A step-up chopper shown in figure is to deliver 3AIn to the 10 Ω...
Calculation:V
o = I
oR
L = 3 × 10 = 30 V
V
b = V
o (1 – D)
On time = T
ON = DT = 0.6 × 20 × 10
-6 = 12 μs
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, the 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 l
o and I
1, as shown in.