Voltage Controlled Oscillator (VCO)

5.1 Dr. Yuri Panarin, DT021/4, Electronics Introduction  Voltage-Controlled Oscillator (VCO)– is a principal part of the PLL.  It determines the ov...
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5.1 Dr. Yuri Panarin, DT021/4, Electronics

Introduction  Voltage-Controlled Oscillator (VCO)– is a principal part of the PLL.  It determines the overall performance of PLL system, i.e. • The operating frequency range, • FM distortion, • center-frequency drift, and • center-frequency, • supply-voltage sensitivity are all determined by of the VCO.  Integrated-circuit VCOs most often are simply R-C multivibrators in which the charging current in the capacitor is varied in response to the control input

Voltage–Controlled Oscillator (VCO) Recommended Text: Gray, P.R. & Meyer. R.G., Analysis and Design of Analog Integrated Circuits (3rd Edition), Wiley (1992) pp. 695-707

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

DT021/4 Electronics – 8 Voltage Controlled Oscillator

Emitter-Coupled Multivibrator as VCO

Notes

 Consider the emitter-coupled multivibrator, which is typical in this application VCC

Assuming that Q1 is turned off and Q2 is turned on The Q4 base voltage (VB4) is one diodes drop (0.6V) and emitter voltage (VE4) is two diodes drops below VCC Base voltage of Q3 (VB3) is VCC and emitter voltage (VE3) is one diodes drops below VCC Thus the emitter of Q2 is two diode drops below Vcc.

R

Q5

VCC

Q6 Q3

Q1

Q4



R

VCC-0.6V

VOUT

Q2

VCC-1.2V V -0.6V CC

IB

IB C

VIN

2

I1 RE

DT021/4 Electronics – 8 Voltage Controlled Oscillator

VCC-1.2V

I1 RE

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

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5.2 Dr. Yuri Panarin, DT021/4, Electronics

Notes

VCO Analysis 

Since Q1 is off, the current I1 is charging the capacitor so that the emitter of Q1 is becoming more negative. Q1 will turn on when VE1 becomes equal to three diode drops below Vcc As a result, the base and emitter of Q3 (and the base of Q2) moves in the negative direction by one diode drop

VCC

R

Q5

Q6 Q3

VCC-0.6V

Q4

VCC-1.2V

R

2I1

VCC-0.6V

VCC-1.2V

Q1

VOUT

Q2

IB

IB

VCC-1.8V

VCC-1.2V

C I1

I1

VIN

I1

RE

RE

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

DT021/4 Electronics – 8 Voltage Controlled Oscillator

VCO Analysis (cont.) Q2 will turn off causing the base of Q4 and Q1 to move positive by one diode drop because Q6 will turns off. As a result, the emitter-base junction of Q2 is reverse biased by one diode drop because the voltage on C cannot change instantaneously. Current I1 must now charge the capacitor voltage in the negative direction by an amount equal to two diode drops before the circuit will switch back again.

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Notes 

VCC

R

Q5

VCC-0.6V

Q6 Q3

Q1VCC-1.2V

Q4

VCC

VOUT

VCC-0.6V

Q2

IB

VCC-1.2V

R

IB C

VCC-0.6V

I1

VIN

I1 RE

DT021/4 Electronics – 8 Voltage Controlled Oscillator

I1 RE

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

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5.3 Dr. Yuri Panarin, DT021/4, Electronics

Notes

Voltage waveforms 

Since the circuit is symmetrical, the half period is given by the time required to charge the capacitor and is T/2=Q/I1 , where Q = C·∆V = 2·C·VBE(on) is the charge on the capacitor. The frequency of the oscillator is thus f=1/T = I1/4CVBE(on)

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

DT021/4 Electronics – 8 Voltage Controlled Oscillator

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Notes

Temperature coefficient  The emitter-coupled configuration is capable of high operating speed, up to approximately 100 MHz .  It displays considerable sensitivity of center frequency to temperature even at low frequencies, since the period is dependent on VBE(on)  The temperature coefficient of the period can be calculated (assuming a base-emitter voltage temperature coefficient of -2 mV/°C) as:



1 dωosc 1 dVBE ( on ) 2 mV / o C =− = = 3.3 × 10 −3 / C VBE ( on ) dT 600 mV ωosc dT

 This temperature sensitivity of center frequency can be compensated by causing current I1 to be temperature sensitive in opposite way DT021/4 Electronics – 8 Voltage Controlled Oscillator

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

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5.4 Dr. Yuri Panarin, DT021/4, Electronics

Typical Problem

Notes

 A typical structure for a PLL voltage-controlled oscillator (VCO) is shown in Fig. 1. VCC

The VCO contains matched transistors with VBE(ON) values of 0.6 V and is operated at a supply voltage (Vcc) of 12 V and a freerunning quiescent current (I1) of 2.5 mA.

R

Q5

Q6 Q3



R

Q4 VOUT

Q1

Q2 IB

IB C

VIN

I1 RE

I1 RE

DT021/4 Electronics – 8 Voltage Controlled Oscillator

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

Voltage waveforms

Typical Problem  Describe the performance of this circuit and sketch, to a common time-scale, the anticipated voltage waveforms at the emitter of Q2 (VE2) and of the capacitor voltage (VC =VE1 - VE2) over one cycle of oscillation.  Assuming a base-emitter voltage temperature coefficient of -2 mV/°C and a symmetrical input voltage range of 0.8 V VBE(on) and that I1 is temperature invariant.

 As a result, the emitter-base junction of Q2 is reverse biased by one diode drop because the voltage on C cannot change instantaneously.  Current I1 must now charge the capacitor voltage in the negative direction by an amount equal to two diode drops before the circuit will switch back again.  VCC VCC -VBE(on)

12V 11.4V 10.8V 10.2V

VE2 0.6V

VBE(on) VC

-0.6V

DT021/4 Electronics – 8 Voltage Controlled Oscillator

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

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5.5 Dr. Yuri Panarin, DT021/4, Electronics

VCO performance

Solution

 Assuming that Q1 is turned off and Q2 is turned on. The Q4 base voltage (VB4) is one diodes drop (0.6V) and emitter voltage (VE4) is two diodes drops below VCC  Base voltage of Q3 (VB3) is VCC and emitter voltage (VE3) is one diodes drop below VCC. Thus the emitter of Q2 is two diode drops below VCC.  Since Q1 is off, the current I1 is charging the capacitor so that the emitter of Q1 is becoming more negative.  Q1 will turn on when VE1 becomes equal to three diode drops below VCC  As a result, the base and emitter of Q3 (and the base of Q2) moves in the negative direction by one diode drop. Q2 will turn off causing the base of Q4 and Q1 to move positive by one diode drop because Q6 will turns off. DT021/4 Electronics – 8 Voltage Controlled Oscillator

(i) The value of capacitance required to attain a free-running oscillation frequency of 1 MHz. Since the circuit is symmetrical, the half period is given by the time required to charge the capacitor and is T/2=Q/I1 ,where Q = C·∆V = 2·C·VBE(on) is the charge on the capacitor. The frequency of the oscillator is thus: 1 I1 I1 2.5 ⋅ 10 −3 f = = C= = ≈ 1nF T 4 ⋅ C ⋅VBE ( on ) 4 f ⋅V 4 ⋅ 10 6 ⋅ 0.6 BE ( on )

(ii) The upper and lower limits of the oscillation frequency range; For the range : 0.8 V < VS < 1.4 V the central voltage is VS = 1.1 V: I1 =

and therefore RE = 0.5 V / 2.5 mA =200Ω.

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

Solution

RE

RE

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Notes

 (ii) The frequency is proportional to the current I1. Therefore the upper frequency limit is corresponds to upper voltage range V − 0.6 1.4 − 0.6 0.8 V=1.4V I 1max = S max = = ≈ 4mA f max =

1 .1 − 0 .6 0 .5 = = 2.5mA RE RE



200Ω

I1 max 4 ⋅ 10 −3 = ≈ 1.66 MHz 4C ⋅ VBE ( on ) 4 ⋅ 10 −9 ⋅ 0.6

 Similar for the lower limit I1 min =

0.8 − 0.6 0.2 = ≈ 1mA RE 200Ω

f min =

10 −3 ≈ 0.417 MHz 4 ⋅10 −9 ⋅ 0.6

(iii) The oscillation frequency temperature coefficient. Assuming base-emitter voltage temperature coefficient (dVBE/dT)of -2 mV/°C the temperature coefficient of the period can be calculated as

1 dωosc 1 dVBE ( on ) 2 mV / o C =− = = 3.3 × 10 −3 / C ωosc dT VBE ( on ) dT 600 mV DT021/4 Electronics – 8 Voltage Controlled Oscillator

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DT021/4 Electronics – 8 Voltage Controlled Oscillator

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