Where and why are static relays used?
Static relays are used to control (switch) without risk the consumers of AC power supply to the 230Vac network, in dangerous environments (with danger of explosion, for example, due to the electric arc produced by the mechanical contacts), static relays are used.
The order is made at low currents, but also has the disadvantage of high power consumption that mechanical contacts do not have.
How does the electronic scheme work?
For control, the mains voltage (via R2-C1) is rectified with the 4 diodes D2-D5: the obtained voltage (signaled by D6), stabilized at 9V (with D1) controls the optotriac LED when pressing switch S1.
The advantage: the S1 switch has a much smaller size than if the order had been made classic. The optocoupler in the optocoupler controls the power triac grid, Q1, which feeds the consumer (when S1 is pressed).
List of required components (with equivalents)
R1 = 220 Ohms;
R2 = 68 Ohms;
R3 = 30 Ohms;
R4 = 560 Ohms;
C1 = 100 nF (polyester capacitor, working voltage of at least 275Vac);
D1 = Zener diode 9V
D2 - D5 = 1N4007;
Q1 = BT136 (or BTA08-600B);
Q2 = MOC3040 (or MOC3041);
S1 = normal switch;
1 x terminal contact (for AC power supply);
Bibliography:
Conex Club Magazine - no.7-8 - 2006
Electronic Practice: Electus Distribution
https://www.findchips.com/
https://components101.com/
5,307 total hits, 1 hits today
It looks like an optocoupler and a thyristor that simulates the relay and gives voltage to the control.
I don't know how this scheme works, I see that the triac is the consumer !!
In the absence of the control pulse on the grid, the triac is blocked, and behaves, between the main terminals T1 and T2, like an open switch (the anodic current is negligible in both directions).
When the terminal T2 is positive in relation to T1, it triggers the thyristor T1; When the T2 terminal becomes negative and the T1 terminal acts as an anode, the thyristor T2 enters the conduction, which initiates when the gate is negative in relation to T1 (if the potential of the G gate is positive in relation to that of the A1 terminal).
BT136-TRIAC-Pinout
Thanks for the info, I knew how the triac works, its position in the assembly confused me, it's basically the consumer.
I don't see why you say it's the consumer, because basically the triac is made not to be. That is his purpose. Controls current in both directions.
We found a similar scheme, simpler using the same main components: MOC3021 and BT136 -> https://youtu.be/svYsNfAOtyw - See here, maybe you will better understand how the assembly works.
Here it appears in parallel with the consumer, not in series, that's what I meant.
triac-in-parallel-with-the-consumer
If you connect the consumer as shown in the attached picture, you will not have any problem.
how-to-make-the-right-consumer-triac-connection
And once it opens, does it short-circuit its own opening voltage? At the next alternation, it will close, voltage will appear again, and it will open, etc., but it will always open after a while (from the beginning of the alternation) until it has enough current on the gate.
As long as S1 is activated, the optotriac Q2 is continuously activated and commands the gate Q1. As long as S1 is activated, Q1 is activated. I don't see much of a short chance here.
The rectified voltage disappears almost instantly with the opening of the triac (luckily with C1 it gives another drop). The voltage on the gate branch also disappears, the triac still remaining open if the maintenance time of the current in the gate has been ensured.
He is an eighth deacon and commands a triac, we are talking about a static electronic switch in alternating current.
According to catalog data, the BT136 has 1,4V when it is "on", from where the voltage on the LED in the optocoupler is provided when the BT136 is "on" (on-state). In my opinion, the assembly should oscillate at the S1 coupling. The functional diagram is with 3 wires for this assembly.
Once open it remains in conduction on that alternation until the current drops below the maintenance value but at the next alternation it will open at a voltage above that of the deacon + something else until it reaches the required current in the gate .. at this value the rectified voltage is sufficient and for optocoupler… A small capacitor may be needed to ensure the gate current is maintained.
Only the oscillation range is increased with one capacitor. To be "connected" the optocoupler needs 3Vdc / 30mA after the rectifier bridge and 3Vac before the bridge. How is 30mA provided by the 100nF capacitor when the triac is "ON"?
They were insured before "on" .. I don't know how much the deacon has, maybe tens of volts, so until the triac opens the respective alternation (including the rectified one) reaches a sufficient value .. the transient phenomenon of the conduction of the triac and that of the entry into the conduction at the next alternation .. namely the angle of entry which is not zero .. and thus does not benefit the consumer from full power.
The ON operating diagram must be tested with an oscilloscope. On a light bulb, I don't think you would see the oscillations needed to connect the optocoupler, only if a capacitor of at least 100uF is connected to the bridge. I remain in my opinion that this would work correctly:
220V-static-switch-variant-with-3-wires
There are commercially available Solid State Relays of different voltages, ie 220-400 V and currents from 40-90 amps.
This does not work, they have not connected the load and a capacitor of approximately 330nF is missing on the main thyristor grid, in order to eliminate the false controls.
If you connect the consumer as shown in one of the comments above, you will not have any problems. In fact, the problems of false commands are eliminated by C1.
Honestly, I don't like it because electrical safety is at risk.
There is no galvanic separation between the switch and the mains. 9V there to him, but the grounding girl ..
The article specifies the following: "the order is made at low currents, but it also has the disadvantage of a high power consumption that mechanical contacts do not have." This means that the assembly controls consumers in pulses. Well maybe we should add a radiator for Q1 and probably a 330 nF capacitor on the main thyristor grid. The part with the galvanic separation ... I didn't really understand it 🙂
Dangerous voltages can easily occur between the ends of the S1 and the null / phase of the network. Some, perhaps seeing 9V there may be tempted to let it soften with some stuff. The assembly must be done carefully and not to put your gloves on when it is plugged in. I'm not a fan of food without a transformer.
But it's very simple. The confusion appears through the notation of the CON which does NOT mean the consumer but the connection of the "switch" so at the CON the consumer connects the series with the network and the series with the CON that is the switch! I think it's clear enough! The diagram does not show the power supply to the consumer, it is just a substitute for a switch!
"..in hazardous environments (with danger of explosion, for example, due to electric arc produced by mechanical contacts), static relays are used."
Reed relays no longer manufactured?
There is also the option of using reed relays. What is presented here is something else.
Interesting debates, I somehow summarize and add.
1. It is more of an electronic test montage, I would not put one in production. As the colleague pointed out, obtaining a 9 volt voltage through that method is risky and unstable.
2. The control / protection when the triac is switched off / on is done mimalistically / does not exist.
3. Lack of galvanic separation from the grid and the control part of the scheme.
4. Lack of correct notations, this does not take into account the correct phasing of the entire assembly.
5. Yes, for lower, continuous voltages, it would make sense, even with a thyristor.
6. A sincere reed relay would solve the problem, simpler, practically more secure than electrical safety, endurance, cost.
7. In the case of an inductive consumer, the present assembly has absolutely no protection, as George Marinescu said, a self-induced voltage can make the cabbage and the zener and the eventual touch of the metal parts of the S1 becomes extremely dangerous.
8. C1, has too low working voltage, preferably at least 400Vac.
Alex, no offense, the scheme is uninspired, working with triacs and thyristors is an art in electronics, when they are ordered in phase, especially when working at high AC voltages and especially for AntiEX conditions.
I see it as a good scheme for low voltages, higher currents, if the triac is cooled properly. Of course, with the recalculation of the values to correctly start the triac or thyristor, used only in CC.