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NNTP-Posting-Date: Sat, 08 Mar 2025 22:06:55 +0000
From: john larkin <jlArbor.com>
Newsgroups: sci.electronics.design
Subject: Re: dumper circuit
Date: Sat, 08 Mar 2025 14:06:50 -0800
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On Sat, 8 Mar 2025 16:46:14 -0500, bitrex <user@example.net> wrote:

>On 3/8/2025 3:03 PM, john larkin wrote:
>> On Sat, 08 Mar 2025 18:57:11 +0000, JM
>> <sunaecoNoChoppedPork@gmail.com> wrote:
>> 
>>> On Sat, 08 Mar 2025 08:57:43 -0800, john larkin <jl@650pot.com> wrote:
>>>
>>>> I'll have an isolated class-D amplifier powered by a dc/dc converter
>>> >from my big 48 volt supply. Simple push-pull, transformer, rectifier,
>>>> making maybe isolated 46.
>>>>
>>>> In one situation, the class-D amp can push current uphill into the
>>>> isolated supply, and we need to dump it. This circuit acts sort of
>>>> like a 25-watt zener diode.
>>>>
>>>> https://www.dropbox.com/scl/fi/98n07mnf3qokvbewuv1oy/Dumper_1.jpg?rlkey=poiep8hzhz33qecswdj50sdne&raw=1
>>>>
>>>> It didn't work in simulation. It was driving me to despair. No amount
>>>> of chocolate would help.
>>>>
>>>> The problem was of course that the Spice initial condition simulation
>>>> perfectly biased everything, balanced the pencil on its point.
>>>>
>>>> So how can the initial conditions sim ignore the positive feedback?
>>>
>>> Use .ic together with uic.
>> 
>> It sims fine if I click the box to skip the initial condition
>> solution. That's faster too. Delaying the power supplies breaks the
>> tie too.
>> 
>> What I don't understand is how negative feedbacks work during the ic
>> solution, but positive feedbacks don't.
>> 
>
>
>Strongly non-linear ODEs and systems of ODEs are more likely to have 
>singular solutions i.e. parts of the domain where the initial value 
>problem fails to return a unique answer...I'm not an expert in how this 
>translates to discretized/numerical solutions but I believe the same 
>sort of thing applies, a system of finite difference equations 
>describing a positive-feedback circuit is more likely to be strongly 
>non-linear than one for negative feedback.
>
>Sometimes an initial value problem is only troublesome at a single point 
>like t = 0 and if you move off that point the problem is well-posed.

Version 4
SHEET 1 880 680
WIRE 272 -16 208 -16
WIRE 336 -16 272 -16
WIRE 336 0 336 -16
WIRE 336 96 336 80
WIRE 208 112 208 -16
WIRE 80 128 32 128
WIRE 176 128 80 128
WIRE 272 144 240 144
WIRE 288 144 272 144
WIRE 304 144 288 144
WIRE 176 160 128 160
WIRE 208 208 208 176
WIRE 128 240 128 160
WIRE 272 240 272 144
WIRE 272 240 128 240
WIRE 32 288 32 128
WIRE 208 288 32 288
WIRE 272 288 208 288
WIRE 32 320 32 288
WIRE 272 320 272 288
WIRE 32 432 32 400
WIRE 224 448 208 448
WIRE 272 448 272 400
WIRE 272 448 224 448
FLAG 208 208 0
FLAG 336 96 0
FLAG 32 432 0
FLAG 272 -16 +10
FLAG 224 448 +10
FLAG 80 128 N
FLAG 208 288 P
FLAG 288 144 U
SYMBOL OpAmps\\UniversalOpAmp2 208 144 R0
WINDOW 0 -63 48 Left 2
SYMATTR InstName U1
SYMBOL voltage 336 -16 R0
SYMATTR InstName V1
SYMATTR Value 10
SYMBOL res 48 416 R180
WINDOW 0 -52 67 Left 2
WINDOW 3 -58 38 Left 2
SYMATTR InstName R2
SYMATTR Value 10K
SYMBOL res 288 416 R180
WINDOW 0 -46 69 Left 2
WINDOW 3 -54 39 Left 2
SYMATTR InstName R4
SYMATTR Value 10K
TEXT -176 304 Left 2 !.tran 1
TEXT -224 208 Left 2 ;Opamp Startup
TEXT -224 248 Left 2 ;JL  Mar 8  2025


Run that for 1 second and look at node U.

Repeat but run for 10 seconds.

Then run for 1000 seconds.