Re: 2018A33 step up converter

On Tue, 17 Jun 2025 17:24:32 +0200, albert@spenarnc.xs4all.nl wrote:

In article <nnd$5972e5da$6d05f277@13e635c803c3f1a9>,
<albert@spenarnc.xs4all.nl> wrote:

In article <nnd$1620c1ab$7a0594d5@1745475ac2689b3d>,
Arie de Muijnck  <noreply@ademu.nl> wrote:

On 2025-06-05 14:15, albert@spenarnc.xs4all.nl wrote:

In article <p50p3khhstse6vfn0u4hepdagvv83c5enq@4ax.com>,
john larkin  <jl@glen--canyon.com> wrote:

On Sun, 01 Jun 2025 17:25:16 +0200, albert@spenarnc.xs4all.nl wrote:
>

I am experimenting with the 2018 step up converter.
The circuit is available from aliexpress , most everywhere.
>
There is a inductance, a Skottky diode, and capacitance, and the
circuit. The 2018 cuts the coil short to earth. The diode fill the
capacitor and the output is measured for a feedback.
Textbook step up. Rock solid 5 Volt.
>
Now I have this (apparently) wild idea. I add another diode reversed
and a capacitor, to prey on the output. The circuit is not phased at all,
solid 5.08 V as far is it original function.
But I get a couple of negative millivolt out.
>
What am I doing wrong? Or should this work?
I checked this over and over again.
>
Groetjes Albert

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

>
     The IC is not drawn. (it connects A B and ground)
>
     Vi                               Vo
    --------UUUUUU-------I>|----------+
                     A         B      |
                                      |
                                      |
                                      =
                                      |
                                      |
     ---------------------------------+
       |
       _
       -
       .
>
    Vi and Vo are 1.5 V and 5 V.
    The switch is connected to A (connect to ground)
    and the voltage is sensed at B.
    Now I want to add a negative voltage.
>
Now I want a negative voltage.
On second thought I came up with this addition:
>
    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 --------------------------

>
Thanks. I'll try it.

>
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.