2018A33 step up converter

If you are trying to add a charge pump, be sure that there is a charging path into the capacitor as well as a discharge path into the load.

As long as the power drawn in the main converter is steady, a single-ended switch should be able to support a low power parasitic charge pump.

An integrated switch might be affected adversely if the resulting current spikes trip internal protection.

RL

That's a lot of words. Can you post schematics?

Too late to cancel this post. I discovered that I rectified the negative swing. Soo embarassed to leave out the charge capacitor.

The IC is not drawn. (it connects A B and ground)

Vi Vo --------UUUUUU-------I>|----------+ A B | | | = | | ---------------------------------+ | _ - .

A --------II-----------|<I------+----+ | | | | - | A = | | | | -----------------------------+----+ | _ - .

Can this work? P.S. I have 10 of those tiny boards. The funny thing is I can remove the ic from one board, reverse the diode and replace the coil with a capacitor. What remain is add the extra diode over the capacitor and connect grounds.

Groetjes Albert

No, it won't work since A never becomes negative. And a cap in series with a diode never works. Try to find a copy of "The art of electronics" and do some studying.

This might work as long as the normal outputs are loaded (or else the drive to A disappears). It obtains a negative voltage from any waveform with enough amplitude.

A ---||-------|<------------- out | _|_ _V_ ___ | | GND --------------------------

Arie

Yikes. Ascii art.

Given the ususal non-synchronous buck switcher, series switch with a catch diode, you can get a modest negative voltage as I think you suggest, namely a series cap and a DC-restore circuit, sometimes called a "half wave voltage doubler."

Or you can add a winding to the inductor and rectify that.

One picture is worth a thousand words.

Thanks. I'll try it.

Groetjes Albert

I tried this and it generates a -4,3 voltage and less than 100 mV ripple. Combined with the original 5 V output. The 9.3 volt is ideal to simulate a 9V battery and indeed it worked to power a voltmeter.

Then it started to draw more and more current. Have I made a short circuit? I experiment on.

It gets interesting when the load is from V+ to V-.

It might help to put a bit of dummy load on the V+ side.

I rebuilt the circuit with other components, 10 uF electrolytic capacitor and 1N4001 diode. The symptoms are the same and diminish after a dummy load was applied. of 10 mA. Then the negative voltage is ca -4V.

Can you give a rule of thumb? If the current from -5 to +5 is X mA, how much should the dummy load be? (I do not understand much of the circuit, apparently.)

Groetjes Albert

I now use 1K (5 mA) summy load of the 5 V circuit. The goal was to use a 1.5 AA cells instead of a 9V battery. Using a voltmeter as a load:

The goal was limited, my voltmeter works. Thanks for the help.

This may not seem practical at first sight, however. There is 30 cm of 0.35 mm wire. Replace that with two times