MULLARD BATTERY VALVES (iii)

I thought that today, we would conclude my review of the Mullard battery D series valves with the arrangement of the filament circuit in which they may be used.  The filament current for all valves in the D series is 50mA for a nominal filament voltage of 1.4V. The output pentodes however, have two  filament sections which may be connected in either series or paralllel configuration, a useful facility where one can either use a single dry cell with parallel heater connection or conversely, a series chain.

To use series operation, precautions need to be taken to ensure that the voltage acrosss any single 1.4V nominal filament device does not exceed the range 1.3 - 1.6 V and this may be done by shunting resistors being fitted across the individual 1.4V filaments as shown in the example below: -

A good rule of thumb is to design the filament circuit such that the voltage across each filament section has a nominal votage of 1.3V and a maintenence range of 1.2 - 1.4V whether or not the supply voltage is the 1.5V battery, the 2V accumulator or a mains PSU.  

The big question then is how to calculate the values of the shunt resistors needed?   When operated at 1.3V the filament current is 47.4mA with no cathode current flow, however, in actual operation, cathode current flows through the filament in a non linear manner. Another consideration is that in starting the filament from the positive end of the chain then each filament has to bear the cathode current of each of the preceeding valves.  Hence the value of the shunt for each individual section must be calculated separately on the basis of the cumulative value of the cathode current at that particular point in the chain.  

Going back to the non linear flow, we need to consider that the current at the positive end of each filament is less than the average filament current by an amount equal to one third of the cathode current.  Similarly the current at the negative end of each filament is greater than the average filament current by an amount equal to two third of the cathode current.

Still with me, not to worry, let's try some practical calculations using the filament chain diagram shown previously:-

If the cathode current at output valve, V4 is 6mA then 2.4mA will flow to the positve end and 3.6mA to the negative end.    

The filament current at 1.3 V is 47.4mA exclusive of cathode current so: -

the total current at the positive  end  of the positive limb is 47.4 - (1/3 x 2.4)  = 46.6mA.

the total current at the negative end  of the positive limb is 47.4 + (2/3 x 2.4)  = 49mA.

the total current at the positive end of the negative limb is 47.4 - (1/3 x 3.6)  = 46.2mA.

From this little lot then, the current, I  in R1 must be 49 - 46.2 = 2.8mA for a V of 1.3V therefore R of R1 must be (1.3 x 1000)/2.8 = 464 Ohms.

Whew, in these blog entries we stumble from chemistry to higher maths, well, that's enough for me so if you want to complete the calculations for V3, V2, V1 then please ava go by yourselves in a darkened room with no gibbering and gnashing of teeth!

At least you now know how to design filament circuits to suit the operation of these smashing little valves should you so wish to.