Is my circuit correct?

I checked the Circuit 1 and Circuit 2 and the SCR was working, inplace of LOAD, I had a small 12V torch bulb.Both of these circuits are working fine.
By looking at the way Circuit 1 and Circuit 2 are working, I am willing to make the Circuit 3, is this circuit ok.I'm using 2N2222 low power general purpose NPN transistor.I want to know is this good design(Circuit 3)?

I'm doubtful about Circuit 2 also, although it's working fine, maybe it only works fine in low voltage like 12V AC and will blow up in high voltage like 220V AC.I'm thinking about this, because when I see Cathode of SCR.I know it's just 0.7Volts lesser than the Anode.I measure 220V AC, so the peak voltage is 220*1.414=311.08V AC, this voltage will appear at anode and 311.08-0.7V = 310.38V AC will appear at the cathode.So now Power supply is Gnd is connected after the load and I know potential would have droped to 0V .Now when the AC supply changes it polarity maybe the higher potential voltage would than flow directly to the GND of power supply.Should I connect a diode in there, so as to block current flowing into it.This configuration worked fine at 12V AC Rms, but don't know what the consequences would be at 220VAC Rms.Actually all I want is to make a good triggering circuit for SCR.The load current is all about 20A, so the circuit 1 containing the MOC3011 will blow up at high current.

Hello,

circuit 1 is (almost) correct, circuit 2 and 3 will backfire to trigger circuit and (surely in case of 230 V supply) blow it up. Trigger voltage should always be applied to gate-cathode.

Furthermore, you should always have a series current limiting resistor at the gate. This is also usual with circuit 1, otherwise the optotriac and/or gate circuit may be possibly damaged if the main switch doesn't trigger fast enough.

I guess you are using a triac rather than a thyristor shown in the circuit. With a thyristor, there must be a means to block trigger current during negative halfperiod, otherwise the trigger circuit and the thyristor would be destroyd.

Regards,
Frank

P.S.: I see BT152 is a thyristor, so circuit 1 should have a 1N4007 in series with optotriac.

Yes, In circuit 1 I had proper current limiting resistance, I forget to mention them in the diagram.

Well SCR and thyristor means the same thing.I made circuit 1, when I was using 12V DC motor.I now have 220V DC,16A currentof 5HP motor.I have designed the gate triggering circuitry with 12V motor.As how the gate signal would be generated.Now I'm confident about my triggering circuit.Now triggering SCR is making me worried in 220V AC.As in Circuit 1(12V motor) on every pulse, anode of SCR was getting connected to the Gate there by triggering it, also the current which was flowing through the triac of MOC3011 was also very low and there by MOC3011 was safe to operate.It was less than 1A.In 220V operation, instead of <1A current about 16-20A of current would flow through the triac of MOC3011.There is going to be no mercy on that opto-triac.Although there are opto-le which can handle 20A of current, but that would be an expensive solution for this small project.I would only go for this if I don't get any other option.Infact I had seen some SCR triggering design too in the past, that were empolying the cheap low current opto-isolators(less than 1A) for triggering SCR that were passing on about 10A of current, that was a clever design.I unfortunately forget that triggering circuit design.I now have to make one.
Tell me how circuit 3 can be improved?
Thanks

Hello,

if also have MOC30xx in a lot of circuits, low as well as mains voltage. Did you understand what happens, when you use a thyristor (SCR) instead of a triac. The Opto triac will also trigger during negative periods, but the main switch can't take over the current. Very nasty.

Circuit 3 would be o.k., when the cathode is connected to control circuit ground. This would be possible by changing places of motor and thyristor. Also a current limiting resistor should be used. However, the control circuit is connected to mains supply and must have the same safe isolation as mains voltage itself.

Regards,
Frank

THERE WAS A BOOM!!!
AND SOME SPARKING IN MY CIRCUIT....

I then quickly remove the power cable from the socket, before anything worse happens.When I looked at the circuit, 2/3rd of the opto-coupler IC disappeared, only few pieces was in the circuit and ash.I don't know how much damage has occured to the remaming of my circuit.This is what I have done.

This has been said:

circuit 2 and 3 will backfire to trigger circuit and (surely in case of 230 V supply) blow it up. Trigger voltage should always be applied to gate-cathode

Click to expand...

Circuit 3 would be o.k., when the cathode is connected to control circuit ground

Click to expand...

This is neither Circuit 2 and nor Circuit 3.It's a new circuit.I think I have connected SCR gate properly, there is no problem.If you see bridge circuit, in top of it I have connected Ground.I think this is also OK, because there's no potential in there, so I make it as Ground.

Trigger voltage should always be applied to gate-cathode

Click to expand...

I think, if I apply trigger voltage between gate-cathode, then only back firing will occur, because SCR Cathode will be at a higer potential than the emitter of the BJT.Instead of current getting into the gate of the SCR and comming out of the Cathode of the SCR.The current will go into the emitter of the BJT from the Cathode of the SCR and then blow it up.I then connect the BJT resistor one end to the top of the Bridge circuit.Because there is very low potential over there and as far as bridge rectifier supplies are conerned people make that point as a Ground.

I then add the two grounds together.One ground of +5V power supply and other ground of SCR bridge, because I think ground is a ground indeed, every where I go.Ground has the same meaning, so it doesn't matter if I connect the two grounds together.

Circuit 3 should be corrected like this:

Here the trigger voltage is applied between gate and cathode terminal. As an additional remark: It is also assumed here, that the control circuit has no galvanic connection to mains supply, the ground symbol can't be but a local ground without external connection.

Your last circuit under test is basically identical to circuit 3 in so far, as trigger voltage is applied between gate and the other load terminal (opposite to cathode). The fact, that you have another SCR and two diodes doesn't count in this respect.

It is better to have isolation. It will help to avoid risk to life while handling mains voltages.
Sarma

THIS TIME MY CIRCUIT CAUGHT FIRE AND FLAME WERE COMING FROM COMPONENTS.BUT CIRCUIT WAS WORKING FINE.




The configuration of the circuit is correct.There is no problem, SCR can be triggered in this way.It's a valid method and it works.

I read in the datasheet of opto-coupler that to find the value of the resistor, you need to use this formula.

Resistance=Vpeak/1.2(Constant)

So, 220V rms=220*1.414=311.12Vpeak
Resistance=311.120/1.2=259Ohms

So after calculating this value at home, the next day I went to the shop-keeper and asked him to give me 330Ohms value resistance instead, because I don't want to use the minimum rated value 259Ohms, so he gave me 1/4 watt resistances, as I was leaving the shop.I met my university teacher right at the shop on time of my leaving.I tell him what I was doing, he told me not to use 1/4 watt.So I bought another 330Ohms resistances of 2/3 watt.I soldered them into my circuit.I turned on the supply.My circuit start to work as it should.Then resistance caught on fire and after few seconds it was burning like a candle flame, but no SPARKING sound or no short circuit sound was coming...I had attached the lamp as a load, so I was able to vary the average voltage sent to the load by varying firing angle(I was doing this by pontentiometer in my circuit).So I think that 2/3 watt resistor is not good, they catch fire in my circuit, but they work for less than 10sec and then burnt up completely.So can I use this formula to know how much wattage resistance I need?
P=V^2/R
and then
P=(220*220)/330=146.66W
Do I need to buy the 146.66W resistance?

About one thing I'm sure, that the value 330Ohm is correct.

It's also beeing said.

With a thyristor, there must be a means to block trigger current during negative halfperiod, otherwise the trigger circuit and the thyristor would be destroyd.

Click to expand...

P.S.: I see BT152 is a thyristor, so circuit 1 should have a 1N4007 in series with optotriac.

Click to expand...

With a thyristor, there must be a means to block trigger current during negative halfperiod, otherwise the trigger circuit and the thyristor would be destroyd.

Click to expand...

I don't think there's any need to block trigger current during negative half cycle.Current doesn't go in there.These are the two ways in which current would flow in this bridge circuit.How could a negative current would flow back into the trigger circuit and destroy it?

have you tried your last schema and has it worked like in your pict??

Sorry I don't remember anything today.