Re: any comments on this circuit?

From: John Popelish (jpopelish_at_rica.net)
Date: 07/22/04


Date: Thu, 22 Jul 2004 17:45:43 -0400

andy wrote:
>
> On Thu, 22 Jul 2004 18:41:48 +0100, andy wrote:
>
> > It's meant to send a 1 second pulse through an electromagnet at the end of
> > every day, to start a plant watering system, as i said in an earlier post.
> > I've tested it on breadboard, but with only one transistor and a single
> > coil winding. I'm trying splitting the coil into 3 parts to get more
> > current through it. It would be good to know what people think of this
> > circuit before i build the final version - it's only the second circuit
> > i've built.
> >
> > http://www.niftybits.ukfsn.org/electronics/daily-water.png
> > http://www.niftybits.ukfsn.org/electronics/daily-water.gif
> >
> > andy.
>
> I've been through it again, and worked out the component values for the
> second half of the circuit again. The new version is at:
>
> http://www.niftybits.ukfsn.org/electronics/daily-water2.png
> http://www.niftybits.ukfsn.org/electronics/daily-water2.gif
>
> If there's something wrong with the basic idea of how i've laid it out,
> i'd rather be told why than just have it written off.
>
> --
> http://www.niftybits.ukfsn.org/
>
> remove 'n-u-l-l' to email me. html mail or attachments will go in the spam
> bin unless notified with [html] or [attachment] in the subject line.

Some basic comments:

Observe the orientations of the 3 output transistors.

The output transistors are NPN types, but a PNP symbol is used.

If you use NPN output transistors, they are operating as emitter
followers, so the emitter voltage will always be at least .6 volts
more negative then the base voltage.

The base current for the output transistors comes entirely from the
base pull up resistor, which provides no current unless there is
voltage across it. This uses up some more of the 12 volt supply.

The BD139 must sink lots more current through the output drive
resistor than that resistor ever supplies to the output transistors.

There is little reason to expect that what output drive current there
is will divide into three equal parts where it meets 3 bases.

I don't want to ruin all your fun by designing the circuit for you,
but think about these comments and show me what you come up with as a
result.

-- 
John Popelish


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