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Simple Light Sensor (because I am lazy) - Capacitor Drop Supply
Posted by paul_g_787 on 16 Jan, 2022 20:50 -
I am working on a little light sensor, it is a very simple project to detect when a light comes on in a rack cabinet at work.
(This way I can hear it from the bottom of the stairs and don't have to run all the way up just to check one dang light all the time).
I have designed it out of parts I have knocking about.
I am not sure what value the LDR is though, it is unlabelled, but I just used trial and error to find a working voltage divider combination.
The LDR measures over 200MΩ in the dark and 0Ω in the light...
The breadboard version is working great so far on a 9V battery!
It uses 14mA when idle and 39mA when the buzzer and LED come on! (practically nothing).
Next is to put it in a case and make a power supply.
I have an old faulty mains battery charger which I would like to use as a project case. It is a wall-wart type device that has been sitting in my box of broken junk for ages.
It already has 3 see-through plastic windows in the front of the casing which I can use for the LDR and 2 indicator LEDs and the hole from the original wire can be used to let the sound out from the buzzer.
I think this would be a great casing as once it is assembled it will be impossible to actually touch any of the components unless you unplug it and unscrew the casing! so pretty safe (safer than some of the LED light bulbs I have seen).
So I have settled on this as the casing as a winner!
Next I need to supply the circuit with 9V. I would like to use a capacitor dropper supply for this for cheapness and simplicity.
So I re-used my design from my LED light bulb project, however this is still experimental and there are a few bits I would like people's opinions on:
C1 (Class X2 dropper cap) - I need to pick a value but when I calculated this the same as I did for my LED project I ended up having to use the next value up. is 470nF enough?
C2 - is 47µF large enough here to smooth the ripple?
ZD1 - Is a 9.1V 0.5W zener going to be sufficient here? (1N4739A is the one I have)
R3 - I thought this would be good as a load resistor? Is this even required?
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You don't need a bridge rectifier, a single diode for half wave rectification would be plenty for this project. You can bump the smoothing capacitor up to something like 220uF 25v or even higher... they're not much bigger in volume.
I'd encourage you to find a buzzer that works at lower levels like 5v because you're wasting a lot of current on the resistors limiting current to leds, you could make it much more efficient with a lower voltage.
ex: https://www.digikey.com/en/products/detail/pui-audio-inc/AI-1027-TWT-5V-R/1745429
I nearly said "go get a greeting card with some melody" and run everything at 1.5-3v but the greeting card is probably expensive and you already have a buzzer.
Microchip has an application note about transformerless power supplies: http://ww1.microchip.com/downloads/en/AppNotes/00954A.pdf
It's a bit controversial as the formulas are not quite correct, but eh, good enough and the schematics can be helpful : -
You don't need a bridge rectifier, a single diode for half wave rectification would be plenty for this project. You can bump the smoothing capacitor up to something like 220uF 25v or even higher... they're not much bigger in volume.
I will give that a try to see however I may keep the rectifier as I could still see the LED flickering on my LED Lamp circuit I made before. I just try it and see what it looks like.
I have a few other caps to try. I will measure the ripple and see what it looks like.I'd encourage you to find a buzzer that works at lower levels like 5v because you're wasting a lot of current on the resistors limiting current to leds, you could make it much more efficient with a lower voltage.
I could run it on 5V yes as I actually have a spare 6V buzzer! I will make a new version that runs on 5-6V and see what the current difference is. -
Well, it's simple math ... let's say the green led has a 2.2v voltage drop and you feed led+resistor 9v ... that means the 470 ohm resistor has to drop 9v-2.2v = 6.8v ... V = I x R => 6.8v = I x 470 -> I = 5.8 /470 = 0.0144 A or 14 mA and the power wasted in the resistor is P = IxIxR = 0.014 x 0.014 x 470 = 0.09 watts
Drop the voltage to 5..5.5v (ex if you use a 5.6v zener diode) and use the same 470 ohm resistor ... now your green led will do 5mA and your resistor will only dissipate 0.01 watts
The green led will still be visible at only 5mA
With a full bridge rectifier, you can sort of estimate the minimum capacitance you'd need for a current amount with the formula C = Current / [ 2 x ac frequency x ( Vdc peak- Vdc min desired) ]
So let's say for 50mA and minimum 5v all the time and a peak voltage of 9v (if 9v zener is used) then C = 0.05A / (2 x 50 hz x (9v -5v) ) = 0.05 / 400 = 0.000125 F or 125uF
Of course with half wave rectification, you'd need a higher capacitance value ... but 220/330/470uF capacitors should be very common and easy to get or you should have one around. -
It's all kind of questions that a small simulation could answer in no time.
IIRC, I found 470nF to be adequate for average current of ~7mA on 230V mains.
I don't think full bridge buys you anything except smoother current delivery (twice denser pulses, half charge per pulse).
14mA idle doesn't seem like "practically nothing", where are you burning it? The green LED?
Regarding zeners, they have a somewhat hard life in this circuit because their current is not steady state but pulsed. -
You don't need a bridge rectifier, a single diode for half wave rectification would be plenty for this project. You can bump the smoothing capacitor up to something like 220uF 25v or even higher... they're not much bigger in volume.
I will give that a try to see however I may keep the rectifier as I could still see the LED flickering on my LED Lamp circuit I made before. I just try it and see what it looks like.
I have a few other caps to try. I will measure the ripple and see what it looks like.I'd encourage you to find a buzzer that works at lower levels like 5v because you're wasting a lot of current on the resistors limiting current to leds, you could make it much more efficient with a lower voltage.
I could run it on 5V yes as I actually have a spare 6V buzzer! I will make a new version that runs on 5-6V and see what the current difference is.
and then you can power it from a USB port -
Which simulation software would you reccomend I use? I used a piece of software called Livewire at school for windows XP but I can't find this anymore. I don't think they make it anymore.
I use linux mint and I use kicad and eeschema to draw my schematics.
The 13mA is in the green LED yes. It is a regular 20mA LED and I am running it slightly lower current as it doesnt need to be full brightness. 10mA would be acceptable but in my testing lower than that it looks too dim. The LED is visible as low as about 6.5mA but very dim. -
So I re-built this circuit today to work on 5V. I first used a 5V USB power bank to create the circuit. It is working great like this.
I lowered the LED resistors to 330R as they were too dim to see properly at 5V with 470R.
The power LED uses 10mA. The buzzer and warning LED use approx 15mA. (25mA total approx).
I changed the C2 capacitor to 100µF which gives me approx 0.09V ripple with no load.
So I re-designed the power supply with the suggestions made in mind and this is great, and has 5.09V with no load!
However when I connect my circuit to it, it drops to 4.25V and this is causing the circuit to not work properly. Why such a huge drop in Voltage?
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ZD1 should be to the left of D1. I can't see how it would work at all as drawn.
Current flows to the left through the cap: ZD1 sources it from neutral, D1 blocks negative voltage from the load and prevents discharging of storage capacitor.
Current flows to the right through the cap: ZD1 limits voltage, D1 permits current to the load and storage capacitor.
As for simulation, I use LTspice. It will run on Windows, it will run on Linux with wine. It's reasonably easy to operate and you draw the schematic graphically.
I think Kicad was supposed to have some simulation functions too? Never used it. -
ZD1 should be to the left of D1. I can't see how it would work at all as drawn.
Current flows to the left through the cap: ZD1 sources it from neutral, D1 blocks negative voltage from the load and prevents discharging of storage capacitor.
Current flows to the right through the cap: ZD1 limits voltage, D1 provides current to the load and storage capacitor.
As for simulation, I use LTspice. It will run on Windows, it will run on Linux with wine. It's reasonably easy to operate and you draw the schematic graphically.
I think Kicad was supposed to have some simulation functions too? Never used it.
Ah OK thank you. Yes I see you are quite right! I moved the zener diode, however I am now getting 4.7V without the load.
Is this because it is clamping the peak voltage at 5.1V?
Do you think I need to use a larger Zener to increase the voltage to 5V? Maybe 6.2V?
Will do some more testing tomorrow. -
Output is less than nominal ZD1 voltage because of forward loss in D1. A higher zener will compensate for that.
Another thing is that the 330nF cap is essentially a current source. Its right side is more or less at 0V all the time, its left side varies ~650Vpp → there is 650V·330nF of charge injected into the load per mains cycle (20ms). Or 650V·330nF·50Hz of average current. So the cap needs to be sized for the expected load current.
Load voltage is less important, it will work about the same as long as it's much less than mains voltage. -
So I re-built this circuit today to work on 5V. I first used a 5V USB power bank to create the circuit. It is working great like this.
I lowered the LED resistors to 330R as they were too dim to see properly at 5V with 470R.
The power LED uses 10mA. The buzzer and warning LED use approx 15mA. (25mA total approx).
I changed the C2 capacitor to 100µF which gives me approx 0.09V ripple with no load.
So I re-designed the power supply with the suggestions made in mind and this is great, and has 5.09V with no load!
However when I connect my circuit to it, it drops to 4.25V and this is causing the circuit to not work properly. Why such a huge drop in Voltage?
if it works from USb why not power it from a random charger from the junk pile?
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Output is less than nominal ZD1 voltage because of forward loss in D1. A higher zener will compensate for that.
Ah yes I see!! I didn't even think of that. How silly of me, it is obvious now you pointed it out!Another thing is that the 330nF cap is essentially a current source. Its right side is more or less at 0V all the time, its left side varies ~650Vpp → there is 650V·330nF of charge injected into the load per mains cycle (20ms). Or 650V·330nF·50Hz of average current. So the cap needs to be sized for the expected load current.
Load voltage is less important, it will work about the same as long as it's much less than mains voltage.
What do I need to look for in the capacitor datasheet to see if it it can handle the load current?
I just did the maths for different values of capacitors and now I understand that this is where the rest of my voltage drop is coming from as the 330nF is barely providing enough current! I think a 470nF would be more suitable to provide a little more. -
if it works from USb why not power it from a random charger from the junk pile?
I could very well do that yes, however I saw this as a perfect oppurtunity for a practical project to use what I have been learning about capacitive dropper supplies.
Theory is one thing and I find that I learn much better putting things into practice. I have already learned a few things from my design mistakes already!
I think the capacitive dropper supply is an important topic to understand due to the potential dangers in such a design if done incorrectly. -
What do I need to look for in the capacitor datasheet to see if it it can handle the load current?
I only mean that capacitance must be sufficient.
The caps have some maximum current ratings too, but they are many amps, no problem. -
What do I need to look for in the capacitor datasheet to see if it it can handle the load current?
I only mean that capacitance must be sufficient.
The caps have some maximum current ratings too, but they are many amps, no problem.
Ok I misunderstood, I thought you meant there was another rating I was not aware of. -
Why would you go with a capacitive dropper circuit though? They're used inside PIR relays and smoke alarms because they are dirt cheap, but they have terrible efficiency (and power factor, though I guess the effect is negligible).
I wouldn't make a project like this in a work setting and leave it connected to the mains 24/7 unsupervised. If you use an off the shelf adapter, the isolation is taken care of by an approved device. I wouldn't automatically assume that means you're free from liability, but a DIY mains connected device would certainly seem to open you to liability on multiple fronts. -
Why would you go with a capacitive dropper circuit though? They're used inside PIR relays and smoke alarms because they are dirt cheap, but they have terrible efficiency (and power factor, though I guess the effect is negligible).
I wouldn't make a project like this in a work setting and leave it connected to the mains 24/7 unsupervised. If you use an off the shelf adapter, the isolation is taken care of by an approved device. I wouldn't automatically assume that means you're free from liability, but a DIY mains connected device would certainly seem to open you to liability on multiple fronts.
yeh, there is a reason why manufacturers use wall-warts/AC-adapters when ever they can -
I have built capacitive dropper powered circuits before and they work pretty well for low powers.
I usually use it in places where i am doing something with mains, like switching things with a triac, so there is already dangerous mains potentials on the board anyway. The capacitive dropper lets me make a tiny, simple and reliable power supply for it. But you want your power consumption to be under 10mA average or your dropper cap gets rather big physically(negating the point of this tiny PSU type).
How i typically do it is Mains->Fuse->Inrush resistor->Dropper cap-> Halfwave rectifier->Zenner 10V->Filtering cap->Small 5V linear regulator. Gives a nice clean stable 5V even during short surges of current draw (that's why the zenner is set at 10V). The ones i built have been running for many years and survived trough quite a few thunder storms just fine (where lightning blew up other electronics). The efficiency is not that bad with these really. A transformer at these powers is worse, only well designed small SMPS supplies are better. Yes these cap droppers have a lot of apparent power draw, but very little of it is actual real power. That reactive power that they do draw is capacitive and nicely sinewave, so it helps balance out the mostly inductive reactive power in the mains network.
As for your circuit. That always on LED is a huge waste of energy. If you need that much current trough a LED for it to be visible, then you have a really crappy low efficiency LED. A good modern high brightness green LED at 10mA will be so bright that its uncomfortable to look at indoors at a distance. These modern LEDs only need about 0.5mA to be plenty bright enough for use as an indicator.
However you should NOT be messing with these if you are not experienced with mains electronics. They are very dangerous things. -
You can save one diode by using the "zener to ground followed by forward diode" configuration. It also means that if the zener goes open circuit, the supply will simply stop working rather than blow up the load.
OTOH, small 0.5A USB chargers are getting really small these days. Particularly if you don't need the safety and are willing to crack their case and extract bare PCB. -
Yeah my designs used a separate diode for rectification. Once of them actually went swimming in water because it was mounted outdoors and a corner of the plastic box got damaged, giving it a crack that let rain start collecting inside. There was supposedly a bang inside and it stopped working. What i found was a black mess around the PSU cirucit. I think the fuse was blown, inrush resistor blown, the zenner blown and perhaps a diode here or there. But i do remember that the 7805 regulator and all the circuitry afterwards was fine, so it was an easy fix to clean up the board, replace a fuse and some cheep common parts.
I have also use taken apart 5V chargers and soldered them down to the board as a "PSU module"(I think you can actually buy them as potted bricks these days too). The one downside is that these cheep little chargers is that they get killed by lighting the easiest of all mains powered devices. I tried to make them a bit more robust by adding a series resistor on the mains input and it worked so far. The current draw on the mains side is really small so it doesn't get very hot while it helps absorb some of the surges.
In terms of long term reliability oldschool iron transformers are probably the best, but very rare these days (SMPS is both smaller and cheaper) -
Thanks for all the useful information everyone, I am now understanding this all a lot better thanks to all the help!
I now understand the concept behind the fusible resistor and also now understand the concept of constant current source and how the Zener diode is 'regulating' the voltage.
I have been having a look at LEDs to find some lower current ones as suggested, I think this would be a great idea. Although almost all of the LEDs I can get are nominal current 10mA, max current 20mA.
I have found some 3mm LEDs that are 5mA nominal current, max current 10mA but these would be too small for my project and it would look a bit odd. 5mm Seems to be the size I need.
Am I looking at the right thing here?
Also, I have recently come across this neat little calculator! This is quite helpful for double checking my calculations.
http://www.nomad.ee/micros/transformerless/index.shtml
Those that are worried, I appreciate the concern and yes I do understand the dangers of this type of circuit and that the circuit is always at mains voltage (with respect to ground and therefore me as well). I do know that working with mains voltages can kill me if I am not careful. If I was not confident I could do so safely I would not work on this.
If it will put your minds at rest, while I have been working on this project I have taken some safety precautions:
I have been using a plug-in RCD GFCI type adapter to connect to the mains voltage.
I have been running the live via a 1A fuse before connecting my circuit as a back-up fuse.
I have been un-plugging from the outlet before touching the circuit.
I have been checking with my volt meter that the capacitors are discharged every time before touching the circuit.
I have also been wearing rubber gloves and my rubber-soled slippers and sitting on a wooden chair. -
The test current will be 10mA or 20mA but the amount of light emitted will be different at various currents.
For example :
1. https://www.digikey.com/en/products/detail/inolux/INL-3AG30/7604619
datasheet: https://www.inolux-corp.com/datasheet/Display/Through-Hole-Display/Round-Lamp/INL-3AX30%20series.pdf
min 11000 mcd, max 18000 mcd at 20mA test current, but at page 6 you have current vs relative luminous intensity so if you run at 10mA you can expect around 80% of intensity and at 5mA around 40-50%
2. https://www.digikey.com/en/products/detail/w%C3%BCrth-elektronik/151033GS03000/4490000
datasheet: https://www.we-online.com/katalog/datasheet/151033GS03000.pdf
They're nice enough to give you mcd vs current on page 4 ... so as an idea this led claims to do 4000mcd at 5mA
and now you can compare it with cheaper led :
1. https://www.digikey.com/en/products/detail/w%C3%BCrth-elektronik/151031VS06000/4489988
datasheet : https://www.we-online.com/katalog/datasheet/151031VS06000.pdf
See same graph again at page showing 15 mcd at 10mA ... far from 4000mA at 5mA , or around 5000 mcd at 5mA for the first.
or the one available in biggest quantity on digikey
https://www.digikey.com/en/products/detail/tt-electronics-optek-technology/OVLBG4C7/827110
datasheet: https://www.ttelectronics.com/TTElectronics/media/ProductFiles/Datasheets/OVLBx4C7.pdf
50% luminous intensity at 10mA , around 25% at 5mA ... and the intensity is claimed to be minimum 4360, max 8400 ....
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I have been having a look at LEDs to find some lower current ones as suggested, I think this would be a great idea. Although almost all of the LEDs I can get are nominal current 10mA, max current 20mA.
I have found some 3mm LEDs that are 5mA nominal current, max current 10mA but these would be too small for my project and it would look a bit odd. 5mm Seems to be the size I need.
Am I looking at the right thing here?
Yes small LEDs are typically rated for up to 10 to 20mA of current. But the efficiency of cheep LEDs is way worse, so it means the good quality high brightness LED might produce 10x to 100x more light at the same 20mA current. This means if you run a 100x more efficient 20mA rated LED at only ~0.2mA you will get the same amount of light out of it as the cheap one at 20mA. The spec to look for is the milicandelas of light coming out since the light is what you want from a LED. So just buy a really bright led and reduce the current until the LED is at the desired brightness. Tho the highest candela LEDs are typically clear lens narrow angle ones, those are not the best indicator lights because they are hard to see at an angle(all the light if focused straight forwards), so the best ones for indicators are actually the non clear, milky plastic LEDs with a wide beam angle.
The highest efficiently LEDs at 20mA can actually be painfully bright to look at up close. So you don't even want to run those at full rated current unless you like seeing colored blobs in your vision after looking at the LED.