Go Down

Topic: Too many ohms?? (Read 934 times) previous topic - next topic

Wawa

The 220ohm resistor for the primary LED is electrically connected in series with the two 25k resistors,
So is doing nothing useful there. Can be removed.

Still the other two problems left (3/4 swapped and 150ohm load).
Leo..

LandonW

Thanks for the responses.
Wawa, I considered putting caps in the circuit after it was working on a basic level.
I will make changes and redraw my schematic as requested.  However, my work day is about to start. So you guys should see it in about 10 hours

adwsystems

Hi,


Tom... :)
What about using a 0.047uF 250V capacitor in place of the two resistors? In AC circuits, capacitors supply impedence (0.047uF at 60Hz = 56kohms) but they don't dissipate or generate heat like the resistors do.

LandonW

Ok, the circuit in the attached is the new circuit...  and it works.  Thank you all for the help
The rectifier is a D3SBA20
The orange highlighter has not been added to the circuit. But I'm curious on what ratings should be used. (I haven't used caps a lot, I have some but not a huge variety)

Wawa

#19
Oct 18, 2018, 02:10 am Last Edit: Oct 18, 2018, 02:13 am by Wawa
See post#13 (47uF >=6.3v).
A 10-100uF, 6.3-35volt, whatever electrolytic cap will do. Just mind the polarity.
Max voltage on the cap is the two LEDs in series (~3.3volt).
Same for the rectifier. Not a problem if a bridge with a low voltage/current rating is used.
Leo..

JohnRob

HI,


I think the cap will be ineffective.  There is no voltage "compliance".  the capacitor as stated by "WAWA" will only charge to the two diode drops, as soon as the charging voltage goes low, the capacitor will nearly immediately not be able to supply the diodes (LED's) with current.


Perhaps you can take some of the 50k from the AC side of the bridge and put it between the capacitor and the diodes.   Capacitor voltage rating will have to increase as well.

I can't calculate a recommendation right now but consider the relationship  I = C dV/dt  where I is the minimum LED & Opto current and dt ~ 10 ms

You can't move too much resistance because your rectifier is rated for 200V and the peak reverse voltage is (120 + 120 * 10% ) *  squareroot of 2


John

Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

Wawa

I think the cap will be ineffective.
I know what you mean, but it's not that bad.
You can have a smoother LED current if you also add that 220ohm resistor between bridge and LED (post#17).
Leo..

JohnRob

I know what you mean, but it's not that bad.
You can have a smoother LED current if you also add that 220ohm resistor between bridge and LED (post#17).
Leo..
Yes but you will still need some dv/dt to get any energy out of the capacitor.  I'm assuming an average of 2 ma current through LED A.   If I'm correct 0.002 * 220 = 0.44 Volts for dv.  Not much to work with.

Of course if the application is only for LED's there isn't much need for a capacitor as one cannot see 120 Hz (very much).  The only negative will be a dimmer LED.   However if the OP is using this as a base for other things we need to know what is planned to offer applicable suggestions.

In my 120VAC to opto application the output was going into a high impedance circuit.  With the higher impedance it was better for me to filter on the opto output rather than the input.
Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

Wawa

#23
Oct 18, 2018, 06:14 am Last Edit: Oct 18, 2018, 06:16 am by Wawa
+1
Can always increase that 220ohm to 1k or so.
If this is to drive an Arduino pin, then I would just connect the opto transistor between pin and ground,
with pull up enabled in pinMode.
And a small capacitor from pin to ground, to bridge the zero-crossings.
With maybe only an indicator LED on the primary side of the opto.
Leo..

LandonW

Good morning.  The function of LED B is to set a digitalRead pin HIGH.
Im building a project that switches mains power. The circuit in this post is the feedback/sensor circuit so the controller knows that the switch happened and mains circuit is active

JohnRob

I agree with Wawa regarding the input of the Arduino.

If you have an extra digital output you can put LED B on that output and energize it in software.  This will allow you to keep the opto to Arduino input high impedance and easy to filter reliably.


Please do not PM me with thread based messages.  If your thoughts are worth responding,  the group should benefit from your insight.

Wawa

The function of LED B is to set a digitalRead pin HIGH.
Then it might be easier to set a pin LOW.

Post#23.
Opto transistor between pin and ground...
No LED on the Arduino side.
Leo..

ReverseEMF

What about using a 0.047uF 250V capacitor in place of the two resistors? In AC circuits, capacitors supply impedence (0.047uF at 60Hz = 56kohms) but they don't dissipate or generate heat like the resistors do.
And, be sure to include a series fuse, in case that cap shorts out.  Around 1 Amp should do it.  You want the fuse to blow if a short happend, but not for any of the normal operation currents.  The normal operation currents for this circuit will be very low, in this case.
"It's a big galaxy, Mr. Scott"

Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement!  Let it ALL hang out!!

ReverseEMF

And, again, so the OP's new schematic can be seen without having to download the file:



And, OP--if you'd like to learn how to do this, follow the following link:

https://forum.arduino.cc/index.php?topic=364156.msg3749026#msg3749026
"It's a big galaxy, Mr. Scott"

Please DON'T Private Message to me, what should be part of the Public Conversation -- especially if it's to correct a mistake, or contradict a statement!  Let it ALL hang out!!

LandonW

Thank you. I've been wondering how to do that

Go Up