Calculating time when SCR starts up

[Femme_physics]

Homework Statement

Given a circuit to control the volume of an electric bell P. Potentiometer T is set to 600K ohms. The breakout voltage of the SCR (at the SCR gate) is 20 micro-amperes. And the max gate that will start up the SCR is 0.7 Volts


http://img41.imageshack.us/img41/4322/yest2.jpg



Calculate the time when the SCR starts up.

The Attempt at a Solution

http://img205.imageshack.us/img205/1268/yest1.jpg

 

[I like Serena]

Hmm, how did you get U(diac)?
It does not look right.

Btw, the quotient would not be in degrees.
You would need to take the inverse sine to find alpha.
Or did you already do so?
Either way, that does not look like the right angle.

 

[Femme_physics]

I thought Udiac was given as 0.7V, but perhaps I misunderstood.

And you're right, it should be a radians.

I admit, I glanced at another person's solution in my class. Problem is that it confuses me even more!

http://img703.imageshack.us/img703/2155/vinvm.jpg

To my understanding, V(t) is Vin's voltage during the set-off of the SCR.

I failed to understand the equation: Vt(t) = Vin(t) - Vd - Vp

Because if it uses Kirchhoff's law on the loop there should be 5 voltages. 1 source and 4 drops.

Source: Vin

Drops:
Vt (potentiometer), Vd, Vp and the SCR.

Can this be clarified for me?

 

[I like Serena]

I you take the loop through the potentiometer, your drops are:
Vt, Vd, Vs (SCR gate).

Apparently the drop over the gate has been marked as Vp, but I find that confusing, since I'd expect Vp to be the drop over the electric bell.


If I understand your problem statement correctly, the SCR is triggered when Vs (SCR gate) is 0.7 V.
And when the current at the SCR gate is at least 20 μA.
Btw, the current at the SCR gate is the current that flows through the potentiometer, the diode and the SCR gate.

Can you calculate what Vin has to be, if the current through the potentiometer is 20 μA, and if the voltage drop over the SCR gate is 0.7 V?


Btw, in the other person's calcution, you first get an expression for the voltage across the potentio meter, and then an expression for the current through the potentiometer.
If this current reaches 20 μA, the gate opens.
You can solve this equation mathematically to find the corresponding time.

However, there is a mistake in it.
I'm missing a factor 2 in the 3rd and 4th lines.

 

[Femme_physics]

I you take the loop through the potentiometer, your drops are:
Vt, Vd, Vs (SCR gate).

Apparently the drop over the gate has been marked as Vp, but I find that confusing, since I'd expect Vp to be the drop over the electric bell.

Ah..true

If I understand your problem statement correctly, the SCR is triggered when Vs (SCR gate) is 0.7 V.
And when the current at the SCR gate is at least 20 μA.
Btw, the current at the SCR gate is the current that flows through the potentiometer, the diode and the SCR gate.

How so? After the firing angel been set-off I see 3 different currents

http://img96.imageshack.us/img96/1683/i1i2i3.jpg

Btw, in the other person's calcution, you first get an expression for the voltage across the potentio meter, and then an expression for the current through the potentiometer.
If this current reaches 20 μA, the gate opens.
You can solve this equation mathematically to find the corresponding time.

However, there is a mistake in it.
I'm missing a factor 2 in the 3rd and 4th lines.

I can't see it, but I rather try my own calculations anyway

 

[I like Serena]

How so? After the firing angel been set-off I see 3 different currents

Yep.
I3 is the current through the resistor, the diode, and the SCR gate.
I2 is the current through the SCR anode.
I1 is the current through the SCR cathode.

 

[Femme_physics]

I'm glad this is sorted out. The original calculations are kinda mumbo jumbo to me-- IMO it should be "just that easy"->

Using pure math

http://img684.imageshack.us/img684/3808/vptpp.jpg

 

[I like Serena]

If only it were that simple... ;)

Your equation applies to a straight line, but your voltage is a sine and not a straight line.
A different formula is needed here.

You're alos leaving some stuff out, in particular the voltage drop over the resistor and the voltage drop over the diode...

 

[Femme_physics]

I had a feeling it's not that easy, but give me some points for general sneakiness ;)

At any rate, I suppose that by "different formula" you mean the

Vin(t)=Vmax sine omega t

I get that. It relates the voltage to the angle and the time... That's logical.

Why need the voltage drop over the resistor and diode is a bit beyond me because the same voltage drops on all the components between the point where the current split to when it reunites.

 

[I like Serena]

I'll give you 0.4125 point for sneakiness! ;D

And yes, I meant that formula.

In your problem statement it says that you need 20 micro amperes flowing through the SCR gate (I3=20 μA).
If I3=20 μA, then you get a voltage drop over the resistor.
If Vin is not high enough, this will mean the the voltage at the SCR gate is not high enough to trigger it.

You need to find the Vin that is high enough to trigger the SCR gate.

 

[Femme_physics]

I'll give you 0.4125 point for sneakiness! ;D

Hehe

In your problem statement it says that you need 20 micro amperes flowing through the SCR gate (I3=20 μA).
If I3=20 μA, then you get a voltage drop over the resistor.

But the SCR is not in series to the potentiometer. We need 20 micoamperes through the bell and the SCR. I guess the potentiometer and the diode are for... Actually, I'm not sure entirely what is the use of them..

 

[I like Serena]

It says "The breakout voltage of the SCR (at the SCR gate) is 20 micro-amperes."

Since it says "at the SCR gate", I interpret that to mean I3.

Here's a picture of an SCR:

"At the gate" should mean the connection on the left side.

 

[Femme_physics]

Ah, true :)

http://img18.imageshack.us/img18/4859/scragain.jpg

OK, so it means Vin is 12.6 [V] because:

Sigma V = Vin - Vp - Vscr = 0
Vin = 12.6 [V]

So that's the value of Vin during SCR's set-off.

Now do I use the

Vin (t) = Vmax Sin ωt

Formula? Whereas ω is 2pi, Vin is the value I found at set-off, and Vmax is just 12√2

Is that right? And the T I find is the T during SCR's set-off that they ask me to find.

 

[NascentOxygen]

Hi FP! Are you challenging the laws of physics again?

The triggering current flows gate→cathode, and is largely independent of the SCR's main anode→cathode current, so you are not at this stage involving current through the bell.

For a gate current of ≥20uA, and the variable resistor set at 600kΩ, the required instantaneous level of the AC supply voltage level is going to be:

≥ 0.7 + 0.6 + 12v

Vmax is just 12√2

Correct.

 

[I like Serena]

Close.
The Vscr voltage is between the gate and the cathode, and not between the gate and the anode (that's also what NascO said).And yes, the "t" you find is the t during SCR's set-off.
But ω is not 2pi, but ω = 2 pi f, where f is the frequency 50 Hz.

 

[Femme_physics]

Hi FP! Are you challenging the laws of physics again?

Why my favorite pastime uncle Nascent! :)

The triggering current flows gate→cathode, and is largely independent of the SCR's main anode→cathode current, so you are not at this stage involving current through the bell.

Oh

Close.
The Vscr voltage is between the gate and the cathode, and not between the gate and the anode (that's also what NascO said).

Ohx2 :)

In this case:

-Vp +Vscr +Vd + Vt

-Vp +0.7 + 0.6 +12

Vp = 13.3 [V]

And then,

Vin (t) = Vmax Sin ωt

13.3 = 12√2 x Sin 2pif x t

Sin2pif = sin(2pi x 50 )= -0.71713.3 = -12.174t

t = 1.09 [secs] (since time can't be negative it's swapped to positive)

The result doesn't make much sense to me...these things suppose to take mili if not micro secs to set off

 

[NascentOxygen]

Vin (t) = Vmax Sin ωt

13.3 = 12√2 x Sin 2pif x t

13.3 = 12√2 sin_([/size][/color]250t_)[/size][/color]
Parentheses and brackets are cheap; I recommend that you make good use of them. ☺[/size][/color]

 

[Femme_physics]

Weird. When I use my calculator equation-solver I get X = 0.164 secs

http://img259.imageshack.us/img259/9134/degdeg.jpg

Yet again a senesless result :-/

 

[NascentOxygen]

Okay, so you included some parentheses but then did something wrong.

Actually, I can't figure out what you are doing, but it doesn't look right. ☹[/size][/color]

13.3 = 12√2 sin_([/size][/color]250t_)[/size][/color]

Following on from what I wrote above, the next step is:

13.3 / 12√2 = sin(250t)

sin⁻¹ (13.3 / (12√2)) = 250t

∴ t = https://www.physicsforums.com/images/icons/icon5.gif

 

[Femme_physics]

I think the parenthesis are to indicate which angle belong to the sine function.

Regardless, if I use your method I get the same result as when I did originally when I used my calculator function

i.e.

http://img29.imageshack.us/img29/933/93218109.jpg

 

[I like Serena]

You need the inverse sine in radians instead of degrees.

 

[NascentOxygen]

Regardless, if I use your method I get the same result as when I did originally when I used my calculator function

Before hitting the sin⁻¹ key you should switch your calculator to radians instead of degrees. https://www.physicsforums.com/images/icons/icon6.gif

 

[Femme_physics]

Ehm...getting error now

http://img696.imageshack.us/img696/8158/raddude.jpg

 

[Femme_physics]

I prefer to work with degrees and just use the formula to convert between them

 

[I like Serena]

You liked calculating with angles didn't you?
Care to improve your skills with sines, radians, and degrees?

 

[NascentOxygen]

Whatever method you use, it needs to be done correctly.

So, repeat your calculation, this time with 51.6° converted to radians.

 

[Femme_physics]

You liked calculating with angles didn't you?
Care to improve your skills with sines, radians, and degrees?

I just forgot how to convert radians to degrees mode in my calculator, but I looked up the manual just now so I got it figured out

So, result is

t = 2.866 millisecs

Make sense, I think, right?

 

[NascentOxygen]

That looks right.

 

[Femme_physics]

Thanks :)

I'm weirded by your angel result. Shouldn't it be->

http://img140.imageshack.us/img140/4467/1333p.jpg Also, do those graphs look accurate? I'm worrying about the Vmax for Vp. I made it 12√2 - 0.7 V

http://img19.imageshack.us/img19/8913/vinvptp.jpg

 

[NascentOxygen]

your angel result

It's nice of you to say that!

You are doing it correctly, I hope? Using sin⁻¹ (13.3 / (12√2)) = ...

None of these FP 'shortcuts' that are based on wistful thinking!

Also, do those graphs look accurate? I'm worrying about the Vmax for Vp. I made it 12√2 - 0.7 V

That looks right.