Feedback To Source

[nul-points]

hi Jesus

Puerto Rico must be a very nice place to live, i'm very envious!


the circuit i posted above is just one suggestion to try - none of the values should be critical, so it would be good to see if you have any components already which might work in it, just to try it out - then, if it looks promising,  get some different values to see if you can improve performance

the inductor could be wound with, say, around a  hundred turns of 0.45mm magnet wire on a 6mm ferrite core

the diodes could be any type, eg 1N914, 1N4148, etc, but schottky will give lower losses

the decoupling cap could be in the range 0.1 to 0.47uF, say (higher value is better for lower losses)

the storage cap value also is not critical for a test, but is best chosen to suit the current drive required by the 1.5-to-12V inverter circuit (100uF might be sufficient, but i think its possible that 1000uF, 4700uF, or even 10,000uF would be better - this could be checked in different test runs)

since the storage cap will be operating at around 1.5V then a 3V or higher rating should be ok, but if the spike voltages are several tens of volts then it would be better to use say 20 or 30V WKG (or higher, if available & not too expensive)

higher value/rating storage caps should have lower internal losses but the physical size becomes an issue

hope this gives some guidance on options for using components you may already have - ideally you want to see how much difference the feedback to source makes to the battery draw, before spending too much money on new components

let us know how the testing goes

all the best
np


http://docsfreelunch.blogspot.com

[nievesoliveras]

A question @nul-points.

Does the circuit you posted gives a signal and stops? Because the problem is not that the motor stops. The problem is that the transistors stay sending the signal for too long.

If we use the fact that just by touching the positive pole with a wire from the charging diode, it charges the battery and create a circuit that do that for us, it will be runing and doing some work and keeping itself charged with the automatic touch the circuit provides.

I repeat, a circuit that mimics the wire touching to the positive AA pole is what we need.
Saddly I need to send for the LT ic again because when I left the wire connected accidentally, the coil began to give a high pitch sound and the neon was very bright for fractions of a second and the LT ic is gone.
I have had a lot of losses on my feedback to source quest.

Jesus

[nievesoliveras]

Is it this what you mean?

Jesus

[nul-points]

<<double post removed>>

[nul-points]

hi Jesus

>> "Is this what you mean?"

no, that's not quite correct...

instead of your previous method of connecting the transistor collector to the battery via the 1N4007, my earlier post is suggesting that you use the extra components

if you have smaller caps in the range i suggested for C2, try them first before trying largest value - 10,000uF might be too big physically (& in cost)

the circuit i posted above only replaces the 1N4007 between the collector of the transistor and the 1.5V battery

i included your existing transistor/motor connection on the left, and your existing 1.5V battery on the right to show you where to connect the new circuit - they should still be connected to the rest of your circuit as you showed earlier

all the rest of your circuit should stay exactly as it was in your previous post

<<EDIT>>
i've just seen your earlier question asking if this new circuit makes connection & then stops (you want this because the signal is on too long)

my first post tried to answer that question - in your original circuit, with the 1N4007 diode, the 12V line gets connected to the 1.5V battery- AND the LT device (via the 1N4007) when you touch that diode to the battery (or leave it connected)

this is true whether any coil-field collapse spikes occur or not - and it causes the damage to your circuit

the new circuit will only allow feedback spikes (not the 12V DC) to pass through C1, so only the spike energy will get fed back to the 1.5V battery (having been filtered first by C2 & L2)

the LT device will now be protected from the voltage of the feedback spikes

i hope this helps

regards
np


http://docsfreelunch.blogspot.com