# RL circuit short circuit issue

Was trying to solve the below circuit. But couldn't get any thought process going on.

Voltage source is square pulse alternating between 0 V and 5 V with period of 3 seconds.

So tried to simulate the circuit and i got the error as in figure. What might be the issue in this circuit?

• Do your inductors have series resistance? It might be complaining because DC resistance to ground is zero (just a guess, I'm not familiar with Spectre). – uint128_t Mar 6 '16 at 7:09

I assume you are using the inductors which are ideal I (i.e. from analogLib). For a DC solutions the inductors are replaced by shorts, therefore the simulator can't find a solution because you are shorting your source.

As pointed out by Neil_UK another problem is the loop of inductors by itself. Since the current is not determined this leads to another error. So there are actually two problems. Splitting the circuit one gets two error messages.

Fatal error found by spectre during topology check.
FATAL: The following branches form a loop of rigid branches (shorts) when added to the circuit:
V0:p (from 0 to 0)
L2:1 (from net4 to 0)


• it's not so much 'shorting your source', but note that the source is also zero impedance, so it's a loop of zero impedance. The solution is the same, put a sniff of resistance into any zero impedance loop, 1m$\Omega$ is plenty. – Neil_UK Mar 6 '16 at 9:31
• A loop of zero impedance wouldn't be a problem. But as soon as you short a source, you force two different conditions (voltage equal to zero and voltage equal to the source voltage). Therefore you get an error. – Mario Mar 6 '16 at 9:39
• Yes i used ideal inductors. Do inductors get short circuited under dc voltage also? What will be the nature of the circuit under t=0+ and at infinite time? Will the inductor get short circuited at t=0+? – Anjana Verma Mar 6 '16 at 10:21
• Even for a transient simulation you need a starting point, which is the DC simulation. So you can't get around that problem. But as Neil_UK suggested, just add a small resistor in series. There are no ideal inductors anyway. Less advanced SPICE implementations have even more restrictions so you should avoid too ideal circuits. – Mario Mar 6 '16 at 10:47
• @Mario no, I beg to differ, but I may be wrong, and an experiment on your favourite simulator would give the answer. It's the loop of zero impedance that's the problem, not one of them being a source. The reason is that such a loop makes the matrix that needs to be solved degenerate, therefore uninvertable. The experiment would be to connect a delta of inductors to a source with a resistor, and see what happens during the DC solve step. – Neil_UK Mar 6 '16 at 13:52