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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