🚀 go-pugleaf

Bitsavers just posted a book on the internal logic and circuits
of the IBM 700 series.  It goes into a great deal of detail, such
as the voltage and resistance of each component.

If you're into hardware this is a must read.

[sample]
The binary input is a negative shift input to a trigger and allows it to be flipped regardless of its previous state (unless it is being held by another circuit). A binary input is indicated when the same signal is applied to both the left and right grid (Figure A12c). For example, if the trigger is off and it is fed a pulse through a binary input, the trigger is flipped on. If another pulse is then applied to the binary input, the trigger is flipped off. 

One grid is initially at ground, and the other is about -12 volts. The negative shift begins at ground and pulls the conducting grid negative. The first twelve volts of the negative shift cannot affect the cut-off grid because the binary input diode connected to that grid is cut off. By the time this diode is ready to conduct, the regenerative action initiated by the negative shift on the opposite grid produces a positive shift at the formerly cut-off grid. 

If the differentiated spike is narrow enough and not excessively large in amplitude, it does not interfere with the positive shift at the formerly cut-off grid. However, if it is too wide or too large in amplitude, it may interfere with the positive shift, causing the trigger to flip back to its initial state. For this reason, the binary input should not be fed a pulse of very great amplitude. 

see:

http://www.bitsavers.org/pdf/ibm/logic/223-6746-1_700_Series_Data_Processing_Systems_Component_Circuits_Apr1959.pdf