# Mono 555 Connected to Photo-interrupter

#### kaufmanda

Joined Mar 5, 2020
8
Hi all,

I am pretty new at dealing with ICs, transistors, etc. so forgive me for any inexperience that might come out. I'll explain the operation of the circuit before I get to the question.

1. I am using a 555 in monostable mode to delay a signal from a photo-interrupter on an existing piece of equipment then delivering that signal through some logic gates back to the PCB on the equipment. All power and grounds come from that PCB over a wiring harness.

2. The input pulse comes from 5V being supplied from 2 photo-interrupters which then go low when the flags block the beam.

3. I have bench tested this and tested it on the equipment with a breadboard. The circuit works as intended except for after being left for ~3-4 hours with the trigger pulse at 5V. After this, the photo-interrupters give a reading of 600mV high, instead of 5V. It seems to me as if the reverse current from the trigger pin is somehow harming the transistor etc. in the photo-interrupter? Just a guess there. Unfortunately, there is no way to distinguish any markings on the interrupters to pull a datasheet on them. The current at pin 2 is around 50mA

I've attached a schema of the circuit. U1, U2 are the logic gates. That part works as intended. C1 is 180microFarad, C2 10microfarad, R2 10kOhm, R1 100Ohm.

Any ideas are why the photo-interrupters are breaking?

Thanks

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#### ericgibbs

Joined Jan 29, 2010
11,122
hi k,
Welcome to AAC,
The 555 Trig input has no pull up resistor to +Vcc.
What are the U2 and U3 device types.?
Show the photo interrupter circuitry and component values.

E

#### Alec_t

Joined Sep 17, 2013
11,743
Depending on the characteristics of the interrupters, you may need pull-up resistors on both sides of C2.

#### dl324

Joined Mar 30, 2015
11,890
Welcome to AAC!

I am pretty new at dealing with ICs, transistors, etc. so forgive me for any inexperience that might come out.
You have the trigger input to the timer floating. It needs to be a pullup resistor. R1 is unnecessary. C2 doesn't need to be so large.

It would be helpful if you told us what the logic gates were.

Here's an approximation of your circuit to show a better way to draw it:

I added the missing resistor on the trigger input.

The connection from C3 C1 to U3 is unwise. That's a slowly changing voltage.

EDIT: corrected component designator.

Why are you using 3 input gates when you could use 2 input? I guessed at the function; NOR could also be used.
EDIT: Or NAND.

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#### kaufmanda

Joined Mar 5, 2020
8
I appreciate the information from all, thank you!

What are the U2 and U3 device types.?
The logic gates are TI SN74LVC1G58DCKR Configurable Multi-Function gates. Both have 2 inputs. U1 is set up as an inverted A input (IN1) AND gate. U2 is the same gate but in an OR configuration instead.

The connection from C3 to U3 is unwise. That's a slowly changing voltage.
The connection between C3 and U3 is unwise? I believe you mean C1? I know that connection is a bit funky, but I'm trying to capture a LOW value from threshold and discharge to make my logic work. The purpose of U3 is to make sure the output of U3 is LOW if pin 2 on the 555 remains LOW. The changing voltage does give an unwanted 100ms "blip" (HIGH) on U3, but with the RC timing, it isn't really an issue, just not the "correct" way? Is there a better way to capture that LOW?

If R1 is unnecessary, I can eliminate that, then a 10k pull-up resistor is appropriate after C2 (your R3)? What about possibly needing a resistor before and after C2? I'm guessing the lack of a pullup resistor is damaging the photo interrupters?

I can try and get some images of the circuit, resistors, and ICs on the interrupters, but there is no way the company that puts them in this equipment will give me the information to pull a data sheet on them, it's a terrible company to work with, and the interrupters have a severe lack of markings to look up any sort of specific data on them.

Once again, thanks for all the information, diagrams, etc.

#### kaufmanda

Joined Mar 5, 2020
8
The changing voltage does give an unwanted 100ms "blip" (HIGH) on U3
I couldn't find how to edit. This number is inaccurate for my new RC values. It looks to be around .74s.

#### Alec_t

Joined Sep 17, 2013
11,743
I couldn't find how to edit.
There's an Edit button at the foot of your post.

#### dl324

Joined Mar 30, 2015
11,890
The connection between C3 and U3 is unwise? I believe you mean C1? I know that connection is a bit funky, but I'm trying to capture a LOW value from threshold and discharge to make my logic work.
Yes, I meant C1.

The connection is less "funky" now that we can see that the gate has Schmitt inputs. That will prevent excessive power dissipation and problems with a slow rise time at the threshold voltage. Schmitt inputs have different positive and negative thresholds, so you won't have issues with noise.

#### dl324

Joined Mar 30, 2015
11,890
The logic gates are TI SN74LVC1G58DCKR Configurable Multi-Function gates. Both have 2 inputs. U1 is set up as an inverted A input (IN1) AND gate. U2 is the same gate but in an OR configuration instead.
I was going to try to draw an updated schematic, but you started using different component designaters. U1 was the 555 timer in your original schematic.

Edited version:

#### kaufmanda

Joined Mar 5, 2020
8
U1 is set up as an inverted A input (IN1) AND gate. U2 is the same gate but in an OR configuration instead.
My apologies, this should have read as U2 and U3 respectively.

You have the trigger input to the timer floating. It needs to be a pullup resistor. R1 is unnecessary. C2 doesn't need to be so large.
I did put a 750Ohm resistor I had laying around as a pull-up after C2 on my breadboard, I can now see the RC curve clearly in both positive and negative directions, unlike before. As for C2 would 10nF or so capacitor be a better choice so the RC time isn't longer than need be?

As for the interrupters, I am going to try with some new pieces at the beginning of next week with the pullup added. How hard is it to tell if the floating input was taking out the interrupters without being able to pull any data on them?

Thanks,

#### dl324

Joined Mar 30, 2015
11,890
I did put a 750Ohm resistor I had laying around as a pull-up after C2 on my breadboard, I can now see the RC curve clearly in both positive and negative directions, unlike before.
750 ohms is smaller than it needs to be. You just need the trigger input to be above Vcc/3 before the timer times out.
As for C2 would 10nF or so capacitor be a better choice so the RC time isn't longer than need be?
We still don't know the characteristics of whatever is triggering the one shot. You only need AC coupling if the trigger signal doesn't go above Vcc/3 before the timer times out.

Did you remove the 100 ohm resistor?

#### dl324

Joined Mar 30, 2015
11,890
This is the modified schematic. I added Schmitt inverters so it would be clear that the gates you're using have Schmitt inputs:

If this schematic is correct, the signal on I2 would have to be a negative pulse with a HIGH steady state. Otherwise, the output of U2 would always be LOW. So you wouldn't need C2 or R3.
The purpose of U3 is to make sure the output of U3 is LOW if pin 2 on the 555 remains LOW.
It won't do that; it would have to be an AND gate to be able to make O1 LOW. The timing cap voltage isn't dependent on the voltage on the trigger input. Once the oneshot is triggered, the voltage on C1 will rise to 0.67Vcc. At that time, the capacitor will be discharged and the timer output will go low; provided that the trigger input voltage is above 0.33Vcc.

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#### kaufmanda

Joined Mar 5, 2020
8
Dennis- I think I need to explain the details of the entire circuit more clearly than I have been. Here is the breakdown:

1. The interrupters are activated by a "manual" method where if a metal bar literally runs into an object it will rotate, block the interrupter beam, and send a LOW signal (without the 555) through a PCB to what appears to go "directly" to an Altera Cyclone IV chipset. This is turn activates code/instructions that tell the equipment (a printer in this case) to move up and away from the object. This only needs to see a LOW for this action to happen, any high signal after this is doesn't matter as the printer takes about 5 minutes to finish that process.

#### kaufmanda

Joined Mar 5, 2020
8
After some painstaking work, I believe I have found the correct data sheet on the interrupters. Does it provide any useful information as to why the interrupters are getting fried?

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#### dl324

Joined Mar 30, 2015
11,890
After some painstaking work, I believe I have found the correct data sheet on the interrupters. Does it provide any useful information as to why the interrupters are getting fried?
No.

You'd need to be exceeding specs or switching an inductive load.