Page 36 - Autumn 2024

The memory stores the carry from the position of the quite often referred to as a flip-flop. This consists of
If the two registers are pulsed again ten times the adding register to the next position-agrand carry two similar valves supplied with h.t. from a common
number 325 will simply circulate round both of them, which requires a time delay as against an incident carry source through anode resistors of the same value, and
and therefore a device is needed to enable one number that requires no delay. with the grid of one valve connected to the anode of
to be added to another. this device is called an “Adder.” the other. Negative bias is applied to the grids.
The 10 or more detector is a half-adder connected to
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
Fig. 23. Shifting Register Layout showing simple -thedotted line from the box C.M. The carry has a of A negative will result in A reducing current flow,
explanation of Shift. 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
The Multiplier, Multiplicand and Product Registers value of which is not known. Therefore a means must become positive because it is connected to the anode
each possess this shifting quality, so that when a pulse be provided for adding back to the existing digit that of A.
is applied to all their triggers simultaneously, the pat- unknown amount by which the carry exceeds 10.
tern changes by being shifted at each pulse. If we Fig. 25. Schematic arrangement of Half Adder A will now be in the state that B was previously and
and Full Adder
consider the register shown, each square represents a This is achieved by inserting another bank of adders will be cut off.
trigger, the slide shows a capacity of eight digits, where- The adder comprises two elements called half-adders. in the 8, 4 and 2 sum lines and whenever the carry buss
as on the machine it may be 9 or 14, it is assumed that It will be seen from the table that with no input to A bar is energised a 4 and 2 filler are entered into these
the register holds the figure 325. or B, there will be no sum or carry, whereas an input adders which causes an addition of 6 to the total of the
A single pulse applied to all the triggers simultane- on line A only will give a sum and no carry. Similarly, sum lines.
ously will result in the pattern shown on the centre with an input on line B only. With an input on line An example would demonstrate more clearly what
register, and the number would now read 32. If, how- A and B, there will be no sum, but there will be a takes place. Referring to the lines marked with an X
ever, four connections were made from the bottom to carry. in Fig. 26:-
the top of the register as shown, the units figure 5 would Whilst a half-adder can deal with A and B, if two Take 8 + 4 = 12.
not have been lost, but would have been fed back into half-adders are connected together to give a complete This sets the 8 + 4 adders which give out a sum on
the top of the register. Thus, if the same number of adder, this will then deal with A + B + C. Reference each line, these sums set the C.M. which gives out the
pulses are applied as there are register positions, the to the table will show the output for all variations. 4 and 2 fillers-thus,reading from the right, we have
figure 325 would move right round and return to its 2 followed by in the 4 position a sum on both inputs
original position. 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.
Fig. 26. Adder-Toadd Fig. in Reg. A to Fig. in Reg. B. With the trigger in this state, the right-hand anode of
Fig. 24. Shifting Register Layout showing how figure It will be seen that the method of connecting the full VI will be at a low potential, whilst the left hand is at a
can be retained. adders is to add the contents of register A to register B. high potential. Therefore, the left-hand side of the
diodes D1 and D2 will also be respectively at a low and
To transfer a number from one register to another, all The A and B inputs to the four half-adders first re- high potential. The actual values will be D2 at 135
that needs to be done is to inter-connect the loops of ceive the digits from the units position of the registers, volts and D1 at approximately 50 volts +.
both registers. Now, if both registers are pulsed ten sum them, and set a Carry if the sum exceeds 10. This Fig. 27. The Eccles-Jordon Trigger. Now bearing in mind that it is desired that V2 will
times the number standing in register 1 (325) will be carry is held in a Carry Memory which is shown in the This shows the well-known Eccles-Jordon Trigger, assume the state of V1, and that we shall consider at the
transferred to register 2 and still retained in register 1. diagram as the Box marked C.M.
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