Re: Help needed for simple circuit troubleshoot

Bob Monsen wrote:

If you can use a double-throw switch instead of a single-throw switch, you
can simply connect the other throw to ground. When you switch it from the
ground throw to the Vcc throw, it'll charge up in 3 seconds. Then, when
you switch it back, it'll come down in the same amount of time.

The N-MOSFET is hooked up as a follower. This means two things. First,
it'll hold the load at between 2 and 4 volts below the battery voltage.
That means your load is going to get less voltage, and thus less power
with a given current flow. Second, if the load is drawing an amp, that
means you are wasting between 2 and 4 W in the N-MOSFET, so it'll get hot
and waste your battery time.

If you are willing to build a somewhat more complex circuit, you can
avoid this power loss, and thus improve the life of your battery. Here is
an example circuit:

                                                        Vcc
                                                      .------.
                                                      |      |
                                                      |      |
                                                      |     ---
                                                      |     / \ 1N4001
  Vcc-----.                                           |     ---
          |                                         [LOAD]   |
          o                                           |      |
         \                              |\            o------'
   SPDT   o---[4K7]--o--[330K]---o------| \       ||--'
                     |           |      |  \      ||<-.
          o          |           |      |   ------||--|
          |          |           |      |  /          |
          |   470uF ---         [R]  .-o| /           |
          |         ---          |   |  |/            |
          |          |           |   |                |
          |          |           |   '----------------o
          |          |           |                    |
          |          |           |                  [0.1R]
          |          |           |                    |
          |          |           |                    |
 GND------o----------o-----------o--------------------'


When the switch is thrown to Vcc, the current through the load will
gradually increase over a few seconds to whatever maximum you set with R.
I don't know how much current your load will draw, so I can't set it up
exactly. You'll have to adjust the value of the resistor marked R in the
picture to ensure that the MOSFET turns on all the way, but doesn't get
too much above the voltage required. Measure the current when your load is
connected directly across the battery, using a multimeter. Then, assuming
the current is I amps, you want to set the voltage at the non-inverting
input to be I/10 when the SPDT switch is connected to Vcc. That will be

R = (334.7k*I)/(10*Vcc - I)

So, for example, if your max current is 2A, and Vcc is 9.6 then 

R = (334.7k*2)/(96-2) = 7121 ohms

You can use any cheap single supply opamp. I added in the diode to prevent
the mosfet from getting fried by back-emf from the motor. With this, you
don't have to use a super high voltage mosfet like the guy specifies; you
can use one that is about 30V ds without a problem.

Regards,
 Bob Monsen


Thanks Bob :-)

I really appreciate your help. I'm learning electronics in my old age by doing ;-), since I somehow managed to avoid learning any electronic theory when I was younger! Your circuit looks very interesting, I think I understand most of it ;-) so I'll build it and see if I can add a bit more to my knowledge base.....

--
James McInerney

My G gauge garden homage to the now long gone railways of Tasmania's west coast, the "Rurr Valley Railway" http://www.cia.com.au/bullack/rvrtitle.html

also http://www.cia.com.au/bullack/ , where the steam era NSWGR secondary lines live on in HO at bucolic "Lambing Flat"

and http://members.optusnet.com.au/lambingflat/ for the family stuff!

.

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