Re: Help needed for simple circuit troubleshoot

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


.

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