NOTES ON RM 96 FINAL PA TANK COIL
By: S, Rama Mohan Rao, VU2RM

    After several qso's on the7 Mhz band with new hams who have built RM 96 upto BD139 driver and not able to get power from IRF stages, the often asked question is "how many turns should I use on a very good imported toroid. I have used but it is not giving more than 2 to 3 watts. What should I do? Well, it is a very pertinent question. Yes, so let us go a bit deep to get an answer.

    The theory part of it tells us that there are certain Parameters, which are required to calculate this data. Having a very good-looking imported toroid alone will not help us. We need to know either the permeability or the AL value of the core and its maximum frequency it can be used. Next, it is to be known weather it is iron core or ferrite core. As per little knowledge about these cores, my idea is that iron dust core toroids are light in weight and ferrite cores are a bit heavy. To calculate the number of primary turns & secondary turns the following formulae are used at this end. 

    Firstly: we should calculate the drain load resistance into which it has to deliver power output. Let us assume that our PA stage employs IRF 830 TMOS FET. Say we are applying about 36 Volts DC to the drain and it draws about 1 to 1.5 Amps of current. (This assumption is based on the practical experience on these sets and BD139 drive is capable of giving that much gate voltage.). We measured up to 8 V with RF diode probe. 

    Therefore, power input is P=:ID x VD i.e. 1.5 A X 36 V = 54 watts DC input. Let us take the efficiency of this stage with all sorts of cheap components as 50%. Therefore the power RF output at the drain is 54/2=27 watts or say 25 watts. 

    The remaining 29 watts (54-25) is to be dissipated as heat, so use a reasonable heat sink with vertical fins.

     Now the drain load RD=VCC2/2 PO. Where Vcc is the drain supply voltage. Here it is 36 V. PO is the Power output (25 watts). 

Therefore RO=(36X36)/(2X25)=25.9 or 26 q.

    These 26 ohms is the load resistance into which the IRF 830 will deliver these 25 watts of RF output power. Now to proceed further we should know the AL value of the core, and weather it is a ferrite or a dust core, which is necessary.

    Here we have taken the cores manufactured by Ferrita Enterprise, Belgaum, Nickel Zinc Ferrite meant for RF use. They manufacture cores with material HF-A group going up to 40 MHz and HF-B group going upto 80 MHz. For our PA tank choice was made for T20-HFA core, considering cost...etc. This HFA, T20 core is having an AL value of 60. 

Now further calculations are as follows: 

    Therefore the primary inductance of our tank coil should have an inductance of 2.95 uh or say 3 uh. Now, with our Belgaum HFA.T-20 core we have to calculate the turns required on the toroid. 

The number of turns as follows: 
N=1000 x sq. root of (L mh)/AL) for ferrite core, 
N=..00 x sq. root of (L mh)/AL) for dust core. 

    Here note the inductance value in milli Henries where as we had the answer for L in micro Henries. So, convert pH to mil. Now we have to wind 7 turns on this core in the primary. However, remember the PA tank coil is an impedance matching transformer i.e. your antenna feeder cables impedance of 75 or 50 g are to be transformed into the required drain load of 25 . 

    If all goes on like this, We have to calculate the secondary turns, for inverted V antenna where we use 50 cable. The turn's ratio between the primary and the secondary is all as far as winding is concerned. Use a gauge of wire which carries that current in the primary i.e. about :!.8 SWG is OK, for secondary it can be around 22 SWG. First wind the secondary turns as it contains more turns, then wind primary turns over the secondary. 

    Here P1 goes to drain and P2 goes to (+) supply. Then S1 goes to ground and S2 goes to antenna change over relay. This sequence is to be followed for proper coupling. 

    Further Note: If the primary inductance is more, let it be due to more primary turns or due to more AL value of the toroid core. The drain current will decrease and will not give the estimated power output. In such cases reduce primary turns maintaining the same turn ratio between primary and secondary. 

I think this much gossip is enough for our PA tank. 

Wish you good luck. 
73 de VU2RM