What happens if the Forward voltage of a diode is not met?
The following is an example of a relaxation oscillator using the 2N6028 PUJT:







simulate this circuit &ndash; Schematic created using CircuitLab

In this circuit, \$R_4\$ and \$C_1\$ are the timing pair. I think these are doing what you hoped for. Current flows through \$R_4\$ and charges up \$C_1\$, over time. \$R_2\$ and \$R_3\$ form a voltage divider that sets the "trigger voltage" that \$C_1\$ needs to reach before "things happen." As \$C_1\$ charges up, it eventually reaches that trigger voltage and when it does the PUJT "fires" and dumps the charge sitting on \$C_1\$ into \$R_1\$ as fast as it can. In doing so, the current caused by this charge-dump causes a voltage to develop across \$R_1\$, which is then used to turn \$Q_1\$ on and saturate it -- just as you'd hoped.

No Arduino. No microcontroller. Just a basic timer. You can change the timing of the pulses by adjusting the value of \$C_1\$. At it is, it runs somewhere between \$3\:\textrm{Hz}\$ and \$4\:\textrm{Hz}\$. I've set \$R_{LOAD}\$ to a particular value. If you discuss what your load actually is, I might be able to help adjust the above circuit. But at least this gives you an idea.