Will both of these circuits yield the same results?  I know a single 20K would be better but I had several 10K on hand.

Yup, same total resistance in circuit.

 

What's the load and the desired effect?

 

It may be better to use X rated capacitors to reduce the power dissipation.

13 minutes ago, Crossy said:

Yup, same total resistance in circuit.

 

What's the load and the desired effect?

 

It may be better to use X rated capacitors to reduce the power dissipation.

It is actually part of a larger circuit that I am trying to fit on a circuit board and the top schematic works much nicer.  Thanks so much for the help.

It will be same total resistance in the circuit, but it will depend on what the load is.

.

Edited July 4, 20205 yr by GregTN

The actual circuit uses a 64K 1 watt resister (I had several 33k 2 watts on hand), a couple of diodes and the purpose is to light an LED from 220 ac.  I just used the above pictures to simplify the question.  Don’t laugh to hard at the completed project as I am definitely a rookie but love the hobby.

Totally with you about the fun hobby side of things but your little circuit dissipates around 2.4W from the two 10K resistors and less than 4mW of power for light (they are only ON for half of each mains cycle).

 

It would be MUCH safer to use a 5v USB supply and a 470 Ohm resistor - and the LED would be on continuously.

 

Edited July 4, 20205 yr by sandbox

3 hours ago, GregTN said:

The actual circuit uses a 64K 1 watt resister (I had several 33k 2 watts on hand), a couple of diodes and the purpose is to light an LED from 220 ac.  I just used the above pictures to simplify the question.  Don’t laugh to hard at the completed project as I am definitely a rookie but love the hobby.

Do you have capacitor 100n (marked 104) on hand?  Use that in series with one resistor and the led, and the led in parallel with diode other way around.

 

2 hours ago, Metropolitian said:

Do you have capacitor 100n (marked 104) on hand?  Use that in series with one resistor and the led, and the led in parallel with diode other way around.

 

You mean, do you have a 100n capacitor of suitable voltage rating and be careful of voltages left on capacitors after disconnection.

Eg 100n-400V in series with 500Ω inrush limit + diodes

Ah yes, direct power from mains, a great way to electrocute yourself AND your friends ????

 

Pure resistive works but as you might have noticed it's a much better heater than a light source.

If you do insist on living on the edge, here is a great app note from Microchip regarding resistive and capacitative power circuits, including the formulas to calculate your components:

http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf

Thanks so much for all the suggestions.  Considering all the good advice I will search out some small neon lamps and replace the LED’s.  Space is limited and I had some spare parts and a lot of time on hand.  Here are a couple of pictures of the almost finished product.

I would keep the LEDs but go for a capacitative (reactance) voltage dropper, as per the APP note.

 

2 hours ago, Jan Dietz said:

I would keep the LEDs but go for a capacitative (reactance) voltage dropper, as per the APP note.

 

Thank you very much for the information and link.

Do it like this:-

 

 

Get an "X" rated capacitor suitable for AC mains use anything around 0.1 - 0.22u will do the trick. 

https://www.aliexpress.com/item/32947949184.html

 

0.1u will give you about 32k + the 4.7k => about 6mA through the LED

0.22u will give you about 14k + the 4.7k => about12mA through the LED

 

The 4.7k will dissipate 0.2W or 0.6W respectively.

 

Capacitive reactance calculator here if you want to try other capacitors / currents  

http://www.sengpielaudio.com/calculator-RC.htm

 

5 hours ago, Crossy said:

Do it like this:-

 

 

Get an "X" rated capacitor suitable for AC mains use anything around 0.1 - 0.22u will do the trick. 

https://www.aliexpress.com/item/32947949184.html

 

0.1u will give you about 32k + the 4.7k => about 6mA through the LED

0.22u will give you about 14k + the 4.7k => about12mA through the LED

 

The 4.7k will dissipate 0.2W or 0.6W respectively.

 

Capacitive reactance calculator here if you want to try other capacitors / currents  

http://www.sengpielaudio.com/calculator-RC.htm

 

Thank you Crossy.

On 7/5/2020 at 1:19 AM, maxpower said:

You mean, do you have a 100n capacitor of suitable voltage rating and be careful of voltages left on capacitors after disconnection.

No, I did not mean to say that so I didn't.

It's common sense to use one of suitable voltage (it's more important that it's an X2 type) -and- we are talking about capacitve dropper, not a elco (dc) or ladder structure which holds strong charge.

It would not always much, but yes it can, hold a charge.

 

You can shunt it with a bleeding resistor of 1Mohm, which empties such a fully charged cap of 0.1u in 0.6 seconds.

 

As indicator light connected in parallel it will eventually bleed out fast caused by the connected electronics at the mains side.

 

Used just as one thing, only the led and nothing else, where you pull it from the mains then you need the shunted resistor for bleeding the cap.

 

On 7/5/2020 at 1:19 AM, maxpower said:

Eg 100n-400V in series with 500Ω inrush limit + diodes

A capacitor in this setup (X2 , 100n) has 32Kohm resistance and a no-load voltage of around 4V and 8mA when put in series on the mains.

Add the 500 Ohm (470Ohm) is good for using one led.

 

Most leds are happy with 20mA and around 2 volt (Green 2.2 and red 1.8)

 

I would use the three leds with a 1.2K in series each led, with  1uF cap which gives you no-load voltage of 24 volt and 40mA.

 

 

 

On 7/4/2020 at 4:57 PM, Crossy said:

Yup, same total resistance in circuit.

 

What's the load and the desired effect?

 

It may be better to use X rated capacitors to reduce the power dissipation.

Really a 'must' to use X2 rated capacitor opposite to other kinds of capacitors.

It's a feed that gets beating 50 times in a second continuously and X caps are more suitable for such task.

Failing caps are dead shorts and that would pass the 230 mains to the led, poof.

 

Leds cost near to nothing.

 

Thinking about using cap drops for the ESP boards, would be better with zener (5.6v) and fuse.

Or 'bigger cap ' like the 225K and little transformer for 24v to... 5v.  Not sure yet. What would you do?

 

 

On 7/4/2020 at 10:27 PM, Metropolitian said:

Do you have capacitor 100n (marked 104) on hand?  Use that in series with one resistor and the led, and the led in parallel with diode other way around.

On 7/6/2020 at 5:22 AM, Crossy said:

Do it like this:-

 

Just what I had imagined ????

On 7/4/2020 at 7:06 PM, GregTN said:

The actual circuit uses a 64K 1 watt resister (I had several 33k 2 watts on hand), a couple of diodes and the purpose is to light an LED from 220 ac.  I just used the above pictures to simplify the question.  Don’t laugh to hard at the completed project as I am definitely a rookie but love the hobby.

 

 

1 hour ago, Metropolitian said:

No, I did not mean to say that so I didn't.

It's common sense to use one of suitable voltage (it's more important that it's an X2 type) -and- we are talking about capacitve dropper, not a elco (dc) or ladder structure

 

I like to spare a little thought for those on a learning curve before flying the knowledge flag or playing the common sense card. One of the reasons why I seldom post in one-up·man·ship frenzies.

If you are eager to spread the word and crunch some numbers, join us over at the EEVblog Electronics Community. That's of course if you're not already a member.

Edited July 7, 20205 yr by maxpower

10 hours ago, Metropolitian said:

Thinking about using cap drops for the ESP boards, would be better with zener (5.6v) and fuse.

Or 'bigger cap ' like the 225K and little transformer for 24v to... 5v.  Not sure yet. What would you do?

 

If I didn't need mains isolation then C/R series combo, bridge, filter cap, zener.

 

Calculate the C/R series and filter cap to suit your load current. Ensure your zener won't over-dissipate when there's no load (all your anticipated load current goes through the zener).

 

Some quick sums, adjust the numbers as required. It's been a while (I think I was about 16 last time I designed a zener regulator) so you may want to check the maths ????

 

Required output 5V @ 50mA 
5mA minimum zener current to maintain regulation at full load

 

Zener needs to be 5.1V @ 1W so it doesn't fry at zero load

 

C/R combo needs an impedance of about 3.5k, use a 1.0uF (250V AC X2) and 330R 1.5W .

 

470u 25V filter cap.

 

IMPORTANT NOTE - The 0V (-) output isn't referenced to anything in particular and will bounce around at about 50% mains voltage. Don't connect a grounded scope probe or anything that needs a pukka ground reference. Should power an ESP just fine.

 

It's also worth noting that there are loads of isolated baby switchers available on AliExpress and Lazada for minimal cost, hardly worth messing about with non-isolated power.

 

5V @ 700mA for 45Baht! 

https://www.lazada.co.th/products/i339218668-s657506548.html

It's probably smaller than you could make one too ????

 

 

 

 

On 7/6/2020 at 5:22 AM, Crossy said:

Do it like this:-

 

 

Get an "X" rated capacitor suitable for AC mains use anything around 0.1 - 0.22u will do the trick. 

https://www.aliexpress.com/item/32947949184.html

 

0.1u will give you about 32k + the 4.7k => about 6mA through the LED

0.22u will give you about 14k + the 4.7k => about12mA through the LED

 

The 4.7k will dissipate 0.2W or 0.6W respectively.

 

Capacitive reactance calculator here if you want to try other capacitors / currents  

http://www.sengpielaudio.com/calculator-RC.htm

 

I received all my parts today and put it together and it lights up nicely however the resister gets very hot.  I used the .22u capacitor.  Thanks for the idea.

8 minutes ago, GregTN said:

I received all my parts today and put it together and it lights up nicely however the resister gets very hot.  I used the .22u capacitor.  Thanks for the idea.

 

Replace the resistor with another of your 0.22u capacitors. Should still give adequate brightness.

 

Or just add the extra cap in series and retain the resistor (which will still get warm).