Using TIMM pot with 2 pins

[anishkgt]

Hey All,

Not sure if a trimm pot can be used with only two pins as a variable resistor. I tried as per schematic but i don't see any change in voltage with a DMM at pins 1 and 2 of the optocoupler.

Would appreciate if somebody could advise.

[JS]

Using a trim with 2 terminals as a variabke reaistor is fine, you usually want to connect the unused pin to the wiper so it doesn't go open if wiper fails or skips but no problem in your app.

I've used an H11AA1 in an aplication similar to yours with a fixed 1W 100k resistor to detect 220V mains active. No external diodes. Success and reliable so far, I have 15 of them working for a few years. It does take some power but this is an industrial application so a few mW isn't a real concern for efficency sensing a 50kW installation. Kind of hard to work it to archive any sort of threshold like it looks like you want with the variable trim, if you want to know if there is under or over voltage you should probably go a different way. I've also got it to work with 3 resistors as a zero crossing detection.

Maybe you can get it to work with a resistive divider and a zenner, then to the optocoupler. Also load on the other side of the optocoupler would change your results as current in both sides has a relationship and you will get some little current even if your voltage is much lower.

JS

[vk6zgo]

Hey All,

Not sure if a trimm pot can be used with only two pins as a variable resistor. I tried as per schematic but i don't see any change in voltage with a DMM at pins 1 and 2 of the optocoupler.

Would appreciate if somebody could advise.

The voltage across pins 1 & 2 will be clamped at the forward voltage of the two back to back diodes.

[anishkgt]

Tried without the diodes as well, nothing happened.


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[JS]

I don't know what the soldering iron has to do with this project 

As I said, I wouldn't expect the voltage to change much at the input of the optocoupler, the current will, they are diodes after all, use a transformer to insulate from mains and have a look with your scope so you understand what's going on, I know what you will see... If you don't have a mains insulation transformer you can take two 9V transformers, hook the 9V windings together, plug the mains side of the first to mains, do your measurements using the second one as insulated mains.

JS

[anishkgt]

i don't see  me mentioning about a soldering iron other than those mentioned in my signature.

The input voltage is 31vac, i thought it was about the floating pin i had but i connected the wiper to the floating pin, that did not change the voltage either.

[anishkgt]

The diode is to clamp voltages above 10vac, as the voltage would increase to 240vac in short pulses. The application here is to detect a voltage drop at the primary when the secondary are shorted during a spot weld. This circuit would go in parallel with the primary of a Microwave Oven Transformer.

[JS]

i don't see  me mentioning about a soldering iron other than those mentioned in my signature.

Yes, it wasn't on the first post, I thought you where telling what you could use...

The input voltage is 31vac, i thought it was about the floating pin i had but i connected the wiper to the floating pin, that did not change the voltage either.

That shouldn't make anything other than improved reliability, as I said, quite important when you are trimming the feedback resistor of an opamp, if it goes open the feedback loop opens and can bring some troubles, in your case isn't needed, and maybe you could detect a failure easier without connecting it, if the pot starts to act up without it connected it will be quite evident.

What is you reading? you should probably be reading about 1.2V, at least with a true RMS meter as yours, as you will have something like a 1.2V square wave. Voltage shouldn't change much while you change that resistor. You should probably put a limiting resistor of at least 600Ω as that will make the maximum current the 50mA the optocoupler is rated for, if you don't have it and you set your pot to minimum it will blow the optocoupler, probably open, making your reading that 31V you have from the source.

If you measure the current at pin 1 or 2 you will see some changes while turning the pot. fig 5 of this datasheet shows the transfer ratio of that current, so after some multi point calibration considering the final resistance you can do some measurements in the output current and determine the input voltage.
http://www.mouser.com/ds/2/239/S_110_LTV-814%20824%20844%20(M,%20S,%20S-TA,%20S-TA1,%20S-TP)%20Se-337200.pdf

  You should use a much higher resistor, at least 5k6 would keep the current low enough for the optocoupler to manage by it self, you could go even higher. If you use 10k resistor you will have at the output 30mA at 240V, (that's over 5W! on the resistor) and about 0.7mA at 10V. 5W would fry a small resistor or potentiometer even if it lasts a few ms, if you connected this and made it run you had likely burned the pot.
  The datasheet only shows currents down to 1mA, it should work a bit under that but the drop could be quite steep, so you should set your minimum useful sense voltage to that point. I'd probably would put the transistor from VCC to a resistor to ground and measure the voltage across the resistor, as the transistor is floating there's nothing wrong doing that.

JS

Sorry, I hadn't seen the last schematic, I saw the last post while posting, you have the optocoupler input shorted in that schematic, you should have it between there and the neutral. Also, the fuse should be right into the live input, in case something goes wrong with the triac, like it shorting to ground while a screw driver went in contact with the PCB.

[anishkgt]

connected the neutral to pin 2. So probably the pot may have blown or the optocoupler. i could see 32vac at the pins 1 and 2. But do these trimm pot fail short ?

[anishkgt]

attached schematic

[krish2487]

As I understand it, the entire section to the left, starting from the fuse is what is your power switch. A triac based static switch, driven by the VOM IC to connect and disconnect the primary of your transformer..


The optocoupler section to the right of the fuse is serving as a opto zero crossing detector, nothing more. when the line voltage, in either cycle goes beyond the diode forward drop voltage, the diodes conduct and the opto secondary transistor conducts.


To answer your question,
1. Yes, a trim resistor can be used with only two terminals as shown. Yes, they do fail open. There are also fakes of bourns trim resistors that are spectacularly horrible


And I don't understand the objective of your exercise in varying the trim resistor and expecting the voltage across the opto input to change. the resistor is for only limiting the current through the opto input diodes.
Irrespective of what voltage you apply at the input of the resistor-opto combination, the voltage you see across the opto inputs will be the same ALWAYS.


Why, you ask.. because the input of the opto is nothing but a pair of anti parallel diodes. The rest of the voltage is dropped across the trim resistor, which, depending on the wiper position decides the opto diodes current.


Set the trim wiper to too low a resistance and you risk blowing out the diodes in the optocoupler and/or the trim resistor itself. Thats what JS meant by including a current limiting resistor.


2. The opto circuit in question cannot be used to detect a sag in the input voltage. its a zcd circuit. plain and simple.


If you are seeing 32VAC at the input terminals of the opto, chances are high you set the wiper position to short and thus most likely blew out your opto input diodes.


<rant>
So, the question now is, what is it exactly that you want??
Please be coherent, simple and succinct in what your end objective is and how you plan to achieve it. Open ended questions do not help us, nor you.
If you need someone to critique your work, then don't paste the entire work and ask the other members to spend time sieving through the rest of non important parts. highlight the area in question and be specific.
</rant>

[anishkgt]

What am trying to acquire is to detect a voltage drop at the primary when the secondary is shorted. The opocoupler is just an isolation for the microcontroller.

A resistor in series with the optocoupler is all that i would need but not sure of what value. The resistance should be low enough for the optocoupler to light up and activate the transistor hence the microcontroller at pin 4 would read a "HIGH" and large enough to drop the voltage when the secondary is shorted which would read "LOW" continuously at pin 4.

The circuit was in its simplest form in my first post. Thought i could give a clearer picture if i posted the remaining.

[krish2487]

The opto data sheet is your best friend. Look under the absolute maximum ratings and try to decide a resistors values which does not violate the parameters for the diode current ratings.


Voltage drop of what value?? do you want to detect a 5v drop or a 50v drop??
And after how long since the drop do you want to trigger an output??


As I mentioned, the circuit is a zero crossing detector. it will not be useful for anything otherwise without substantial changes..


Please go through this excellent link


http://www.bristolwatch.com/ele2/zero_crossing.htm


Your original circuit is the second image in that page.
Now, do you understand how the circuit works?? and why it is not an appropriate solution for the problem what you are trying to solve??


Please look at comparator applications, specifically a threshold detector which is more appropriate in your case.
And then see how you can isolate the mains from the low voltage circuits section and still achieve the same result.
(Assuming you actually mean that you need to detect when the primary voltage sags during the spot welder on time..)

[JS]

connected the neutral to pin 2. So probably the pot may have blown or the optocoupler. i could see 32vac at the pins 1 and 2. But do these trimm pot fail short ?

No, it wouldn't fail short, check with your DMM for resistance... In this case your optocoupler's magic smoke went first.

JS

[anishkgt]

Please look at comparator applications, specifically a threshold detector which is more appropriate in your case.
And then see how you can isolate the mains from the low voltage circuits section and still achieve the same result.
(Assuming you actually mean that you need to detect when the primary voltage sags during the spot welder on time..)

Yes i first thought of a peak detector then i was lost with no help on it. Second was the comparator again lost with no help. I guess i would need to give the peak detector a go again.

[JS]

Why not to measure in the secondary, very little voltage of course but you are insulated, pot an opamp and go, you can even measure at the electrodes making the highest dynamic range to measure from.

JS

[anishkgt]

Not sure how it could be done at the secondary. The secondary would have large current flow as well how can it be bypassed ? I was thinking of the below peak detector.

[JS]

Not bypassed, in parallel. Measure the voltage at the secondary.

JS

[anishkgt]

I mean there would be a huge current during a spot weld so wouldn't that damage the opamp and those underlying circuitry ?

[JS]

No, it would connected in parallel, current wont touch it, not much protection required, the usual to ensure reliability but even without any it should work.

JS

[anishkgt]

Thank you. I am not sure how to go about it with am opamp. So it would compare the voltage from the electrodes at the invtering and non-inverting pins when the electrodes are open. So how would the output look like from the opamp when the electrodes are open and closed ? I don't have any at the moment. Which one would i need to test this on a bread board and what other components would i require ?

[anishkgt]

I was thinking of getting this Diff amp (https://www.digikey.com/product-detail/en/on-semiconductor/LM393N/LM393NFS-ND/458695) Would any resistors be required ? i've got couple of 1k, 10k 39R, 360R bunch of 0.1uf ceramic caps.

[JS]

Ok, I see where this go. Build a non inverting stage, I don't know the voltage at the point where you measure, but that dividing the reference voltage would be the desired gain and define resistance values.

Now, 393 aren't differential amplifiers, are comparators.

If you don't really know what you are doing this isn't a good project, high energy and low knowledge are a bad mix, some of the questions seems like that. To starts, it's always recomended to use an insulation transformer while working at mains projects, you are not using one for what I've seen, there are a ton of safety issues with this and I can't make them into a post, you should learn that starting with small projects and build up confidence and experience befor going to something like this... Sory dude, but that's the way it is.

JS

[anishkgt]

Well thanks for the heads-up. I am pretty sure about what am doing but just don't how to use opamps. Comparators are something i can understand. I tried using a configuration at the primary to set a dc reference and compared it against the voltage at the primary when the secondary are shorted. Well it did not work actually they were beyond my knowledge about comparators.

If you could quide me i could get it done or suggest me an opamp to test on a bread board and i will try learn as i go building it. Would a general pupose opamp like MCP6002-I/P (http://ww1.microchip.com/downloads/en/DeviceDoc/21733j.pdf) or the LM358 (http://www.ti.com/lit/ds/symlink/lm358.pdf)help me begin with ?

[JS]

I would say the lm358 should do, they are quite situable for working ih micro cobtrollers projects. What micro are you using? In particular... What adc? What voltage range?

From a single rail you should ac coupke your signal to an inverting stage and feed 0.5 Vcc to the non inverting input. The gain will depend on other factors. What voltage do you see on the secondary for the 3 states you are interested?

JS