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6.3.2 Clapp oscillator
A variation of the oscillator Colpitts by connecting a capacitor C in series with the inductor L gives the oscillator Clapp. The Sch.6.5 shows such an oscillator with a BJT. The addition of capacitor improves the stability of oscillation. This stability can be further improved if we use a piezoelectric crystal. When the capacitive resistance of the capacitors C1 and C2 are high on the capacitor is added, the oscillation frequency is determined by the relationship we saw in the oscillator Colpitts. H capacity C may be variable in order to have variable resonant frequency at a given frequency range. The resonance frequency, the parallel resistance of series LC circuit is
The oscillation frequency is given again by:
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minimum so that the resonance frequency is practically independent of the parameters of the transistor.
Figure 6.5. Clapp oscillator with a BJT
(6.3.11)
but where now we
1 = _L + _L +1
Ct C1 C2 C
(6.3.12)
Also, according to Eq. (6.3.2) proves that to ensure the maintenance of oscillations, must apply again the same relation to the oscillator Colpltts, ie Eq. (6.3.5).
If C1 << C2, why anasyzefxis given by:
(6.3.13)
and the strengthening of the amplifier:
(6.3.14)
Example 6-3
O above oscillator is L = 33 mH, C1 = C2 = 100 nF, C = 1200 nF, hfe = 100, RC = 470 and Z hie = 1 kQ. Find the frequency of oscillation, the ratio anasyzefxis maintained if the oscillations.
Solution
From Eq. (3.6.10) we have
So
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C = 48 nF
The Eq. (6.3.9) gives
O ratio is anasyzefxis
To maintain the oscillations we have
The voltage gain of the amplifier transistor is:
which is much greater than 2. Therefore we maintain the oscillations.
The Sch.6.6 shows Clapp oscillator with TE. The oscillator that all of the above relations with the only difference is the voltage gain of the amplifier is given by
known relationship with a reversal of the amplifier, ie
(6.3.15)
Example 6-4
We calculate oscillator Clapp gives oscillations with resonance 50 kHz, using the TE 741.
Solution
We accept as a reasonable value L = 2.2 mH. Therefore, from Eq. (6.3.9),
we
We choose the standard value b = 0.1. Therefore,
Analogue Electronics
Figure 6.6. Clapp oscillator with TE
Assuming a reasonable value Ct = 4.7 nF and C2 = 100 nF
Therefore,
To have a dependent oscillations should apply Eq (6.3.5),
ie
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So if dechoume, eg, R1 = 10 kQ, we have
R2> 110 KW
6.3.3 Hartley oscillator
O oscillator Hartley, Sch.6.7, like the oscillator Colpitts, except that the position and role of capacitors and coils have been reversed. That is, to create the anasyzefxi used two coils L1 and L2 (in place of the capacitors C1 and C2) and a capacitor C (in place of L). O resistors R1, R2, Re RC and create the proper polarity of the transistor. The resonant frequency of oscillation is determined by the lattice coordination (L1, L2, C). O capacitor C can be variable and therefore the circuit to selectively coordinate within a given frequency range. O capacitor C2 is capacitor barrier (the dc voltage to not appear at the output) and the capacitor Ce is capacitor leakage.
Repeating the reasoning we did for the oscillator Colpitts, but now putting X1 =] h L1, X2 =] oh and L2 X3 = 1 / joC in Eq. (6.3.1) and Eq. (6.3.2), we find that the oscillation frequency is given by
(6.3.16)
where,
Lt = L + L
(6.3.17)
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Analogue Electronics
Figure 6.7. BJT Harley oscillator with
Also, it turns out, that to be dependent oscillations, we must apply the formula:
where, Av is the strengthening of the amplifier in this case (transistor CE), are:
Example 6-5
O above Hartley oscillator has the following component values: L1 = L2 = 4.7 mH, C = 22 nF, RC = 2.2 kQ, and the transistor has hie = 1 kQ and hfe = 50. Find the resonant frequency of oscillation and to check whether the oscillations are dependent.
(6.3.18)
(6.3.19)
Solution
Because
Lt = L1 + L2 = 4.7 + 4.7 = 9.4 mH
Besides, from Eq. (3.6.19), we
Therefore satisfied the inequality of Eq. (3.6.18) so we maintain the oscillations.
The Sch.6.8 shows the Hartley oscillator with TE. The resonant frequency is given again by Eq. (3.6.16). It remains to apply the Treaty of Eq. (6.3.18) for the maintenance of oscillations, but the aid If it is known from the type of reverse amplifier with TE
(6.3.20)
R1
Because the input impedance of the TE is great, why are anasyzefxis,
L2
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Therefore
L1
(6.3.21)
Figure 6.8. Harley oscillator with TE
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L1 = (1 -,) Lt = 0,9 x 1 = 0.9 mH
and
L2 = Lt - L1 = 1 - 0.9 = 0.1 mH
Finally, according to Eq. (6.3.18), the aid to give the amplifier to have a dependent oscillations are:
Therefore, in Eq. (6.3.20) should
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Analogue Electronics
Example 6-6
We compute Hartley oscillator with TE (eg the 741) for the resonance frequency 30 kHz.
Solution
We accept the logical value Lt = 1 mH, so Eq. (06/03/16)
Then we, as usual, b = 0.1. Therefore, if you connect the Eq. (3.6.17) and
Eq. (3.6.21), we get
So if we accept the R1 = 10 KW, so we find that R2> 10 x R1 = 100 KW.
