Re: Help needed for simple circuit troubleshoot
- From: "Jim Douglas" <james.douglas@xxxxxxxxxxxxxxxxxxxx>
- Date: Sat, 16 Apr 2005 08:25:29 -0500
Your web site is incredible! I have spend the last 45 minutes checking out
all that work! This stuff is beautiful and I admire the work and research
you have done! I have this site in my favorites and will check back often
to see what's new.
"Bob Monsen" <rcsurname@xxxxxxxxxx> wrote in message
news:pan.2005.04.15.07.05.41.684401@xxxxxxxxxxxxx
> On Thu, 14 Apr 2005 17:03:23 +1000, Lambing Flat wrote:
>
> > G'day all :-)
> >
> > I am very much a beginner at this electronics lurk so I am hoping that
> > someone can help me figure out what is wrong with my version of the
> > circuit shown in the following diag.
> > http://www.trainweb.org/toenailridge/circuit.jpg
> >
> > The circuit's function is to smoothly accelerate a G gauge battery
> > powered locomotive to running speed once the circuit is turned on, and
> > then to smoothly decelerate it back to stop once the circuit is turned
off.
> >
> > My problem is that it accelerates very nicely to speed in about 3 secs
> > once the switch is turned on, but when the switch is turned off it takes
> > about 2 minutes to stop! I have obviously done something wrong, as the
> > deceleration is supposed to be at the same rate as the acceleration, but
> > I can't figure out what my mistake is :-(
> >
> > Here are photos of both sides of my version of the circuit, (apologies
> > for my poor standards of soldering ;-)
> > http://www.cia.com.au/bullack/Circuit1.jpg
> > http://www.cia.com.au/bullack/circuit2.jpg
> >
> > The only difference my version has to the circuit diagram is that I have
> > temporarily moved the connection to the base tag of the MOSFET from the
> > "power always" side of the on/off switch to the other side, as part of
> > ongoing experiments to try to find out why it won't slow down! Made no
> > difference at all, I'm afraid......
> >
> > I have carefully checked all other connections and components, and as
> > far as I can tell, (other than as mentioned above) everything is
> > connected and in the correct place and there are no shorts.
> >
> > As part of my careful checking of everything I have established that the
> > batteries produce 7.5v and that is the voltage present at the collector
> > leg of the MOSFET once the switch is turned on and the circuit powers
> > up. The voltage drops to 5v on the emitter leg of the MOSFET.
> >
> > Hopefully the fine brains trust here will be able to point me in the
> > right direction!
>
> 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
>
>
.
- References:
- Help needed for simple circuit troubleshoot
- From: Lambing Flat
- Re: Help needed for simple circuit troubleshoot
- From: Bob Monsen
- Help needed for simple circuit troubleshoot
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