# Need help with SCR datasheet

Discussion in 'General Electronics Chat' started by chipwitch, Mar 13, 2014.

1. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
SCR datasheets are even more difficult to decipher than most.

What are the more critical characteristics to look for?

I was trying to find a "voltage drop" thinking that is important to determine the power to dissipate from the SCR. Aside from Peak voltage and RMS current, I'm not sure what I'm looking at.

No specific purpose in mind. I'm just trying to understand them better and wanted to buy a few to experiment with.

http://pdf1.alldatasheet.com/datasheet-pdf/view/172172/ONSEMI/C122B1.html

2. ### #12 Expert

Nov 30, 2010
16,655
7,294
Watch for dv/dt. Snubbers quench the fast rising anode voltage when the thyristor turns off so as to avid false triggering.

You can use the graphs to derive Von or just look at the graphs to find the power to dissipate at any given current.

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3. ### crutschow Expert

Mar 14, 2008
13,466
3,353
The voltage drop is shown as the Peak On-State Voltage in the table.
Figures 3 and 4 show the power dissipation. From the DC curve in Figure 3 you can determine the voltage drop for any current using Ohm's law.

4. ### #12 Expert

Nov 30, 2010
16,655
7,294
VTM
Missed it.
I was looking for Vf

senior moment.

5. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
Ah... thanks.

These are pretty cool, if I'm understanding them right. So, the only power they need to dissipate is determined by 1.83V*I? Compared to a linear voltage regulator where the power dissipated is the based on the voltage dropped which is the delta of source and output, these have a huge advantage. Any reason why these wouldn't be ideal for narrowing that gap? I'm thinking any voltage regulator circuit where I've got 20 or even 10 volts difference between Vin and Vout, I should be using an SCR to reduce the Vin. Almost seems like getting something for nothing. Am I missing something?

6. ### crutschow Expert

Mar 14, 2008
13,466
3,353
The power at other than the maximum current is shown in Figure 3. The forward drop reduces somewhat as the current is reduced.

An SCR only reduces the AC voltage significantly if it is used as a part of a phase-control circuit, which is a common application for SCRs. For typically home applications, a TRIAC (bilaterial SCR) is often used since it requires only one device rather than two to perform full-wave operation.

7. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
I knew I should have mentioned the Vin was AC!

Though I imagine DC Vin could work too? Use an RC circuit to operate the SCR gate at intervals creating a PWM? Filter cap for smoothing SCR output? Any frequency issues from it? Or am I all wet?

8. ### #12 Expert

Nov 30, 2010
16,655
7,294
A thyristor won't shut off unless the current through it approaches zero. That won't naturally happen in a DC circuit.

9. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
That I know. My last post, I was asking if a simple RC circuit couldn't provide the trigger... charge, discharge (to zero), charge, discharge, etc. Or a transistor shunt to ground around the SCR?

<edited> By "Or a transistor shunt," I mean in addition to the RC circuit. The shunt would be enabled at some point after the discharge cycle began and ground the SCR anode, effectively turning it off.

Last edited: Mar 13, 2014
10. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
I'm looking at another datasheet:
http://www.mouser.com/ds/2/302/BT151-500RT-111779.pdf

In the following chart, Fig. 3. "Total power dissipation as a function of average on-state current; maximum values" appears to be saying that the temperature of the mounting base is inversely related to the conduction angle. The more the device conducts, the cooler it operates? This is counter-intuitive. Am I reading it right? When a relay is off, it runs cooler. It also appears that if I use a conduction angle of 30, the allowable amperage is drastically reduced from the devices maximum. I would think this too is the opposite of how it would work.

11. ### crutschow Expert

Mar 14, 2008
13,466
3,353
You are reading it wrong. Fig. 3 shows the Power Dissipated (left axis) versus the Average Current (bottom axis). This is the power that must be dissipated by the heat sink. This shows the power increases with average average current and is greater for the same average current at a smaller conduction angle, as would be expected. That's because the peak current is higher for the same average current (averaged over one cycle) for a smaller conduction angle.

Make sense?

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12. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
...

When I look at it through the lens of calculus, the area under the wave, it makes more sense. A 30 degree conduction angle will result in a lower current (as well as a directly proportional average voltage?) when compared to 180 degrees, assuming the input source remains the same. Yes?

So, if I use an SCR to drop voltage, a consequence will be a loss in current from the max rating. Temperature seems to be a key part. What is the mounting base temperature on the right of that figure conveying? Is it a "do not exceed" temperature?

The last thing... form factor. Looked it up. I get the basic concept (barely), but the math is discouraging. Do I need to use form factor for anything?

13. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
Form factor * Iaverage = peak I?

14. ### crutschow Expert

Mar 14, 2008
13,466
3,353
Lower average current, yes. For a resistive load the peak current will stay the same.
Yes, the right side of the graph is labeled Tmb(max).
I've never used it myself.
No. The formula at the bottom of the graph gives Form factor * Iaverage = Irms

15. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
AARrrghHH! I appreciate your patience.

I didn't see the formula at the bottom. I think why this has been so hard is due to what I'm identifying as cause and effect, taking one for the other.

The little sine wave showing how alpha is defined shows a the full sine wave (partly with dotted line). But, that's misleading. I took it that the conduction angle essentially masked the rest of the sine wave. But it wouldn't work like that. I'm not accustomed to thinking of current in a wave form. Sine waves are an icon used to indicate AC voltage. Intellectually, I recognize sine waves are everywhere, not exclusively voltage, however, I have a bit of a blind spot when their applied to something else. Like, current in this case.

I've also been (and am still) assuming that the conduction angle will always be measured back from where the wave becomes negative. Otherwise, the current would keep the scr in conduction mode.

So, now I'm looking clearly at the current sine wave segment as a function of voltage/resistance. Resistance is fixed, naturally. What is the voltage output of the SCR? If we have a voltage source of say a fixed Vrms of 10 VAC, the peak will be 14V. In an ideal SCR with a conduction angle of 180 degrees, won't the output be 14Vpeak/π= 4.45VDC? And a conduction angle of 90 degrees will be even less (I don't know how to calculate that). Given a circuit of a fixed load and source, varying only the conduction angle, you're saying the peak current will remain the same? How is that possible? Voltage is lower @ 30 degrees than 180 degrees, resistance is fixed.

... I think I'm going to go take up pottery

16. ### crutschow Expert

Mar 14, 2008
13,466
3,353
The conduction is "a" on the graph. For a resistive load, both the current and voltage follow the sold black part of the sine wave based upon Ohm's law, I = V / R.
True. When the voltage goes to zero, so does the current and the SCR stops conducting.
I misspoke. The peak output voltage will be less then the peak sine wave voltage if the conduction angle is less than 90°. The average voltage, of course, will depend upon the phase angle and is indeed 4.45V for 180°.

17. ### chipwitch Thread Starter Member

Mar 29, 2013
48
4
Okay, so I AM starting to get it.

You said, "conduction is 'a' on the graph." I think you meant to say, "α" (alpha) is conduction angle. "a" is form factor.

Thanks for your help. I'm ordering a few to experiment with today.

18. ### crutschow Expert

Mar 14, 2008
13,466
3,353
That is correct.