555 light detector circuit with self holding output

Thread Starter

jenom

Joined Nov 15, 2019
11
I have a simple 555 light detector (sensor) circuit, attached is the diagram.(not showing pin 5 connected to GND with a 10nF capacitor)
Want to make output self holding, so once input triggered on pin 2 & 6 and output pin 3 level changes...... then output will remain the same regardless what is happening at input.
Very likely, pin 2 (trigger) and pin 6 (threshold) needs to be separated, and some kind of feedback coming from output pin 3 needs to be created,
Feedback maybe to pin 6 threshold ? or perhaps to pin 5 control ?
I am looking for a solution with minimal additional components, 1-2 resistors
Thank You!
555-detector.png
 

WBahn

Joined Mar 31, 2012
29,978
Your description of what you are trying to do is not very well spelled out.

When you say that "then output will remain the same regardless of what is happening at the input," for how long? For ever? You don't want any way to reset the output (other than cycling power)? If so, then just use a latch that is triggered by the LDR sensor.

If you want something more, then you need to clearly state what that is.
 

Thread Starter

jenom

Joined Nov 15, 2019
11
Your description of what you are trying to do is not very well spelled out.

When you say that "then output will remain the same regardless of what is happening at the input," for how long? For ever? You don't want any way to reset the output (other than cycling power)? If so, then just use a latch that is triggered by the LDR sensor.

If you want something more, then you need to clearly state what that is.
Yes, I want the output remain at the same level after triggered......until power cycled.
Circuit is used in an "alarm" setup, so only one time trigger activation is all that needed.
How exactly I need to modify circuit to do it?
Thanks.
 

MrSalts

Joined Apr 2, 2020
2,767
You just need a small NPN transistor and a resistor. r1 can be any size from 1k to 100k. It doesn't really matter.

the problem with a 555 is that it tends totrigger when you power it up. So you can add a delay to the reset pin (pin4). This allows a small delay to bring the voltage up to pin8 voltage (or close to it). R2 and the capacitor are the delay, R3 automatically discharges the cap once you power off to reset the system. You could also replace R3 with a small momentary switch to discharge the 10uF capacitor. Note that the capacitor can hold charge for some time after you power off so it needs to be discharged with the resistor or the momentary switch.
A good value for R2 is 10k and R3 is 220k. The higher R3 is, the longer you'll have to keep it turned off to reset the unit. If you use a momentary switch instead of R3, it will reset immediately. .
D976B885-8E7C-4521-BC5E-D09E5AF79DAE.jpeg
 
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Thread Starter

jenom

Joined Nov 15, 2019
11
A simple latch would do:
View attachment 280069
Latch or toggle switch (flipp-flopp) changes output to opposite each time when triggered.
I need the output to stay "forever" on the same level as after the FIRST trigger.
Experimenting now by using only pin2 trigger as a sensory input. Pin6 threshold is connected to a middle slider of potentiometer between ground and Pin3 Output.
 

crutschow

Joined Mar 14, 2008
34,284
Here's the LTspice simulation of my take on a 555 circuit:
The output (green trace) goes high when the input (yellow trace) goes below the TRIG threshold, and remains high after the TRIG input goes back high (or whatever the TRIG input does) until power is removed (red trace) and reapplied to reset the circuit.
V2 simulates the voltage from the LDR network.
C1 and R1 generate a momentary positive pulse at the THRES input to set the output low when power is applied to the circuit.

1667752882631.png
 
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Thread Starter

jenom

Joined Nov 15, 2019
11
Here's the LTspice simulation of my take on a 555 circuit:
The output (green trace) goes high when the input (yellow trace) goes below the TRIG threshold, and remains high after the TRIG input goes back high (or whatever the TRIG input does) until power is removed (red trace) and reapplied to reset the circuit.
V2 simulates the voltage from the LDR network.
C1 and R1 generate a momentary positive pulse at the THRES input to set the output low when power is applied to the circuit.

View attachment 280082
In reality it does not like that. I have 2/3 power voltage on pin6 threshold, and triggering happens when input pin2 goes below 1/3 of power voltage level and output goes high.However, when input goes above trigger level, output will turn low.
Now I am considering to use a thyristor at output, which supposed to stay permanently on after triggered . (unless power supply interrupted or load current goes too low)
 

MrSalts

Joined Apr 2, 2020
2,767
In reality it does not like that. I have 2/3 power voltage on pin6 threshold, and triggering happens when input pin2 goes below 1/3 of power voltage level and output goes high.However, when input goes above trigger level, output will turn low.
Now I am considering to use a thyristor at output, which supposed to stay permanently on after triggered . (unless power supply interrupted or load current goes too low)
See my post #4. It works well.
 

crutschow

Joined Mar 14, 2008
34,284
In reality it does not like that. I have 2/3 power voltage on pin6 threshold
Why do you have that voltage applied to pin 6?
You need less the 2/3 power on pin 6 to keep if from resetting to zero (that's why I have 0V).
 
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crutschow

Joined Mar 14, 2008
34,284
Now I am considering to use a thyristor at output, which supposed to stay permanently on after triggered . (unless power supply interrupted or load current goes too low)
That would work also, but a thyristor does not have a precise trigger level, as the 555 does.
 

Thread Starter

jenom

Joined Nov 15, 2019
11
See my post #4. It works well.
Circuit looks like when light detected, trigger input goes low, output goes high, opens transistor and brings trigger input to ground.... so output will continue to stay high.This should work as intended.
My additional issue is that I want to detect very low level of light, so I need to increase circuit sensitivity to maximum.Already changed shown 15 kohm pot to 5 Mohm resistor, not sure what else can be done ?
 

crutschow

Joined Mar 14, 2008
34,284
I think pin6 voltage coming from some internal voltage divider ?
There is an internal voltage divider that sets the pin 6 threshold at 2/3 of the supply voltage, so any pin 6 voltage above that will reset the output to zero.
So if you keep the pin 6 voltage below 2/3 of the supply, it won't reset.
 

crutschow

Joined Mar 14, 2008
34,284
Here's the basic 555 Operation description---

It is a level-triggered latch (flip-flop) with an added DIScharge output to reset the timing capacitor.
When the TRIG voltage goes below 1/3 Vcc, the latch is set (OUT high and DIS open).
If the TRIG voltage is back above 1/3 Vcc, then when the THRS voltage goes above 2/3 Vcc, the latch is reset (OUT low, and DIS connected to ground to discharge the timing capacitor).
It stays low until a low on the TRIG input again sets the latch (if the THRS voltage is below 2/3 Vcc).

Edit: Sim below illustrates the trigger levels:

1667773056439.png
 
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crutschow

Joined Mar 14, 2008
34,284
I need to increase circuit sensitivity to maximum.Already changed shown 15 kohm pot to 5 Mohm resistor, not sure what else can be done ?
So even 5 Mohm does not give you sufficient sensitivity?
If so, then you could add an op amp amplifier to increase the sensitivity.
 

Thread Starter

jenom

Joined Nov 15, 2019
11
So even 5 Mohm does not give you sufficient sensitivity?
If so, then you could add an op amp amplifier to increase the sensitivity.
I attached a potentiometer to +/- to apply voltage to pin6, you are correct about reset/no reset voltage levels.
My LDR has a big Mohm resistance in dark, so I ended up without any resistor between + voltage and pin2 trigger.
Now circuit works according to plan, thank you everybody for their help!
 

MrSalts

Joined Apr 2, 2020
2,767
Circuit looks like when light detected, trigger input goes low,
that is true

output goes high, opens transistor and brings trigger input to ground.... so output will continue to stay high.This should work as intended.
that is correct as well

My additional issue is that I want to detect very low level of light, so I need to increase circuit sensitivity to maximum.Already changed shown 15 kohm pot to 5 Mohm resistor, not sure what else can be done ?
Can use a proper comparator as a latch instead of a 555 timer. A 555 is not exactly a precision device and you are using a slightly off-label circuit.
You can get some comparitor that are extremely low current input and raise your input resistance quite high. Then you can make a separate reverence voltage voltage divider.
Finally, you can add some light blocking to the back side of your LDR to minimumze stray light that can allow current to leak through.
Lastly, you create a bit of a paradox for yourself with very high resistance resistors to make your sensor input. You can do the measurements and the math at various very low current levels (for example of ten micro amps leaks through the LDR and and the 5Mohm resistor). Note that your meter is also likely like adding 10M ohms (or 30M ohms) in parallel with any measurement you make so you'll need to account for that for high resistance inputs.

my recommendation would be to go with a comparitor with two voltage dividers as input. One being a potentiometer (possibly with additional divider resistors) and one being your LDR (back side shielded with black electrical tape (punch the leads through the black tape) and then add a second resistor in series (like 330k ohms). I know it is much less than your 5M but, since you have a comparison input as a reference, you can tune as needed.
 

Thread Starter

jenom

Joined Nov 15, 2019
11
that is true


that is correct as well


Can use a proper comparator as a latch instead of a 555 timer. A 555 is not exactly a precision device and you are using a slightly off-label circuit.
You can get some comparitor that are extremely low current input and raise your input resistance quite high. Then you can make a separate reverence voltage voltage divider.
Finally, you can add some light blocking to the back side of your LDR to minimumze stray light that can allow current to leak through.
Lastly, you create a bit of a paradox for yourself with very high resistance resistors to make your sensor input. You can do the measurements and the math at various very low current levels (for example of ten micro amps leaks through the LDR and and the 5Mohm resistor). Note that your meter is also likely like adding 10M ohms (or 30M ohms) in parallel with any measurement you make so you'll need to account for that for high resistance inputs.

my recommendation would be to go with a comparitor with two voltage dividers as input. One being a potentiometer (possibly with additional divider resistors) and one being your LDR (back side shielded with black electrical tape (punch the leads through the black tape) and then add a second resistor in series (like 330k ohms). I know it is much less than your 5M but, since you have a comparison input as a reference, you can tune as needed.
Thanks for your advise, but I needed to keep it small and simple.All it does to detect presence of natural gas flame in a totally dark environment.
 
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