Question

A thyristor operating with a peak supply voltage of 400 V has the following specifications. Repetitive peak current IP = 200A, (di/dt)max = 50 A/μs, (dv/dt)max = 200 V/μs. With a factor of safety 2 for the above specifications, design a suitable snubber circuit for the thyristor. Assume minimum value of load resistance to be R = 10Ω. Damping ratio may be taken as 0.65.

Answer 1

Answer:

$\frac{d v}{d t}=74.07 \mathrm{v} / \mu_s$

Explanation:

Step 1: For a factor of safety of 2 , the permitted Values are $I_p=\frac{200}{2}=100 \mathrm{~A}$

$\left(\frac{d i}{d t}\right)_{\max }=\frac{50}{2}=25 \mathrm{~A} / \mathrm{us},\left(\frac{d v}{d r}\right)_{\max }=\frac{200}{2}=100 \mathrm{v} / \mu \mathrm{s}$

Step 2: Inductor is required in series with thyristor to restrict rate of rise of current, beyond specified limit,

$L=\frac{V_s}{(d i / d t)_{\text {max }}}=\frac{400 \times 10^{-6}}{25}=16 \mu \mathrm{H}$

$R_s=\frac{L}{V_s}\left(\frac{d V}{d t}\right)_{\max }=\frac{16 \times 10^{-6}}{400} \times \frac{100}{10^{-6}}=4 \Omega$

Step 3: $C_S$ is charged to$400 \mathrm{~V}$, before thyristor turned on. when thyristor is on then peak current through thyristor is

$\frac{400}{10}+\frac{400}{4}=140 \mathrm{~A} > 100 \mathrm{~A} \text { (permissible) }$

Take. $R_S$ as $8 \Omega$, Now, Peak current

$\Rightarrow \frac{400}{10}+\frac{400}{8}=90 \mathrm{~A} \text {, so } P_S=8 \Omega \text { should }$

be chooses.

Also, $\quad C_S=\left(\frac{2}{R_S}\right)^2 L=\left(\frac{1.3}{8}\right)^2 \times 16 \times 10^{-6}=0.4225 \mu \mathrm{F}$

S may be taken as$0.30 \mu \mathrm{F}$

$\begin{aligned}& \left(C_s \frac{d v}{d t}=\frac{v_s}{R_s+R_L}\right) \\& 0.3 \times 10^{-6} \frac{d v}{d t}=\frac{400}{10+8} \Rightarrow \frac{d v}{d t}=\frac{400}{18} \times \frac{1}{0.3 \times 10^{-6}} \\& \frac{d v}{d t}=74.07 \mathrm{v} / \mu_s\end{aligned}$

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