Ok, can you explain it to me ? My amazement is that this all rather predates solid state transistors. — unenlightened
Are you looking at the 9th canon where he constructs an ever deepening series of nested a's and b's? Page 55 in my version?
If so, you just take the whole right hand expression of a & b as = r. and use J2 in reverse. — unenlightened
Wow, if someone implemented something like that we could have computers and an internet!
Sorry, couldn't resist. — wonderer1
I'm sure I don't understand how a circuit has a memory, still. — Moliere
I didn't mean feedback necessarily, just the view that process might be seen as fundemental, not substance. — Count Timothy von Icarus
The story of a hole in a state of flow with an innumerable number of other holes towards ~Q: We start at 5 V and move through R1 to TR1 because the voltage at Q is lower than the voltage at ~Q (assuming we're already in a steady state), then we go through the unmarked resistor on the other side of the transistor, up through R3 and out ~Q. If you touch "Set" to the zero volts line than you ground the flow causing the voltage to switch over to R4-T2-R2-Q. — Moliere
[tl;dr]More accurately than looking at it as a switch, we might look at the path from collector to emitter as a resistor with a resistance of about 42 Ohms when the transistor is in the 'on state', and as an open circuit when the transistor is in the 'off state'. In that case, if we suppose the resistance of R1 and R4 to be 1000 Ohms, then we have an explanation for why the flip-flop schematic shows a voltage at Q of 0.2 volts for the blue state. In the blue state the 5 volts of the flip-flop power supply gets divided between the 1000 Ohm resistance of R1 and the 42 Ohm resistance of the 'on stated' TR1. In the red state the voltage at Q simply is the +5 Volts of the power supply. (Ignoring for the sake of simplicity, the the relatively low base currents flowing through R3 and R4. The ambitious reader who is into that sort of thing can assume that R3 and R4 have a resistance of 100000 Ohms, and look up linear circuit analysis, and calculate voltages out to more decimal places. However, for pragmatic purposes we can ignore current through R3 and R4, and just consider whether the voltage at the transistor bases are above or below 0.7 Volts to know whether a transistor is off or on.)
And for this discussion we can ignore the resistance of the unlabeled resistors altogether and treat them as open circuits. They are for practical details engineers need to worry about but, not of any help in looking at things in the simple voltage focused model of the flip-flop schematic.[/tl;dr]
That helps me understand the feedback part very well -- so thank you again for taking the time. When Set is grounded the voltage from R3 no longer gives the voltage necessary for the transistor to be in the "on" state, but the parallel circuit through R2 does so the circuit flips over to Tr2. Since Tr1 is now off that means 5V goes to Q as the path of least resistance. The same holds for reset and the blue state. — Moliere
Based on the website I linked it looks like Q and ~Q are out of phase with one another. So the memory comes from being able to output an electrical current at inverse phases of one another? How do we get from these circuits to a logic? And the phase shift is perhaps caused by subtle manipulations of the transistor? — Moliere
At this point it is pragmatic to jump up a level in abstraction and think in terms of logic gates instead of transistor circuits. — wonderer1
I think what I'm wanting to settle, for myself, is whether or not the circuits are in turn being interpreted by us, or if they are performing logical operations. — Moliere
What makes Q and ~Q different other than one is on the left side, and the other on the right side? Do we just arbitrarily choose one side to be zero and the other side to be 1? Or do the logical circuits which have a threshhold for counting do it differently?
To my mind the circuit still doesn't really have a logical structure anymore than a stop light has the logical structure of Stop/Go without an interpretation to say "red means stop, green means go". So are we saying "Q means 1, and ~Q means 0"? — Moliere
The SR f!ip-flop circuit is symmetrical, so it is somewhat arbitrary which output is chosen to be Q and ~Q. However, the Set pin is defined as the input that can cause Q to produce a 1 (5V) output. So one could swap Q and ~Q, but to be consistent with the conventions for SR flip-flops one would also need to swap which input is labeled S and which R. So like the stoplight it is a matter of convention. — wonderer1
Also, flip-flops themselves don't perform logical operations. They just serve as memories that can be used to provide inputs to logic gates (or combinations thereof), and store outputs from logic gates.
Latching relays require only a single pulse of control power to operate the switch persistently. Another pulse applied to a second set of control terminals, or a pulse with opposite polarity, resets the switch, while repeated pulses of the same kind have no effects. Magnetic latching relays are useful in applications when interrupted power should not affect the circuits that the relay is controlling. — wiki
Wait, I think i see what you are doing - treating each line as an equation, and then substituting the right back in for f. — unenlightened
You don't want to do that! Each line is a result for a combination of an and b. There is no working shown, and almost none to do. so for (2):– — unenlightened
and the re-entered f can be ignored.
We can now find, by the rule of dominance, the values which f may take in each possible case of a,b.
You're making the very simple, complicated. these are not equations, but evaluations of f for all the possible values of a & b. the right hand side in each case is the result of simplifying the left — unenlightened
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