Page 37 - Autumn 2024

The memory stores the carry from the position of the quite often referred to as a flip-flop. This consists of
adding register to the next position-agrand carry two similar valves supplied with h.t. from a common
which requires a time delay as against an incident carry source through anode resistors of the same value, and
that requires no delay. with the grid of one valve connected to the anode of
The 10 or more detector is a half-adder connected to the other. Negative bias is applied to the grids.
the 8, 4 and 2 sum lines, so that whenever the digits Such an arrangement has two stable states-ifvalve
received at the A and B inputs contain an 8 and 2, an A is conducting, valve B is not, and vice versa.
8 and 4, or an 8, 4 and 2 (all of these combinations When valve A is conducting its anode potential will
being ten or more), the carry buss bar is stimulated and be low, the grid of valve B will also be low, thus pre-
this will set the Carry Memory. venting it from conducting; its anode will therefore be
It should be remembered here that the number in the at a high voltage and because of its cross connection to
register will not exceed nine, the units 8, 4, 2, 1 are the grid of valve A this will tend to cause valve A to
arranged to give any figure between 0 and 9. pass more current.
When a carry of 1 has been obtained it will be held Consider now a negative pulse applied to both grids.
until later when it will be added back into the 1 adder As the grid of B is already negative, the pulse will have
for inclusion into the next position of the adding register no direct effect on it, but the effect of making the grid
-thedotted line from the box C.M. The carry has a of A negative will result in A reducing current flow,
value of ten by stipulation, but it is known to have been with a consequent rise in its anode voltage. This will
stipulated by any number greater than ten, the exact cause a swing in the other direction, the grid of B will
value of which is not known. Therefore a means must become positive because it is connected to the anode
be provided for adding back to the existing digit that of A.
unknown amount by which the carry exceeds 10. A will now be in the state that B was previously and
This is achieved by inserting another bank of adders will be cut off.
in the 8, 4 and 2 sum lines and whenever the carry buss
bar is energised a 4 and 2 filler are entered into these
adders which causes an addition of 6 to the total of the
sum lines.
An example would demonstrate more clearly what
takes place. Referring to the lines marked with an X
in Fig. 26:-
Take 8 + 4 = 12.
This sets the 8 + 4 adders which give out a sum on
each line, these sums set the C.M. which gives out the
4 and 2 fillers-thus,reading from the right, we have
2 followed by in the 4 position a sum on both inputs
which creates a carry to the 8 position leaving zero at 4.
The 8 position will therefore have two inputs which will
create a carry which is not used and leaving a zero at 8. Fig. 28. The Shifting Register.
The adders will therefore give out the figure 2 at the We shall now look at the circuit of the shifting regis-
units position, but because the C.M. was set a 1 carry ter, and the slide shows two positions where VI, a
will be given out at the tens position-thusgiving the double triode valve, represents one Eccles-Jordon Trig-
correct answer of 12. ger, and was represented on the previous shifting regis-
ter by one square. This then can hold one digit,
representing either an 8, 4, 2 or 1, depending on its
position in the register, and whether or not the trigger
is set.
For the purpose of explanation, let us assume that
VI is set, i.e. it holds a digit. This then means that the
right-hand portion of the valve is conducting and this
of course prevents the left-hand side from conducting.
With the trigger in this state, the right-hand anode of
VI will be at a low potential, whilst the left hand is at a
high potential. Therefore, the left-hand side of the
diodes D1 and D2 will also be respectively at a low and
high potential. The actual values will be D2 at 135
volts and D1 at approximately 50 volts +.
Fig. 27. The Eccles-Jordon Trigger. Now bearing in mind that it is desired that V2 will
This shows the well-known Eccles-Jordon Trigger, assume the state of V1, and that we shall consider at the

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