Calculation of SCR Power Loss, Gate Resistor

Example 1:

An SCR has a V_g – I_g characteristics as V_g = 1.5 + 8 I_g . in a certain application, the gate voltage consists of rectangular pulses of 12 V and of duration 50μs with duty cycle 0.2.

a) find the value of Rag series resistor in gate circuit to limit the peak power dissipation in the gate 5 watts.

b) calculate average power dissipation in the gate.

Solution

During conduction

V_gs = R_g I_g + V_g = R_g I _g + 1.5 + 8 I_g

12 = ( Rag + 8 ) I_g + 1.5

Peak power loss = V_g I_g = 5

5 = (1.5 + 8I_g ) I_g

I_g = -1.5 ± √(1.5)² – 160 / 16 = 0.7A

I₂ = (R_g + 8 ) 0.7 + 1.5 R_g = 7Ω

Average power loss

Average power loss = Peak power loss * duty cycle

= 5 * 0.2

= 1W

Calculation of SCR Gate Resistor:

If the V_g – I_g characteristics of an SCR is assumed to be a straight line passing through the origin with a gradient of 3 * 10 ³ , calculate the required gate source resistance. Given E_gs = 10V and allowable P_g = 0.012Ω .

Solution

The allowable power

P_g = v_{g *} I_g

= 0.012

Also gradient = v_g / I_g = 3 * 10³ v_g = 3 * 10 ³I_g

( 3 * 10³ * I_g * I_g ) = 0.012 , I_g = 2mA. V_g = 3 * 10³ * 2 * 10^-3 = 6V

Example 3:

A thyristor has a forward characteristic which may be approximated over its normal working range to the straight line below fig. determine the mean power loss for

a) a continues on state current of 23 A

b) a half – sine wave of mean value 18A

c) a level current of 39.6A for one – half cycle

d) a level current of 48.5 A for one third cycle

solution

a)power loss

V_T = 1 + 23 * 1.1 / 60 = 1.42 V

Power loss = V_T I_T = 1.42 * 23 = 32.7W

b) The maximum value of the sine wave

=18πA

from the figure at any current I , voltage V = (1 + 1.1 /60) I

over one cycle , the total base length is 2π from o to π , I = 18 π sin x ,and from π to 2 π , I =0.

c) Mean power

the mean power loss will be half the instantaneous power loss over the half cycle when the current is flowing

Mean power = (39.6 (1+ (1.11/60) * 39.6 )) /2 = 34.2W

SCR Triggering Frequency Calculation:

For an SCR the gate cathode characteristics is given by a straight line with a gradient of 16volts per amp passing through the origin, the maximum turn – on time is 4μs and the maximum gate current required to obtain this quick turn – on is 500 mA. If the gate source voltage is 15V.

a) calculate the resistance to be connected in series with the SCR gate.

b) compute the gate power dissipation , given that the pulse width is equal to the turn – on time and that the average gate power dissipation is 0.3W. also compute the maximum triggering frequency that will be possible when pulse firing is used.

Solution

a) I_{g min} = 500mA = 0.5A , V_g / I_g = 16 V/A , V_g = 16 * 0.5 = 8V

R_s = E_gs – V_g / I_g = (15-8) / 0.5 = R_s = 14Ω

b) Power dissipation

P_g = v_g I_g = 8 * 0.5 = 4W

P_gmax = P_gav / f . T_on = 4= 0.3 * 10⁶ / f * 4

f = 18.75 kHZ , f=19kHZ

Commutating Capacitances Calculation:

For the class c commutation. The dc source voltage E_dc = 120V and current through R₁ and R₂ = 20A. the turn off time of both the SCR is 60 μs. Determine the value of commutating capacitances C for successful commutation.

Solution

The resistance R₁ = R₂ = E_dc / I = 120 /20 =60W

Now we have the relation for C for successful commutation as

C = 1.44 * t_off / t_on = 1.44 * 60 *10^-6 / 6 = 14.4μf

Example5:

Compute the value of commutations capacitor C and commutating inductor L for the following data

E_dc = 50V , I­_L (max) = 50A , t_off of SCR₁ = 30 μs chopping frequency f = 500Hz and the load voltage variation required is 10 to 100%

Solution

For reliable operation, let us assume 50% tolerance on turn off time of SCR₁

t_off = (30 + (50/100) * 30 ) = 45 μs

now, we have the relation for the commutating capacitor C as

C = I_L t_off / E_dc = 50 *45 * 10^-6 / 50 = C = 45 μf

Therefore the minimum load voltage available is given by

V_o = t₁ – t₂ / T E_dc

Where t is chopping time period

V_o (min) = (π√LC / T) * E_dc

L ≤ ( V_{o (min)} / E_dc )² (T/ π²c)

V_o (min) = 10 % (50) = 5V

T = 1 / f = 2* 10^-3 s

L ≤ (2* 10^-3 )² /( π² * 45 * 10^-6) * (5/50)²

Or

L ≤ 90 μH

Also we have the relation

L ≥ C (E_dc / I_{L (max)} ) or L ≥ 45 * 10 ^-6 (50 /50 )²

L ≥ 45 μH

Example 6:

An SCR with I_g (min ) = 0.1mA and v_g = 0.5V. the diode is silicon and the peak amplitude of the input is 24volts. Discuss the trigger angle α for R_V = 100KΩ and R_min = 10 KΩ.

Solution

e_s = I_g (R_V + R_min) + V_D + V_g

at the trigger point

e_{s (trigger)} = 0.1mA (110 KΩ) + 0.7V + 0.5V = 12.2V

since e_s is a sine – wave , it obeys the expression

e_s = E_max * sin ωt = E_max * sin (2πft)

where  2πft is the phase angle at any instant of time. For our purposes this angle is α. Thus E_max = 24V

e_s = 24 sin α 12.2 = 24 sin α

sin α = 12.2 / 24 α =30.6⁰