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From: albert@spenarnc.xs4all.nl
Subject: Re: 2018A33 step up converter
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Date: Thu, 19 Jun 2025 16:46:40 +0200
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In article <o0435k51sa6bk58ppkt58d1o2qsmtq7n0t@4ax.com>,
john larkin  <jl@glen--canyon.com> wrote:
>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.

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.
1 mA is not sufficient.

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
>
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