When the simulator lies

Recognise the runs that return a confident wrong number (missing DC paths, coarse time steps, ideal sources, more than one answer) and fix each one.

Lesson: E1.6
Time: 25 min
Runs on: ngspice
Check: 5 questions

A bench drove an ideal 1 mA current source into a node whose only other part was a 1 pF capacitor. The operating point printed v(a) = 9999.95 V, ten kilovolts on a 1.8 V chip. The log had "singular matrix" warnings, the run finished without an error, and the next check used the number.

An ideal current source into a capacitorI1 pushes 1 mA into node a. The only other part on node a is C₁ = 1 pF, which carries no DC current.

Nowhere for the current to go
In an operating point the capacitor is open. Kirchhoff's law at node a says 1 mA = 0, which no voltage satisfies.
1 mA

Read the circuit as text

I1 current source (1 mA): + → 0, − → a
C1 capacitor (1 pF): pin 1 → a, pin 2 → 0

a = 9999.95 V

The pill is what op printed on ngspice 42.

SPICE does not refuse a circuit that makes no sense: it often returns a number anyway, and the warnings are part of the result.

No DC path: where the number comes from

A node with no DC path is joined to the rest of the circuit only through capacitors or a current source. In an operating point the capacitors are open, so nothing fixes that node's voltage. The equations have no single solution, and ngspice reports a singular matrix.

It does not stop there. It tries gmin stepping, which adds small conductances and removes them step by step, then source stepping, which ramps the sources up from zero. When both fail, it runs a short transient, the transient operating point, and takes its end as the answer.

That transient lasts 10 µs. A 1 mA current into 1 pF for 10 µs charges it to 1 mA × 10 µs / 1 pF = 10 kV. The missing 0.05 V is a leak equal to gmin, the solver's 10⁻¹² S minimum conductance: change gmin and the number moves exactly as that leak predicts. So 9999.95 V describes ngspice's fallback, and none of it describes your circuit.

Run a DC sweep on the same netlist and you get a different answer, va_dc = 1.000000e+09: 1 mA through the gmin conductance alone, 1 GV. Two analyses, two confident numbers, neither real.

Bench	What the log says	What prints	Fix
1 mA into 1 pF, `op`	singular matrix; gmin and source stepping failed	9999.95 V	a DC path: `R1 a 0 1meg` gives 1000 V
two 1 nF in series from a 1 V source, middle node	singular matrix; transient op	0.5000 V	a resistor to a defined voltage
NMOS with its gate left open	singular matrix; transient op	gate at 0.5005 V	drive the gate; `.options rshunt=1e12` shows 0 V
1 mA and 2 mA current sources in series	nothing at all	v(a) = −1.000 mV	real sources have resistance; add it
two voltage sources (1 V and 2 V) in parallel	Timestep too small; op aborted	nothing	one source, or a series resistor

Each line measured on ngspice 42. Only the last one stops; the other four print a number with no error.

The fix is always the same idea: give every node a DC path to a defined voltage. That can be a bias resistor, a real driver, or the output resistance a real source has. The middle-node voltage of two capacitors, 0.5000 V, looks plausible, which makes it the most dangerous number in the table.

A waveform is samples

TransientA 1 V sine, stepped every 2 µsA 100 kHz sine of 1 V amplitude, simulated with tran 2u 100u. Straight lines join the computed points.

Time 56.56 µsv(in) = -830.6 mV
The lowest computed point is −0.8306 V. meas MIN reports it, because meas only sees the points the solver computed.
Time 92.56 µsv(in) = 999.3 mV
The highest point, 0.9993 V, lies close to a crest by luck. Five points per period cannot draw a sine.

Read the plot as a table

Time	v(in)
0 s	0 V
20 µs	27.98 mV
40 µs	27.98 mV
56.56 µs	-830.6 mV
60 µs	27.98 mV
80 µs	27.98 mV
92.56 µs	999.3 mV
100 µs	-2.449 fV

Measured on ngspice 42: 59 time points in 100 µs.

A transient result is a list of samples joined by straight lines. Each sample is right: −0.8306 V is the true value of the sine at 56.56 µs. What is missing is the points in between. With tran 2u 100u the solver may step up to 2 µs at a time, which is a fifth of the 10 µs period. The run reports vmin = -8.305959e-01, and nothing in the log warns you.

Ask for a shorter step and the peaks return. tran 10n 100u gives −0.9999985 V and 0.9999985 V. So does keeping the 2 µs print step and capping the internal step with a fourth number, tran 2u 100u 0 10n, which gives ±0.9999968 V. Resolve the fastest thing you care about with ten to twenty points.

Aids that steer, and circuits with two answers

gmin, 10⁻¹² S by default, is also what ngspice places across device junctions to help convergence. It is invisible until a current is as small as gmin times a voltage; E2.6 shows it dominating a transistor's off current.

.nodeset gives the solver a starting guess for some nodes, then releases them, so it cannot change the answer unless the circuit has more than one. .ic is stronger. It holds nodes at the given values while the transient's starting point is solved, then lets go, so the run starts from a state you chose.

A latch, plain op

XP1 q qb vdd vdd sky130_fd_pr__pfet_01v8 W=1 L=0.15
XN1 q qb 0 0 sky130_fd_pr__nfet_01v8 W=0.5 L=0.15
XP2 qb q vdd vdd sky130_fd_pr__pfet_01v8 W=1 L=0.15
XN2 qb q 0 0 sky130_fd_pr__nfet_01v8 W=0.5 L=0.15

op prints v(q) = v(qb) = 0.8819 V, with no warning: the balanced point where neither inverter wins.

The same latch with a guess

.nodeset v(q)=1.8 v(qb)=0
* or the opposite guess:
* .nodeset v(q)=0 v(qb)=1.8

v(q) = 1.8 V and v(qb) = 0.62 µV; the opposite guess gives the opposite state. All three are valid solutions.

Two cross-coupled SKY130 inverters at 1.8 V, measured on ngspice 42.

A latch has two stable states and one balanced, unstable one. op returns whichever one its start leads to, here the balanced one, which a real latch never stays in. E4.5 shows the same trap in a bias reference that the operating point finds working and a power-up transient finds dead.

The rule for every case on this page: read the log, not only the numbers. A run that finishes is not a run that is right.

Asked in interviews

What does "singular matrix" mean in SPICE, and how do you fix it?
The circuit equations have no unique solution, usually because a node has no DC path to ground (only capacitors or current sources) or voltage sources form a loop. Give every node a DC path, add series resistance to ideal sources, and treat any number printed after the warning as meaningless.
Why can a transient waveform miss a peak?
The result is samples at the solver's time points, and measurements only see those points. A step that is long compared with the signal skips the peaks. Shorten tstep or set tmax so the fastest edge or period gets 10-20 points.
Can a circuit have more than one DC operating point?
Yes: latches, flip-flops and self-biased references have several. The operating point returns the one its starting guess leads to. Use .nodeset to pick a state, and a power-up transient to see which one the real circuit reaches.

Try it

Run it and open the log: the singular-matrix warnings, the failed stepping, then the transient operating point. The transient's starting point is 9999.95 V; the DC sweep's point is 1 GV.

Now add R1 a 0 1meg and rerun: both numbers become 1000 V, and the warnings vanish. Then change C1 to 1n without the resistor and predict va_tran from the 10 µs rule.

SPICEngspice 42 · simulator service

Edit the netlist, then run it

Expected measurements: va_tran = 9.999950e+03 va_dc = 1.000000e+09

Check yourself

Predict the outputThe same 1 mA source and 1 pF capacitor, now with 470 kΩ from node a to ground. Type the line meas prints (for example va = 9.999950e+03).

* 1 mA into 1 pF with 470 k to ground
I1 0 a 1m
C1 a 0 1p
R1 a 0 470k
.control
tran 1n 2n
meas tran va FIND v(a) AT=0
.endc
.end

ConceptWhat does a "singular matrix" warning say about the circuit?

The circuit has more nodes than the default matrix size allows, so the solver stops earlyA device model is driven outside its valid range, so its results are clipped to the limitsSome node voltage or source current is not fixed by the circuit, so no unique solution existsThe circuit oscillates, so no steady DC value exists

ConceptEach bench drives node b from V1 in 0 1. Which one has a node with no DC path?

C1 in b 1n and R2 b 0 1megR1 in b 1k and R2 b 0 1kR1 in b 1k and C2 b 0 1nC1 in b 1n and C2 b 0 1n

ConceptThe run status is OK and no line says "error". Why is that not enough to trust v(a) = 9999.95 V?

Because ngspice may fall back to aids such as a transient and report that with only warningsBecause ngspice prints at most four correct digits, so a value this large is only roundedBecause the run status only covers the netlist syntax, not the analysis that followsBecause the value has to be checked against a second simulator

ConceptOn the sine plot, the lowest point is −0.8306 V although the source amplitude is 1 V. What does that tell you?

R1 loads the source and clips its negative halfThe solver stepped past the trough: the computed points never land on itngspice limits a sine to ±0.83 V unless the amplitude is given in voltsThe source is asymmetric: its offset is about +0.08 V

Remember

A node with no DC path makes the matrix singular; ngspice falls back to gmin stepping, source stepping and a 10 µs transient, and prints a number anyway.
Read the log: warnings are part of the result, and a run that finishes is not a run that is right.
A transient result is samples; MIN, MAX and peaks are only as good as the time step, so cap it with tstep or tmax.
Ideal sources in impossible combinations either abort or print nonsense; real sources have resistance.
Some circuits have several DC solutions; op returns one, and .nodeset picks which.