Test: Power Electronics- 3

Concept:

Steady state ripple

The peak to peak ripple current has maximum value when duty cycle ∝ = 0.5

In case 4FL ≫ R, then tan 
Calculation

∴ maximum value of Ripple current is inversely proportional to chopping frequency and the circuit Inductance.

We known Boost regulator is a step – up chopper

V_dc = 220V, f = 500 Hz, V_o = 180 V

(2×10⁻³) = 1.636 msec

Current ripple does not depend on the value of E
ΔI_L = V_dc ∝ (1−∝)T

First find rated load current

For Regeneration braking load current should be negative.

-1.8 = ∝230 – 120

230∝ = 118.2

∝ = 0.5139

∴ Duty cycle of chopper = 51.39 %

Concept:

For Single – quadrant (type A) Chopper

Average value V_o = ∝ Vs
R.M.S value V_rms = 
where ∝ = Ton/T
We known the form factor F.F = 
 

Calculation:

Form factor = 

∴ form factor = 1.528

Calculation:

V_o = I_oR_L = 3 × 10 = 30 V

V_b = V_o (1 – D)
1−D = 12/30 = 0.4
D = 0.6

Periodic time 

On time = DT = 0.6 × 20 × 10^-6 = 12 μs 

over an on-time the change in battery current will be

Concept:

When the transistor is switched on, current ramps up in the inductor and energy is stored.

When the transistor is switched off, the inductor voltage reverses and acts together with the battery voltage to forward bias the diode, transferring energy to the capacitor.

When the transistor is switched on again, load current is maintained by the capacitor, energy is stored in the inductor and the cycle can start again.

The value of the load voltage is increased by increasing the duty cycle or the on-time of the transistor.

During the transistor on-period, assuming an ideal inductor and transistor

V_b = V_L = Ldi/dt

V_b/s = (sL) i(s)

i(s) = V_b/s²L

i(t) = (V_b/L)t = kt

Over an on-time of T_ON, the change of battery current will be

Δi = k T_ON

During the off-period

V_o = V_b + V_L = V_b + L(di/dt) = V_b + L(Δi/Δt)

= V_b + L(V_b/L)T_ON /T_off = V_b(1 + T_ON /T_off)

Let the duty cycle

D = T_ON T, and T_off = T – T_ON = T(1 - D). Now

V_o = V_b (1 – DT/T(1 – D))

which simplifies to

V_o = V_b/(1 - D)

In an ideal circuit, V_bI_b = V₀I₀. Therefore, from eq.

I_b = (V_o/V_b)I_b = I_o/(1 – D)

I_b = I_b + Δi/2

I_bo = I_b – Δi/2

When the steady variation of battery current has been reached, the variation will be between l_o and I₁, as shown in.

For Buck Converter
V₀ = αV_dc ⇒ α = 5/12
The peak to peak ripple current is

L = 129.62 μH