Help needed for simple circuit troubleshoot

[Lambing Flat]

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!


--
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!

 

[Fritz Schlunder]

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 :-(

You haven't done anything wrong. The problem is the circuit was designed by
a fellow newbie, so by design it takes approximately 100 times longer to
turn off than to turn on. If you want to make the delays approximately the
same...

Then replace the 470k ohm resistor with a 1k ohm resistor and hook it up
such that the top side of the resistor goes to the right side of the on/off
switch (which also hooks to the left of the 4.7k resistor). The bottom side
of the 1k resistor should go to the negative terminal of the battery pack
just as the current 470k resistor does. If the timing is still not to your
satisfaction after this modification you can play with adjusting the value.
Larger resistance will take longer to stop, lower resistances than 1k will
not make much difference. You can also play with the value of the 4.7k
resistor or the 470uF capacitor if you like.

 

[Lambing Flat]

You haven't done anything wrong. The problem is the circuit was designed by
a fellow newbie, so by design it takes approximately 100 times longer to
turn off than to turn on. If you want to make the delays approximately the
same...

Then replace the 470k ohm resistor with a 1k ohm resistor and hook it up
such that the top side of the resistor goes to the right side of the on/off
switch (which also hooks to the left of the 4.7k resistor). The bottom side
of the 1k resistor should go to the negative terminal of the battery pack
just as the current 470k resistor does. If the timing is still not to your
satisfaction after this modification you can play with adjusting the value.
Larger resistance will take longer to stop, lower resistances than 1k will
not make much difference. You can also play with the value of the 4.7k
resistor or the 470uF capacitor if you like.

Thanks Fritz, I'll give that a try and post my success or failure later

--
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!

 

[Bob Monsen]

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

 

[Lambing Flat]

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!

 

[Jim Douglas]

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.

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