sawtooth wave form generator using Hspice
You have the option of including UIC on the .TRAN card if you don't want the Spice program to perform an &quot;initial transient solution&quot; step (so-called &quot;ITS&quot;) to find the DC solution at \$t=0\$. If you don't use UIC, then Spice will proceed with the ITS step and this will set the initial conditions for the energy storage devices before it starts performing the transient step. On the other hand, when you use UIC, then the initial value of every single energy storage (voltage and current) device is treated as zero, except for those which are explicitly provided using the .IC card.
However, I'd like to use this opportunity to illustrate a more managed design. Sometimes, it helps to see them. The following circuit shows such a design:



simulate this circuit &ndash; Schematic created using CircuitLab
I've changed everything, basically. The current source has been replaced with a more manageable design topology. You can find much more on the topic here, if interested. But it's a better arrangement. Also, the &quot;SCR&quot; has been replaced with a much better and more consistent (managed) SCR behavior, including positive feedback through \$C_2\$ and \$R_7\$ to lock in the event in a reliable fashion. \$D_1\$ replaces a diode-connected BJT, with very different saturation current and emission coefficients, and forms up with \$Q_5\$ to create a current mirror with a current gain very much less than 1 (about 0.01 or less.) This is really all that is needed to drive \$Q_4\$. \$Q_3\$ is the voltage trigger for the SCR.
I've leave it there, but it's worth seeing what it takes to build something that copes with the vagaries of discrete part and temperature variations.
Afterword
I've just placed all this into LTspice, today. Here are the results of a run:

The current source is about \$3.6\:\mu\text{A}\$ (\$Q_2\$'s \$V_\text{BE}\$ is about \$540\:\text{mV}\$.)
I had originally guessed that \$Q_3\$ would trigger at about \$V_\text{BE}\approx 500\:\text{mV}\$ and that \$D_2\$ would have about \$200\:\text{mV}\$ across it at the time. So this worked out to a trigger voltage of about \$3.5\:\text{V}\$. As it appears, it's closer to about \$2.9\:\text{V}\$.
But I'd call it &quot;close enough.&quot;