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    RF Electronics & Analysing the Magnetron Drive Circuit

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    This solution deals with the analysis of a given Magnetron Drive Circuit to derive the circuit characteristics as applied to a resulting Radar pulse. Circuit parameters are given including Delay Line parameters and from this such things as the number of turns on the secondary winding of the drive transformer, the pulse period and duration, DC input power requirements and outgoing RF pulse power are derived.

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    https://brainmass.com/physics/circuits/rf-electronics-analysing-magnetron-drive-circuit-344802

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    Determination of circuit parameters for a Magnetron Drive Circuit - RF Electronics

    PROBLEM:

    A Magnetron Drive Circuit is described by the circuit shown in Figure 1. In addition the said Magnetron has the following circuit parameters defined.

    Magnetron Definition parameters:

    Input impedance of Magnetron (ZMAG ) = 625 Ohms
    DC to RF conversion efficiency of Magnetron (Eff (DC-RF) ) = 75 % = 0.75

    Magnetron Drive circuitry definition:

    {SEE ATTACHMENT FOR CIRCUIT DIAGRAM}

    VS = 1 kV, LCH = 3.12 H, L = 160 nH, C = 260 pF, no. sections in LC delay
    line n = 20

    Figure 1: A Magnetron Drive Circuit

    (Q1) Determine an estimate for the required number of turns (N) on the secondary of
    the transformer.

    (A1) Theory states that the output impedance as seen by the transformer (in this case
    the input impedance of the Magnetron) is given as the input impedance (in this
    case the delay line impedance) multiplied by the transformer turns ratio squared.

    Equation [1] shows this general identity

    N2/n2 = ZMAG / Z0 [1]

    {where N represent the turns on the secondary winding, n represents the turns on the primary transformer winding}.

    Since in this case the primary winding is just represented as one turn, n = 1, we can simplify [1] to [2] as below.

    N2 = ZMAG / Z0 [2]

    Also from standard theory we can say that the characteristic delay line impedance, Z0 , can be calculated from the identity involving the square root of the Inductance to Capacitance ratio shown in [3]

    Z0 = Sqrt(L/C) [3]

    [3] substituted into [2] and square rooted yields [4] as

    N = (ZMAG)^1/2*(C/L)^1/4 [4]

    Putting values given into [4] yields an estimate for number of turns required on the secondary winding of the transformer, thus

    N = (625)^1/2*(260 x 10-12 / 160 x 10-9)^1/4 turns

    N ~ 5 turns

    (Q2) Estimate the amplitude (VP) of the resulting Radar voltage pulse

    (A2) The voltage VS is applied as a discharge to the input of transformer

    The pulse amplitude (VP ) to the Magnetron is simply this voltage multiplied by the turns ratio N and is given by [5] as

    VP = N*VS [5]

    Putting in values given or calculated yields an estimate for the amplitude of the Radar pulse of

    VP = 5 x 1 kV = 5 kV

    (Q3) Calculate the period (T) of the pulse applied to the Magnetron

    (A3) The period of the pulse can be determined from standard theory and is given
    by [6]

    T = pi*Sqrt(LCH*nC) [6]

    Substituting the values given into [6] an estimate for the pulse period can be made as

    T = pi*Sqrt(3.12 x 20 x 260 x 10^-12 ) s

    T = 400 us

    (Q4) Determine an estimate for the duration (τ) of the pulse applied to the Magnetron

    (A4) The duration (t) of the pulse applied to the Magnetron is defined by circuit theory
    and is given by [7]

    t = n*Sqrt(LC) [7]

    Putting in values given to [7] we can estimate the pulse duration as

    t = 20*Sqrt(160 x 10^-9 x 260 x 10^-12) s

    t = 0.129 us

    (Q5) Estimate the pulse power (PDC) given to the Magnetron

    (A5) The DC pulse power is given by [8] via the classical Power = Voltage Squared x
    Impedance.

    As we are looking at the input side to the transformer then

    PDC = (VS )2 Z0 [8]

    Where Z0 is given by [9] (Equation [2] transformed) as

    Z0 = ZMAG / N2 [9]

    [9] in [8] yields [10] as

    PDC = (VS )2 ZMAG / N2 [10]

    Putting in the values given in to [10] gives an estimate of the pulse power input to
    the Magnetron as

    PDC = (1000 )2 x 625 / 25 = 25 MW

    (Q6) Find the RF power (PRF ) transmitted by the Magnetron.

    (A6) The RF power transmitted by the Magnetron is given as the product of the power
    conversion efficiency and the DC power calculated previously in (A5). It is
    shown by the identity [11]

    PRF = Eff(DC-RF) *PDC [11]

    Putting in values given to [11] gives an estimate for the RF pulse power as

    PRF = 0.75 x 25 = 18.75 MW

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