Branching from the comments here, TimWescott gave this example of an open-loop unstable electrical system:

Find the nearest NPN power transistor. Put it in a circuit with a healthy (for it) voltage on the collector, and bias it with a fixed voltage to flow a healthy (for it) current from collector to emitter.

In the absence of a truly massive heat sink, as it flows current, it'll heat up. As it heats up, it'll flow more current. That'll make it heat up more. While the precise dynamic equation is both complicated and nonlinear, if you simplify it and linearize it then the dominant behavior will once again be an unstable 1st-order response of \$e^{+at}\$.

As far as I know, there are two types of stability: BIBO, which considers a bounded input and checks if the output is bounded too; and impulsive, which consider an impulsive input and checks what happens to the output when the impulse is gone. For example, according to what I read a capacitor is BIBO unstable with a direct current input, since the output voltage isn't bounded; but impulse stable, since when the impulse is gone, the output doesn't diverge.

The transistor example seems to me to refer to BIBO, since the bias voltage is kept applied instead of being impulsive, so I was wondering if an open-loop voltage-controlled electrical system could be unstable also with regard to impulsive stability, at least from a mathematical point of view (meaning that - quoting TimWescott - "a strictly mathematically stable system can still be 'unstable' for all practical purposes", if it comes to rest after breaking).

In short, what I'm asking is, can an open-loop voltage-controlled electrical system be unstable with regard to impulsive stability? And, if there are more instability types in addition to BIBO and impulsive, what kind of instability can exist in such a system?

When referring to open and closed loops, I'm referring to what (in my experience) is taught in automatic control classes, i.e. open loop and closed loop. Maybe it'd be better to phrase it as "stability for open/closed loop control systems, as seen in those block diagrams".

  • \$\begingroup\$ It's not actually clear what you're asking. Perhaps edit your question, summing it up with a clear and simple question, or set thereof? I could assume that you're asking "what sort of instabilities can exist in electronic circuits?", or perhaps "can either BIBO or impulsive instability exist in an electronic circuit?" -- but it's not clear to me. \$\endgroup\$
    – TimWescott
    Commented Nov 17, 2021 at 2:46
  • \$\begingroup\$ @TimWescott I edited the question, thanks. \$\endgroup\$
    – Mauro
    Commented Nov 17, 2021 at 9:00
  • \$\begingroup\$ @Mauro , see if I understand the very question: Is BIBO only criteria for impulsive stability of "open-loop voltage-controlled electrical system" ? What are other stability criteria for "open-loop voltage-controlled electrical system"? \$\endgroup\$
    – jay
    Commented Nov 17, 2021 at 13:55
  • 1
    \$\begingroup\$ @Mauro, so.... I am not answering your question, neither asking you the question, but I was asking you .... did I understand your question there; you want to learn more about stability criteria of "open-loop voltage-controlled electrical system". Right or Wrong? \$\endgroup\$
    – jay
    Commented Nov 17, 2021 at 14:13
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    \$\begingroup\$ @sarthak: whether a circuit's model can disobey the law of conservation of energy depends on the rules one follows in composing it. For an abstract circuit, I can put in dependent voltage or current sources at will, and the energy is just magically there (and energy conservation is violated). For an actual circuit, I can either take the power rail as just being magically there (and again, energy conservation is violated), or I can take every wire connected to the circuit to be an input -- and energy is conserved. It's a matter of definition, which is my point below. \$\endgroup\$
    – TimWescott
    Commented Nov 18, 2021 at 19:46

3 Answers 3


The basic problem here is that your question, as stated, is unanswerable. So rather than try to answer it, I'm going to show you why.

The reason it's unanswerable is because you're trying to map abstract mathematical concepts onto real physical systems, and they just don't fit.

So I'm going to rephrase your question in a variety of ways and show you how one can, simply by redefining the mapping from abstract to real, that one can get whatever answer one wants.

Are there any systems in the known universe that can be experimentally proven to be unstable for either the impulsive or BIBO instability cases?

No. Because the universe is finite in both physical extent and energy content. So there are no systems that have or can have infinite extent, energy, luminosity, or whatever. Nothing physical matches the abstract mathematical definitions.

This leads to an interesting and very practical problem -- how do you define "instability" for practical purposes? Engineers don't usually hang out at the water heater discussing this, because for practical purposes if it grows without any bound except that it breaks itself, or puts itself into an unusable operating mode, it's "unstable".

So a motor whose position grows without bound until it smacks into a stop, breaks the stop and destroys the thing its in is part of an "unstable system", even though once the smoke clears and the bits stop bouncing on the floor, everything is bounded in a mathematical sense.

Ditto, an amplifier that has a sinusoidal mode that grows without bound until the amplifier overloads and then the system sits there in a perfectly stable in nonlinear dynamics terms limit cycle is, for practical purposes "an unstable amplifier".

OK, smart@**, limit your answers to open-loop systems.

"Open loop" is a mathematical abstraction. Any physical object in the universe is held together by particle interactions, and these interactions operate by a form of feedback. So if you drill down far enough, there is nothing made of matter that doesn't, at some level, operate in "closed loop". Electrical systems are made of matter, ergo, they all have some sort of feedback.

Even limiting the question to electrical feedback leaves us with a full, practical expression for something mundane like, say, an actual resistor -- an actual resistor has capacitance, inductance, resistance and feedback between these effects (or "elements", if you don't like the word "effects"). Any object that interacts with electrons has the same property, if you drill down far enough. Electrical systems, by definition, interact with electrons, so they have some sort of feedback.

So, for practical purposes, you have to define what you mean by open-loop. If you're designing a control system at the block diagram level and you have a block with a transfer function in it, then at that level of the diagram that block, in itself, is "open loop". It may stand for a physical part that has a dozen intentional feedback loops in it (and is held together physically by another practical infinity of "feedback loops), but for that block diagram it is open loop until you wrap a loop around it.

You linked to two block diagrams that exemplify "open-loop" and "closed-loop". Remember, block diagram designs can be hierarchical. So if I am buying a motor with a built-in controller from you, and all I care about is getting a certain RPM out of it, then my block diagram is "open loop" -- I issue commands, the motor does what it does. On the other hand, your block diagram, as the designer of the motor/controller system, is closed-loop, because your part of the design has feedback (probably multiple feedback paths, in fact). So, when I tell my boss that "the system is open loop" I'm right, in a certain sense -- and when you tell your boss "the system is closed loop" -- you're right.

Which boils down to: the terms "open loop" and "closed loop" only have local meaning. You simply cannot point to a physical system and make a global statement "that's open loop" (although arguably you can always say "that's closed-loop"). You can only say "I can accurately analyse this system as if it were open loop", or "in order to accurately analyze this system I must take into account its feedback".

Yes, but I meant open loop and voltage controlled electrical systems.

What do you allow inside this electrical system? How do you define what is a control, and what is a power connection? Unless you're designing a power supply, your circuit analysis will take the various power supplies as given, not as "inputs".

So in the example you cite, the bias to a transistor isn't a control input, it's a bias supply that's derived from the power supply -- it's just a given.

If you exclude exterior power supplies, what about batteries? Put a battery inside of something, put a latching relay on the interior power line, and an impulse on the "on" lead will turn the thing on, and then whatever is inside will be stable -- or not.

If you define an "electrical system" as something that starts with no stored energy (i.e., no batteries), and whose power feeds are designated as inputs, then by that definition the system must be stable to impulses. Further, for any real system it'll be BIBO stable, first, because see discussion about the finite nature of the universe, above, and second, because capacitors leak.

If you define an "electrical system" as something for which the power inputs are assumed, it may be autonomously unstable, without needing an input at all.

My real point is you must define how you are mapping from the abstract to the real. Moreover, you must define what you mean by "unstable". For each combination of those mappings you get different answers. In the original question you asked two different engineers, and got two different, and opposite, answers. That's because each one of those engineers had a set of mappings in their head, and considered that mapping to be the one and true mapping.

  • \$\begingroup\$ Thanks, I think the part about "something that starts with no stored energy" is what I was looking for; just to be sure, and not sure if this helps, about "open loop" and "closed loop" I was thinking at what (in my experience) is taught in automatic control classes, i.e. open loop (electronicscoach.com/wp-content/uploads/2019/11/…) and closed loop (elprocus.com/wp-content/uploads/…). Maybe it'd be better to phrase it as "stability for open/closed loop control systems, as seen in those block diagrams"? \$\endgroup\$
    – Mauro
    Commented Nov 17, 2021 at 16:38
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    \$\begingroup\$ @Mauro: this is stackexchange, where if you want to change your question you edit your question. If you feel that I missed the point of your question, put those references into the question (and don't feel you need to call out that you've edited the question -- Stackexchenge = not a normal forum). \$\endgroup\$
    – TimWescott
    Commented Nov 18, 2021 at 18:31
  • \$\begingroup\$ @TimWescott I edited as per your suggestion, thanks. I think you could have answered me, specifically in the part about defining an electrical system as something with no stored energy at the start, but I noticed that I never said I was referring to those control systems and block diagrams, so I added those reference to be sure. \$\endgroup\$
    – Mauro
    Commented Nov 18, 2021 at 19:19

As far as stability goes, there are many ways to analyze physical systems to understand their stabilty. There are linear methods and nonlinear methods. Either way you need to come up with a set of equations or model to describe what the physical system will do over time.

With linear analysis methods we want to come up with a set of linear equations (or linear differential equations) because they are easy to analyse. Usually this also means using Fourier transformations to analyse the systems properties like gain and phase to test for stability.

BIBO analysis depends on the system able to be modeled as a linear system* and is not a way to linearize a system. Since a transistor itself is not linear, BIBO could not be used on only a transistor. We can however create circuits that function linearly across a range of inputs and outputs (like when we put a transistor into saturation). First we must show that the circuit behaves linearly across a range, then we can assume that we can apply math that is linear (almost no physical systems are linear across their entire range, most all analog systems saturate at some point).

*In addtion to linearity the system has bounds \$ \lvert y(t)\rvert \leq B \$ notice that the bounds will be centered around zero.

But you don't have to use linear methods to prove stability, there are non-linear methods that are much harder mathematical tools to use to prove that the circuit is stable and you can use nonlinear functions or differential equations and still prove stability.
But what Tim is talking about is temperature on transistors and runaway thermal effects and not much to do with open loop gain. Temperature would be more of an error factor in the equation So on to the question.

Edit: what I'm asking is, can an open-loop voltage-controlled electrical system be unstable with regard to impulsive stability? And, if there are more instability types in addition to BIBO and impulsive, what kind of instability can exist in such a system?

An ideal voltage controlled voltage source or voltage controlled current source? No, they are impulse stable because they are linear. You put an impulse in and you get a scaled impulse out.

Or if the system is some kind of filter then you could put an impulse in and see that it decays.

As far as the transistor goes thermal runaway is unstable in the sense that the physical plant or controller is changing. Typically we want a controller or plant that does not change or has minimal change during operation, and including an exponential term in a linear equation will make it nonlinear. There are a few options: One is to linearize the exponential. The other is to make the physical system not be subject to temperature effects so the temperature effects don't even need to be included in the analysis.

If you are an engineer, you can either increase the complexity of the model OR you can make the physical system fit the model.


As I understand, you are interested in finding out if an open loop system which is unstable with respect to impulse stability exists in electric circuit theory. You are not interested in going deep into the Physics or the question of your model accurately representing the real world.

The answer is yes, there do exist systems that become unstable when subject to an impulse. In my hometown, this happens sometimes when lightning strikes the electric pole that supplies power to the apartment I live in. For our purpose, the electric power supply network can be treated as an open loop system because it supplies power in one direction, and does not take feedback. When lightning strikes, the power gets cut. One might wonder what could be the reason for the power getting cut. This is a separate question that fortunately, has some interesting answers here: Why do thunderstorms knock out power?.

One more example, consider the case of a circuit breaker. It prevents a device from blowing up. In the absence of a circuit breaker, an impulse input such as high current or high voltage destroys an open loop system such as a toaster, making it unstable.

How do we write the mathematical model of such a system?

$$F(t,x) = H(t,x), \forall x < x_0$$

$$F(t,x) = Not\hspace{0.1cm} Defined, \forall x \geq x_0$$ where,

  • \$F(t,x)\$ is the transfer function of the system
  • \$H(t,x)\$ is an open loop transfer function
  • \$t\$ represents time
  • \$x\$ represents input voltage/current
  • \$x_0\$ represents a threshold value of the voltage/current

Even if \$x > x_0\$ for a very short duration, i.e. an impulse input, the system will become unstable.


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