In a type A chopper, the maximum value of Ripple current is
Steady state ripple
The peak to peak ripple current has maximum value when duty cycle ∝ = 0.5
In case 4FL ≫ R, then tan
∴ maximum value of Ripple current is inversely proportional to chopping frequency and the circuit Inductance.
A boost – regulator has an Input voltage 8V and the average output voltage of 24 V the duty cycle is _
We known Boost regulator is a step – up chopper
A DC chopper operates on 220 V dc and frequency of 500 Hz, feeds on RL load of the output voltage of chopper is 180 V the ON time of the chopper is –(in msec)
Vdc = 220V, f = 500 Hz, Vo = 180 V
(2×10−3) = 1.636 msec
DC chopper feed an RLE load. If the value of E increased by 20%, the current ripple
Current ripple does not depend on the value of E
ΔIL = Vdc ∝ (1−∝)T
A class C chopper is operated from 230 V battery the load is a dc motor with R = 0.15 Ω, L = 12 mH and Eb = 120 V. The duty cycle of the chopper to achieve regenerative braking at the rated current of 12 ampere would be equal to
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 %
A single - quadrant chopper is operating with the following specification: Ideal battery of 240 V; on time (Ton) = 1.5 ms, OFF time = 2 ms. The Form factor factor will be –
For Single – quadrant (type A) Chopper
Average value Vo = ∝ Vs
R.M.S value Vrms =
where ∝ = Ton/T
We known the form factor F.F =
Form factor =
∴ form factor = 1.528
A class C chopper is operated from a 220 V battery. The load is a dc motor with R = 0.1 Ω, L = 10 mH and Eb = 100 V. The rated current is 10 A. The duty cycle in motoring mode is
A buck converter is to be designed where Vs = 10V, Vo = 5V,R = 500Ω. The switching frequency is decided to be 20 kHz. Allowable current ripple is 5 mA. The optimum value of inductor is
A step-up chopper shown in figure is to deliver 3A
In to the 10 Ω load. The battery voltage is 12 V, L = 20 μH, C = 100 μF and chopper frequency is 50 kHz. Determine the battery current variation in ampere
Vo = IoRL = 3 × 10 = 30 V
Vb = Vo (1 – D)
1−D = 12/30 = 0.4
D = 0.6
On time = DT = 0.6 × 20 × 10-6 = 12 μs
over an on-time the change in battery current will be
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
Vb = VL = Ldi/dt
Vb/s = (sL) i(s)
i(s) = Vb/s2L
i(t) = (Vb/L)t = kt
Over an on-time of TON, the change of battery current will be
Δi = k TON
During the off-period
Vo = Vb + VL = Vb + L(di/dt) = Vb + L(Δi/Δt)
= Vb + L(Vb/L)TON /Toff = Vb(1 + TON /Toff)
Let the duty cycle
D = TON T, and Toff = T – TON = T(1 - D). Now
Vo = Vb (1 – DT/T(1 – D))
which simplifies to
Vo = Vb/(1 - D)
In an ideal circuit, VbIb = V0I0. Therefore, from eq.
Ib = (Vo/Vb)Ib = Io/(1 – D)
Ib = Ib + Δi/2
Ibo = Ib – Δi/2
When the steady variation of battery current has been reached, the variation will be between lo and I1, as shown in.
In the circuit shown below, if load R = 500 Ω, switching frequency is 25 kHz and peak to peak ripple current of inductor is limited to 0.9 A then the filter inductance L is"
For Buck Converter
V0 = αVdc ⇒ α = 5/12
The peak to peak ripple current is
L = 129.62 μH