EDITOR: B. SOMANATHAN
NAIR
1. INTRODUCTION
Voltage-controlled
oscillator (VCO) is a circuit that produces an output frequency which is
proportional to the input voltage of the circuit. We can construct a VCO by
disconnecting the base resistors RB1
and RB2 of an astable
multivibrator from VCC and
reconnect them to a new variable power supply VBB as shown in Fig.1.
2. WORKING PRINCIPLES OF VCO
By
disconnecting RB1 and RB2
from VCC, and
reconnecting them to a different supply +VBB,
we notice that capacitors charge from this value to +VCC. The waveforms for this situation are shown in Figs.
2 and 3, respectively.
Now, consider the capacitor-charging
equation
vC = VF ‒ (VF‒ VI)
exp(‒t/RBC) (1)
From
Figs. 2 and 3, we find that vC
≈ 0, VF = VCC, and VI = −VBB. Substituting these in Eq. (1) yields
0 = VCC
‒ (VCC‒ VBB) exp(‒t/RBC)
(2)
Rearranging
and solving Eq. (2), we get the half-period of oscillation
T = RBC ln [(VCC‒ VBB)/ VCC]
= RBC ln [1+(VBB/VCC)] (3)
From
Eq. (3) we find that by varying VBB,
we can change the period and hence the frequency of oscillation. Now, full period
of oscillation
2T = 2RBC ln [1+(VBB/VCC)] (4)
And hence frequency of
oscillation
fo = 1/2T = 1/2RBC ln [1+(VBB/VCC)] (5)
3. WAVEFORMS OF VCO
As
stated earlier, the waveforms associated with VCO are shown in Figs. 2 and 3,
respectively. It can be seen that the waveforms of VCO are the same as those of
the astable multivibrator shown in a previous blog with the exception that the
capacitor charging voltage rises from ‒VBB
to +VCC. Varying VBB then varies the output
frequency.
4. SPECIFICATIONS
Variations
required in frequency : 1 kHz to 10 kHz
5. DESIGN STEPS
The
initial design for the astable multi follows the steps we have given in the
earlier blog on Design of Astable Multi.
We now have to find out the value of VBB
that will yield the vales of capacitors to produce the desired frequency range.
Now, for a frequency
of 1 kHz, we find from Eq. (5)
1 kHz = 1/2T = 1/2RBC ln [1+(VBB1/VCC)] (6)
and for a frequency
of 10 kHz, we find from Eq. (5)
10 kHz = 1/2T = 1/2RBC ln [1+(VBB1/VCC)] (7)
Substituting for VCC = 10 V, RB = 1 MΩ, and choosing C = 0.001 μF (or any other suitable
value), in Eq. (6), we find
1/1 kHz = 2x106 x 0.001x10‒6 ln [1+(VBB1/VCC)]
ln [1+(VBB1/10]
= 0.5
Taking
antilog and after some rearangements, we get
VBB1 = 6.5 V
Similarly, for 10
kHz, we find from Eq. (7)
1/10
kHz = 2x106 x 0.001x10‒6 ln
[1+(VBB2/VCC)]
ln [1+(VBB2/10]
= 0.05
Taking
antilog and after some rearangements, we get
VBB2 = 0.51 V
Thus we must have a supply voltage
that can be from about 0.5 volt to 6.5 volts for producing a variation in
frequency of 1 kHz to 10 kHz. The fully designed VCO is shown in Fig. 4.
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