Ecad Labmanual

Electronics and Telecommunication Department

SVERI’S

COLLEGE OF ENGINEERING, PANDHARPUR

ELECTRONICS AND TELECOMMUNICATIONDEPARTMENT

ELECTRONIC CIRCUIT ANALYSIS & DESIGN -I

LABORATORY MANUAL

PREPARED BY,

Mr.S.A. Inamdar &

Mr.A.D. Shinde.

Electronic Circuit Analysis & Design -I1


ELECTRONIC CIRCUIT ANALYSIS & DESIGN -I

INDEX

SR. NO. TITLE OF THE EXPERIMENT

PAGE NO.

1

To Study of Full Wave Rectifier

2

To Study of Clipper & Clamper

3

To study of Voltage Multiplier

4

To Study of V-I Characteristics of Zener diode

& its application as Regulator

5

Design & Implementation of unregulated

power supply using FWR & Capacitor Filter

6

Input & Output Characteristics of CB

configuration

7

Input & Output Characteristics of CE

configuration

8

Design & implementation of relay driver circuit

9

Frequency response of single stage CE

amplifier

10

To study of Astable Multivibrator



EXPERIMENT NO. 1

Performance of Full Wave Rectifier



Aim: To check performance of full wave rectifier.

To verify output voltage & ripple factor of FWR.

Equipments: CRO

Components:

Sr. No. Item Quantity1 IN 4007 diode 22 9-0-9 v Transformer 13 RL (1M) 1

Theory:

The circuit diagram of full wave rectifier is as shown in fig. It

consist of one step down transformer & two PN junction diode which

are connected as shown in fig.

During +ve half cycle of input signal diode D1 get +ve signal,

hence conducts & current flows through diode D1 & intern through

RL. At the same time because of tapping of secondary winding diode

D2 gets –ve voltage & D2 gets reverse biased & no current flows

through diode D2.(Reverse saturation current will be there but as it is

very small in value it is neglected).

Similarly during –ve half cycle of ac input ,diode D1 gets

reverse biased & does not allow current to flows where as diode D2

gets +ve voltage because of tapping of secondary windings & hence it

conducts & allow current to flow through it & intern once again

current through D2 passes through RL in the same direction as like

previous. Therefore voltage drop across RL is available in both +ve & -

ve half’s of input. Hence we can say entire waveform is rectified. Hence

called as Full Wave Rectifier.

Ripple factor:

The output of the rectifier consist of both ac & dc

component.



The ac component present in output of rectifier is called as

ripple & ratio of r.m.s. value of ac component present in output to the

DC component is called as ripple factor.

r=r .m. s . value of ac component present∈the outputThe dccoponent present∈output

¿Vr , rmsV dc

¿Ir ,rmsIdc

r=√[ Irms ²Idc ² ]−1

DC output voltage: (Vdc)

It is the dc component present in output of full wave rectifier.

It is the average of values over one entire cycle & it can be given as

follows

Vdc =2Vmπ

Procedure:

1. Do the connections as shown in circuit diagram2. Switch on power supply3. Observe waveform on CRO4. Note down values of required parameter using multimeter &

CRO.

Conclusion:

Thus we checked performance of FWR & also it is concluded that theoretical & practical values of dc output voltage & ripple factor are approximately same.



Circuit diagram & input output waveform:



EXPERIMENT NO. 2

Clipper & Clamper



Experiment No. 2

Aim:

1) To verify the performance of clipping circuit & observe it waveform on CRO.

2) To verify the performance of clipping circuit & observe it waveform on CRO

Equipments:

Sr No Decription Specification Qty Req1 Signal

generator0-1Mhz 1

2 CRO 0-20 Mhz 13 Power Supply 0-32V 1

Components:

Sr No Item Quantity1 PN Jnction

diode (IN4007]

1

2 Resistance R=2.2.KΩ

1

3 Load resistance RL=1MΩ

1

4 Capacitor(470µf)

1

Theory:

Clipping circuit:

There are verities of network that have ability of cliff off a portion of input signal without distorting remaining part of input signal called



clipper. Clipping circuits are also referred to as voltage limiters ,amplitude selectors or slicers. Depending upon the orientation of diode and polarity of reference voltage the input signal will clip.

A Clipping circuit requires a minimum of one diode & one resistor .Power supply is often used to set the various clipping levels.There are two types of clipper circuits

1) Positive Clipping circuit2) Negative Clipping circuit

Let us analyze the following few circuits , where R=1KΩ Diode forward bias drop is assumed to 0.7V .During +ve half cycle of the input 2V dc less than supply is dominant over the input signal. So diode is reverse bias & o/p becomes a faithful reproduction of i/p till it reduces 2.7V.During –ve cycle , diode becomes fully reverse biased.Hence o/p waveform is a faith full reproduction of the i/p. Let us consider ,one more circuit where the diode comes in series with i/p signal sourse.during the +ve half cycle of i/p,less than 2v .DC power supply dominant over the i/p signal .so the diode is reverse biased & o/p comes dc till it reaches 2V.Once i/p signal exceeds 2.7V.diode becomes forward biased & there is a current in the circuit.During –ve half cycle diode becomes fully reverse biased. Hence o/p waveforms equals to the DC supply voltage. The same explanation holds for good for ant clipping circuit with or with out DC supply.

Clamper Circuits:The clamper network is one that will clam on i/p signal to a different dc level. The network consist of capacitor , a diode & resistance. But if can also have an independent supply to introduce an additional dc shift.

The magnitude of R & C must be chosen such that the time constant ,T=R.C is large enough to ensure that the voltage a/c the capacitor does not discharge significantly during the interval when diode is non conducting.

During the +ve half cycle, diode is forward biased & short out the effect of resistor R. This result in small RC time constant so that capacitor will charge to peak value [diode drop is considered very quickly during this interval the o/p voltage is almost short circuited voltage,Vo=0.7V(diode drop alone)].

During the –ve half cycle diode will reverse biased therefore the voltage drop across the resistance

Voltage drop across resistance R=drop across capacitor +i/p Voltage

= -(V-diode drop)-V



=-(-4.3-5)=9.3V

Procedure:1) Built up the circuit as per the circuit diagram.2) Set i/p signal voltage (10V 100Hz) using signal generator.3) Observe the o/p waveform using CRO4) Sketch the observed waveform on graph sheet.

Conclusion:1) The waveform of clipper circuit are observed on CRO & o/p got clipped according to reference level chosen.2) The waveforms of clamper circuit are observed on CRO & we observed that it got clamped to level equal to peak voltage of

i/p



EXPERIMENT NO. 3

To study of Voltage Multiplier



Experiment No. 3

Title: Voltage Multiplier

Aim: To study & verify working of half wave voltage doubler.

Equipments:

Sr No Decription Specification Qty Req1 Signal

generator0-1Mhz 1

2 CRO 0-20 Mhz 13 Power Supply 0-32V 1

Components:

SR No Item Quantity1 Diode [IN4007] 22 Capacitor[470µF] 23 Connection Wires 4-54 CRO probe 1

Theory:

Voltage Multipliers:Voltage multiplier is a circuit ,which produces an output d.c.

voltage whose value is multiple of peak A.C. input voltage[i.e. 2 Vm,3 Vm , 4 Vm & so on].Such circuits are used as power supply for high voltage /low current devices like cathode ray tues used in televisions receiver ,oscilloscope.

Voltage doubler: A voltge multiplier ,whose output d.c. voltage is double the peak

a.c input voltage is called a voltage doubler.



Circuit diagram shows the circuit of a half wave voltage doubler.In this circuit each section of a diode and capacitor (say D1 & C1 or D2 & C2) is called a peak rectifier.

During the +ve half cycle of the input signal , the diode De conducts ( and diode D2 is cut –off charging the capaciot C2 up to peak rectified voltage (i.e. Vm).

During the –ve half cycle diode D1 is cut off and diode D2 conducts charging capacitor c2.It may be noted that ,during negative half cycle, the voltage across capacitor C1 is in series with the input voltage. Therefore the total voltage presented to capacitor c2 is equal to 2Vm.As a result of this ,the capacitor c2 is charged to a voltage of 2Vm during negative half cycle.On the next positive half cycle the diode D2 is non conducting and the capacitors will discharge through load.If no load is connected across capacitor C2,both capacitors stay charged at their full values(i.e. c1 to Vm & c2 to 2Vm).also the both diodes D1 & D2 have a peak inverse voltage of 2Vm each. This is the case when the diodes are ideal diode. But to use ideal diode is not possible, so the output got across o/p terminal given by,

Vdc = 2Vm-2*0.7 = 2Vm -1.4Vdc = 2Vm -1.4

To find Vdc , we have to substract voltage drop across both diodes from 2Vm.

Observation Table:

Parameter DVM CRO

Average load voltage (Vdc)

Procedure :

1) Do connections of circuit on breadboard as shown in circuit diagram.2) Switch on the all supply & instruments.3) Apply i/p voltage as 10V peak to peak to the circuit4) Observe waveforms on CRO.5) Note down o/p voltage also measure the same on DVM.



Conclusion:Thus we studied Half wave voltage Doubler & verified practically

o/p of half wave doubler is approximately double of peak of ac input voltage & it is dc in nature.



EXPERIMENT NO. 4

To Study of V-I Characteristics of Zener

diode & its application as Regulator



Experiment No. 4

Aim:: 1 )Study of VI characteristics of Zener diode 2) Study of zener diode as voltage regulator Introduction to Equipments:

Sr No Decription Specification Qty Req1 Power Supply 0-32V 12 Multimeter

Components:

SR No Item Quantity1 Zener diode2 Resistor

Theory:

VI characteristics of Zener diodeA PN junction diode normally doesn’t conduction in reverse biased is increased at a particular voltage it starts conduction heavily. This voltage is called as breakdown voltage. High current through he diode can permanently damages if once the diode starts conducting .A Zener diode is PN junction diode specially made to work in breakdown region. It is used as voltage regulator in order to maximize the loop equivalent for this equation is given by

(I1+I2)Rs-(Vin-V2)Rs=Vin(min)+ V2 (max) /I2(min)+I1(max)

Zener Diode as Voltage regulatorRegulator is measure of cicuit ability to maintain constant o/p

voltage even when i/p voltage varies or load varies.a zener diode when working in breakdown rregion can seve as voltage regulator.

As shown fig i.e. i/p DC voltage whose variation are tobe regulated the zener diode do reverse connected across Vin when potential difference across diode is greater than V2.The large current passes through the series resistance Rs so we get the constant voltage



across load resistance Rl. The total current passes through Rs equal to sum of diode current & load current.

i.e. I=I1+I2; Under all condition Vout=V2; Vin= IR+Vout =IR+V2 =IR+V2The value of resistance which is connected in series with Rs.It

must be of such value so that Zener circuit willnot drop below minimum value of I1[min].This minimum circuit is necerray to keep the device in breakdown region .

Line Rgulation:Suppose RL is fixed but input voltage Vin increases slightly it

will increase current I.The increase in Vin will be dropped across IR so that voltage across the zener diode is constant.Load Regulation

For finding load regulation make the connection as shown in fig.Keep the input voltage constant say 10v vary load resistance value. Note down no load voltage Vn; for maximum load resistance value & full load voltage Vfl for minimum load resistance value.

Calculate load resistance using % load regulation =Vnl-Vfl *100

Vfl

Conclusion: 1) Thus we have studied VI characteristics of Zener diode.2) It can be calculated that the value of o/p voltage nearly same

either by changing supply voltage or by changing load.

Observations tables1) VI characteristics of Zener Diode

Sr No Vin(V) Iz(ma) Vz(v)

2) Load RegulationInput Voltage =

Sr No Rl(Ω) Vout(V)

3) Line Regulation

Rl= Ω


Sr No Vin (V) Vout (V)




EXPERIMENT NO. 5

Design & Implementation of unregulated power supply using FWR & Capacitor

Filter



Experiment No. 5

Aim: To design an unregulated power supply using full wave rectifier (Center Tap) with capacitor filter with

following specifications. Vdc=8V ,Idc=80mA , Rf =6%

Equipments: Sr No Decription Specification Qty Req1 Power Supply 0-32V 12 CRO 0-20 Mhz 1

Components:

SR No Item Quantity1 Diodes [IN 4007 ] 22 Resistance [100Ω] 13 Capacitor [470 µF) 14 Transformer (9-0-

9)V1

Theory:In an unregulated power supply AC mains is given to step down transformer where high voltage is converted to low voltage & then rectifier rectifier converted to low AC signal to pulsating DC signal which feel to filter to get pure DC o/p.Almost all electric circuit requires DC source of power for portable low power system batteries may be used. Batteries have the advantages of AC component in the o/p but there is danger leakage.

Hence o/p is not constant so it is known as unregulated power supply. The transformer is used to step down the AC mains the required voltage rectifier converts AC into unidirectional current(AC to DC). However impure DC called ripple can be filtered out by filter circuit. The o/p is unregulated i.e. not a constant level. Regulator regulates the varying o/p suitable value then constant is fed to load.

Types of filter circuits1) L-filter 2) C-filter 3) LC-filter 4) π−filter



Expression for the C filter :The total amount of the charge lost by capacitor during this interval is given by as Q(discharge)=Iav * T2

This charge is replaced during short interval T1 during this voltage across capacitor charge by amount equal to peak to peak voltage of the ripple Vp-p using relation

Q=VcQcharge=Vrp-p * CBut Qchrge=QDischargeVrp-p*C=Idc*T2Vrp-p=Idc * T2/CAssuming load is light ripple is small & time for

recharging capacitor is small compared with the time for which is discharges

T1<<T2T2=T/2=1/2fVrp-p = Idc/2fcWith the assumption the ripple

Vrrms= Vrp-p/2√3So Idc = Vdc/R2Vr-rms= Idc / 4√3 fCr = Vr-rms/Vdc = ¼√3RL f C

Vdc=Vm-Vr-p-p/√2

Hence some important mathematical expression for full wave rectifier1) Vavg = Vdc = 2Vm/ π

Iavg=Idc =2Im/ π2) Vrms=Vm/√2

Irms = Im/√23) r = ripple voltage/dc voltage

r= Vr-rms/Vdc4) Vr-rms= Idc/fc5) r = ¼√3RL f C

Design statement - Design unregulated power supply with full wave rectifier to give o/p of 10v with load current of 100mA. The ripple should not exceed 1%.Given vo = 10V = Vdc

Idc = IL=100mAr =0.01

1) Selection of RL= Vdc/Idc = 10/100=100Ω2) Selection of capacitance

C=2900/r*Rl=2900/(0.01)*100 =2900µF3) Selection of transformer



Vrms= Vm/√2 =10.89/√2 =7.68VVdc=Vm=500Idc/C

=500+1000+0.001/2900 =10.72V

Selected transformer =(9-0-9)V4) Selection of diode

PIV =2Vm=21.44Im=Ic/2 +0.25*Ic=75mA

5) Component test Load resistor Rl=100 ΩCapacitor ,C =2200 µF ,25VTransformer =(9-0-9)V ,100mADiode = 1N4007.

Procedure:1) Choose approximate component from design value

obtain2) Buit ckt using those component on breadboard3) Remove capacitor & measure Vm4) Now connect filter ‘c’ & measure Vr-p-p i.e. peak ripple

voltage on CRO5) Calculate Vrrms= Vrp-p/2√36) Calculate r = Vr-rms/Vdc7) Calculate Vdc= Vm-Vrp-p/2√38) Plot graph

Conclusion : The amount of ripple content in the o/p waveform depends upon ‘c’ .If ‘c’ increased ripple

to decreases.Practically observed value of ripple factor using i) r = 0.00947ii) CRO e.g. r= 0.005%





EXPERIMENT NO. 6

Input & Output Characteristics of CB configuration



Experiment No. 6Title: Common Base characteristics of TransistorAim:: To Study the input & output characteristics of transistor in

common base configuration.

Equipments: Sr No Decription Specification Qty Req1 Power Supply 0-32V 12 Multimeter

Components:

Theory:Circuit Diagram:

Procedure Part IInput characteristics

1) Connect the circuit for CB configuration as per the circuit diagram

2) Fix the collector base voltage (VCB)3) Vary the base-emitter voltage (VEB) and note the base current (IE)4) Plot the graph VEB v/s IE from the curve calculate input

resistance Ri

Output characteristics1) Connect the circuit for CB configuration as per the circuit

diagram2) Fix the Base Current (IE)



3) Vary the collector-base voltage (VCB) and note the collector current (IC)

4) Plot the graph VCB v/s IC from the curve calculate input resistance Ro

Observation Table:Input CharacteristicsVCB= 5V (Constant)

Obs. No VEB (V) IE (mA)12

Output Characteristics:Obs. No IB1= µA IB2= µA IB3= µA

VCB (V) IC (mA) VCB (V) IC (mA) VCB (V) IC (mA)12

Calculations:1) Input resistance(Ri) = Change in input voltage

Change in input current (At constant output voltage VCE)

2)Input resistance(Ro) = Change in input voltage

Change in input current(At constant output voltage VCE)

= VCE

IC

Results:1) Input Resistance (Ri)=2) Output Resistance(Ro)=

Theory:Introduction of Transistor:

A transistor consist of two P-N junction formed by sandwitching either P-type or

N-type semiconductor between a pair of opposite type material. Accordingly there are two types of transistor namely

1) N-P-N transistor2) P-N-P transistor



An n-p-n transistor is composed of two n-type semiconducting material, separated by thin section of p-type .However , a p-n-p transistor is formed by two p-section separated by thin section of n-type semiconductor.A transistor is having two junctions and is a three terminal semiconductor device mainly used as an amplifier. The three terminals are emitter, base and collector. The emitter region is heavily doped so that it can inject large no of charge carriers in to the base region . The base region is lightly doped and is very thin. It passes most of the emitter injected charge carries to the collector. The collector is moderately doped.

IE=IC+IB

Common Base Configuration

In this circuit arrangement, input is applied between emitter& base and output is taken from collector & base. Here base of the transistor is common to both input & output circuit hence the name common base configuration.

Input characteristics:

It is curve between emitter current IE and base emitter voltage VBE at constant collector-base voltage VCB. The following points may be noted from characteristics1) The emitter current (IE) increase rapidly with small

increase in emitter-base voltage (VCB) . It means that input resistance is very small.

2) The emitter current is almost independent of collector-base voltage. This leads to the conclusion that emitter current & hence collector current is almost independent of collector voltage.

Output characteristics:

It is the curve between collector current IC and collector-base voltage VCB at constant emitter current IE . The following points may be noted from the characteristics1. The collector current IC various with VCB only at very low

voltages.2. Above a voltage the collector current becomes constant. It

means that IC is independent of VCB and depends on IE only.



3. A very large change in collector-base voltage produces only tiny change in collector current. This means that output resistance is collector current. This means that output resistance is very high.

EXPERIMENT NO. 8



I Input & Output Characteristics of CE configuration

Experiment No. 8

Title : Common Emitter characteristics of Transistor



Aim: Introduction to Computer Communication Network.

Equipments: Sr No Decription Specification Qty Req1 Power Supply 0-32V 12 Multimeter

Components:

Theory:

Procedure Part IInput characteristics

5) Connect the circuit for CE configuration as per the circuit diagram

6) Fix the collector emitter voltage (VCE)7) Vary the base-emitter voltage (VBE) and note the base current (IB)



8) Plot the graph VBE v/s IB from the curve calculate input resistance Ri

Output characteristics5) Connect the circuit for CE configuration as per the circuit

diagram6) Fix the Base Current (IB)7) Vary the collector-emitter voltage (VCE) and note the collector

current (IC)8) Plot the graph VCE v/s IC from the curve calculate input

resistance Ro

Observation Table:Input CharacteristicsVCE= 5V (Constant)

Obs. No VBE (V) I IB (µA)12

Output Characteristics:Obs. No IB1= µA IB2= µA IB3= µA

VCE (V) I IC (mA) VCE (V) IC (mA) VCE (V) IC (mA)12

Calculations:1) Input resistance(Ri) = Change in input voltage

Change in input current(At constant output voltage VCE)

= VBE

IB

2) Input resistance(Ro) = Change in input voltage Change in input current

(At constant output voltage VCE)

Results:3) Input Resistance (Ri)=4) Output Resistance(Ro)=

Theory:Introduction of Transistor:



A transistor consist of two P-N junction formed by sandwitching either P-type or

N-type semiconductor between a pair of opposite type material. Accordingly there are two types of transistor namely

3) N-P-N transistor4) P-N-P transistor

An n-p-n transistor is composed of two n-type semiconducting material, separated by thin section of p-type .However , a p-n-p transistor is formed by two p-section separated by thin section of n-type semiconductor.A transistor is having two junctions and is a three terminal semiconductor device mainly used as an amplifier. The three terminals are emitter ,base and collector. The emitter region is heavily doped so that it can inject large no of charge carriers in to the base region . The base region is lightly doped and is very thin. It passes most of the emitter injected charge carries to the collector. The collector is moderately doped.

IE=IC+IB

Common Emitter Configuration

In this circuit arrangement , input is applied between base & emitter and output is taken from collector & emitter. Here emitter of the transistor is common to both input & output circuit hence the name common emitter configuration.

Input characteristics:It is curve between base current IB and base emitter

voltage VBE at constant collector-emitter voltage VCE. The following points may be noted from characteristics3) The characteristics resemble that of forward biased diode

curve. This is expected since the base emitter section of the transistor is diode & it is forward biased.

4) As compared to CB arrangement IB increases less rapidly with VBE.

Output characteristics: It is the curve between collector current IC and collector-emitter voltage VCE at constant base current IB . The following points may be noted from the characteristics4. The collector current IC various with VCE between 0 & 1

volts only. After this collector current becomes almost constant and independent of VCE.This value of VCE above



which collector current IC is independent of VCE is called the knee voltage.

5. Above the Knee voltage C almost constant.

EXPERIMENT NO. 8



Design & implementation of relay driver circuit

Experiment No. 8

Title: Relay driver circuit.AIM : Design & implementation of relay driver circuit

Equipments: Sr No Decription Specification Qty Req



1 Power Supply 0-32V 12 Multimeter

Components:

SR No Item Quantity1 Resistor2 Diode3 Transistor4 Relay

Theory:Relay device which function as an electrically operated switch. In response to an electrical signal known as control signal. The relay open or close if switch contacts in some prearranged & fixed combination . The contacts may be in same circuit as operating signal in another circuit. The relays are widely used in industries as control device.Application

Control of high power , load circuits ,low voltage control of remote equipments, Isolation of control circuits from load circuits.ClassificationsElectromagnetic relayCrystal relayDry reed relayMercury wetted relaySolid state relayTime delay relayElectromagnetic RelayIt consist of an iron core surrounding coil wire, Iron yoke & set of contacts , they arrange as shown in fig.The iron yoke provides a low reactance path for the magnetic flux. It is shaped in such a way that a magnetic circuit can be closed by a armature & set of contacts armature is movable piece of iron.It is hinged to the yoke & held a spring in such way that there is air gap in magnetic circuit. There are three contacts in electromagnetic relay normally closed or break contact normally open or place on mounting bracket ,NC & No contacts placed on armature.When voltage is applied at coil i/p of relay it magnetic Yoke. The armature to its original position this operation takes in few milliseconds the dc electromagnetic relays are faster than ac relays.Procedure:

1. Connect the circuit as shown in fig on breadboard2. Check the connections apply Vin.3. Observe the state of LED at Vin=0



4. Change value of Vin & observe state of LED.

Conclusion : Thus we are designed & implementation a relay circuit using transistor as a switch & observe its working



EXPERIMENT NO. 9

Frequency response of single stage CE Amplifier



Experiment No. 1

Aim:: Introduction to Computer Communication Network.

Equipments: Components:Theory:



EXPERIMENT NO. 10

To study of Astable Multivibrator

Experiment No. 10

Tile: Transistor Astable multivibrator



Aim : To study of Transistor as Astable multivibrator.

Theory:

When power is applied to the circuit one might expect both transistors to switch on because both of them have paths to their bases for current to flow (R2 & R3). If the supply voltage is not applied quickly, i.e. it increases from zero slowly, this can in fact happen but there is a technique to overcome this. However, in most cases the supply is established quickly enough and operation is as follows.

Because of inevitable differences in nominally similar components, and in particular the transistors, one of the transistors will conduct before, or more rapidly than, the other.

Unless stated otherwise any voltage mentioned in the following description implies that it is measured with respect to the negative pole of the supply (0V) and therefore the transistor emitters.

Lets suppose that TR1 conducts first.

It is held on by the current from the supply through R3 and its base-emitter junction.The voltage at its collector will be around +100 mV. Since both capacitors will not have any charge on them at switch on they will both have zero voltage across them.With +100 mV at the left hand plate of C1 and zero volts across it there will also be +100 mV at its right hand plate and therefore at the base of TR2. This is not sufficient to cause TR2 to conduct.

Since TR1 is conducting there will be +0.6V at its base. C2 has zero volts across it so TR2 collector will also be at +0.6V.

Two things now start to happen.

Both C1 and C2 start to charge. The right hand plate of C1 starts charging, through R2, from +100mV towards +9V, and the right hand plate of C2, through R4, from +0.6V towards +9V. The next significant thing which happens is when the voltage on the right hand plate of C1, and therefore TR2s base, approaches +0.6V.

TR2 begins to conduct.

The collector voltage of TR2, which by now will have increased in value as C2 started to charge through R4, will fall and go down to +100mV.



The voltage across C2 cant change instantaneously so when the voltage at its right hand plate goes down the voltage at its left hand plate goes down by the same magnitude. If TR2s collector voltage had risen to, say, +3V, then when it reduces to +100mV (a negative-going change of 2.9V) the left hand plate of C2, and the base of TR1, will also go 2.9V more negative, but starting from +0.6V.

They will end up at 2.3V. This will cut TR1 off.

TR1s collector voltage will rise towards +9V as C1 initially discharges (left hand plate rising from +0.1V to +0.6V) and then charges (left hand plate rising to +9V while its right hand plate stays at +0.6V).

C2, with its right hand plate at +0.1V and left hand plate initially at 2.3V, will now start to discharge through R3. Its left hand plate rises from 2.3V towards +9V, its right hand plate being held at 0.1V. It will discharge (+0.1V on each plate) and then charge as the voltage at its left hand plate continues to rise.

When this voltage reaches +0.6V TR1 starts to conduct. TR1s collector voltage then falls from +9V to +100mV. As the voltage across C1 cant change instantaneously, the same magnitude of voltage change on its left hand plate will also appear on its right hand one. The voltage at TR2 base will therefore go from +0.6V to 8.3V cutting TR2 off.

TR2s collector voltage rises to +9V as C2 charges through R4. C2s left hand plate staying at +0.6V.

C1 will now begin to discharge through R2, the voltage at its right hand plate rising from 8.3V towards +9V, reaching +0.1V (C1 discharged), continuing to rise (C1 charging) until it reaches +0.6V. At this point TR2 starts to conduct again.

By this time the operation should begin to sound familiar to you. From now on, however, the negative excursions of voltage at both bases will be to 8.3V.

The above sequence repeats. As each collector voltage alternately changes from +9V to about 0V, the other transistor's base voltage changes from about 0V (+0.6V) to about -9V.

******** Note ********

The above is all very well for the purposes of explaining circuit operation, but it has one serious flaw.The max reverse base-emitter voltage of most transistors is around 5 to7. Which means that operating it with a 9V supply in the circuit as shown isn't a very clever thing to do.



In order to preserve the life of the transistors if the supply is greater than 5V we need to put a diode, a common or garden signal variety is adequate, in series with the base of each transistor, i.e. between C1/R2 junction and TR2 base and between C2/R3 and TR1 base.

**********************

In applications where the supply voltage rises slowly at switch-on the following modification will ensure correct operation without lock-up.

Circuit diagram:


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