

Example


Step1: Find the maximum ID. VD = 0V→ ID = IR= E/RStep 2: Find the maximum VD. ID =0A → E = VD + IDRStep 3: Plot the load lineStep 4 : Find the intersection between the load line and thecharacteristic curve. This is the Q-pointStep 5: Checking !!!!

Solution


Determine ID, VD2 and Vo for the circuit.

Remember, the combination ofshort circuit in series with anopen circuit always results inan open circuit and ID=0A.

Example


Determine I, V1, V2 and Vo

Example


Series – Parallel ConfigurationsSolve this circuit like any Series/Parallelcircuit, knowing VD = 0.7V (or up to 0.7V) inforward bias and as an open in reverse bias.

VD1 = VD2 = Vo =0 .7VVR = 9.3V

Diodes in parallel are used to limit current:IR = E – VD = 10V -0 .7V = 28mA

R 0.33kID1 = ID2 = 28mA/2 = 14mA


Determine the resistance R for the network whenI=200mA.

Example

Si Si


Determine the currents I1, I2, and ID2 for the networkExample


Diodes in AC Circuits

•Inputs: -Sinusoidal waveform-Square wave

•This circuit is called half-wave rectifier, which generatewaveform vo that will have an average value of particularuse in the ac-to-dc conversion process.

•The diode only conducts when it is in forward bias,therefore only half of the AC cycle passes through thediode.


Half-Wave Rectification

•The diode that employed in the rectification process is typicallyreferred to rectifier.

•The diode only conducts for one-half of the AC cycle. The remaininghalf is either all positive or all negative. This is a crude AC to DCconversion.

•The DC Voltage out of the diode :Vdc = 0.318Vm

where Vm = the peak voltage


Peak Inverse Voltage (PIV)• Because the diode is only forward biased for one-half of the

AC cycle, it is then also off for one-half of the AC cycle. It isimportant that the reverse breakdown voltage rating of thediode be high enough to withstand the peak AC voltage.

– PIV (PRV) > Vm

• PIV = Peak Inverse Voltage; PRV = Peak Reverse Voltage• Vm = Peak AC Voltage


Full-Wave Rectification:Bridge Network

• The dc level obtained from a sinusoidal input can beimproved 100% using a process called full-waverectification.

• The most familiar network is bridge configuration with 4diodes.

Vdc = 0.636 Vm


Operation of the Bridge RectifierCircuit

For the positive half of the AC cycle:

For the negative half of the AC cycle:


Determining Vo for silicon diodes in the bridgeconfiguration

The effect of using a silicon diode with VD=0.7 isdemonstrated in below figure. The dc level has change to:


Determine the output waveform for the network below andcalculate the output dc level.

Example


Conduction path for the +ve region

Solution


Conduction path for the -ve region

Solution


Full-Wave Rectification:Center-Tapped Transformer

A second popular full-wave rectifier with only two diodes butrequiring a center-tapped transformer to establish the inputsignal across each section of the secondary of the transformer.

Two diodes and a center-tapped transformer are required.

VDC = 0.636(Vm) for ideal diode

Note that Vm here is the transformer secondary voltage to the tap.


Operation of the Center–Tapped TransformerRectifier Circuit

For the positive half of the AC cycle

During the positive cycle of vi applied to the primary of theTransformer the network will appear as shown in figure. D1assumes the short-circuit equivalent and D2 the open-circuitequivalent, as determined by the secondary voltages and theresulting current directions.


For the negative half of the AC cycle

During the negative cycle of vi, reversing the roles of the diodes(D2 is short-circuit) but maintaining the same polarity for thevoltage across the load resistor R.


Animation of center-tappedtransformer rectifier


Show the voltage waveform across the secondary windingand across R when an input sinusoidal is applied to theprimary winding.

Example


The transformer turns ratio = 0.5.The total peak secondary voltageis,Vp(sec) = nVp(pri) =0.5(100)=50V.There is a 25 V peak across each ofthe secondary with respect toground.

Solution


Note: Vm = peak of the AC voltage. Be careful, in the center tappedtransformer rectifier circuit the peak AC voltage is the transformersecondary voltage to the tap.

Rectifier Circuit Summary


Clippers• Clippers or diode limiting is a diode network that have the

ability to “clip”(cut short/crop) off a portion on the inputsignal without distorting the remaining part of thealternating waveform.

• Clippers are used to eliminate amplitude noise or tofabricate new waveforms from an existing signal.

• Simplest form of diode clipper- one resistor and a diode• Depending on the orientation of the diode, the positive or

negative region of the applied signal is clipped off.• 2 general of clippers:

a) Series clippersb) Parallel clippers

• Series Clippers– The series configuration is defined as one where the diode is in

series with the load.– A half-wave rectifier is the simplest form of diode -clipper-one

resistor and diode.• Parallel Clippers

– The parallel configuration has the diode in a branch parallel tothe load.


Series Clipper• Diodes “clip” a portion of the AC wave.• The diode “clips” any voltage that does not put it in forward

bias. That would be a reverse biasing polarity and a voltageless than 0.7V for a silicon diode.

Any type of signals canbe applied to a clipper


Analysis steps for series clippersThere is no general procedure for analyzing series clippersnetwork but there are some things one can do to give theanalysis some direction.

1. Take careful note of where the output voltage is defined.2. Try to develop an overall sense of the response by simply

noting the “pressure” established by each supply and theeffect it will have on the conventional current directionthrough the diode.

3. Determine the applied voltage (transition voltage) that willresult in a change of state for the diode from the “off” to the“on” state.

4. It is often helpful to draw the output waveform directlybelow the applied voltage using the same scales for thehorizontal axis and the vertical axis.

Series clipper with dc supply examples


Series clipper with dc supply

By adding a DC source to the circuit, the voltage requiredto forward bias the diode can be changed.


Series clipper example

• Positive region of Vi - turn the diode ON.

• Negative region of Vi - turn the diode OFF.• Vi > V to turn ON the diode.• In general, diode is open circuit (OFF state) and short circuit (ON state)• For Vi > V the Vo = Vi – V• For Vi = V the Vo= 0 V


Example 1Determine the output waveform for the network below:


Solution (continued)


Repeat previous example for the square-wave input.

Example 2


Solution (continued):- ve region OFF state


Parallel Clipper• By taking the output across the diode, the output is now the

voltage when the diode is not conducting.• A DC source can also be added to change the diode’s required

forward bias voltage.

Parallel clipper exampleMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN

Determine the Vo and sketch the output waveform for thebelow network

Example 2


Solution


Repeat the previous example using a silicon diode withVD=0.7 V

Example 2

Solution


For input voltages greater than 3.3 V the diode opencircuit and Vo=Vi.For input voltages less than 3.3 V the diode short circuitand the network result as shown below



Series Clippers (Ideal Diode) Summary


Parallel Clippers (Ideal Diode) Summary


Clampers• A clamper is a network constructed of a diode, resistor,

and a capacitor that shifts a waveform to a different dclevel without changing the appearance of the appliedsignal.

• Clamping networks have a capacitor connected directlyfrom input to output with a resistive element in parallelwith the output signal. The diode is also in parallel withthe output signal but may or may not have a series dcsupply as an added element.


A diode in conjunction witha capacitor can be used to“clamp” an AC signal to aspecific DC level.

Element of the clamper circuit• Magnitude of R and C must

be appropriate to ensureг=RC where the timeconstant is large enoughand capacitor may notdischarge during the timeinterval while diode is notconducting.

• We will assume that allpractical purposes thediode will fully charge ordischarge in 5 timeconstant.


•The input signal can be any type of waveform:- sine, square, triangle wave, etc.

•You can adjust the DC camping level with a DC source.

Clampers example


There is a sequence of steps that can be applied tohelp make the analysis straight forward.1. Start the analysis by examining the response of the

portion of the input signal that will forward bias thediode. If the diode is reverse bias, skip the analysis forthat interval time, and start analysis for the next intervaltime.

2. During the period that the diode is in the “on” state,assume that the capacitor will charge up instantaneouslyto a voltage level determined by the surroundingnetwork.

3. Assume that during the period when the diode is in the“off” state the capacitor holds on to its establishedvoltage level.

4. Throughout the analysis, maintain a continual awarenessof the location and defined polarity for vo to ensure thatthe proper levels are obtained.

5. Check that the total swing of the output matches that offthe input.


Clampers Summary


Determine Vo for the below network.

Example


Solution• f=1000Hz, so a period of 1ms or

interval 0.5ms between each level.

• Define the period that the diode isstart to conduct (t1~t2), which isthe V0=5V.

• Determine VC from the Kirchoff’sLaw– VC=20V+5V=25V

• When in the positive input, we willfind V0=35V(outside loop)

• Time constant, г =RC = 10ms,total discharge time = 50mswhere is large enough before thecapacitor is discharge duringinterval t2~t3


Output waveform


Zener Diode

The state of the diode must be determined followed by asubstitution of the appropriate model and a determination ofthe unknown quantities of the network. For the off state as adefined by a voltage less than Vz but greater than 0V. TheZener equivalent is the open circuit.


Vi and R FixedThe applied dc voltage is fixed, as the load resistor.The analysis :1. Determine the state of the Zener diode by removing it

from the network and calculating the voltage across theresulting open circuit.


2. Substitute the appropriate equivalent circuit and solve forthe desired unknowns.- For the on state diode, the voltages across parallelelements must be the same.

VL=VZThe Zener diode current is determined by KCL:

IZ = IR – ILThe power dissipated by the Zener diode is determined by:

PZ = VZ IZ- For the off state diode, the equivalent circuit is open-circuit.


Fixed Vi, Variable RL• Due to the offset voltage Vz, there is a specific range of

resistor values (and therefore load current) which will ensurethat the Zener is in the on state.

• Too small RL VL < Vz Zener diode will be in the off state• To determine the min RL that will turn the Zener diode on :

• Any load resistance value greater than the RL min will ensurethat the Zener diode is in the on state and the diode can bereplaced by its Vz source equivalent.

• The max IL


• Once the diode is in the on state, the voltage across Rremains fixed at:

• Iz is limited to IZM as provided on the data sheet, it doesaffect the range of RL and therefore IL.


Fixed RL, Variable Vi• For fixed values of RL, the voltage Vi must be sufficiently

large to turn the Zener diode on. The min turn-on voltageVi=Vi min :

• The max value of Vi is limited by the max Zener current IZM.IRmax=IZM+IL

• Since IL is fixed at VZ/RL and IZM is the max value of IZ, themax Vi is defined by:

Vi max =VRmax+Vz

Vi max=IRmaxR+VzMDM KASUMAWATI BT LIAS DR THELAHA BIN MASRI DR WAN AZLAN BIN WAN ZAINAL ABIDIN

Zener Diode Examples


Determine the network to find the range of RL and IL tomaintained VRL at 10V.

Example


Solution:


Voltage Multiplier Circuit

Half Wave voltage doubler



Full Wave voltage doubler



Voltage Tripler and Quadrupler


Practical Application

• Rectification• Protective Configuration• Polarity Insurance• Controlled Battery-Powered Backup• Polarity Detector• AC Regulator and Square Wave

Generator


End of Chapter 2

Thank you


C hapter 2 diode applications

Documents

Transcript of C hapter 2 diode applications