# [SOLVED] Resistor Change for ET-OPTO-ACIN4 to Detect 24vac instead of 110-220vac

Hello,
I am using an ET-OPTO-ACIN4 (PDF attached) to Detect 110vac in the US. It works great and lets my microcontroller know if 110vac is flowing through a circuit.

However, I would like to change a couple of the channels to detect 24vac instead. I plugged in 24vac and the Indication LED lit very dimly but did not send a “0” to the controller. The Board sends a TTL 0 if the Circuit is active and and a TTL 1 if the circuit is inactive.

I’m assuming this will just be a resistor change but am not well versed in how to modify this circuit to support the 24vac.

http://www.micro4you.com/files/ETT/ET-OPTO-ACIN4.pdf

ET-OPTO-ACIN4.pdf (195 KB)

I don't have a calculator to hand but basically

The 0.1 uf capacitor is acting as a resistor by capacitive reactance

Xc = 1 / ( 2 x PI x Freq x Capacitance )

Not sure if freq is 50 or 60 hz for you, but you can do the maths

I think that the cap is an equivalent resistance of 26k

So even if you shorted out the 100k the capacitive reactance would be too high to trigger the led in the opto

I think you will need to short out the resistor and replace the capacitor with another resistor or perhaps 10k or lower, perhaps as low as 2k

However, removing the capacitor removes a certain degree of insulation from your circuit and the overall load on the 24v circuit is proportionally higher than on the 110v

So I'd get a second opinion before making any changes

Ie you make changes at your own risk

Thank you for the response. I'm not familiar with the formula you showed or how to use it.

Xc = 1 / ( 2 x PI x Freq x Capacitance )

Also, If it helps It is 60hz here. I want to be sure (if possible) before altering the board that it will work for the lower voltage.

Once again, thank you for taking the time to reply

See

Bascially at 60Hz

Assuming the 0.1 capacitor is 0.1uF (the units are not listed in the diagram you posted)

This gives

Xc (effective resistance) = 26525 ohms

Total resistance in mains side is 100k + 26.525k

i.e around 126k

So current in the circuit at 110V RMS would be

I = V /R
I = 110/126,000

I = 0.87mA

but the smoothed voltage (or current is root 2 times as much as this, so the value would be 1.2mA

However that seems too low to trigger the LED in the opto. (though not impossible)

So if the opto does need just 1.2mA, it looks like it would need just 27k at 24V to give the same current, in which case you can just replace the 100k with something very low, e.g. 1k or less

However, I’d recommend that perhaps you ask the moderator to move this to the general electronics part of the forums so that other people could comment.

https://www.futurlec.com/LED/PC817.shtml

However I’m not sure

Thank you for the suggestion. The moderator kindly moved the post.

Replace the 100K resistor with a 2.2K, 1/2 W and short the capacitor. That will get you around 9 mA which will make SURE the opto comes on regardless of temperature. You may or may not want to lift the 50uF capacitor off the board, although it will be OK for things slower than about 1/10 second.

rogerClark:
I = V /R
I = 110/126,000

I = 0.87mA

but the smoothed voltage (or current is root 2 times as much as this, so the value would be 1.2mA

However that seems too low to trigger the LED in the opto. (though not impossible)

So if the opto does need just 1.2mA, it looks like it would need just 27k at 24V to give the same current, in which case you can just replace the 100k with something very low, e.g. 1k or less

So if I understand this correctly, I actually may be able to just remove the resistor and bridge the gap on the board and it would detect 24vac? This system is rated at 220vac 50hz which means I = 1.67mA.

If I pulled the 100k resistor and ran 24v it would then be I = 0.93mA correct? or would this cause to much power to be dissipated by the capacitor?

rmetzner49:
Replace the 100K resistor with a 2.2K, 1/2 W and short the capacitor. That will get you around 9 mA which will make SURE the opto comes on regardless of temperature. You may or may not want to lift the 50uF capacitor off the board, although it will be OK for things slower than about 1/10 second.

By "Short the capacitor", do you mean to solder a connection from each lead of the capacitor to the apposing lead? Should I just remove it and bridge the connection?

Whichever you prefer.

rogerClark:
Total resistance in mains side is 100k + 26.525k
i.e around 126k

Wrong!

Not that it makes a lot of difference, but you did not read your reference, did you?
Total impedance in mains side is 100k + i26.525k, around 103.45k in fact.

The circuit or at least the choice of values is in fact, wrong, the resistor should be more like 4k7 in the original design; its only function is to limit the inrush current when the AC is switched on abruptly, and at that value, it would dissipate negligible power (OK, about one tenth of a watt) while the opto-isolator current would be about 4 mA.

It follows that if the present circuit is functioning with about 1 mA to the opto-isolator, then this modification would permit it to operate adequately at 24V (as well as 110V). With a 10k pull-up, you only require half a milliamp to pull the output low, so that if the opto-isolator has a CTR (current transfer ratio) of 100% or better, it should do this quite well.

Take-home message - the use of a capacitor as dropper is eminently sensible, you do not want to remove it, but the original circuit is not properly designed to utilise it.

Thank you all for the help.

However, I have three very different answers. Can someone help me definitively know which one will work.

Here are the condensed versions:

1. Replace 100k resistor with 1K resistor
2. Replace 100k resistor with 2k2 1/2 watt resistor and short the capacitor
3. Replace 100k resistor with 4k7 resistor

Also, what wattage does the resistor need to be. The current resistor is 1 watt.

However, I would like to change a couple of the channels to detect 24vac instead. I plugged in 24vac and the Indication LED lit very dimly but did not send a "0" to the controller. The Board sends a TTL 0 if the Circuit is active and and a TTL 1 if the circuit is inactive.

Have you tried changing this:

to this:

Then in your code, don't use INPUT_PULLUP, just use INPUT.
This circuit is 10X more sensitive than the original.

EDIT: You could make the opto-isolator's output 5X more sensitive by using 47K and 0.022µF if available.

dlloyd, Thank you for the suggestion, however, I would rather adjust the input to do more like 24v to 110v (no need for 220v) instead of changing the way the collector is working.

Do you have a suggestion on how to make that happen?

dlloyd, Thank you for the suggestion, however, I would rather adjust the input to do more like 24v to 110v (no need for 220v) instead of changing the way the collector is working.

Do you have a suggestion on how to make that happen?

It all comes down to IF, the forward current through the IRLED of the photocoupler. At 24VAC, IF will drop 1/9 th the value it would be at 220VAC (the pdf shows a 220VAC circuit design). If you change the 100K to 10K you would get approximately the same current through IF at 24VAC as the original design at 220VAC.

The new RMS current (using 10K) at 120VAC is 120/10,000 = 12mA (IF max = 50mA)
However, the power dissipated by the resistor would greatly increase. Using V2/R, the power rating of the resistor would need to be 1.44W minimum, but I wouldn’t use anything less than a 3W resistor. You would have to ensure 220VAC is never connnected, as this will increase the resistor’s power dissipation by another 4X. EDIT: I neglected the effect of the series capacitor (see other replies for calculations).

You could most likely achieve the same results (untested) by boosting the sensitivity on the low level isolated side, as per my previous post, and achieve operational range of 24VAC to 220VAC.

jgiebler:
However, I have three very different answers.

Such is life.

Well, I have explained mine; I can do no more.

Thank you all for your responses!

after considering the increased power consumed if I were to lower the resistance on the input, I decided to attempt to make the Collector more sensitive.

dlloyd, I did not have a .01uF capacitor but I did have a 100k 1/2 watt resistor (Little over kill) I traded it with the 10k resistor and viola, I received a TTL 0 when I activated the 24v circuit.

Do I still need to change out the capacitor beings it worked? Does the capacitor just affect the speed at which it responds? If so, Speed isn't a huge deal, This is simply going to check if a solenoid is on or off in software.

dlloyd, I did not have a .01uF capacitor but I did have a 100k 1/2 watt resistor (Little over kill) I traded it with the 10k resistor and viola, I received a TTL 0 when I activated the 24v circuit.

Do I still need to change out the capacitor beings it worked? Does the capacitor just affect the speed at which it responds? If so, Speed isn't a huge deal, This is simply going to check if a solenoid is on or off in software.

It just affects the speed. The RC time constant (response time) is now 0.01 seconds.

Thank you all for the help. In the End, I'm going to change out the 10k resistor on each channel with 100k. This will accommodate 24vac and 110vac which suites both of my needs and does not require specific attention as to which channel I purpose.

Thank you again!

I was able to use the ET-OPTO-ACIN4 board on a 24 VAC system without modifying the board. I attached the ACIN4 board to an arduino Analog input port (rather than a arduino digital port. I then used the following code: