EEVblog 1482 - Mains Capacitor Zener Regulator Circuit - Page 1
Forum section: electronics / general

author: EEVblog
author_rank: Unknown
timestamp: June 23, 2022, 10:28:12 pm
content: A follow up to the previous video on repairing the heater.A viewer asked how the capacitor diode rectifier gave a 24V output. The key is in the zener regulator, so this vidoe looks at how mains powered zener voltage regulators work, and their limitations.X class capacitor and self healing.Zener diode tutorial video:

author: Kleinstein
author_rank: Super Contributor
timestamp: June 24, 2022, 05:11:02 am
content: The calculated 16 mA is the RMS current. The average current is a bit (some 20%) smaller.With the 3.3 V zener in series the current for the µC and LEDs does not add to the current for the relay. It only adds to the voltage. So this is a clever (cost saving) design as with a capacitive dropper supply the current is limited, but a higher voltage come essentially for free. This is why they did choose the 24 V relay and not the 5 V version.There is an even simpler version of the capacitive droper circuit, using only a zener and 1 extra diode for rectification. This version however uses only half the current and thus needs twice the capacitance. So it's only attractive for very low power. One sometimes still finds this even for a comparable power level - maybe tradition from old times when a diode was more expensive than a capacitor.The reactive power is usually not a big issue. From mains transformers and induction motors the tendency is to have excess inductive load. So the capacitive load from the dropper supplies does not really hurt and oftzen acts as part of the power factor compensation. Most of the reactive power does not have to travel all the way to the generators. Even if it does the generators don't need much true power to provide the reactive power. Moving the reactive power through the grid adds to the grid loss - though not very much.

author: Peabody
author_rank: Super Contributor
timestamp: June 24, 2022, 03:46:27 pm
content: So the power used by such a circuit is constant. If the relay coil isn't energized, that current will instead flow through the zener. In fact the maximum zener current is when the load current is zero.In my limited experience, when these circuits die, it has never been the dropping capacitor that went bad. It's always been the output smoothing capacitor. Probably because it's always an electrolytic. In theory they shouldn't fail - they're never exposed to more than 5VDC. You would think it would always be the cap that's exposed directly to the mains that fails. But in front of me is a Disney coffee mug warmer with this circuit, and the three I've bought have always failed after a couple years, and in all cases it was the electrolytic that failed - actually, very high ESR. No problems at all with the dropping capacitor. So far. Of course the electrolytics aren't exactly name brand parts, so that may be the explanation for my Disney experience. Mickey Mouse capacitors.

author: SeanB
author_rank: Super Contributor
timestamp: June 24, 2022, 04:15:36 pm
content: Lovely vidoe Dave. Seen a simpler circuit, for industrial logic where the designers decided to go even easier. They used a 555 timer, and use pin 3 to operate the relay, 150R coil, and also have a 180R 2W resistor to the other power rail. Set of pads which are drilled out during manufacture, and a second place for the resistor, so that you can decide if the relay is to be normally energised or normally deenergised on power application. Power on some is via a tiny mains transformer, 110VAC primary, or 24VAC primary on some, to a 12VAC secondary, and a simple bridge rectifier and 220uF 25V capacitor for smoothing. There are also 24VDC versions, where they use another 180R resistor as a dropper from 24VDC, with a single diode in the bridge to provide reverse polarity protection. Same PCB for all the dozen or so timer variants, you just cut tracks for relay operation, and use either transformer and bridge, or resistor and diode, as the pin spacing matches on all those options, and where you connect power.Modules all date from the late 1970's, it was a job lot I got on auction for around $5, for around 500 modules, so they have been supplying all my relay needs for a long time, provided I can use the 12V or 24V relays, and get to look for whatever is needed, generally DPDT with 5A contact ratings, though some are SPDT as well. Just have to bear with the 150R coils on 12V, and 300R coils on 24V. A lot of vintage 555 and 741 IC's in there, all well aged, but pretty much all do work when tested. Even used some in industrial control as well, as they were around, and I had the right 8 pin or 11 pin bases.

author: SeanB
author_rank: Super Contributor
timestamp: June 24, 2022, 04:24:05 pm
content: Quote from: Peabody on June 24, 2022, 03:46:27 pmOn 115VAC the dropper capacitor is very much under run voltage wise, so does not degrade from spikes much. The same is not true on 230VAC mains, where the capacitor has full working voltage applied, plus the transients are much higher in voltage. Yes that electrolytic does fail, though more often than not because it is overheated and cooked, probably by sitting next to a heater controlled by the circuit, so it will dry out. Better quality capacitor, and higher voltage rating, will help, but really it just needs to run cooler. If you put in say 1000uF 35V into the application Dave has, it will still turn on the relay even with the class X capacitor having self healed down to 50nF, simply because the 1000uF capacitor will charge up to 24VDC, and have enough stored energy to pull in the relay, which will hold in with much lower current, often a 24V relay will still hold closed on 6VDC across it, simply because of the closed magnetic circuit, and this often is exploited to reduce current draw, here because the relay will probably have 20VDC across it when closed, but still have full voltage to pull in before dropping down. The relay would likely have been pulling in with a bit of mains hum for a while before failing, not something you would notice unless you actually listened, but a fair indication it was losing pull in power.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 24, 2022, 04:40:30 pm
content: The circuit as shown takes an about constant power. There is however a rather counter-intuitive way to reduce the power when the relay is off, by shortening the 24 V supply in this case.

author: robbak
author_rank: Contributor
timestamp: June 25, 2022, 11:18:03 am
content: That relay has a 15mA coil current, a 75% of rated voltage for reliable switch on, and a min release voltage of 5%. What I wonder is what the reliable hold current is for such a relay? Once that capacitor has pushed current through the coil to energize the relay, how low could they let the voltage drop before they risk having the relay turn off?

author: ticktok
author_rank: Newbie
timestamp: June 25, 2022, 12:48:25 pm
content: Wouldn't the cap actually be assisting the power generators of the grid because it would be assisting in power factor correction for the grid?

author: tunk
author_rank: Super Contributor
timestamp: June 25, 2022, 03:26:53 pm
content: Quote from: ticktok on June 25, 2022, 12:48:25 pmA couple of comments on the youtube video:Worf:"While capacitive droppers make poor power factors it really helps the power company. Most load anywhere is generally inductive - think electric motors used for practically everything in industry and residential. Fans used in ventilation, or motors driving compressors in HVAC and refrigeration or motors pumping liquids and such. Thus power companies know the power factor is generally inductive and they have to add capacitive reactance to balance it. There are capacitor banks on distribution lines and at substations to compensate for the inductive loads. You using something with a capacitive dropper helps bring the inductive load down a bit. Though modern LED lights are back to being resistive loads as they are using linear regulation driver ICs. But since the LED array drops about 90% of the line voltage the linear regulator doesn't waste much power. So I wouldn't worry too much about capacitive droppers at all. "Terry Smithwick:"....I once worked for a company with many large assembly lines. They had many large motors each and we used heaters with some inductance themselves. In order to prevent extra fees because we were inductive-loading the power factor out of tolerance, we had a Capacitor Farm. All those magnetic fields growing and shrinking in our motors were feeding power into our own capacitors locally, so the generators saw a normal load and the long-distance wires to the power company could stay cool.So if you plug a capacitive load into your house receptacle, rejoice in knowing that you're just balancing out a bit of your refrigerator's inductive load, and making those generators a bit less loaded."

author: golden_labels
author_rank: Super Contributor
timestamp: June 25, 2022, 06:05:11 pm
content: I have deleted an earlier comment of mine to provide an example. I’m reposting it now and adding a comment regarding what tunk said above. Since those two parts were written separately, they may duplicate some things.edit: by accident I have given maximum instantaneous power, not average power, as the values in the example. The specific numbers are, however, not important for the idea itself.Quote from: ticktok on June 25, 2022, 12:48:25 pmNo — the device lacks power factor correction.That capacitor is storing energy and then dumping it back on the other half of a cycle. Which means that energy is moved back and forth, doing no useful work and just heating up wires.Seepower losses comparisonin Falstad’s circuit simulator to get the feel of what’s going on. The left circuit has 100Ω load and a capacitor to drop voltage, so the load gets about 1W. The right circuit has only a resistive load that also gets 1W, but from much higher voltage. Bottom charts are showing power consumed by loads. Then there is 1Ω wire to power the device, with graphs showing power lost in that wire. Do you see the difference?Regarding tunk’s comment: that is true. But that applies to a scenario where power-corrected load is inductive, the capacitance is properly chosen,(1)and it can actually match the inductive load. Any contribution from that dropper will be swamped by your washing machine or vacuum cleaner. But that’s also why such solutions are not a problem: they are so small that it doesn’t matter they waste power. If you waste 90% of 1W power used by a controller for a 1kW heater, it’s still less than a permille.(1)See any of the magical “energy saver” debunking videos with a random capacitor inside to understand why, in particular in home setting, that is not the case.Simulator code:Code:$ 13 0.0000049999999999999996 19.867427341514983 50 340 50 5e-11c 368 272 368 320 2 0.000001 187.6082321017914 0.001g 368 320 368 336 0 0r 368 208 368 272 0 100r 304 176 368 176 0 1w 368 176 368 208 0R 304 176 272 176 0 1 50 330 0 0 0.5R 512 176 480 176 0 1 50 330 0 0 0.5w 576 208 576 176 0r 512 176 576 176 0 1g 576 320 576 336 0 0r 576 208 576 320 0 101500403 224 208 352 272 0 2_64_7_12291_0.0001_0.0001_-1_1_1.073664970659974403 432 208 560 272 0 10_64_7_12291_0.0001_0.0001_-1_1_1.0728852632394683403 240 80 368 144 0 3_64_7_12291_0.0001_0.0001_-1_1_0.010736649706599702403 448 80 576 144 0 8_64_7_12291_0.0001_0.0001_-1_1_0.0001

author: ticktok
author_rank: Newbie
timestamp: June 26, 2022, 04:09:01 am
content: Thanks Golden_labels.. Yes, but this cap is included in the circuit for practical reasons, and even though the losses are negligible, this minute amount would be offsetting a naturally inductive grid right? Everyone is a winner when capacitors are used as load droppers in these grid connected domestic appliances? Dave seemed to suggest that there would be some sort of cost that is passed on to the generators, so I am a bit confused.

author: Kleinstein
author_rank: Super Contributor
timestamp: June 26, 2022, 06:33:45 am
content: For this application wasting a little power is not a problem - it is a heater after all and supposed to convert electricity to heat. So heat lost in the electronics is not even really lost.

author: golden_labels
author_rank: Super Contributor
timestamp: June 26, 2022, 01:57:16 pm
content: Quote from: ticktok on June 26, 2022, 04:09:01 amNo. The current still goes back and forth along the wire. Just in this case to the inductive load elsewhere in the grid.Power factor must be corrected at the spot where it is needed. Between the load and the device used for power correction, whether it is a capacitor, an inductor or an active power correction circuit, the losses are still present and unavoidable. This is why the capacitor banks mentioned by tunk are near the motors and not simply rented from some company 40km away.This is not a physically correct analogy, but imagine a tall pole on top of which some person swings back and forth. Your goal is to keep the pole stable. A hard task, right? You are the electric grid, the person on the pole is the load. I believe we will both agree that giving you a task of stabilizing a second pole with another person swinging is not making that task any easier. What you need is both persons to be on the same pole, swinging in opposite direction.

author: Peabody
author_rank: Super Contributor
timestamp: June 26, 2022, 02:04:23 pm
content: Do things like air conditioning compressors have capacitors built in to correct the power factor?

author: Kleinstein
author_rank: Super Contributor
timestamp: June 26, 2022, 03:17:43 pm
content: 3 phase motors usually don't have a capacitor for PFC correction build in. With variable speed motors the power may go though a rectifier and inverter. In this case they should have PFC included for more than just the capacitive / inductive part but also the harmonics ( a simple rectifier gives pulsed current and thus harminics a simple capacitor does not correct very well).With a 1 phase induction motor there normally is a capacitor as part of the normal motor function and this capacitor may very well also do some PFC correction.Quite often the phase angle is noted on the name plate - at least there is often a field for this, though not always filled in. So one can see how bad it is. For normal private home use there is no much to worry about as the reactive power is not charged extra and the utilities take care of the phase shift by adding more or less capacitors at the distribution transformers (at least in Europe - the small ones in the US may not have this).The inductive reactive power is mainly an issue with industrial use from multiple large motors and also in a 230 V country with old style balasts for fluorecent lamps. With the lamps it was common to have some (e.g. every 3rd or 2nd) run with a series capacitor and do some compensation (with additional flicker reduction). Those old style lamps are a thing of the past anyway.There is no need to get perfect local compensation - as long as the reactive power is small the losses are also small as the resistive loss goes with the square of the current. So not much to worry about 10% reactive power: that would increas the wire losses by some 1% (1% of the loss not 1% of the total power).

author: f4eru
author_rank: Super Contributor
timestamp: June 26, 2022, 03:18:22 pm
content: One very good way to save power in sleep mode (relay not activated) would be to short circuit the 24V zener with the transistor to open the relay.