SCR Bridge Control - Is my Circuit Correct ?

For MOC3021 input circuit I have choosen 220E resistor.

I took Vf = 1.15, If = 20mA, v = 5V

So,

(V - Vf) / If = (5 - 1.15) / 20mA = 192.5

so, 220E is used

Now I want to know how to choose MOC3021 ouput circuit series resistance value. I also have a 1N4007 between MOC3021 ouput and SCR gate

SCR is TYN612

For TYN612 according to datasheet gate current max is 15mA and so if I take gate current 10mA then 10mA has to flowthrough MOC3021 output circuit. If I take AC as 230V then all the voltage drop across the series resistor in the MOC3021 output circuit. so, I don't have to worry about the voltage drops across MOC3021 triac, 1N4007 and SCR gate junction. Right ?

So, R = V/I = 230V/10mA = 23k

so, can I choose 24k ?

If I am wrong then please explain how to connect MOC3021 to TYM612's gate using resistor and 1N4007 for 230V AC.

In the attached paper, the Scientific Journal has hidden the values of R and L.
However, with a little bit of math, I have found them even though the claimed values in the Scientific Journal's paper are impossible to achieve with that input voltage (120 V AC), so, I have used 240 V AC (240*sqrt(2) Volts peak).

DATA:
Vin peak = 240*sqrt(2) Volts
With alpha=90º, Average Voltage=60.68 V. From those values I get phi~55.83º which means, if you want your circuit to perform as their circuit, the L/R ratio must be 4.69*10^-3.

A simple example would be using R=100 ohms, L=469 mH resulting in Average Voltage = 70 Volts (it is not 60.68 Volts because the circuit's phi is not the assumed due to the influence of the transient response), I_RMS=0.9 Amps.

L/R = 4.69 * 10^-3

V = 240V

If R = 100R then

L = 100 * 4.69 * 10^-3 = 469mH

but if I choose lower reistance then how it will handle high currents ?

Please tell me how you calculated 4.69 * 10^-3 value ?

Also tell me for the above values of R and L what will be the back emf released into the load during negative half cycle.

How to calculate the amount of back emf or energy stored in RL circuit during positive half cycle. How to get more back emf so that it can be clearly shown on the scope. If L increased bck emf increases. right ?

but if I choose lower reistance then how it will handle high currents ?

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Chose suitable power ratings for the resistor.

Please tell me how you calculated 4.69 * 10^-3 value ?

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I have calculated in order to get similar results as in the paper you have attached repeatedly. You said you want to have the same as in that paper. Then stick to those values.
1)I calculated the DC voltage's expression in terms of phi, alpha and input peak voltage.
2) Found out phi knowing the input Data
3) Figured out L/R ratio.

How to calculate the amount of back emf or energy stored in RL circuit during positive half cycle. How to get more back emf so that it can be clearly shown on the scope.

Click to expand...

Do you like what you see in the paper you have repeatedly attached ? If so, it will be visible on the scope. Also.. can Proteus simulate such circuit ?

The circuit mentioned that document doesn't simulate in Proteus and thats why I made a separate simulation circuit. The simulation and practical circuit are different.

Yes, the project that has to be demonstarted by my client who is a student is the same project. He has to show output on scope for R and RL load. Code I have written but I can't post it until he completed the project submission. Here I am attaching the proteus simulation. For simulation I have used a different circuit but that circuit will not be built in real hardware.

The PDF circuii that I attached will be the one that will be built on hardware.

Actually at first he only asked me to write the code but later asked to design the circuit as mentioned in that document and show the simulation.

Actually that was the only document given to me for writing the code.

Here I am attaching another two documents which contains a badly written code which doesn't use timer interrupts to generate the firing delay. I don't know the french project is working in simulation and hardware. It has lcd and gate triggering delay code in the main loop.

In my code I have used timer interrupt. When zc is detected timer is started for required firing delay. after delay expires proper scr pair is fired. adc and lcd code are in main loop. adc value is converted to timer reload value to get firing delay and when zc is detected timer is reloaded with this delay value and started. After timer expires gate of required scr pair are triggered.
After his project completion I will post my code other wise he will not get degree.

In that document which has to be followed the author shows the 2 PCBs they used with simple circuit for demonstartion. They have shown the scope signals for that circuit.

I convinced my client to use 24V transformer and a LM317T to get 5V DC for PIC circuit. 1A bridge is used and also to use 24V AC from transformer output as input for SCR. So, he will not be using 230V AC for the output side circuit. So, no high voltages.

Now please provide inductor value for RL load. Not for motor. Current will be kept minimum. Just tell me how to calculate RL component values. SCR output DC max will be 24V * 1.4142 DC.

Also I want to know can TYN612 SCR can be used with 24V AC ?

I reviewed the thread to understand why you want an inductive load, but I didn't find a reasoning.

- - - Updated - - -

If the purpose is simply to demonstrate behavior with inductive load, it's your job to specify the parameters. Or use a common inductive load like a DC solenoid valve.

What is the difference between why I want an inductive load and reasoning. My client has to show the waveforms for RL load where negative voltage appears at the output during negative half cycles of AC.

That is the purpose of the project.

Regarding opto triac circuit design. I am now referring this document. Please tell me in page 2 from where they got 1.2A in the calculation and also from where they got VTM. I am sure if somebody answeres from where they got 1.2A and VTM then I will be able to design MOC3021 circuit.

Is 1.2A max surge current for MOC3011 ?

If yes, where is it mentioned ?

Also as now I am using 24V AC for SCr should I really use MOC3021 for firing the SCR ? Do I really need isolation because the voltages are low.

My client asked to use resistor and inductor for RL load. His instructor has mentioned to use it. And also he has to show calculation. So, if inductance is more than backmf will be more and so he has to show it with calculations and hence I am asking for the equation to desing RL values and calculate back emf value.

At this point, I give up asking about specifications because seems your client does not care.
Do you want it to perform as in the paper you have repeatedly showed ? (i.e. with phi~55º?)
If not do this:
1) choose a phi according to the DC ouput voltage desired
2) figure out L/R ratio
3) choose R and L values according to availability or current etc..

Also as now I am using 24V AC for SCr should I really use MOC3021 for firing the SCR ? Do I really need isolation because the voltages are low.

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Not strictly isolation, but floating trigger sources referred to each SCR cathode. You have previously demonstrated failure of ground referred trigger sources, e.g. in post #21.

@CataM

Yes, I want it to perform like in the referred document that is ijsrp_p4937.pdf but for 24V AC.

1. I will choose max conduction angle 180 degrees

2. Then how to calculate ?/R ratio ?


@FvM

Forget about post #21. it was used for simulation. In real hardware MOC3021 will be used.


This is my calculation for MOC3021 circuit done using AN-3003.pdf attached in post #48. Will this work for TYN612.

For MOC3021

R1(min) = Vpk/(peak repetative surge current)

R1(min) = 24V / 1A

R1(min) = 24E

1. This will allow max 1A current through MOC3021. Right ?

2. So, can I increase R1 so that I will be equal to Igt for SCR ?

Igt for TYN612RG = 15mA

For TYN612

Igt = 15mA
Vgt = 1.3V
Vtm = 1.8V

VinT = R1(min) * Igt + Vgt + Vtm

VinT = 24 * 15mA + 1.3V + 1.8V

VinT = (0.36 + 1.3 + 1.8) V

VinT = 3.46V


Edit:

VinT for my condition is


VinT = R1(min) * Igt + Vgt + Vtm + V(diode) (I have used 1N4007 at output of MOC3021)

So, VinT = 3.46 + 0.7 = 4.16V.

Are my calculations correct ?

Not strictly isolation, but floating trigger sources referred to each SCR cathode.

Click to expand...

But in the attached document they show PCB which has only resistor and diode in MOC3021 output circuit. They have shown the output for it.


Here they have used 12V and thay have got output. They have only used resistor and diode in MOC3021 output circuit.

**broken link removed**

https://www.youtube.com/watch?v=odx9V9DIjgg

Forget about post #21. it was used for simulation. In real hardware MOC3021 will be used.

Click to expand...

But in the attached document they show PCB which has only resistor and diode in MOC3021 output circuit. They have shown the output for it.

Click to expand...

Don't know what you mean. I tried to answer your question

should I really use MOC3021 for firing the SCR ? Do I really need isolation because the voltages are low.

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If you use MOCxx, it can be connected with resistor and diode. If you want to use a non-isolated trigger source, it must provide a voltage between cathode and gate. MOCxx effectively does the same in a different way.

Now as you decided to use MOC3021 with resistor and diode, everything is fine.

R1(min) = Vpk/(peak repetative surge current)

R1(min) = 24V / 1A

R1(min) = 24E

1. This will allow max 1A current through MOC3021. Right ?

Click to expand...

Vpk of 24 VAC is about 34V in my calculation.
But generally, you want to make the series resistor low to allow maximum duty cycle.

@FvM

If 24V AC is converted to dc then it becomes apprx 39V. How you got 34V ?

Vpk = Vrms * 1.4142

How you got 34V ? If you meant 39V then why 39V has to be used ? because we are feeding AC. Should we still consider Vrms * 1.4142 for the calculation ?

If I am wrong then correct me.

Edit:

Sorry FvM. I did some wrong calculation. Indeed Vpk is 33.9V.

So,

R1(min) = 33.9V / 1A = 33.9R

So, shall I use 22R ?

If yes, then

VinT = R1(min) * Igt + Vgt + Vtm + V(diode)

= 22 * 15mA + 1.3V + 1.8V + 0.7V

VinT = 4.13V

Then how to calculate ?/R ratio ?

Click to expand...

For phi=55.83, the L/R ratio must be 4.69*10^-3. Do you know what is phi ?

Vpk = Vrms * 1.4142

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Sure. How do you get 39V?

I would use 39 or 47 ohm series resistor.

So, shall I use 22R ?

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R(min) gives a minimum value to be used. You can not use lower than the minimum value.

@CataM

tau = L/R but without knowing L and R how can I calculate tau ?

Okada said:

tau = L/R but without knowing L and R how can I calculate tau ?

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That is the point. You have to have some specs...

Are you using

theta = phi = arctan(VL/VR) ?

Okada said:

Are you using

theta = phi = arctan(VL/VR) ?

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I am not sure what does VL and VR mean but I am using φ=arctan(ωL/R)