# The Silicon-Controlled Rectifier (SCR)

Discussion in 'Feedback and Suggestions' started by arjunaditya, Feb 7, 2010.

In phase-shifting capacitor the voltage across the capacitor before the scr is triggered
is phase shifting when once it is triggered the scr gets short circuitted and the voltage across it decreases exponentially.
but in the wave form it is given the whole thing as phase shifted voltage

(I am if there is any mistake in my explanation)

2. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
Where is this in the book?

volume 3
semiconductors
thyristors
scr

4. ### Wendy Moderator

Mar 24, 2008
21,470
2,959

This appears correct. Are you pointing out an error, or wanting to discuss this circuit?

If you want to discuss this circuit please register. It is free, and easy to do.

5. ### jpanhalt Expert

Jan 18, 2008
6,775
1,393
I think he is saying that during SCR conduction, the capacitor is effectively short-circuited, and therefore its voltage would approach zero fairly quickly depending on the value of the variable resistor and the rate of discharge through the gate-cathode PN junction.

Therefore, capacitor voltage should be shown as dropping to zero until zero crossing causes the SCR to cease conducting. After which time, the capacitor will again begin to charge. I don't know if that is right or wrong, but it does seem to be a pretty thoughtful question.

John

6. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
Actually, the PN junction stays on either way, it is the BE of a transistor. What happens on the gate cathode is not affected by the anode cathode, much like the voltage drop of a transistor on the CE has very little effect on the BE junction (we have had similar misconceptions there too). Like a transistor, it is current, not voltage, that turns a SCR on (though we have had some lively debates about BJT being voltage devices elsewhere on the forum).

It is important to remember that an SCR is two transistors.

A quick search seemed to show the book is basically correct. I'm no expert on these circuits, so it is still up for discussion from my point of view. It also seems to match what I was taught.

7. ### jpanhalt Expert

Jan 18, 2008
6,775
1,393
I am almost completely naive with regard to theory. That's why I said I didn't know whether that analysis was correct. If I had the time right now, I would do the experiment with a dual channel scope, but I don't.

The simple question in my mind is whether the voltage across the capacitor effectively flat lines for any appreciable period during the cycle.

John

8. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
Given it is a true PN junction I suspect it does. It doesn't matter though, it doesn't conduct as a switch does, shorting the capacitor. The phase shift is what is important though, it is what makes it work. The comparison to a transistor is more than an analogy, it is accurate.

9. ### Jony130 AAC Fanatic!

Feb 17, 2009
4,667
1,311
Buy how can a voltage on the capacitor be larger then "trigger" voltage?
And if you compare to BJT so you have to keep in mind that Vce(sat) is much smaller then Vbe.
And during SCR conduction the left site of "pot" is shor by SCR anode-cathode junction.
And capacitor also will be discharge through gate-cathode because there is non resistor on series with gate.

• ###### 03217.png
File size:
14.1 KB
Views:
33
Last edited: Feb 17, 2010
10. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
The trigger voltage in this case is actually current, remember that we are dealing with a transistor BE junction. As such once the 0.7V BE junction voltage has been achieved then it is how much current flows through the gate. The voltage waveform will be flattened severely.

Once an SCR is triggered it doesn't matter, the capacitor has done its job. The SCR will stay triggered until current reverses direction, at which time the SCR becomes reverse biased and turns off. It would be more correct to show the capacitor voltage also going to zero, but the idea is to show how the phase angle is triggering the SCR, not the actual waveform, which is already highly distorted from the BE junction.

It is not meant to be a completely accurate diagram , but a conceptual one. I've seen something like this diagram in several books discussing SCRs.

The same argument could be used for this drawing, but again, it is a conceptual drawing.

My personal verdict is it is OK as is. It is debatable though, so other opinions might rule.

11. ### eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
104
The following is a simulation of the SCR firing circuit. It is done using the free LTSpiceIV. I have chosen the breakdown voltage of the Diac to be 30V.

File size:
21.5 KB
Views:
123
12. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
If you read what I said, that pretty much matches. The diac boosts the voltage and the BE drop of the Gate/Cathode.

But consider this, would someone new to the idea be able to understand the gate trace as shown? Or would the other drawing be more obvious? This is what I mean by a conceptual drawing.

As I have said, this isn't the only text book to show it this way, for a good reason.

13. ### jpanhalt Expert

Jan 18, 2008
6,775
1,393
If by gate trace you mean C1 voltage, then it is understandable and shows the expected discharge through the resistor. A simple sentence in the description could point that out and make the e-book correct.

The fact that some other texts show it as originally shown here is testament only to the fact that textbooks copy each other without giving credit. I grew up in an era when the number of human chromosomes was incorrectly stated in virtually all textbooks because of an original error. Correcting that error helped rather than hindered understanding.

John

edit:http://www.nature.com/scitable/topicpage/Human-Chromosome-Number-294

Last edited: Feb 19, 2010
sbombs likes this.
14. ### eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
104
The diagrams in the ebook are in-correct.

Just a section earlier in the ebook a correct symbol of the diac is shown but then in the next section on a SCR circuit a diode symbol is used to mean "diac". This is evident in all the images posted so far.

A silicon diode has only ~0.6V across its terminal and thus is no good in the case for phase control depending on the charge voltage of the capacitor.

Please amend the diagram to show a proper DIAC symbol.

15. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
I think I have to disagree with the diac. It is an improvement, but not necessary for correct operation.

The diode is required to prevent the BE junction from taking the full brunt of back bias. However, most SCRs have a resistor between the Gate to Cathode (aka, Base to Emitter). Coupled with the variable resistor shown in the original illustration (R1 in yours) this forms a variable voltage divider. This divider will have much the same effect.

The Gate-Cathode resistor was described in the AAC book. Wikipedia shows a zener in a similar circuit, this area is pretty flexible in the exact construction used. The concept is important, not the actual design used.

...............Wikipedia's Illustration.

It might be a good idea to modify the text in the area to explain the sine wave shown is for illustration, and not an actual representation of the waveform. Since I don't feel it is a major issue I'll let someone else pick up the torch on this one, I currently have several other articles in progress that are much closer to my priorities.

The OP has brought up a valid point, though it took me a little while to reach the same conclusion.

• ###### temp.GIF
File size:
3 KB
Views:
89
Last edited: Feb 19, 2010
16. ### eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
104
Are you saying the author is meant to use a DIODE instead of a diac in the SCR firing circuit?

If so, case closed.

17. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
Two sources, betcha I can find another in my old industrial electronics text books. The diode is meant for a different function than what you are thinking.

You are ignoring a component that is invisible, but incorporated in most SCRs, an extra resistor.

18. ### eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
104
Only high power SCRs have these internal resistors built in and normal SCRs don't.

I have re-read the ebook article and now knows the exact thinking of the author in using a diode in the SCR triggering circuit. That is a continuation of the previous example, by adding a capacitor.

Anyway, relying on the gradually increasing magnitude of the gate current to trigger a SCR remains doubtful as a good practice.

The example serve well as an illustration inside the ebook but in practice most likely one would need to use a diac instead to have proper phase control from a few ten degrees upto 180 degree.

Last edited: Feb 19, 2010
19. ### Wendy Moderator

Mar 24, 2008
21,470
2,959
Other sources by way of example...

Though it is interesting to note they have the load reversed. They also show the resistor I was talking about.

Googling for more...

Side note, a triac is two SCRs back to back, with the gates connected.

Which brings us to this one.

http://pcbheaven.com/wikipages/Dimmer_Theory/

The diac you are advocating is shown here, along with the explanation why they are used.

I suspect you are mixing SCRs (which are ½ of a TRIAC and TRIAC circuits together).

Something to keep in mind, this is a text book, with simplified schematics for illustration. Just as you wouldn't use a simple BJT with two resistors, but would show it in a text book as the first example of bias. I think it is reasonable to assume that you would do something similar with SCR circuits. If you were to simulate a SCR with a 10MΩ variable resistor, and include the Gate to Cathode resistor, I think you would find you could do the full 90° conduction. Larger resistors also mean smaller caps too.

Last edited: Feb 19, 2010
20. ### eblc1388 AAC Fanatic!

Nov 28, 2008
1,543
104
If you still can't see why using the diode is a bad idea, may be the following simulation can shed some light on how the circuit will behave.

See what happens to the SCR current if one changes R1 by a very small amount.

File size:
30.2 KB
Views:
84