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EE1/ISE1 Analogue Electronics 2008/2009 Part 5 9

Example 2 (1996 exam)

= 20v

1 k

vOUT8 k

+10 V

100 mH

i 2 k

vOUT

i

vA

vBvCvD

t ( sec)

t ( sec)

0 20

iA

iBiC

V has been at 8.7 V for a long time. It then falls to -2 V at t = 0, remains at this levelfor 20 sec, and then rises to +2.7 V. Assign values to v A, vB, vC, vD and i A, iB, iC.


Example 3

Both transistors have = 100. Sketch the time variation of V OUT when an input pulseis applied as shown. Assume V IN was at zero volts for a long time before t = 0.

V (V)IN

5

0 t (msec)0 200

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Switching Oscillators -1Ring Oscillator

Odd number of inverters connected in ringis ASTABLE i.e. has no stable state(check for yourself!)

Simple circuit suitable for very highoscillation frequencies

Oscillation period (=1/frequency) is sumof inverter propagation delays:

iPLHiPHLi ttT

Astable Multivibrator

Two MONOSTABLE multivibratorsconnected in loop

T = kR 1C1 + kR 2C2

where k( 1) depends on detailed designof monostables


Switching Oscillators - 2Discrete Astable Multivibrator (historical/lab interest only!)

During Phase A, current flows in C 1via R 1 and Q 2. VB1 rises from initialvalue of (0.9 V CC) until Q 1switches on, so T 1 given by:

0.7 VCC + (0.9 2V CC)e -T1/(R 1C1)

7.0V9.0V2

lnk withCkR TCC

CC 111

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Switching Oscillators 3Relaxation Type

Characterised by single RC network with comparator(s) to control switchingof source

T1 given by VTH+

Vomax

+ (VTH-

Vomax

)exp{-T 1/(RC)}with similar relation for T 2

For comparator version, thresholds are set by resistivefeedback network, with V TH+ = Vo max R B/(R A+R B) etc

Comparator version:

Schmitt version:

More complex circuit also possible, with two comparators and latch, asin 555 timer (see lab notes)


Sinusoidal Oscillators - 1Transistor amplifiers with feedback can be made to oscillate under certain conditions sometimes this even happens when we dont want it to!

To see how oscillation can arise, consider following system:

G(s)x-

K

y

)s(KG1)s(G

xy

)s(H

H(s) is the closed loop transfer function, expressed in terms of complex frequency ,s = + j . This tells us about the response of the system to inputs of the form exp(st).Signals of this form with , 0 correspond to oscillations with growing ( > 0) or decaying ( < 0) amplitude. H(s) can be obtained by normal nodal analysis, usingZC = 1/sC and Z L = sL in place of the normal impedance expressions for capacitors andinductors.

Note that H(s) at (complex) frequencies where the loop gain KG(s) -1. At suchfrequencies it is possible to have a finite output with zero input i.e. the circuit can generatean output signal without an input signal

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Sinusoidal Oscillators - 2

G(s)x-

K

y

)s(KG1)s(G

xy

)s(H

The equation 1 + KG(s) = 0 is referred to as the characteristic equation . Later in thecourse (year 2) you will find that the stability of a feedback system depends on where theroots of the characteristic equation lie in the complex plane. For our purposes, we only needto know that:

roots with < 0: oscillations will die away over time in absence of inputroots with > 0: oscillation amplitude will grow with timeroots with = 0: oscillation amplitude will be stable

The key to making a sinusoidal oscillator is to construct a circuit with feedback for whichthe characteristic equation has one (complex conjugate) pair of imaginary roots, and noroots with > 0.

An alternative way of stating the first requirement is that there should be a unique frequency0 for which -KG(j 0) = 1 i.e. for which the total gain around the loop (including the

sign on the input combiner) is unity. This is the Barkhausen Criterion .

what we want for an oscillator!


Sinusoidal Oscillators - 3 Typical oscillator structure:

Passive frequency-selectivenetwork (RC, RL or RLC)

G(s) -K

Amplifier (K real & >0)

Stable oscillations will occur at some frequency 0 provided:

Arg{G(j )} = and K = 1/|G(j )| = K 0 Note that K has to be controlled precisely - too large and oscillations will grow; toosmall and they will decay away. In practice, the amplitude of oscillation is controlled byhaving a second feedback mechanism that reduces K as the amplitude increases. At turn-on, oscillations build up until amplitude is reached where K = K 0.

Traditional methods for amplitude control include:

exploiting natural o/p swing limitations of amplifier (leads to distortion)

introducing clamp diodes to limit signal amplitude (ditto)

incorporating temperature-dependent resistors (e.g. thermistors or filaments) ingain-defining network so that gain decreases as signal power rises

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Sinusoidal Oscillators RC typePhase-Shift Oscillator Wien Bridge Oscillator

Transfer function of RC network is:

G(s) = = u3

u3 + 6u 2 + 5u + 1vivo

where u = sRC. Also note K = R F/R

Barkhausen criterion (KG = -1 withs = j 0) gives:

29K ; RC61

0

Transfer function of RC network is:

G(s) = sR 2C1sR 2C1 + (1 + sR 1C1)(1 + sR 2C2)

and K = -(1+R 3/R 4) (NB amp is non-inverting!)

Barkhausen criterion gives:

1

2

2

1

4

3

22110 C

CR R

R R

; CR CR

1


Sinusoidal Oscillators LC typee.g. Single Transistor Colpitts Oscillator

Oscillation frequency determined by resonance of LCnetwork.

Transistor provides gain to compensate for losses(represented in SSEC by r o , although in practice other components will also be lossy!)

Oscillation amplitude controlled by transistor non-linearity (transistor enters triode region)

Conditions for oscillation obtained by nodal analysis of SSEC, which yields characteristic equation directly:

s3LC1C2 +s2LC2/r o + s(C 1 + C 2) + (g m + 1/r o) = 0

Imposing the condition s = j 0 we obtain:

1

2om

21

210 C

Cr g ;

CCCLC

1

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EE1/ISE1 Analogue Electronics 2008/2009 - Problems 5 ASH 1

ANALOGUE ELECTRONICS

PROBLEMS 5

1. In the circuit below, V IN = -1 V for t < 0 and +1 V for t 0. Derive an expression forVOUT (t) and calculate the time at which it reaches +0.5 V. (Ans: t=139 sec)

1 k

100 mH

VIN

VOUT

2. For the simple CR network below, write down a general expression for V OUT (t), t 0, if the value of V OUT (0) is known, and V IN is constant. Starting from this result, show that if a square wave of amplitude V 0 and frequency f is applied to the input of the network,then the peak-to-peak amplitude of the waveform at the output will be given by:

fRC41

tanhV2 0

Sketch the output waveforms for the cases f = 1/10RC, f = 1/RC and f = 10/RC.

VIN V OUT

R

C

Hint: make t = 0 correspond to a transition in the input waveform, and apply thecondition V OUT (1/2f) = -V OUT (0) over the following half-cycle.

3. In the circuit of Figure Q3 overleaf, the switch S has been closed for a long time, and isopened at t = 0. Draw dimensioned sketches showing the time variations of the voltagesVE, V R and V C, and also of the voltage V CAP across the capacitor, starting just before t =0 and extending to t = 100 msec. Indicate any points at which the transistor moves fromone mode of operation to another. (from 1995 Exam)

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+10 V

5.7 V5 k

20 F

3 k

VE

VC

VCAP

VR

S

Figure Q3

4. In the circuit below, the voltage V IN has been zero for a long time. At t = 0 it is raisedinstantaneously to 5 V, held at this value for 500 sec, and then returned to zero. Derivean expression for the voltage V(t), starting just before t = 0, and continuing until t = 2msec.

+10 V

V

V IN

4.3 k

= 100

200 mH

500

1 k

Figure Q4

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5. Show by nodal analysis that the transfer function of the passive network in Figure Q5can be expressed as:

1u5u6u

u)s(G

23

3

where u = sRC and s is complex frequency. Hence, by applying the Barkhausen criterion,derive the conditions for oscillation and the oscillation frequency of the phase-shiftoscillator shown on page 8 of the lecture notes (Part 5b).

Figure Q5

6. (tricky) Figure Q6 below shows the basic configuration of a Hartley oscillator (NB nobias components shown!). Draw a small-signal equivalent circuit of the oscillator,ignoring the transistors output resistance, and show by nodal analysis that thecharacteristic equation is of the form:

0rsLsCr)LL(s)1(CLL be12

be21321

Hence show that the condition for oscillation and the oscillation frequency are given by:

C)LL(

1 ;

LL

210

2

1

Figure Q6

0809 part 5

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