Electronic – Blowing resistors in parallel, in an order-controlled way


I'm supposed to make some questions, to gain some badges (just the excuse), so I'll throw one I've always been very curious about.

Imagine I want a set of resistors arranged in parallel to blow in a sequential and order-controlled way, just for me to admire the show, or to share it with someone else.

Look at this schematic:


I want to blow as many resistors in the set {R1, R2,… RN} as possible, as I said, in an order-controlled way. First R1, then R2, etc. I don't want to blow Rs. We can choose the values for Vs, Rs, R1, R2,… RN, the power ratings for each resistor (let's call them Psmax, P1max, P2max,… PNmax), and the maximum current Ismax that the source is able to provide. Also, assume that a blown resistor is always an open circuit.

Let's call M to the number of resistors (out of those N) that will eventually be blown.

Question: How would you choose those values, to maximize M?

I see two cases:

1) Mathematical "world", with unbounded parameters, and even making unreal assumptions such that a resistor does not blow for P < Pmax, and blows for P >= Pmax. I'm not interested in this one (because it is clear that there are infinite solutions, and with M=infinity).

2) Real-world case, with feasible values for all those parameters, and with the real thermal behavior for the resistors. This is what I'm interested in.

I know that this is a relatively complex question, and with little practical use, but I'm still curious about it, as a mathematical/engineering challenge. Aren't you? Just take your time.

Edited: Actually, let's bound Vs, so that we don't end up with HV generators. Since Olin already used 12 V in his example, let's fix Vs=12 V for all of us. Also assume a value of Ismax=100 A.

Best Answer

If the resistors are all the same package and wattage, they should blow in order of high to low abuse. In this case abuse would be dumping too much power thru them. The power dissipated by a resistor is V**2 / R. Since the resistors are in parallel and V therefore the same for all, those with smaller R will suffer proportionately higher abuse.

So, arrange them in order from low to high resistance. The existance of Rs will cause the voltage accross the resistors to go up each time one pops, hastening the demise of the next one in line. This also means you should calculate each value so that it dissipates the necessary power to pop with all previous resistors open. Note that Rs needs to be quite beefy so as to not itself pop.

Let's say you have determined that 1 W dissipation will cause the desirable popping in the types of resistors you plan to use and that Vs is 12 V (a car battery would work well as it is a good voltage and can easily handle the power). Let's also say that when only the last resistor is left, Rs drops 1 V.

To calculate the cannon-fodder resistors, work backwards from the last. When only the last resistor is left, it will have 11 V applied to it. Since we want 1 W dissipation, the resistance in Ohms will be the square of the Volts applied to it, which is 121 Ω for the last one. This also tells you that Rs must be 11 Ω.

Now you can calculate the value for the second to last resistor. The Thevenin equivalent it sees is 10.08 Ω and 11 V. So the question is what resistance connected to that Thevenin source dissipates 1 W? The equation is a quadratic, which I'll leave for you to solve. Once you have that resistance, you can calculate the Thevenin source the next resistor sees and repeat the process as far as you like.