I want to try this circuit from https://www.electronics-tutorials.ws/waveforms/555-circuits-part-2.html

This simple 555 voltage doubler circuit consists of a 555 oscillator and a single capacitor-diode voltage doubler network formed by C3, D1, D2 and C4. This voltage doubler circuit multiplies the supply voltage and produces an output that is approximately twice the voltage value of the input voltage minus the diode voltage drops.

When the output at pin 3 is LOW, the 50uF capacitor (C3) charges up to the supply voltage through diode, D1 with diode D2 off. When the output from the 555 goes HIGH, the voltage across C3 discharges through diode D2, as D1 is reverse biased, adding its voltage to the source voltage as VCC and C3 are now like two voltage sources in series.

The timing cycle from the 555 changes state again from HIGH to LOW and the cycle repeats once again, thus producing a DC load voltage which is twice the original input voltage, that is a multiplication factor of two (voltage doubler). Then a 555 voltage doubler circuit can produce an output voltage from about 10 to 30 volts at very low current.

Another point to note is that the frequency of oscillation of the 555 astable multivibrator used to generate the square-wave input signal will determine the value of the capacitors used, as they along with the connected load value create a RC charging/discharging circuit to filter the output voltage. Too low a capacitance value, or too low the frequency of oscillation will produce ripples in the output voltage waveform and therefore a lower average DC output voltage.

With no-load connected, the output voltage will be twice the 555's original supply voltage. The actual output voltage will depend on the value of the connected load, RL and the load current, IL. As given, the 555 voltage doubler circuit above can supply about 30mA at the rated voltage.

There are many variations of the voltage doubler circuit above, but each one uses two diode/capacitor pairs to provide the x2 multiplication factor. By adding or cascading more diode/capacitor networks to the voltage doubler, we can create circuits which can create voltage multiplication ratios as high as we want.

So for example, by adding half a diode/capacitor combination to the 555 voltage doubler circuit creates a voltage tripler circuit with a multiplication factor of x3, and adding a second full diode/capacitor section to the 555 voltage doubler circuit will create a voltage quadrupler circuit with a multiplication factor of x4, and so on as shown.

Problem: The highest value bipolar capacitors in my inventory is 10 uF. In another circuit I see electrolytic capacitors being used (with the + as in the above image) so I guess I will use those.

The above image has an error; C2 should be 10 nF.

After building only the left part of the circuit I measured the frequency at 1.5 kHz.

After finishing the circuit I measured the voltage over the load resistor (R3), but got a steady 8 V DC...Nope! the C3 to the 555 was loose. After connecting the output is a 14.8 V DC.

Even better, when decreasing the supply voltage to 5 V the output is around 7.3 V which is near perfect for a project where I power a radio from a 5 V USB powerbank.