Where to get schematics about SCRs full-wave triggering circuits ?

[eagle1109]

Hi,

I modified the circuit a bit.


1. Measuring the signal at the gate, is negative with one scope probe. Why ?

My understanding here, is that the AC has two branches to run in. Path 1 is the path that has the scr, path 2 is the one runs down the pot to the capacitor. So the signal which is I2 should be positive one. Or what exactly ?

a) During the +ve signal, path 1 scr is off so I1 won't run, path 2 the voltage is -18V on oscilloscope and 13V rms, but I2 should be positive.
b) During the -ve signal, path 1 scr is in reverse bias so no current flow, path 2 is also a possible path for the -ve current but I don't know if it's blocked by the capacitor assuming it's an electrolytic one.

So I think I should get both the +ve / -ve pulses, but I only get a negative one ! And I don't know which one is appearing on the scope.

2. The scr doesn't conduct when the pot is at 1% which is 10k ohms + 1k ==> Ig = 220V / 11k ohms = 20mA. That's also shown on the AC multimeter. Also another interesting thing is that I guess the values are in rms.
It's good that I could calculate the gate current thanks Mr Klaus

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OK, I did the same trick of combining two scope signals and get the invert of that signal of the gate voltage and put a pot on the gate to get the exact gate current easily.

But this scr gets triggered at 50mA and I measured the voltage signal at the gate in this picture and divided the voltage which is 76V on the actual resistance which is adjusted at 500 ohms and the current was 152mA but the scr didn't get triggered.



Could anyone tell me why ?

 

[KlausST]

Hi,

1. Measuring the signal at the gate, is negative with one scope probe. Why ?

From post#55:

Thus we need to invert ONE input.

But you inverted both!

Regarding the resistor to measure the current.
The resistor value should be low, low enough not to influence the function of the circuit.
A usual current DVM has a value down to the low milliohms.

In your case - especially when you use a diac - you should not go above 10 ohms.
In a real circuit one would used a differential amplifier to get a useful voltage value proportional to the current.

Klaus

 

[eagle1109]

OK, I think I learned something.



I simplified the circuit more, no need for all those voltmeters, because they give rms value. I can now know what's going on with the scope.

In this photo, the gate voltage is peaking 257V, but this is another issue I want to learn about, is that why the voltage is so high ? In real circuit won't that just destroy the scr immediately ?

So, when the gate triggers, I learned that as soon as the current hit 50mA, the scr conducts.

But now the question is how can I achieve the full phase control ? Because I can now trigger the control just a portion of the pulse phase.

Thanks,

- - - Updated - - -

Alright, this one is much better.

I was able to get a full phase control with one pot, but I had to increase the value of the capacitor to 220uF and a fixed resistor of 100 ohms. I don't know how I did it, I guess it was a hit of luck :smile:

 

[KlausST]

Hi,

I guess it was a hit of luck

Yes, luck. Sadly luck, because it seemed you forgot all we tried to teach you.

Designing electronics has nothing to do with luck. It´s learning, understanding, reading and calculating.
Again you used random part values. In a real circuit there is a good changce to cause explosion and fire.
Not only the too high gate voltage (the gate voltage limits are shown in the datasheet)
but also
* negative voltage on an electrolytics capacitor
* too high voltage on a 40V capacitor
* too much power dissipation on a resistor
* ...

In my eyes the current discussion is about "SCR triggering". I´ve tried to tell you that the gate triggers with current, not voltage.
You still insist in voltage measurement... you never showed a valid gate current measurement.

I´m sad that I didn´t find the right words to teach you. Sorry for this. I don´t know how to go on..

Klaus

 

[eagle1109]

Hi,


Yes, luck. Sadly luck, because it seemed you forgot all we tried to teach you.

Of course not, I actually learned a lot. Like I thought about the last information you told me that the resistance has to be very low, so I removed the unnecessary resistors and kept the overall resistance lower than before.

Designing electronics has nothing to do with luck. It´s learning, understanding, reading and calculating.

Of course I know, it's based on theories and demonstrations.


Again you used random part values. In a real circuit there is a good changce to cause explosion and fire.
Not only the too high gate voltage (the gate voltage limits are shown in the datasheet)
but also

* negative voltage on an electrolytics capacitor

Now this is the issue I want to learn about, I didn't know what to use, a ceramic or electrolytic capacitor. But it seems that I should use an un-polarized capacitor because it's an AC circuit basically.

* too high voltage on a 40V capacitor
* too much power dissipation on a resistor
* ...

What should I do to lower the voltages at trigger circuit ? Should I use a transformer.

In my eyes the current discussion is about "SCR triggering". I´ve tried to tell you that the gate triggers with current, not voltage.
You still insist in voltage measurement... you never showed a valid gate current measurement.

I know the goal you're trying to tell me, but I hope you appreciate that it takes me time to realize this point. I reached the level now to know in which case I can control the phase of the output, and since I got to some point.

Now, I have to fix the issue of high voltage on trigger circuit. Now I have to lower the voltages and then lower the resistance to match the same phase control.

I´m sad that I didn´t find the right words to teach you. Sorry for this. I don´t know how to go on..

I think it's the opposite, you taught me a lot, I'm actually following you assisting all this week.

I think I learned something, I won't reach the full learning curve very fast.


The issue I want to work on now is the high voltage on triggering circuit, how to lower it ? What to use ? A transformer or resistors ?

 

[betwixt]

Back to basics eagle1109 - if you charge up a capacitor, it doesn't matter what value or what the voltage is, then short it out, how much current flows out of it?

Try this simulation: use nothing but a 10K resistor, a 100nF capacitor and a Diac. Connect the diac across the capacitor and ground one end of them. Wire the resistor to their other end and apply a DC voltage of say 100V. Connect the oscilloscope across the capacitor and show what you see.

Brian.

 

[KlausST]

Hi,

Of course not, I actually learned a lot. Like I thought about the last information you told me that the resistance has to be very low, so I removed the unnecessary resistors and kept the overall resistance lower than before.

I clearly talked about the "resistor to measure the current" and wrote it should be not higher than 10 Ohms.
* but you removed all the other resistors
* and the value of the current measrement resistor still is 100 Ohms
* and it is not used as current measurement at all. --> thus no need to use it at all

I know the goal you're trying to tell me, but I hope you appreciate that it takes me time to realize this point

I don´t understand how you want to understand the "gate current" when you never measure the "gate current"... (At least I can not see this in your posts)
It takes 10 seconds of time to do a gate current measurement in the simulation...then you get a scope picture... and a saying is: "a picture is worth a thousand words"

I reached the level now to know in which case I can control the phase of the output, and since I got to some point.

Here is my problem. In my eyes learning is step by step.
* first learn / understand / know how to calculate to trigger the SCR correctly
* then go the next step to adjust firing angle.
In your posts I can´t verify the first step..

Klaus

@Brian:
Yes, good experiment...for voltage, for current, for phase shift, for diac function...

 

[eagle1109]

Back to basics eagle1109 - if you charge up a capacitor, it doesn't matter what value or what the voltage is, then short it out, how much current flows out of it?

Highest current rate possible, if the voltage is so high it could cause a high current spark.
When I work with PC power supplies, I use like 1k ohm/1W resistor to discharge the high voltage capacitor.

So you mean with this analogy, is that when the scr triggers, it's actually taking up the voltage of the discharging capacitor.

Try this simulation: use nothing but a 10K resistor, a 100nF capacitor and a Diac. Connect the diac across the capacitor and ground one end of them. Wire the resistor to their other end and apply a DC voltage of say 100V. Connect the oscilloscope across the capacitor and show what you see.

I'm sorry I don't know if I have done it right. But it didn't work, this is the simulation.

 

[betwixt]

So you mean with this analogy, is that when the scr triggers, it's actually taking up the voltage of the discharging capacitor.

You are not reading what Klaus explained - it is current that triggers the SCR but it does come from the discharging capacitor.

Your schematic is correct but the scope settings are not. Switch off the A+B and show channel B instead of C, at the moment we have no idea where zero volts is on the scope display. Try increasing the resistor to 100K and see what happens.

Brian.

 

[eagle1109]

You are not reading what Klaus explained - it is current that triggers the SCR but it does come from the discharging capacitor.

Yes I understand the scr is a current driven device.

Like BJTs are current driven transistors where MOSFETs are voltage driven. So scr firing is about current amount, and the discharging is the charge of the capacitor reached the required amount of current that is needed to trigger that specific scr in this case it's 50mA.

Your schematic is correct but the scope settings are not. Switch off the A+B and show channel B instead of C, at the moment we have no idea where zero volts is on the scope display. Try increasing the resistor to 100K and see what happens.

I had to change the signal type to DC, then it showed the correct level of the voltage where in AC mode it showed 0V.



with 100k resistor, it's the same.



I've put a switch to show to you what happens when I open/close the switch. When the switch is closed the output is constant 22V, maybe this is the level that keeps the diac on all the time.

But when I open the switch, the capacitor charges up to 28V ! I don't know why this happens, then when I close the switch there is a spike of 42V and voltage drops to 22V again.

 

[KlausST]

Hi,

You still repeat not to strictly follow the recommendations.
You modify the circuit, you modify the part values...
1) May I ask why you used 1k instead of the recommended 10k?
You say you "know" .... so please tell us what you know.
There is Ohm's law. You know?
The datasheet shows an operation point at 10mA.

Δ V Dynamic breakover voltage (1) VBO and VF at 10 mA

100V and 10k ohms gives 10mA.
I'm sure it was not "luck" that makes Brian choose the values. He calculated.
Why did you choose another operation point?

2) the switch is of no help. It makes things more difficult.

3) the scope time should start at zero...the voltage at the capacitor should be zero when starting the simulation...

Klaus

 

[eagle1109]

Hi,

You still repeat not to strictly follow the recommendations.
You modify the circuit, you modify the part values...

1) May I ask why you used 1k instead of the recommended 10k?

I actually did exactly as he said, see 2nd line of #69, he said the schematic is correct, then he also told me to try 100k. So I did two circuits one with 1k (yes it's not 10k because I already did it and because the one with 1k was the last simulation I did so I took a snapshoot of it, and the other one with the recommended 100K), believe me I'm following the instructions and also do more of my own efforts within the scope of the discussion.

You say you "know" .... so please tell us what you know.
There is Ohm's law. You know?
The datasheet shows an operation point at 10mA.
100V and 10k ohms gives 10mA.
I'm sure it was not "luck" that makes Brian choose the values. He calculated.

I downloaded a datasheet for this brand "CHENYI ELECTRONICS"



And I really didn't search for current, I directly took the value of voltage for the DB4 diac and tried to understand the min/max breakover voltages, I didn't think of the current.

Why did you choose another operation point?

2) the switch is of no help. It makes things more difficult.

Yes I know but I wanted to show that there's no charging/discharging, and I wanted to show what happens when I use a switch and I still don't know why the circuit is acting the same as I used three different resistance 1k, 10k and 100k. The same thing, when I close the switch it drops to 22V and when I open the switch it goes up to 28V, so I don't understand why it charges up as there is no flow of current to charge that capacitor !

3) the scope time should start at zero...the voltage at the capacitor should be zero when starting the simulation...

Yes, I mostly don't catch the 0 time because the reading of the oscilloscope is moving constantly and I even can't goes back to the 0 point.

 

[KlausST]

Hi,

Yes, I mostly don't catch the 0 time because the reading of the oscilloscope is moving constantly and I even can't goes back to the 0 point.

Press the "one shot" button at the scope.

Klaus

 

[betwixt]

One shot and try a slower timebase speed.

What you should see, if the simulation is working, is a repeating sawtooth waveform. The capacitor charges through the resistor until enough voltage is across the Diac for it to start conducting (VBO in the data sheet), the current it draws from the capacitor makes it discharge until there isn't enough for the Diac to stay conducting. Then as the Diac 'turns off', the voltage across the capacitor starts to rise again and the cycle repeats.

I'm trying to demonstrate that the current to operate your SCR gate doesn't only flow through the resistors, most of it comes from the capacitor discharging. If the simulation works, I was going to suggest adding a small resistor (about 10 Ohms) in the ground side of the Diac and using the oscilloscope to look at the voltage dropped across it. Ohms law will let you convert the voltage across the resistor into the current through it. I suggested 10 Ohms because it is small enough not to limit the current too much while still letting the circuit work. A smaller value will be better but at the same time it will make it harder to monitor the voltage across it.

Again, if the simulation works, you should see short spikes of voltage across the resistor as the capacitor discharges, these would be the trigger pulses to the SCR in a real phase control circuit.

Brian.

 

[eagle1109]

Hi,


Press the "one shot" button at the scope.

Klaus

I tried, it's the same. I put the capacitor voltage at 45V as maximum DB4 VBO is 45V.



- - - Updated - - -

One shot and try a slower timebase speed.

What you should see, if the simulation is working, is a repeating sawtooth waveform. The capacitor charges through the resistor until enough voltage is across the Diac for it to start conducting (VBO in the data sheet), the current it draws from the capacitor makes it discharge until there isn't enough for the Diac to stay conducting. Then as the Diac 'turns off', the voltage across the capacitor starts to rise again and the cycle repeats.

I'm trying to demonstrate that the current to operate your SCR gate doesn't only flow through the resistors, most of it comes from the capacitor discharging. If the simulation works, I was going to suggest adding a small resistor (about 10 Ohms) in the ground side of the Diac and using the oscilloscope to look at the voltage dropped across it. Ohms law will let you convert the voltage across the resistor into the current through it. I suggested 10 Ohms because it is small enough not to limit the current too much while still letting the circuit work. A smaller value will be better but at the same time it will make it harder to monitor the voltage across it.

Again, if the simulation works, you should see short spikes of voltage across the resistor as the capacitor discharges, these would be the trigger pulses to the SCR in a real phase control circuit.

Brian.

I could observe the diac effect as a voltage clipper. In this circuit I applied 230V rms on the diac and it clips the voltage to its rated VBO which is 35V and according to the datasheet it's the minimum voltage break over.

 

[betwixt]

I fear we are going backwards....

A 100nF capacitor has a reactance (Xc) of 26.5K Ohms. So in your bottom circuit the voltage across the Diac will only reach about 68V peak-to peak even if the diac is removed. Your scope shows 68.5V so it is pretty much exactly as expected.

However, I didn't say to use AC, for the experiment you should be using 100V DC! Did you try 100K in the simulation?

Brian.

 

[KlausST]

Hi,

.. and the diac in series to the scope makes no sense at all.

Your conclusion is completely wrong.

Klaus

 

[eagle1109]

I fear we are going backwards....

A 100nF capacitor has a reactance (Xc) of 26.5K Ohms. So in your bottom circuit the voltage across the Diac will only reach about 68V peak-to peak even if the diac is removed. Your scope shows 68.5V so it is pretty much exactly as expected.

Yes because 100 V - 68.5 V = 31.5 V which is less than minimum break over voltage which is 35 V but it is close.

So the diac would break any time the voltage across it is from 35 - 45 V.

However, I didn't say to use AC, for the experiment you should be using 100V DC! Did you try 100K in the simulation?

Yep in #70, and I also put some notification on the photo.

- - - Updated - - -

But there is a problem in the simulation !

The capacitor continues to charge up to 22V as the switch is open ! That's not logical.

 

[betwixt]

So the diac would break any time the voltage across it is from 35 - 45 V. Any voltage more than the max break over voltage would be clipped; for example

NO! Diacs do not 'clip' the voltage, that's what Zener diodes might do. There is a special characteristic of Diacs that once they start to conduct, the voltage across them drops suddenly and they continue to conduct until the current through them is too small. VBO is the voltage that starts them conducting but then the voltage across the ends will typically drop to just a few volts.

Maybe thinking of them as a switch that turns on when the voltage across it reaches a certain level might help. It isn't quite right but it might help you understand their operation.

Brian.

 

[KlausST]

Hi,

The capacitor continues to charge up to 22V as the switch is open ! That's not logical.

When the switch is open, then node may be seen as "floating". Floating means "undefined" .... anything can happen.
That's why I wrote "2) the switch is of no help. It makes things more difficult.".

An "open switch" is not the same as "Zero volts".

Klaus