Basic principle of Step Up Down Chopper

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written 5.8 years ago by

teamques10 ★ 69k

A chopper can also be used both in step-up and step-down modes by continuously varying its duty cycle. The principle of operation is illustrated in Fig. below. As shown, the output, polarity is opposite to that of input voltage $E_{\mathrm{dc}}$ .

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When the chopper is $\mathrm{ON},$ the supply current flows through the path $E_{\mathrm{dc}+}-C H-L-E_{\mathrm{dc-}} .$ Hence, inductor L stores the energy during the $T_{\text { on period. }}$

When the chopper $C H$ is OFF, the inductor current tends to decrease and as a result, the polarity of the emf induced in L is reversed as shown in above figure.

Thus, the inductance energy discharges in the load through the path,

$L_{+}-\text { Load }-D-L_{-}$

During $T_{\text { on }},$ the energy stored in the inductance is given by

$W_{i}=E_{\text { dc }} I_{s} T_{\text { on }}-----(1)$

During $T_{\text { off }},$ the energy fed to the load is

$W_{o}=E_{0} I_{s} T_{\text { off }}-----(2)$

For a lossless system, in steady-state: Input energy, $W_{i}=$ output energy, $W_{o}$

$\therefore \quad E_{\mathrm{dc}} \cdot I_{s} \cdot I_{\mathrm{on}}=E_{0} I_{s} T_{\mathrm{off}}, \ or \ E_{0}=E_{\mathrm{dc}} \cdot \frac{T_{\mathrm{on}}}{T_{\mathrm{off}}}-----(3)$

$\qquad \qquad E_{0}=E_{\mathrm{dc}} \cdot \frac{T_{\mathrm{on}}}{T-T_{\mathrm{on}}}=E_{\mathrm{dc}} \cdot \frac{1}{T / T_{\mathrm{on}}-T_{\mathrm{on}} / T_{\mathrm{on}}}$

Substituting

$\qquad \frac{T_{\text { on }}}{T}=\alpha, \ we \ get, \ E_{0}=E_{\mathrm{dc}} \cdot \frac{1}{1 / \alpha-1}$

$\qquad E_{0}=E_{\mathrm{dc}} \frac{\alpha}{1-\alpha}$

For $0 \lt \alpha \lt 0.5,$ the step-down chopper operation is achieved and for $0.5 \lt \alpha \lt 1,$ step-up chopper operation is obtained.

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