Help needed for simple circuit troubleshoot

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.

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 ;-)

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!

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

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------| \\ ||--' | | | \\ ||

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.....

off.