Volt drop across inverter is activating optocoupler circuit

Thread Starter

JeremyCondie

Joined Jan 5, 2015
12
My circuit controls a humidifier solenoid that runs off 25VAC

The standard humidifier wastes a lot of water so this circuit has sensors in a drain that turns off the solenoid when the water level is high (the humidifier is wet). The solenoid turns back on when the water level drops.

I used a 555 timer as a debounce circuit. When the sensor detects water, the 555 output goes high (12VDC). But that would turn the optocoupler on when I need the reverse. So I placed an inverter NPN so that the optocoupler turns off when water detected. Diagram should help (the components on right are the optocoupler).

My problem is this. When the 555 goes high, the NPN pulls the input to the optocoupler low as expected. Trouble is, I didn't think the small voltage drop across the transistor would be enough to energise the optocoupler, but it seems to. My solenoid pulses slowly (~ 1 per second) and the AC across the optocoupler is less than 1VAC. I didn't expect the solenoid to be that sensitive.

Any ideas how I can turn off the solenoid completely when water is sensed?

Thanks.
 

Papabravo

Joined Feb 24, 2006
13,926
10K is a pretty big base resistor. What is the Vcc for the 555?
A pulldown resistor on the base will make sure tha Q1 is off when the 555 output goes high
 

crutschow

Joined Mar 14, 2008
25,117
330 seems like a small value.
What current does the opto require?
What opto are you using?
The ltl307EE you in the schematic is an LED not an opto.
 
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LesJones

Joined Jan 8, 2017
2,665
I don't see why you need the transistor to invert sense of the drive to the opto coupler. You could just connect the opto coupler in series with its current limiting resistor between the output of the 555 and the positive rail.

Les.
 

AlbertHall

Joined Jun 4, 2014
10,022
I don't see why you need the transistor to invert sense of the drive to the opto coupler. You could just connect the opto coupler in series with its current limiting resistor between the output of the 555 and the positive rail.

Les.
Yes, but I think you would need something to allow for the fact that the '555 output won't fully reach the supply voltage - a potential divider or a series zener perhaps.
 

Thread Starter

JeremyCondie

Joined Jan 5, 2015
12
So I sorted the problem. I started by changing R1 to 1k ohms and R2 to 470 ohms. No difference.

I put a pull down resistor on Q1's base. No difference.

Then put a scope on output of 555 timer and realized that (after the sensors were triggered) it's output went high, but with regular very short off cycles (one per second). So the sensors might be in an on-state for one minute and the 555 output would be high for that long, but with these regular drops to zero.

A small cap across the base of Q1 sorted the problem, but I'd like to know why the 555 is behaving in this way. Any ideas are appreciated.
 

eetech00

Joined Jun 8, 2013
1,938
Hi

The 555 trigger pin is wired incorrectly.

The trigger pin (#2) should be pulled up to 12v thru the 47k resistor and shorted to ground when the NO button is pressed. That way, the output will be low until the button is pressed. Then it will go high for the pulse duration.

Also, the LM317 programming resistor values don't look correct for 12v output.
Should be 120 ohm and 1k ohm.

And....you don't need a transistor at the output of the 555. You can drive the LED directly with the 555. Remove Q1 and resistors, then connect the anode of LED thru 1k resistor to +12 and the cathode of the LED to the 555 output pin. The LED will light when the 555 output is low and turn off when the output is high.

eT
 
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