Help with ramping voltage

[jimmy_20_05]

Hello all,
I have a PWM that I am using to drive a motor in one direction for about a second (something like a pendulum thing)and then power off. I am using the Velleman k8004. This circuit has a soft start and is perfect, although I would like a soft stop also. Currently I am using an interval timer relay to power this and I was thinking that the way to add a soft stop to this would be by gradually reducing the supply voltage to the Velleman. I can not use a relay to do this, too noisy and mechanical. The on/off time is about 1 second each. Obviously I can not come up with a circuit for this and would appreciate your help or suggestions. I do not know how to make a circuit to do this, so please send me a link or a schematic if possible. My voltage max will be 24 volts for the motor (1 amp) and I do have an input voltage of 30Vdc, but I can limit that to whatever is suggested.
Thank You,
Jim

 

[chuckey]

The nicest way would be to modify the PWM generating circuit, so the on/off period changes on switch on and off but in the absence of a circuit diagram this is not possible.
If you put a series control transistor in series with the motor feed, the controlling the base current, you can control the supply to the motor. As the motor takes 1A, the base current would be 1/50 ~20 mA, this sounds a bit high, so I would use a second emitter follower to provide extra current gain , so now the base current of the first transistor would be in the order of .2 mA. So now connect the first base to the incoming supply via the new on/off switch via a 1K resistor. So now the motor will go on and off via this new switch. Connect the junction of the base and 1K to the positive of a 200 MF capacitor, with its negative end going to earth. Switching the new switch on will result in the capacitor charging up and so the output voltage will rise slowly. when you switch the switch off, the voltage on the capacitor will decay slowly so the output voltage will decay also. The disadvantage is that there will be a voltage drop of 1.6 volt across the two transistors, so the motor will get 12-1.6 = 10.4 V. The output transistor will have to be mounted on a heat sink as it will dissipate power during the time the capacitor is charging/discharging.
Frank