Designing of Single section Band Pass Filter

Dear all,

Today we present a simple method for designing a single section band pass filter.

The Impedance of the coil is 75 Ohms = 2πFL. 
Impedance of capacitance = 1/2πFC

Let us calculate value of  L & C for F of 17.820 MHz and R = 75 Ohms.

Capacitance value calculation:

                                                                           R = 1/2πFC

Where                  F = 17.820 MHz

                            R = 75 Ω

                                                                           C = 1/2πRF

                                                                           C = 120 pF

Inductor Value calculation :

                                                                           R = 2πFL

Where                  F = 17.820 MHz

                            R = 75 Ω

                                                                           L = R/2πF

                                                                           L = (75)/(2*π*17.82*10^6)

                                                                           L = 0.7µH

Approx Expected Q of the copper coil is between  200 and 300.

Bandwidth of the circuit is                              = F/Q

Loaded Q is decided as 20 to take care of low insertion loss.

The loaded Q is around 20      =>           Bandwidth is 891 KHz

Impedance of the coil is 75 Ω

For Q = 20 the impedance of the tank circuit is 75 * 20 = 1500 Ω

Loaded Q is 20

Rin = Rout = 50Ω ( input impedance is 50 Ohms )

Equivalent circuit with double loading.

Equivalent circuit with double loaded

 Impedance of the coil is 75 Ω

For Q = 20 the impedance of the tank circuit is 75 * 20 = 1500 Ω
(Here we are neglecting the coil's finite Q)

Series and parallel equivalent of loading circuit

Series and parallel equivalent of loading circuit

parallel to series and series to parallel conversion

Parallel to series and series to parallel impedance conversion equations.

Rp = Rs{1+ (Xs/Rs)^2}

Xp = (Rp*Rs)/Xs          

Rs = Rp/{1+(Rp/Xp)}  

Xs = (Rs*Rp)/Xp         

These equation are from MOTOROLA application note

Now let us calculate Xp and Xs, we know Rp = 1500 and Rs = 25 Ohms

      

Rp = Rs{1+K^2}        

              Where K = Xs/Rs

1 + K^2 = Rp/Rs                           

1 + K^2 = 1500/25                        

1 + k^2 =  60                               

K = 7.6                      

Xs/Rs =  7.5                              

Xs = 7/5 * Rs                    

Xs = 190                           

                      

Xs = 1/2πFC                      

 C = 1/2πXsF                    

       Where  F  = 17.820 MHz

                  Xs = 190

C = 47pF                        

Series equivalent circuit:

Series equivalent circuit

Input/output circuit:

Input/Output circuit

                                                 Xs = Rs*Rp/Xp

                                                 Xp = Rs*Rp/Xs

Where Rs = 25

           Rp = 1500

                                                 Xp = 197

This means for loaded Q of 20 and above Xs and Xp are almost same

Now tuning capacitance Ct = Total capacitance required – coupling capacitor (Xp)

                                                   = 120 – 47 pF

                                                  = 73 pF

Final circuit is 

Final Circuit

      The insertion loss of the circuit is 20Log(D/D-1)

            Where  D = Unloaded Q/desired Q

In our case unloaded Q in between 200 - 300 and the loaded Q is 20.

In our case insertion loss is 1db

There is no circuit which will give zero loss due to finite unloaded Q

If somebody claims he has designed a zero loss band pass filter  then Prof A.ZVEREV Russian scientist would be unhappy man. As he is the designer of insertion loss equation.