Tuned circuits

TUNED CIRCUITS

Tuned circuits are two types - (a) Single tuned, (b) double tuned. In radio frequency range (500 to 1600 kc/s), the application of signals is by these coupled circuits. The coupled tuned circuits are used in radio receivers to produce uniform response to modulated signals over a specified band width. They are very useful in communication systems.

(a) Single Tuned Coupled Circuits

Consider the circuit shown in the fig.5.40. A parallel resonant circuit on the secondary is inductively coupled to coil 1. This coil 1 is excited by a source E_g. Let R_g be the source resistance.

Let R₁, R₂ be the resistances of coils 1 and 2 respectively and let L₁, L₂ be the self-inductances of the coils 1 and 2 respectively.

Assume that R_g >> R₁ >> jωL₁ i.e., Ignore R₁ and jωL₁, in comparison with R_g

Then, the mesh equations are

When the secondary side is tuned i.e., when the values of the frequency o, is such that

From equation (iii) the current I₂ at resonance is obtained by putting ωL₂ = 1/ωC and replacing ω by ω_r.

Therefore I₂ at resonance = jE_g ω_rM / R_g R₂ + ω_r² M² … (vii)

From equations (vi) and (vii) it is observed that at resonance frequency E₀, I² and A depend on M.

The maximum value of E₀ or A depends upon M. To get the condition for maximum E₀, dE₀ / dM = 0

From this, on simplification, we get

M = √R_g R₂ / ω_r ... (viii)

i.e., When M = √R_g R₂ / ω_r , the output voltage is maximum.

Therefore, maximum output voltage

These maximum values are obtained by substituting M = √R_g R₂ / ω_r in expressions E₀, A, and I₂ at resonance.

We know that M = K√L₁L₂. By changing the coupling factor K, we can vary M. The variation of amplification factor or output voltage with the coefficient of the coupling is shown in the fig.5.41

Example 9 In the single tuned resonant circuit, the applied voltage in a primary coil Eg volts (assume R_g = 0), R₁ = R₂ = 5 Ω L₁ =L₂ = 32μH, M = 20 μH

C = (secondary side capacitance) = 0.5 μF

Determine the resonant frequency and the output voltage at this frequency.

Solution: