Zero-crossing Switch Saver - Page 1
Forum section: electronics / beginners

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 28, 2020, 07:51:20 pm
content: So, I had this problem with mechanical load switching depicted in Fig.1. The switched currents were 10s of Amps and were destroying the mechanical switch. The slightly capacitive nature of the load did not help the matter.As a solution, I have constructed a Switch Saver depicted in Fig.2. It was important to preserve thezero energy consumptionof this circuit in its OFF state.I elected to use power MOSFETs as the switching elements because I was operating this circuit in the area where the voltage drop across the source and drain of the MOSFEts is still lower than the voltage drop across the emitter and collector of saturated IGBTs.I also tried using Triacs, but they did not work very well because they have even higher voltage drop than IGBTs and were heating up under load. Additionally, they turned off 2x per AC cycle and did not turn back on immediately after the AC zero-crossing, thus creating harmonics on the supply line which bothered my other electronic equipment. They do this not only when the circuit is initially switched on, but continuously as long as the load is powered on.The MOSFET Switch Saver circuit depicted in Fig.2 works by charging C1 and C3 on alternate halves of the AC cycle and the C1(R1+R2) time constant*being 5x longer than the period of the AC cycle so the gate of the MOSFET is powered continuously.R1 and R2 form a voltage divider that applies approximately 15V to the MOSFET's gate (R3 and R4, too).R5 limits the current flowing through the switch S1, while D2 and D4 protect the gates of the MOSFETs from overvoltage.However, I still suffer from dips on the supply line when I am unlucky enough to turn on the switch at the peak of the AC cycle...not after the circuit is ON, though.QUESTION: How would you improve this circuit so it turns on the MOSFETs at the zero-crossing of the AC cycle, while preserving its zero energy consumption in its OFF state ?NOTE: The minimum acceptable turn-on delay is 300ms.*and C3(R3+R4) time constant being 5x longer than the period of the AC cycle

author: BrianHG
author_rank: Super Contributor
timestamp: August 28, 2020, 08:08:05 pm
content: Are you against a simple triac?The performance would be similar to an IGBT, but only a triac and resistor with the switch plus 1 additional optional pulldown resistor.Having access to the other side of the AC terminal would also eliminate that 0.7v crossover dropout in the case or the triac or that larger drop voltage in the case of the mosfet design.A triac should not consume any current in the off state unless you are have a snubber resistor and cap. However, you dont have any such protection with your mosfet circuit anyways.Note this does not perform any zero-crossing detection.Use a good hefty 1 or 2 watt 100Ohm resistor and it should last.

author: langwadt
author_rank: Super Contributor
timestamp: August 28, 2020, 08:19:08 pm
content: when the switch is on there is no voltage across it so no way to charge C1/C3

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 28, 2020, 08:20:13 pm
content: Quote from: BrianHG on August 28, 2020, 08:08:05 pmYes, Triacs heat up more than MOSFETs and create interference on the line as they turn on AFTER the zero-crossing not only when the circuit is first switched on, but continuously when the load is powered on as well.I discussed this in my original post.Also, I would like to turn on this entire circuit magnetically with a small reed switch and the 1.2A is too much for it (120V/100Ω = 1.2A). Not to mention that 1.2A*120V=144W, which is a lot of wasted power just to keep the Triac on (although a cap or depletion mode MOSFET connected in series with that 100Ω resistor would reduce that power overhead significantly).Originally the reed switch was energizing an electromechanical relay and the relay's contacts were burning out. I tried putting in a bigger relay, but the current requirement of its coil was too much for the little reed switch (yes, a flyback diode was installed across the relay's coil).

author: BrianHG
author_rank: Super Contributor
timestamp: August 28, 2020, 08:22:34 pm
content: But your mosfet circuit creating worse line interference.Unless you gain access to the 'HOT' feeding the other side of the load, there is no way to charge your mosfet gates continuously through the cycle.Now, if you want to insert a 12v lithium primary cell battery in your circuit with something like a 10megaohm pulldown on the gates and change the 2 mosfet connection a little, that would work. But, you would need a shielded cable to the switch or the switch would need to be really close to the circuit.When on, the battery would drain with 1.2uA or 14.4uW if I got my decimal place correct.That would also be the current sent to your switch when turned on. LOL...Though, when the battery drains to a meeger ~4v, the mosfets will cook as they partially switch on.Another advantage is I can imagine additional circuitry for zero crossing detection, but that may draw some finite current when switched off. Or maybe not if you want to draw a little more than 1.2ua from that battery.

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 28, 2020, 08:49:30 pm
content: Quote from: langwadt on August 28, 2020, 08:19:08 pmSorry, I missdrew how the load is connected in Fig.2. I have corrected it now.

author: BrianHG
author_rank: Super Contributor
timestamp: August 28, 2020, 09:12:14 pm
content: Have you though about flipping around the mosfets, IE tying the sources together and gates together?All you need to do is supply 1 single voltage of 15v between the gates and source to switch on, and 0v for off.A bootstrap style supply generated from the AC now that you have access to the other side becomes possible.Also, if you don't mind spending an additional few dollars, you can go really extreme and wait for a zero crossing for both turn on and turn off.

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 28, 2020, 09:48:30 pm
content: Quote from: BrianHG on August 28, 2020, 09:12:14 pmCould you complete your idea how the bootstrap supply would work with a SPST switch ?

author: BrianHG
author_rank: Super Contributor
timestamp: August 28, 2020, 10:09:24 pm
content: Here you go:C1 needs to hold a large enough charge to keep the mosfets at 15v for the duration where the 120v is at negative phase.You can make C1 a 500v cap and place it before the switch. Meaning the switch is now in between R1 and R2. This will give you instant on and instant off control. allowing for a larger C1, like 10uf guaranteeing that the gate will have a full charge during the negative cycle.R3 is optional.If you went with an MCU in between the zener and gate to drive the gate only during a 0v crossover, this is up to you. This change would allow a controlled turn on and off time so long as the MCU cap's power holds out.You might also do this in an analog fashion by muting the gate's charge until the voltage between both drains are 0v, but, that circuit would add a little off current.

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 28, 2020, 11:27:50 pm
content: I was expecting a full-wave rectification in the bootstrap supply circuit, but that is not possible with the sources tied together. However a half-wave rectification scheme will work, too.Additionally, your solution uses less parts for the same functionality when a bigger cap is used to maintain the ripple voltage amplitude in the gate circuit at the same level as with my circuit. So it is better.Now, for the zero-crossing detector circuit, I was thinking about driving the gate from a RS flip-flop made out of NAND gates from a high-voltage CMOS chip, which has low static power requirements, ...like the CD4011.This chip would be powered from the gate circuit bootstrap supply, i.e. from the C1 in your schematic.The RESET pin of this flip-flop (point A) would be kept active (low) until the chip's supply voltage reached e.g. 15V and after that, the flip-flop's SET pin (Point B) would be armed to wait for the low voltage level from the AC line - the zero crossing.What do you think ?P.S.Note the absence of batteries, which require periodic maintenance.

author: BrianHG
author_rank: Super Contributor
timestamp: August 29, 2020, 12:16:12 am
content: I need to think this one through, but, I think point B should detect the rise and fall at 2 points from the mains or only detect the transition point as to not get fooled at the center of a single phase, or, as the 2 mosfet's sources partially float around after a cap charge.For sensing point B, I would be using something like an XOR gate, 1 input series resistor to mains, bidirectional diode protected, with the second input having a series resistor to the first input with a cap to GND to introduce a delay. The output of the XOR should pulse once every half cycle while the 2 inputs aren't equal when the AC 1/2 VDD crosses due to that RC delay.This logic could be replicated with the NAND gates at your disposal with a little re-wiring.

author: BrianHG
author_rank: Super Contributor
timestamp: August 29, 2020, 02:29:18 am
content: Quote from: GonzoTheGreat on August 28, 2020, 11:27:50 pmMy circuit will also work fine with DC so long as the polarity is correct.

author: BrianHG
author_rank: Super Contributor
timestamp: August 29, 2020, 02:39:42 am
content: You can use a simple 4000 series D Flipflop.Tie the CLK to the AC input through series resistor.Tie the Din to +VDD.Tie the RSTn with a cap to VDD with a pulldown and discharge diode so it always powers up with reset low output.Once the reset is released, the VCC will need to go from low to high before the output switches on. This means if the output is already high, the output will stay low until the AC input cycles low, then back to high another clock cycle later.You may use the zener diode trick on the reset input to make sure the VDD goes past a point before the reset is finally released.Now, because of the way mosfet Q1 and Q2 diodes are open to allow the sources to fly around when the switch is open, you will need to confirm this behavior on the AC input as the sources is you logic GND point.Also, the CLK input should have diode protection with a 100pf to 1000pf cap to gnd and a 1meg series resistor to the AC line so the flipflop sees a 60Hz square wave at the CLK input.Your other choice is a binary counter with the last bit feeding an ENAn input and cheap reset circuit as mentioned above. With this circuit, you may make the switch wait for 4, 8 or 16 60Hz counter cycles before it activates. Yes, a real power up timer clocked to the mains. And it doesn't matter is that counter skips a beat or 2, as long as it the increments or skips happen during the 60Hz transition, your turn on time will always be during that 1/2VDD mains transition time which is where you want it.Using a counter of 8 or 16 will virtually guarantee that during power up, no CLK signal glitching will turn on the mosfets unless for some reason, you get 16 consecutive spikes strong enough to overpower the mains driving the CLK input.For the counter, you need one with a reset & !count_enable inputs. The output which drives the mosfets goes to the !count_enable to stop the counter once the mosfets have been turned on. The reset makes sure the counter powers-up in a reset state.Finally, you can use both D-flipflops in the 4013 as a 2 stage binary counter, where the first flipflop !Q output is tied to that stages Din using a 100k series resistor and a 10nf cap to GND at the D-input to filter out multiple cycle during a noisy clock transition. That stages !Qout goes directly to the CLK in of stage 2 meaning 2 cycles of the first stage is required to clock the second stage. The second flipflop Din would need to be wired to VDD so once clocked by the first stage's !Qout so it goes high and stays high from there on.

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 29, 2020, 05:02:46 pm
content: Quote from: BrianHG on August 29, 2020, 02:39:42 amAll of the D Flip-Flops in the 4000 series have reset inputs that are active high.Quote from: BrianHG on August 29, 2020, 02:39:42 amThat is a very valid point since the zero-crossing across the AC line does not have to appear as 0V with respect to these flying sources.I just tried to probe it, which is not easy without blowing up the scope, and it seems that the reverse current of the MOSFET body diodes flaps the tied sources ±30V.

author: BrianHG
author_rank: Super Contributor
timestamp: August 29, 2020, 05:57:49 pm
content: Quote from: GonzoTheGreat on August 29, 2020, 05:02:46 pmAll this means is that you tie the reset 100nf filter cap to the VDD with a pull-down instead of the other way around. A discharge diode, IE 1N914, goes from GND to the reset input to discharge that cap when powered off.You can also use the same logic with a zener diode reset. It's just wired to GND with a pullup on the reset.

author: langwadt
author_rank: Super Contributor
timestamp: August 29, 2020, 08:54:03 pm
content: is zero crossing really needed, if the components are choosen to slow down the turn-on/off ?

author: BrianHG
author_rank: Super Contributor
timestamp: August 30, 2020, 05:31:07 pm
content: Doesn't actually need to be at the 0 cross itself. It may even be worse at the 0 cross. Maybe turning on after the peak, say when the voltage is at ~50V and falling to give a predictable 1/2 voltage jolt first before the next full power cycle.

author: Zero999
author_rank: Super Contributor
timestamp: August 30, 2020, 06:37:56 pm
content: Quote from: GonzoTheGreat on August 28, 2020, 08:20:13 pmThat depends on the MOSFET vs TRIAC and the current. You've also got two MOSFETs in series, which increases on losses further.

author: BrianHG
author_rank: Super Contributor
timestamp: August 30, 2020, 07:19:17 pm
content: Quote from: Zero999 on August 30, 2020, 06:37:56 pmWell, you cannot say. Maybe GonzoTheGreat is using something like a few IXFH220N20X3 in parallel...He can get a 25 amp load to only drop something like 0.1v or less across the 2 mosfet series with 3 parallel mosfets across each side. IE 6 mosfets total for a 100$ mosfet switch...I kinda like a 2 step switch idea, the first stage with a heavy series 10 ohm resistor and the second mosfets directly connect. Sorta like a soft start I used to use to startup a huge 2kva transformer I used to cycle on and off a few times an hour. Completely erased any surge and also the switch 'knock' noise in the transformer itself. Though, at the time, I did it with 2 normal SSRs.

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: August 31, 2020, 08:00:58 pm
content: Yes, I had to parallel the MOSFETs when powering the capacitive load because they were burning out.When using my CD4011 zero-crossing circuit I can get away with single MOSFETs though.I will post some waveform scopeshots soon.

author: Zero999
author_rank: Super Contributor
timestamp: August 31, 2020, 09:45:19 pm
content: Quote from: BrianHG on August 30, 2020, 07:19:17 pmI didn't say which has lower losses, just that it depends on the MOSFETs and we don't know what MOSFETs he's using. At the time of making that post, I missed that the voltage is 120VAC, rather than the 240V, in my location, so only 200V MOSFETs are required, which will probably make them more likely to have lower on losses, than a TRIAC.Quote from: GonzoTheGreat on August 31, 2020, 08:00:58 pmYes, you need zero crossing for a capacitive load. Incidentally, if the load were inductive, you'd want to turn on during the peak voltage, not zero crossing, but that's off-topic.

author: GonzoTheGreat
author_rank: Regular Contributor
timestamp: September 01, 2020, 11:28:32 pm
content: Quote from: Zero999 on August 31, 2020, 09:45:19 pmYes, if this switch used zero-voltage switching during turn-on and zero-current switching during turn-off, then it would be the optimal behavior for the switch and for the load.I might create another thread to pursue such switch after I debug this one.