# SCR question! help plz

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#### ericyeoh

##### Member level 2 Can anyone help me with this question?

Given Vg=1+ 10Ig. The gate source voltage is a rectangular pulse of 15V with 20micro second duration. For an average power dissipation of 0.3W and peak gate drive power of 5W. Determine

i) Resistance to be connected series with the SCR gate.

ii)Triggering frequency

iii) The duty circle.

#### scanman

##### Member level 5 Hi ericyeoh,

I come up with:

i) Resistance to be connected series with the SCR gate = 52 Ohms.

ii)Triggering frequency = 3000 Hz

iii) The duty cycle = 0.06

Let me know if this checks out.

Cheers,
Scanman

### ericyeoh

Points: 2

#### ericyeoh

##### Member level 2 Yes! i;m getting the same answer as yours for ii and iii.

To Where (Pave/ peak gate power) = Duty circle

Duty circle = Ton / (Ton + toff)

Ton given, Duty circle is apply in order to find Frequency by using F = 1/Period.

However i;m still dun get how could u got the answer for i).

Here is my solution for finding the gate resistance.

Pg = VgxIg <-------1
Vg = 1 + 10Ig <------2

since the Pg is given as 5 W, (5/Ig) = Vg <-----3

substitude equation "3" inside equation "2"
u will be able to getting the equation 5 = Ig + 10 (Ig^2)

Ig = 0.65 A and -0.75A (neglect)

Vg = 1 + (10x Ig)
Vg= 7.5V

Rg = (Vamplitude - Vg) / Ig
V amplitude = 15V

Rg = 11.53 ohm

#### wisley

##### Newbie level 2 Given:
Vgg=15V
Vsr=1V
Ig=?

Vgg=Vsr+IgRg

So, we can obtain
Ig=1.4A
and Rg=10ohm

Plz verify the formula is correct.............................................

#### scanman

##### Member level 5 Originally I got the same number as wisley but the power seemed too high.
I made a mistake on the resistor calculation. Ig = 5W / 15V = 0.3333 Amps
R = E/I
Rg = 15V - (1 + 10Ig) / Ig = 15 - 4.3333 / 0.3333 = 32 Ohms

Hope this helps
Scanman

#### ericyeoh

##### Member level 2 Guys! the 15V is amplitude of the gate source voltage right? how come u guys make it as Vg ( gate voltage) ?

Besides, the peak gate power has the formula Pg = Vg x Ig
The Vg is the gate voltage and the Ig is the gate current!

#### wisley

##### Newbie level 2 Pg=5w-0.3w
=4.7w
Pg=IgVg
Ig=0.3133A
Using ohm rule,
Vg=IgRg
Rg=15/0.3133
=47.87ohm[/img]

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