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passive inverter circuit with one short pulse

ratstar

Aug 20, 2018
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I know you guys prefer doing inverters with transistors, but this one is actually pretty good, I can actually invert half the resistance with double the resistance so it could be useful, And its actually because there is a capacitor, the charge only partially turns off, but it gives the capacitor another absolute resting position, and it dolly's back and forth fine, which u see on the back2back LEDS.

The capacitor has a discharging loop, and a diversion wire which activates when the capacitor is fully charged, when the capacitor is charging the diversion wire is off.

So it has 2 outputs, to use from it, one when charging, one when discharging. Ive got a feeling im close to getting this stuff to work soon!!! =)

125465745_1354068558274531_8688128959368009697_n.jpg
 

Harald Kapp

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You don't expect us to decipher your circuit from that blurred image, do you? Please, please learn to draw legible schematics!

it dolly's back and forth fine, which u see on the back2back LEDS.
Nothing to see in your image.

What the heck is a "diverter"? In all my years dealing with electronics I've never heard of such a device.
 

ratstar

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125200952_1354086894939364_3398839055527881232_n.jpg


The diverter is just an extra channel it has that activates when the capacitor is full. Im pretty sure itll be in the finished design for this thing im making, but nothings totally certain yet...

every time you close the switch, you get a discharge then a recharge when u release. (so if the switch was a cap, it would discharge then charge when the cap on the switch blocked up.) [edit] no thats actually wrong, but i hopefully have a trick to get it to happen.[/ed]

[edit] i had to repost a few times... did the schem wrong, should be ok now. [/edit]
 
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Harald Kapp

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A rather good attempt at a schematic, congrats.
However: connection dots are missing. So we cannot see whether crossing lines/wires are connected or just crossing without connection.

You used a switch. That is, you introduced a kind of an active component where the power comes from you when activating the switch. You'll be getting nowhere near the speed range you envisioned in your other posts (MHz to GHz) with a manual switch.

Anyway, I have no clue what this circuit is supposed to do.
 

ratstar

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thanks! I might be in luck soon... just gotta keep on screwin the components together, and im racing around doing a million diagrams, i get them all wrong so much, but im starting to get a bit more correct with the resistances and wire positions, but too early to say.
 

Harald Kapp

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Getting better, but:
upload_2020-11-16_10-18-40.png
Dots should be at every connection. If in the far far future you want to use a CAE program, it is essential that you place dots in these position, otherwise the program typically will assume unconnected nets. The result will not at all resemble what you expect, be it a simulation or a pcb layout.
 

ratstar

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If that dot isnt there on those T-Junctions in the cad program, I can guess that your right there for the simulation results.
If I get my "big thing" Im going to test it out on all the simulators and see if they can take unorthodoxy.

I have actually got a sim to work (In Falstad.), as you wouldnt believe I suppose, for an analogue subtractor with just resistors, And I did get current where I thought it would be, but who knows if its right or not. Only real life will never lie to you, as long as there isnt the supreme ghosts around, then its all nonsense.
 
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73's de Edd

Aug 21, 2015
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The revered members of the Illuminati, know that is supposed to be a 100 ufd E-capacitor, being installed with proper polarity , but you need to define it as such by use of additional proper mark- up.

73's de Edd. . . . .
 

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Thanks 73's de Edd, for the Ed. got the junctions in - is the cap in the right way around? :)
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