A step down chopper has Vs = 220 V and is connected to RLE load. With R = 1 Ω, E = 24 V and L = 5 mH. The chopping period is 2000 μs and the onperiod is 600 μs. Find the value of minimum steady state output current.

To find the minimum steady state output current for the given step-down chopper, we can apply the following steps:

1. Calculate the Duty Cycle (D):

[

D = frac{T_{on}}{T_{chopping}} = frac{600 , mu s}{2000 , mu s} = 0.3

]

2. Calculate the average output voltage (Vout) of the step-down converter:

[

V_{out} = D cdot V_s = 0.3 cdot 220 V = 66 V

]

3. Calculate the output voltage across R, E, and L.

In steady state, the equation can be expressed as:

[

V_{out} = I cdot R + E

]

Where ( I ) is the steady-state output current.

4. Rearranging gives:

[

I = frac{V_{out} – E}{R}

]

Plugging in the values:

[

I = frac{66 V – 24 V}{1 , Omega} = frac{42 V}{1 , Omega} = 42 A

]

Thus, the value of the minimum steady-state output current is 42 A.