Please help me in this project:

+ Using a rotary switch potentiometer to adjust output voltage.
+ After resuming from power failure, the output voltage is auto down to 0V although the pot is still remained.
+ In order to activate the output voltage again, the pot must be turn all the way back till the switch is off and then turns it on again.

I very appreciate your kindly help.

 

Show us what you have done first...

You should consider a microcontroller connected to the potentiometer (using an analog input.) The rest is just programming.

 

Something like a simple latching relay circuit that is only energised by the pot switch being 'off'.

The relay would drop out with a power failure and need the pot turning back to zero to re-latch it.

Use another contact on the relay to feed the reference voltage in to the pot, so the output is held off until the relay is energised.

 

Actually, I want to copy this circuit from a china apparatus. It uses CD4060 as a timer for relay function and dual opamp LM358 for latching function (???!!!). Robert Jenkins, could you give some suggestion from this idea? Thank you so much.

 

+ Using a rotary switch potentiometer to adjust output voltage.

Are you talking about a potentiometer with an on/off switch at one end of it's rotation...like the volume control/power switch on a radio?

Ken

 

You could use a 555 timer to momentarily activate a relay to simulate the pot being returned to zero and removing the power. Eg. After power is restored the timer would wait for say 1 second then switch the relay on and then off to open the the "switch" contacts.

 

Just a mental try with relays.

Operator turns on units power switch.

R1 wiper is at the bottom and R1SW is open.
DC Supply is turned ON
C1 is effectively a short, so RLY1 closes
The NO-Com contacts on RLY1 close, shorting the NO-Com contacts on RLY2
RLY2 is powered
RLY2's NO-COM contacts close and latches RLY2 on.
DC REF voltage is supplied to R1.
C1 charges, the voltage across RLY1 drops, and RLY1 opens
The user closes R1SW as they turn up the wiper on R1

Power is lost!

R1 is in a mid-range position and R1SW is closed.
RLY2 loses power and its contacts open, removing the power source for R1
C1 is discharged through SW1 and R1

Power is restored!

C1 Charges, but R1SW is closed, shorting RLY1's coil, so RLY1 can't close.
RLY1 can't close and latch RLY2, so there is no power to R1.

Operator turns down R1

R1SW is opened.
C1 is already charged so there is no voltage across it RLY1's coil.
RLY1 can't close and latch RLY2 and power R1.

Operator turns off the unit's power.

C1 is discharged through SW1 and R1

Operator turns on unit's power.

Go back the the top.

ken

 

Please help me in this project:

+ Using a rotary switch potentiometer to adjust output voltage.
+ After resuming from power failure, the output voltage is auto down to 0V although the pot is still remained.
+ In order to activate the output voltage again, the pot must be turn all the way back till the switch is off and then turns it on again.

I very appreciate your kindly help.

I think there is some confusion here. Are you trying to restore the circuit to its original working condition before the power was removed AND not have to touch the pot? The way I read your request is that you don't want to have to turn the pot back to zero and operate the switch. You want to to work as soon as power is restored. Could you clarify?

 

Are you talking about a potentiometer with an on/off switch at one end of it's rotation...like the volume control/power switch on a radio?

Ken

yes, that's it

 

So it CLICKS when the knob gets to the OFF point?

 

Wow, i'm sorry, because my english isn't very strong, that's why it makes you confused. i'll try my best to describe my idea:

+ I have made a electronic muscle stimulator, and i need a safety circuit for patient.
+ Using rotary switch potentiometer to adjust output voltage
+ When power is cut off and turned back on, the output voltage is auto down to 0V with the pot is still remained (haven't been turned to 0V yet ).
+ to activate output voltage again, the pot must be turned all the way to 0V till the switch contact is off, and then turn it on again.

Thank you very much for your help.

 

Just a mental try with relays.

Operator turns on units power switch.

R1 wiper is at the bottom and R1SW is open.
DC Supply is turned ON
C1 is effectively a short, so RLY1 closes
The NO-Com contacts on RLY1 close, shorting the NO-Com contacts on RLY2
RLY2 is powered
RLY2's NO-COM contacts close and latches RLY2 on.
DC REF voltage is supplied to R1.
C1 charges, the voltage across RLY1 drops, and RLY1 opens
The user closes R1SW as they turn up the wiper on R1

Power is lost!

R1 is in a mid-range position and R1SW is closed.
RLY2 loses power and its contacts open, removing the power source for R1
C1 is discharged through SW1 and R1

Power is restored!

C1 Charges, but R1SW is closed, shorting RLY1's coil, so RLY1 can't close.
RLY1 can't close and latch RLY2, so there is no power to R1.

Operator turns down R1

R1SW is opened.
C1 is already charged so there is no voltage across it RLY1's coil.
RLY1 can't close and latch RLY2 and power R1.

Operator turns off the unit's power.

C1 is discharged through SW1 and R1

Operator turns on unit's power.

Go back the the top.

ken

Thank you very much

 

Ok. Now under stood.

KMoffett's idea should work fine.

PS. I have had one of these muscle stimulators for years.

OUCH they can hurt but they do wonders for sore muscles.

 

I believe the idea is that a machine of some sort cannot re-start on it's own when the power is applied - the operator must turn the speed to zero before it will start, then bring it back up to the required point manually.

It's like a no-volt safety relay on a table saw or router, to prevent an accidental start when power is applied, but in this case also acting as a soft-start as it must start from zero speed.

As long as you find a pot with a changeover or normally closed (at zero position) switch, then nothing else other than a two pole relay (with flywheel diode or snubber on the coil) is needed.

Just connect power to one side of relay contact 1, the other side of that contact to one side of the coil, and the other side of the coil to power ground/return.

Connect the pot switch NC contact across that relay contact - the relay will activate when the pot is set to zero and stay in due to the latching contact.

Use the second relay contact inline with the feed the 'top' of the pot, so there is no reference voltage until the relay is latched. The resistance of the pot will hold the output at zero when there is no reference connected.

Use appropriate back emf protection on the relay coil.

 

Just a mental try with relays.

Operator turns on units power switch.

R1 wiper is at the bottom and R1SW is open.
DC Supply is turned ON
C1 is effectively a short, so RLY1 closes
The NO-Com contacts on RLY1 close, shorting the NO-Com contacts on RLY2
RLY2 is powered
RLY2's NO-COM contacts close and latches RLY2 on.
DC REF voltage is supplied to R1.
C1 charges, the voltage across RLY1 drops, and RLY1 opens
The user closes R1SW as they turn up the wiper on R1

Power is lost!

R1 is in a mid-range position and R1SW is closed.
RLY2 loses power and its contacts open, removing the power source for R1
C1 is discharged through SW1 and R1

Power is restored!

C1 Charges, but R1SW is closed, shorting RLY1's coil, so RLY1 can't close.
RLY1 can't close and latch RLY2, so there is no power to R1.

Operator turns down R1

R1SW is opened.
C1 is already charged so there is no voltage across it RLY1's coil.
RLY1 can't close and latch RLY2 and power R1.

Operator turns off the unit's power.

C1 is discharged through SW1 and R1

Operator turns on unit's power.

Go back the the top.

ken

Hi KMoffett,

I have tried your circuit but it doesn't work, C1 isolates the relay 1 from the power supply when power on. And when I removed it and close R1SW, RLY1's NO-com contact is opened but RLY2's NO-com contact is also opened, it can't be holding. Could you give me the specified values of components or further suggestion. Thank you indeed.

 

Hi KMoffett,

I have tried your circuit but it doesn't work, C1 isolates the relay 1 from the power supply when power on.

True. C1/RLY1 form a monostable multivibrator that closes its contacts briefly when power is first turned on.

And when I removed it and close R1SW, RLY1's NO-com contact is opened but RLY2's NO-com contact is also opened, it can't be holding.

The brief closure of RLY1's contacts will cause RLY2's contacts to close and latch.

Could you give me the specified values of components or further suggestion. Thank you indeed.

Remember, I said this was a "mental try". It was not bench tested, though I have done similar circuits. Can you give me the specs (voltage, and coil resistance or current) on your RLY1 and RLY2?

Ken

 

True. C1/RLY1 form a monostable multivibrator that closes its contacts briefly when power is first turned on.



The brief closure of RLY1's contacts will cause RLY2's contacts to close and latch.



Remember, I said this was a "mental try". It was not bench tested, though I have done similar circuits. Can you give me the specs (voltage, and coil resistance or current) on your RLY1 and RLY2?

Ken

I uses 12V relays, coil resistance is 240 ohms. Thank you

 

I believe the idea is that a machine of some sort cannot re-start on it's own when the power is applied - the operator must turn the speed to zero before it will start, then bring it back up to the required point manually.

It's like a no-volt safety relay on a table saw or router, to prevent an accidental start when power is applied, but in this case also acting as a soft-start as it must start from zero speed.

As long as you find a pot with a changeover or normally closed (at zero position) switch, then nothing else other than a two pole relay (with flywheel diode or snubber on the coil) is needed.

Just connect power to one side of relay contact 1, the other side of that contact to one side of the coil, and the other side of the coil to power ground/return.

Connect the pot switch NC contact across that relay contact - the relay will activate when the pot is set to zero and stay in due to the latching contact.

Use the second relay contact inline with the feed the 'top' of the pot, so there is no reference voltage until the relay is latched. The resistance of the pot will hold the output at zero when there is no reference connected.

Use appropriate back emf protection on the relay coil.

Thank you very much indeed

 

elecwindy,

With C1 of 2200uF and a 240Ω relay coil the on-time for the monostable should be ~1 second. Lowering the value of C1 will shorten the on-time. With R2 of 1KΩ, the time before the circuit can be retriggered is ~ 3 seconds. Lowering the value of R2 will shorten the minimum time to retrigger...but will continuously draw more current from the power supply.

Ken

 

elecwindy,

With C1 of 2200uF and a 240Ω relay coil the on-time for the monostable should be ~1 second. Lowering the value of C1 will shorten the on-time. With R2 of 1KΩ, the time before the circuit can be retriggered is ~ 3 seconds. Lowering the value of R2 will shorten the minimum time to retrigger...but will continuously draw more current from the power supply.

Ken

Thank you very much, very much