The voltage across a capacitor is the integral of the current through it. If you feed a constant current to a capacitor, its voltage ramps up linearly, which is exactly what you want for a sawtooth waveform generator.
Yes, you're correct that this cannot continue forever; the complete waveform generator circuit will be discharging the capacitor periodically in order to prevent the constant-current circuit from saturating.
In the circuit you show, D1, D2 and R1 provide a voltage reference, in this case, approximately 1.2V, which is the total forward drop across the two diodes. This establishes the voltage across the B-E junction of Q1 and R2. Since the voltage across the B-E junction is also a diode drop, or 0.6V, this means that the remaining voltage, 0.6V appears across R2.
From Ohm's law, we now know the current through R2, 0.6V/2200Ω = 273µA. This is the current that is charging C1.
The voltage across the capacitor is a function of time: V = I×t/C. Let's rewrite this as V/t = I/C, which means that the rate of change of the voltage is the current divided by the capacitance. In this case, 273µA/0.1µF = 2730 V/s, or equivalently, 2.73 V/ms.
The charge time of the capacitor is hardly affected by load resistance at all (except in extremes). The charging of the capacitor is determined by the resistance of the forward conducting diodes in the bridge rectifier - this resistance is likely to be around 1 ohm or less.
The discharge time of the capacitor is determined only by the load resistor which is probably many ohms hence charge and discharge times will be different.
A more exact answer would consider the leakage inductance of the transformer feeding the bridge and the cable resistance feeding the transformer but, in general, the equivalent forward resistance of the diodes in the bridge ensure the cap charges quickly (compared to the discharge due to load resistance.
Here's a picture that shows an applied AC voltage and a single diode charging up a capacitor: -
After a few cycles of AC the capacitor is starting to receive full charge. Remember the picture is a half-wave rectifier so the negative excursions of the AC aren't contributing. With a significant load resistance there will be a droop/decay on the capacitor voltage between conduction periods of the diode but, under normal circumstances the voltage will have a trend that is asymptotic with the peak AC voltage minus 1x diode volt drop of about 0.7 volts.
Here's a picture that shows the droop in the capacitor voltage between diode conduction periods. This is a full-wave picture and assumes the diode is perfect: -
Best Answer
LTspice calculates the DC operating point before starting the transient simulation. If you check the capacitor voltage at the beginning of the simulation you'll probably find that it is 1GV. The capacitor voltage after a very long time would, in theory, increase without limit with a 1mA current source feeding into it and no parallel resistance, so the DC operating point makes no sense in this case.
If you name the node at the top of the capacitor Vcap and add an initial condition: .ic V(Vcap) = 0 as a spice directive you'll get a more sensible answer.
You can also set the initial condition for currents in inductors, which is useful in the analogous situation where a voltage source is connected across an inductor (though LTspice tries to save you from this by inserting a hidden default resistance of 1m\$\Omega\$ in series with the pure inductance). But if you start off with a 1V source across an inductor you'll still end up with a constant 1000A flowing rather than the linear increase of current with time you might have been expecting.