What actually happens when both 1 input is given in RS flip flop circuit (physical change)?

Question

This question is somewhat similar to:

Why is S=1, R=1 state forbidden in RS flip flop?

But I am asking what actually happens if both 1 input is given forcefully in R and S terminal of flip flop circuit. Will there be any physical change? Will the flip flop circuit burn or get damaged? What will happen?

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In the original question, it is written that:

But if you set both R and S to 1 we have that Q = 0 and \$\bar{Q} = 0\$ at the same time. This contradicts the relation \$Q = \bar{Q}\$. In the real world one of the gates will reach the 1 state first and the result will be unpredictable.

This is true theoretically. Both the output will be 0 which is not possible as both the output are complement to each other. So, is it correct to write the truth table of flip flop circuit in the following manner(here):

instead of writing \$Q = \bar{Q} = 0\$ when R = S = 1. This can be written as output cannot be determined as this case is not possible. We cannot say which output will give 0 and which will give 1. This is mentioned here (from the same link):

R = 1 and S = 1: This condition violates the fact that both outputs are complements of each other since each of them tries to go to 0, which is not stable configuration. It is impossible to predict which output will go to 1 and which will stay at 0. In normal operation this condition must be avoided by making sure that 1's are not applied to both inputs simultaneously, thus making it one of the main disadvantages of RS flip-flop.

It is written that this case must be avoided. What worse would happen if it is not avoided? Will it get damaged or burned?

Answer 1

This is an RS flip flop made from NOR gates

^{simulate this circuit – Schematic created using CircuitLab}

We note that both gates are symmetrical, so there's no need to figure out what both gates are doing.

Each gate is basically an OR function, that generates an output TRUE when either or both inputs are TRUE. If R is '1' or TRUE, then the output will be TRUE. It's an inverted output, so output TRUE is '0'.

So there's your answer. With a NOR-based flip-flop, when both R and S are '1', both 'Q' outputs are '0'. Perfectly predictable.

No problem, unless you insist that the Qs are complements of each other. They are, for at least one of R and S being '0', the normal or expected mode of operation of the flip-flop.

If you make the false assumption that the outputs are always complements of each other, then the R,S='1' state violates that assumption. If you had logic following which basically said if Qa==Qb, then ignite Doomsday Bomb, the consequences could be very serious. But the NOR gates are perfectly happy, doing exactly their logic thang without problems.

Where life does get unpredictable is if we take R and S back to '0' simultaneously after they've both been at '1'. They will 'race' back to a '01' output, the winner getting to '1' first, with the other settling for '0'.

In the ideal case of equal delays, it will be unpredictable which will win. Given the likelyhood of small differences in delay, either in the gates themselves, of the logic driving the R and S inputs, one will tend to always win.

In the unlikely case of very closely balanced delays, the outputs may go metastable, which is both outputs go to a mid-rail voltage, and stay that way for an unpredictably long time, which could exceed their rated propagation delays by a factor of several.

Answer 2

As long as R and S are both 1, both Q and Q' will be 0. If one of R or S drops back to 0 before the other, the flip-flop will start acting normally again.

But if R and S fall back to 0 simultaneously, both Q and Q' will be 1. But Q and Q' are fed back into the flip-flop as a second R and S causing the process to repeat.

The flip-flop may start to oscillate between Q = Q' = 0 and Q = Q' = 1 due to the propagation delay until/unless there is some drift that will make it finally latch into a valid state. Or it could find an equilibrium where both the NMOS and PMOS transistors are partially conducting, i.e. it will burn.

^{simulate this circuit – Schematic created using CircuitLab}

Answer 3

The simple answer is that nobody really uses real SR flip-flips when coding HDL designs for FPGA's or ASIC's because its an asynchronous loop and will lockup the logic simulator with an infinite recursion loop. Instead, they emulate an SR-flip-flop by inferring a D-flip-flops connected to a simple state machine. In this case, its up to you how you implement the next state if S & R are asserted at the same time.

Of course the above doesn't apply if you are building a "real" SR-flip-flop out of TTL gates on a bread-board, or directly connecting transistors together to build logic gates.

Answer 4

Nah, it is as simple as results will be unpredictable. That's it. And it must be avoided just because we don't uncertainties in our logic. In fact, J-K latch (not J-K flipflop, despite lots of conflicts between definitions of latch and flipflop, I called unclocked devices as latch) toggles between one and zero continuously if both its J & K inputs are held high with time delay equal to propagation delay of the latch. (I still doubt if JK latch practically exists in real world because I don't see any practical meaning of having it, but one of my books mentioned it (and book is considered quite standard) & also provided output waveforms so I mentioned it.)

As JK latch is just RS latch with feedback, I don't think helding both inputs high causes any physical damage to gate. Thinking otherway, doing so doesn't causes any excessive current flow, so there is no source of energy for heat, so I don't think it will burn the gate. However, diagram showing internal circuitry of particular gate may finalise my answer.