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From: N_Cook on 8 Jul 2010 07:52
Meat Plow <mhywatt(a)yahoo.com> wrote in message
news:3t134h.u5j.19.4(a)news.alt.net...
> On Thu, 8 Jul 2010 08:54:46 +0100, "N_Cook" <diverse(a)tcp.co.uk>wrote:
>
> >If I ever get this thing working properly , how to check the average
current
> >drain?
> >I am thinking 12V supply >- ammeter >- bank of 20V Cs summing to about
0.1F
> >>- fencer unit
> >
>
> Find a cow, sheep or horse willing to test it.

I don't think asking one "hoo moony mooliemoomps dooze this moozapper gizmoo
take" would get very far


From: Franc Zabkar on 8 Jul 2010 19:32
On Thu, 8 Jul 2010 08:24:40 +0100, "N_Cook" <diverse(a)tcp.co.uk> put
finger to keyboard and composed:

>The cap is 1uF , 280V ac rating. From the .5*C*V*V that comes to 630V, so
>something awry there.
>No more than 250V m 60% of likely DC rating, and I would have thought more
>like an equal split between 12V and 1800V so about 150V, transformers are
>much the same size if that is anything to go by

If the device is designed to output its rated energy over a 9V-12V
supply range, then one would expect that the dump capacitor's voltage
would be regulated. Otherwise the variation in the stored energy would
be (12/9)^2 = 1.8X.

Is there any voltage feedback from the dump cap back to its charge
controller? I'd expect to see a resistive potential divider feeding
one input of an error amp (comparator?), and maybe a 5V or 2.5V
reference on the other input. You may be able to compute the voltage
from the resistance values.

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.
From: N_Cook on 9 Jul 2010 03:45
Franc Zabkar <fzabkar(a)iinternode.on.net> wrote in message
news:8fnc361ijlnbv0spgbgljmn7tfdo8nfpe2(a)4ax.com...
> On Thu, 8 Jul 2010 08:24:40 +0100, "N_Cook" <diverse(a)tcp.co.uk> put
> finger to keyboard and composed:
>
> >The cap is 1uF , 280V ac rating. From the .5*C*V*V that comes to 630V, so
> >something awry there.
> >No more than 250V m 60% of likely DC rating, and I would have thought
more
> >like an equal split between 12V and 1800V so about 150V, transformers are
> >much the same size if that is anything to go by
>
> If the device is designed to output its rated energy over a 9V-12V
> supply range, then one would expect that the dump capacitor's voltage
> would be regulated. Otherwise the variation in the stored energy would
> be (12/9)^2 = 1.8X.
>
> Is there any voltage feedback from the dump cap back to its charge
> controller? I'd expect to see a resistive potential divider feeding
> one input of an error amp (comparator?), and maybe a 5V or 2.5V
> reference on the other input. You may be able to compute the voltage
> from the resistance values.
>
> - Franc Zabkar
> --
> Please remove one 'i' from my address when replying by email.


There is a chain of 3.3M resistors for feedback , I suspect it would change
the 1.5KHz multivibrator rather than the 300KHz one. There is also the 1 to
2 second repetition cycling, perhaps 3 multivibrator package, also
unreadable marking


From: Franc Zabkar on 9 Jul 2010 04:12
On Fri, 9 Jul 2010 08:45:56 +0100, "N_Cook" <diverse(a)tcp.co.uk> put
finger to keyboard and composed:

>There is a chain of 3.3M resistors for feedback , I suspect it would change
>the 1.5KHz multivibrator rather than the 300KHz one. There is also the 1 to
>2 second repetition cycling, perhaps 3 multivibrator package, also
>unreadable marking

I would expect that the multivibrator would have an internal reference
voltage which it would compare against the voltage on the resistor at
the bottom end of the potential divider. The IC's reference voltage
may appear on one of its pins.

I would locate where the divider feeds into the IC, and determine the
values of all the resistances in the chain. Then measure the voltages
on the pins on either side of the feedback pin. If you can supply this
information, then perhaps it will help ascertain the capacitor
voltage.

- Franc Zabkar
--
Please remove one 'i' from my address when replying by email.
From: Ian French on 10 Jul 2010 08:47

"N_Cook" <diverse(a)tcp.co.uk> wrote in message
news:i120g0$ofq$1(a)news.eternal-september.org...
>2 stage derivation of (should be) 1800V pulses.
> Fence had failure in the first stage probably due to intermittant corroded
> battery supply contact.
> Anyway the 15A 60V 8N06 powerFET blew along with an associated double
> diode
> BAV23.
> Replaced both and works to some extent. This FET has plenty of drive 5V of
> pulses of repeat 3uS for the gate and about 3V ringing swing on the output
> to the first step up transformer. I assume that is as to be expected,
> never
> having the chance to see hot-side SMPS drive scope traces.
> But rectified output only climbs to about 1V of DC before the final stage
> (unreadable marking) thyristor discharges that supply into the second step
> up transformer, but of course not even enough to get a glow on a 110V
> neon.
> Doubling the C setting the 3uS pulses, to give longer repeat rate and that
> intermediary voltage rises to about 3V but still not enough for any proper
> final output. Adding more C and DC is then much less than 3V.
> What would people expect the intermediary DC to be? obviously higher than
> 12V battery supply
> Next stage would be disconnect the thyristor if that should be working but
> leaky and draining the intermediary DC but any other ideas? Would explain
> why the printing is very indistinct I suppose.
> DC ohms of the airgap type transformers are
> intermediary 0.1R//10.6R
> HV 0.3R//29R which seem reasonable, ie not major shorted turns if any
>
>

Hi,

I found this schematic (circuit diagram) somewhere on the net, is it
anything like your one ?

http://www.electro-tech-online.com/attachments/electronic-projects-design-ideas-reviews/4007d1108193301-need-help-developing-electric-fencing-my-farm-electricfence.jpg

Ian.


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